Genetics Exam #2

Quiz: Pedigree #1 A man affected with the autosomal recessive disorder cystic fibrosis (CF) mates with a normal woman. Which of the following statements about the transmission of CF to his kids is TRUE? (Choose all that are correct) All of his sons will be affected All of his sons will be carriers All of his daughters will be affected All of his daughters will be carriers

All of his sons will be carriers All of his daughters will be carriers Here the cross is dd (affected) x DD (normal). All the kids are Dd

Chapter (2.6/3.5) Advanced Pedigree Analysis Basic Pedigree Analysis 1: Assuming 100% penetrance and no new mutation... I. This pedigree could be showing the inheritance of a recessive trait or disease. II. This pedigree could be showing the inheritance of a dominant trait or disease. A. I = TRUE II = TRUE B. I = TRUE. II = false C. I = false. II = TRUE D. I = false. II = false

B. I = TRUE. II = false

Problem Set #1: Mitosis, Meiosis Which of the following shows a diploid cell in G1 with 2n = 6?

C

Chapter 3: Mitosis, Meiosis and Sex Determination How many double-stranded DNA molecules are shown in this picture? A. 1 B. 2 C. 4 D. 8

C. 4

Quiz: Pedigree #2 The pedigree shows the inheritance of a disease or trait in three generations of a family I. It is possible to rule out autosomal dominant inheritance II. It is possible to rule out autosomal recessive inheritance I = TRUE, II = TRUE I = TRUE, II = false I = false, II = TRUE I = false, II = false

I = false, II = false You cannot rule out AD by Rule 1, nor can you rule out AR by Rule 2.

Chapter 3: Mitosis, Meiosis and Sex Determination During what part of the somatic cell cycle do chromosomes look like this? A. G1 B. S C. G2 D. M phase, prior to metaphase E. M phase, after anaphase

M phase, prior to metaphase

Chapter (2.6/3.5) Advanced Pedigree Analysis ** CONTINUED ** Know the rules that allow particular mechanisms of inheritance to be ruled out, and understand the logic behind these rules. PROBLEM: This pedigree is typical of X-linked recessive inheritance, but... is it also illustrating something else? - i.e. Could it be.... Autosomal recessive? Autosomal dominant? X-linked dominant? * State what the pedigree can be, if anything, what is ruled out, and by what rule.

Yes! It can be autosomal recessive as well as X-linked recessive. It could not have been autosomal dominant nor X-linked dominant since #1 rules them out.

Problem Set #1: Sex Linkage and Sex Determination Alice is a woman whose father suffers from hemophilia. He has been successfully treated with recombinant factor VIII. Alice's mother's family has no history of hemophilia. The likelihood that Alice is carrier of hemophilia A is: A. 100% B. 67% (that is, 2/3) C. 50% D. 25% E. 0%, Alice cannot be a carrier

A. 100% * Hemophilia A: X-linked recessive inheritance * Father: XaY Mother: XAXA Alice: XAXa

Problem Set #2: Pedigree Problem Set 27. Y-linked conditions are very rare and are associated with male sterility. However, assume a man is affected with a Y-linked disease called beergutitis that is not associated with sterility. If he marries a normal woman, then it can be predicted that.. A. All of his male children will be affected with beergutitis B. Half of his male children will be affected C. All of his female children will be affected D. Half of his female children will be carrier E. None of his male children will be affected

A. All of his male children will be affected with beergutitis

Chapter (2.6/3.5) Advanced Pedigree Analysis Basic Pedigree Analysis 2: Assuming 100% penetrance and no new mutation... I. This pedigree could be showing the inheritance of a recessive trait or disease. II. This pedigree could be showing the inheritance of a dominant trait or disease. A. I = TRUE II = TRUE B. I = TRUE. II = false C. I = false. II = TRUE D. I = false. II = false

A. I = TRUE II = TRUE

Problem Set #1: Sex Linkage and Sex Determination Red-green colorblindness is an X-linked recessive trait. Which of the following statements about a woman who is born colorblind must be true? A. Her father and her brother must also be colorblind B. Both her mother and her father must be colorblind C. Her sister and son must be colorblind D. Her daughter must have a 50% chance of being colorblind E. Both her father and son must be colorblind

E. Both her father and son must be colorblind * Women born colorblind: XrXr

Problem Set #1: Sex Linkage and Sex Determination Which of the following statements is FALSE A. If a female carrier of an X-linked recessive disease mates with a normal man, half of her sons will be affected, on average. B. If a female carrier of an X-linked recessive disease mates with a normal man, half of her daughters will be carriers, on average. C. If a male affected with an X-linked recessive disease mates with a normal woman, all of his sons will be normal. D. If a male affected with an X-linked recessive disease mates with a normal woman, all of his daughters willbe carriers. E. None of the statements are false.

E. None of the statements are false. A. XRXr x XRY = XRY, XrY, XRXR, XRXr - true. B. XRXr x XRY = XRY, XrY, XRXR, XRXr - true C. XrY x XRXR = XRXr, XRY - true D. XrY x XRXR = XRXr x XRY - true

Problem Set #1: Pedigrees Examine the accompanying pedigree and determine the inheritance pattern. A. This trait could only be autosomal recessive B. This trait could be autosomal recessive or X-linked dominant C. This trait could be only be X-linked recessive D.This trait could be only be X-linked dominant E. This trait could be autosomal recessive or X-linked recessive

E. This trait could be autosomal recessive or X-linked recessive. Can't be dominant: unaffected parents have affected child. * Autosomal dominant: If the disease was dominant, then that would mean that somehow a dominant genotype was created from dd x dd, not possible. * X-linked dominant: If the disease was dominant, then that would mean that somehow XRY was created from XrXr and XrY, not possible. * Autosomal recessive: Rr x Rr = rr - Disease is rr. * X-linked recessive: XRY x XRXr = XrY

Chapter 3: Mitosis, Meiosis and Sex Determination Features of X-Linked Dominant Inheritance

EX: Hypophosphatemic rickets, Incontinentia pigmenti 1. Phenotype visible in heterozygous/homozygous females and hemizygous males. 2. About equal numbers of males and females show the trait. 3. Heterozygous females mated to wild-type males transmit the dominant allele to half their progeny of each sex. 4. Dominant hemizgous males mated to homozygous recessive females transmit the dominant trait to all their daughters, but none of their sons. 5. The dominant phenotype is equally frequent in males and females.

Chapter 3: Mitosis, Meiosis and Sex Determination (Not a learning objective, but should know) TEXTBOOK EXPLANATION CONTINUED Know the categories that reproduction can be divided into. * Know what occurs during each type of reproduction. * Know the difference between them. Know the difference between reproduction in bacteria and archea, and reproduction in eukaryotes. * For eukaryotes, know different modes of reproduction. - For eukaryotes, know how haploid gametes are formed. * Know differences between bacteria, archea, and eukaryotes (single-celled and multicellular): which is haploid/diploid?

In contrast to single-celled eukaryotes, multicellular eukaryotes reproduce predominantly by sexual means. In most animal species and dioecious plants, males and females carry distinct reproductive tissues and structures. Mating requires the production of haploid gametes from both male structures and female structures. The union of haploid gametes produces diploid progeny. In monoecious plant species, including Pisum sativum, which Mendel worked with, male and female reproductive tissues are present in each plant, and self-fertilization is the common mode of reproduction, although fertilization involving pollen from one plant fertilizing the flower of another also occurs. In sexually reproducing animals, specialized germ-line cells undergo meiosis to produce haploid gametes, or reproductive cells. Female gametes are produced by the ovary in female animals or by the ovule in plants. Male germ-line cells are located in testes in animals, where they produce sperm. In the anthers of flowering plants, pollen containing two sperm cells is produced. These descriptions are broadly true for most plants and animals, but there are many exceptions, including the observation of asexual reproduction in several species of fish, rotifers (small aquatic organisms), and salamanders. In addition, male ants, bees, and wasps have haploid somatic cells, and their processes of gamete production are distinctive.

Chapter (2.6/3.5) Advanced Pedigree Analysis ** CONTINUED ** Analyze pedigrees and determine the mechanism of inheritance of the disease or trait being followed. Recessive Inheritance of disease/trait: * What does a heterozygous genotype give you in terms of phenotype? * Compare the mutant allele with the wild-type allele, is it dominant or recessive over the WT? * If d is the disease allele and D is normal, then what genotypes are affected? * What are typical patterns of recessive traits/diseases when looking at a pedigree? Know what this means. * Can an affected kid be born to unaffected parents? Explain why or why not. Autosomal Recessive Inheritance * What are typical patterns of recessive traits/diseases when looking at a pedigree? Know what this means. * Can an affected kid be born to unaffected parents? Explain why or why not. X-linked Recessive Inheritance * Where is the gene containing the allele(s) in question? * What is the dominant wild-type allele? * What is the recessive disease-causing allele? * What is the genotype of affected males? * What is the genotype of affected females? * What is noticeable about the proportion of M/F infected?

Recessive Inheritance * Heterozygous genotype -> normal phenotype. * The mutant allele is RECESSIVE to the wild-type allele. * If d is the disease allele and D is normal, then only dd genotypes are affected. * Typically NOT seen in every generation ("skips generations") -> not found in every generation, doesn't mean found in every other generation. * Affected kid CAN be born to unaffected parents. Common Characteristics of Autosomal Recessive Inheritance * Typically NOT seen in every generation. * Affected kid CAN be born to unaffected parents. X-linked Recessive Inheritance * Gene in question is on the X-chromosome. - D is dominant wild-type allele. - d is recessive disease-causing allele. * Affected males are XdY (only need one bad X to show disease). * Affected females are XdXd. * Males are affected much more often than females.

Chapter (2.6/3.5) Advanced Pedigree Analysis Know about (and understand the logic behind) the "rare disease clue". * What is the rare disease clue (what is its argument)? * What is a rare disease/trait? * If a problem tells you the disease is rare, what should you assume when considering: - Hypothesis of recessiveness - Hypothesis of dominance

The "rare disease clue In a given pedigree, there will be people whose parents are not shown, either because they are in the oldest generation of the pedigree, because they married into the family, or because the family history is imperfect. If the problem "tells" you the disease is rare, then... * When considering the hypothesis of recessiveness: you can assume that multiple carriers did not marry into the family. * When considering the hypothesis of dominance: you can assume that the founder is a heterozygote. * The rare disease clue is an argument about probability, not logical certainty. * For this class, a rare disease or trait is one with an incidence of 1% or less. * For word problems in this class, you will be told if a disease or trait is rare in the text of the problem, or you will be able to figure it out from the incidence. * MCAT, DAT, PCAT, GRE, etc. don't always tell you that the disease is rare, and you may need to assume that it is rare to solve the problem.

Quiz: Pedigree #2 Now assume that the pedigree shown in question 1 shows the inheritance of a rare genetic disease: The pedigree shows the inheritance of a rare disease or trait in three generations of a family The mechanism of inheritance of this disease is most likely to be... The disease is most likely autosomal dominant The disease is most likely autosomal recessive The disease is equally likely to be either autosomal dominant or autosomal recessive, but cannot be x-linked Cannot be determined from the information given

The disease is most likely autosomal dominant First, we can rule out x-linked dominant and x-linked recessive using Rules 3 and 4 from certain parent-child trios from the pedigree. We can't rule out AD from Rule 1. Strictly speaking, we also can't absolutely rule out AR by Rule 2. However, the question asks about which is the most likely. Thus, using the RARE DISEASE CLUE, we assume that multiple carriers did not marry into the family. Thus AR is very unlikely, since it would require this.

Chapter 3: Mitosis, Meiosis and Sex Determination (Not a learning objective, but should know) TEXTBOOK EXPLANATION * Know the eukaryotic cell cycle. * What principal phases is the cell cycle divided into? - What occurs during each of these phases. - What phases are each of these principle phases divided into? * What happens during these smaller phases?

The eukaryotic cell cycle is divided into two principal phases—M phase, a short segment of the cell cycle during which cells divide, and interphase, the longer period between one M phase and the next (Figure 3.1a). Interphase consists of three successive stages, G1, S, and G2. During interphase the cell expresses genetic information, it replicates its chromosomes, and it prepares for entry into M phase. M phase is divided into multiple substages that correspond to the progress of the cell during its division. When viewed under a light microscope, somatic cells in interphase may appear rather placid, but their outward appearance gives little indication of the complex activity taking place inside. Gene expression occurs continuously throughout the cell cycle, but during the G1 (or Gap 1) phase of interphase, it is particularly high (Figure 3.1b). Cells of different types vary in how many genes they express, in how they function in the body, and in how they interact with other cells. Consequently, the duration of G1 varies among different types of cells in the body. Some types of cells are rapidly dividing and spend only a short time, perhaps as little as a few hours, in G1. Other cells linger in G1 for periods of days, weeks, or more.

Chapter 3: Mitosis, Meiosis and Sex Determination TEXTBOOK EXPLANATION CONTINUED Interpret independent assortment in terms of meiotic chromosome behavior. * Know the steps in which things occur, and what it means. * Know the end result and what it shows.

The mechanistic basis of Mendel's law of independent assortment is illustrated in Figure 3.16 for a GgRr dihybrid pea plant. Recall that this law of heredity predicts that a dihybrid organism should produce four genetically different gametes at a frequency of one-quarter (25%) each. Once again, S phase creates two identical sister chromatids for each chromosome. In metaphase I, however, two equally likely arrangements of the two homologous pairs, shown in Figure 3.15, can occur. In each arrangement, the homologous chromosomes are on opposite sides of the metaphase plate. Obviously, when the cell undergoes meiosis, only one or the other of these alternative arrangements will occur; thus, each cell undergoing metaphase I of meiosis will have either "arrangement I" or "arrangement II." Over a large number of meiotic divisions, arrangement I and arrangement II are equally frequent. Arrangement I in Figure 3.15 has chromosomes carrying dominant alleles on one side of the metaphase plate, and chromosomes carrying recessives on the opposite side. Arrangement II has a dominant-bearing and a recessive-bearing chromosome on each side of the metaphase plate. The first meiotic division segregates G from g and R from r to create the haploid products of meiosis I division. If we now follow each haploid product of meiosis I through the meiosis II division, we see that the four gametes produced by arrangement I have the genotypes GR and gr in equal frequency. In contrast, the four gametes produced by arrangement II have the genotypes Gr and gR in equal frequency. Taking both possible arrangements of these homologous chromosomes at metaphase I into account, eight gametes are generated with four equally frequent genotypes. The four possible gamete genotypes—GR, Gr, gR, and gr—are produced in a frequency of 25% each as predicted by the law of independent assortment.

Chapter (2.6/3.5) Advanced Pedigree Analysis Genetic Analysis 2.3 Problem The pedigree provided here shows a woman (II-3) and a man (II-4) who each have a sibling with an autosomal recessive condition1 (II-2 and II-5). They seek to determine the chance that their first child (III-1) will have the condition. The child has not yet been conceived. Neither the man nor the woman has the condition, nor do the parents of either of them (see generation I). Using only the genetic information given, perform the following tasks: a. Using D to represent the dominant allele and d to represent the recessive allele, assign genotypes to all members of the pedigree. If a complete genotype (showing both alleles) cannot be given, provide the genotype information that is known. Explain your reasoning. b. Calculate the chance that child III-1 will have the recessive condition. Show your work.

The recessive condition occurs when an individual has the genotype dd. Since the parental pairs in generation I have each produced a child with the recessive condition, and since none of those four parents has the recessive condition (i.e., they all have the dominant phenotype), the members of each parental pair must have the heterozygous Dd genotype. All members of generation II are produced from crosses that are Dd × Dd. II-2 and II-5 have the recessive phenotype and must have the genotype dd. All other family members in generation II are either DD or Dd. Because their genotypes are only partly known, the genotypes of II-1, II-3, II-4, and II-6 can be written as D-. Both II-3 and II-4 have the dominant phenotype, so neither can have the dd genotype. The Punnett square shows that for their possible D- genotypes, each of them has a two in three chance of having the Dd genotype and a one in three chance of having the DD genotype. a) The pedigree shown here includes the complete and partial genotypes assigned to members of generations I and II. To have the recessive phenotype, the child of II-3 and II-4 must have the dd genotype. Each of its parents must be Dd for this to occur. If we look at the Punnett square and consider only the genotypes that meet the condition of producing a dominant phenotype, we see that each parent has a (2/3) chance of being heterozygous. The probability that both are heterozygous is (2/3)(2/3)=(4/9). b) The chance that both II-3 and II-4 are heterozygous is (4/9). The cross of these heterozygotes (Dd × Dd) would have a (1/4) chance of producing a child that is d. This probability is (4/9)(1/4)=(4/36)=(1/9). In other words, given the available information, there is a one in nine chance that this couple will have a child with the recessive phenotype.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN Allelic series: Molecular Basis of the C-gene allelic series * Know the characteristics of the tyrosinase enzyme produced by the c^h allele. - What kind of allele is this? - What phenotype does this allele produce? * Know all the genotypes that produce this phenotype. * Define temperature-sensitive alleles. - Understand the siamese cat coat example. * What occurs at warmer extremities of the body in terms of tyrosinase activity of the c^h allele? What phenotype does this lead to in these portions of the body? * What occurs at cooler extremities of the body in terms of tyrosinase activity of the c^h allele? What phenotype does this lead to in these portions of the body? * Know the characteristics of the tyrosinase enzyme produced by the c allele. - What kind of allele is it? - Describe homozygotes for this allele * Phenotype * Biochemical processes. What does this lead to?

The tyrosinase enzyme produced by the c^h hypomorphic (Himalayan) allele is unstable and is inactivated at a temperature very near the normal body temperature of most mammals. This type of gene product is an example of a temperature-sensitive allele. Cats with the Siamese coat-color pattern are familiar examples of the action of this temperature-sensitive allele. The parts of cats that are farthest away from the core of the body (the paws, ears, tail, and tip of the nose) at most times tend to be slightly cooler than the trunk. At these cooler extremities, the temperature-sensitive tyrosinase produced by the c^h allele remains active, producing pigment in the hairs there. However, in the warmer central portion of the body, the slightly higher temperature is enough to cause the tyrosinase produced by the c^h allele to denature, or unravel. This inactivates the enzyme and leads to an absence of pigment in the central portion of the body. Animals that are c^hc^h or c^hc have the Himalayan phenotype. The final allele in the series, c, is a null allele that does not produce functional tyrosinase. Homozygotes for this allele are unable to initiate the catabolism of tyrosine. This leads to an absence of melanin and produces the condition known as albinism.

Chapter 3: Mitosis, Meiosis and Sex Determination TEXTBOOK EXPLANATION (Not a learning objective, but should know) Know Y-linked inheritance, what causes it, what it means, and results of Y-linked inheritance.

Whereas females receive one copy of X-linked alleles from each parent, males receive their X-linked alleles from their mother and their Y chromosome from their father. This means that Y-linked inheritance, the inheritance of genes on the Y chromosome, is an exclusively patrilineal (father to son) pattern of hereditary transmission. The key to Y-linked inheritance is that the Y chromosome is found only in males. This means Y-linked genes are transmitted in a male-to-male pattern. In mammals, fewer than 50 genes are found on the Y chromosome; and like SRY, those genes are likely to play a role in male sex determination or development. The genes on the human Y chromosome do not have counterparts on the X chromosome, although the DNA sequences in the pseudoautosomal regions are shared by the X and Y chromosomes to facilitate synapsis of the chromosomes during meiosis. There is crossing over between the pseudoautosomal regions, but this does not involve expressed genes. Females never carry Y chromosomes, so from an evolutionary perspective it makes sense that the genes carried on a Y chromosome should be male-specific, having either to do with male sex determination or reproduction. Indeed, the most recent genomic evidence suggests that the mammalian Y chromosome has rapidly evolved over the past 300 million to 350 million years, undergoing multiple changes in structure but preserving a handful of genes that are essential to male fertility and survival.

Chapter 3: Mitosis, Meiosis and Sex Determination TEXTBOOK EXPLANATION Explain the inheritance of X-linked traits. * What are the inheritance patterns we see with respect to X-linked genes? - What does this mean for each pattern? * Explain the transfer of genes on the Y chromosome. * What three features of the inheritance patterns we see with respect to X-linked genes make them distinct from inheritance of autosomal traits? - Know what this means, be able to explain it. * Know Y-linked inheritance and what it means.

With respect to X-linked genes in animal species, two inheritance patterns are common. X-linked recessive inheritance: with this mode of inheritance, females homozygous for the recessive allele and hemizygous males whose X chromosome carries the recessive allele display the recessive phenotype. The alternative mode of X-linked transmission is X-linked dominant inheritance, in which heterozygous females and males hemizygous for the dominant allele express the dominant phenotype. Genes on the Y chromosome are exclusively transferred patrilineally (i.e., from father to son), since the Y chromosome is male-specific. Three features of X-linked dominant and X-linked recessive inheritance make them distinct from inheritance of autosomal traits. First, although autosomal dominant and recessive gene expression are generally the same in males and females, the terms recessive and dominant for X-linked gene transmission refer specifically to the expression of traits in females. For X-linked alleles, females can be homozygous or heterozygous, but males are hemizygous and express the allele on their X chromosome regardless of the hereditary pattern in females. Second, the probability of transmission of X-linked alleles to offspring is not the same for the two sexes as it is for autosomal alleles. Female X-linked transmission is identical to autosomal transmission, but hemizygous males always transmit their X chromosome to female offspring and their Y chromosome to male offspring. Lastly, whereas females receive one copy of X-linked alleles from each parent, males receive their X-linked alleles from their mother and their Y chromosome from their father. This means that Y-linked inheritance, the inheritance of genes on the Y chromosome, is an exclusively patrilineal (father to son) pattern of hereditary transmission.

Chapter 3: Meiosis, Mitosis and Sex Determination Based on the cartoon, we have four generations of fathers, all losers. None of the mothers were losers. None of the many daughters of the loser fathers were losers. The most likely mode of inheritance of the "loser" trait is: A. Autosomal dominant B. Autosomal recessive C. X-linked dominant D. X-linked recessive E. Y-linked

X-linked recessive

Chapter 10: Chromosome Aberrations & Disease TEXTBOOK EXPLANATION Know the four major types of chromosome rearrangements. "Skim": * How does gene dosage relate to the growth and development of animals? * What do mutations resulting in loss or gain of whole chromosomes or chromosome segments have to potential to produce? * What does chromosome breakage lead to? * What events lead to the loss or gain of chromosomal segments? * When a chromosome breaks, what has happened? * Describe the structure of the chromosome ends at the break point of a broken chromosome. What can these ends do after breaking? * What can chromosome breakage result in? - Define this. * What does the extent of this determine? - In a case where the extent is large, how is an individual affected? * Define reciprocal recombination. - What is another name for this? What occurs during this process? * What potential dangers are possible from this? - What does this lead to?

"Skim", don't memorize, just understand: In animals, as mentioned earlier, the proper balance of gene dosage is critical for normal growth and development. For this reason, mutations that result in the loss or gain of whole chromosomes or chromosome segments have the potential to produce severe abnormalities. In this section, we examine changes to chromosome structure that occur by chromosome breakage and other events that lead to the loss or gain of chromosomal segments by the partial duplication or the partial deletion of a chromosome. When a chromosome breaks, both strands of DNA are severed at a location called a chromosome break point. The ends at the breakpoint of a broken chromosome retain their chromatin structure, and they can adhere to one another, to other truncated chromosome ends, or to the ends of intact chromosomes. Chromosome breakage can result in partial chromosome deletion by the loss of a portion of a chromosome. The size of the deletion and the specific genes deleted are significant factors in the degree of ensuing phenotypic abnormality. In these larger deletions, many genes are affected, and the likelihood of substantial phenotypic consequences is very high. The process of reciprocal recombination (crossing over) achieves the recombination of alleles on homologous chromosomes without causing a gain or loss of chromosomal material that would result in mutation (see Sections 5.2 and 12.6). Occasionally, however, crossing over between homologs is inaccurate, resulting in chromosome mutations that are due to unequal crossover. These mutations result in the partial duplication and partial deletion of chromosome segments on the resulting recombinant chromosomes.

Chapter 3: Mitosis, Meiosis and Sex Determination Identify each of the things the arrows are pointing towards: Question 1: A. Different Genes B. Same genes but different alleles C. Same genes and same alleles Question 2: A. Different Genes B. Same genes but different alleles C. Same genes and same alleles Question 3: A. Different Genes B. Same genes but different alleles C. Same genes and same alleles

#1: A (different chromosomes) #2: B (paternal vs. maternal homologous) #3: C (sister chromatids from replication)

Chapter (2.6/3.5) Advanced Pedigree Analysis Calculate the probability that the proband has a disease or disease-causing allele. Practice Problem: Assume your grandfather is a carrier (heterozygote) for a rare recessive, disease-causing allele of a given gene. What's the chance that you are also a carrier of this allele? A. (2/3) B. (1/2) C. (1/4) D. (1/6) E. (1/8)

(D) is a carrier for a rare recessive allele. What's the chance that (H) is also a carrier? C. (1/4)

Problem Set #2: Pedigree Problem Set Making all the above assumptions about the pedigre... If #7 mates with an unrelated, unaffected individual, what are the approx. chances that his first child will have the condition? If #8 mates with an unrelated, unaffected individual, what are the approx. chances that her first child will have the condition?

* 0% Mechanism: Autosomal dominant, from last problem. dd x dd = dd (0%) * 50% Dd x dd = Dd, dd (50%)

Chapter 4: Gene Interactions Practice Problem: What type of mutation is Huntington Disease?

* Autosomal dominant. Wild-type HUNTINGTIN protein: PIC * In mutant protein, the poly-Q track expands from 23 Q's in wild-type to 100's of Q's in patients. * The extended glutamine tracts are thought to promote the formation of toxic aggregates, leading to cell death. An example of a gain-of-function mutation. - The mutant allele has gained a disease-causing function (the ability to form toxic aggregates) that the wild-type doesn't have. - Neomorphic

Problem Set #1: Pedigrees Examine the accompanying pedigree and determine the inheritance pattern. Assume that the traits are not the result of a de novo mutation and that penetrance is 100%. A. This trait could only be autosomal recessive B. This trait could only be autosomal dominant C. This trait could only be X-linked recessive D. This trait could only be X-linked dominant E. This trait could be a, b, c, or d

* E. This trait could be a, b, c, or d. * Can be dominant. * Autosomal dominant: Rr x rr = Rr, rr * X-linked Dominant: XRY x XrXr = XRXr, XRXr, XrY, XrY Girls have the trait, boys don't. * Recessive? * Autosomal recessive: rr x Rr = Rr, rr * X-linked recessive: XrY x XRXr = XRXr, XrXr, XRY, XrY Half girls have the trait, half do not. Half boys half the trait, half do not. WORKS! (half and half is a probability, it doesn't mean this is exactly what will be seen in a pedigree). Therefore, you can't reject this inheritance mechanism since it shows that girls have the trait, and boys do not like as seen in the pedigree. The proportions don't have to be exact. OR XrY x XRXR = XRXr, XRY None of the kids will have the disease.

Chapter 10: Chromosome Aberrations & Disease Describe non-disjunction. * What is expected when non-disjunction occurs in exclusively __________? - Meiosis I - Meiosis II * Define nondisjunction. - During what process does this occur? - What gametes does nondisjunction in meiosis I produce? - What gametes does nondisjunction in meiosis II produce? - What characteristics do zygotes of nondisjunction have? * How does nondisjunction cause disease in newborns? What increases the risk of nondisjunction? - What is down syndrome caused by?

* Non-disjunction: Chromosomes or chromatids fail to disjoin properly. - Only Meiosis I non-disjunction: Both homologous pairs of chromosomes end up in the same spermacyte I cell (spermacyte I cell includes an X homologous pair and a Y homologous pair). - Only Meiosis II non-disjunction: The sperm cell gamete includes both X/Y sister chromatids -> Sperm carries two X-chromosomes/sperm carries two Y-chromosomes. Nondisjunction * Chromosomes or chromatids fail to "disjoin" during meiosis. * Nondisjunction in meiosis I produces gametes with a pair of homologous chromosomes (e.g. XY). * Nondisjunction in meiosis II produces gametes with a pair of sister chromatids (e.g. XX or YY). * Fertilization produces a zygotę with monsomy or trisomy. Risk increases with Age: * The risk of pregnancy with a chromosome disorder increases with the age of the pregnant mother, caused by a nondisjunction event. * The risk of nondisjunction increases dramatically with maternal age. - Down syndrome is mostly caused by the nondisjunction of chromosome 21.

Problem Set #2: Pedigree Problem Set 35-38. The pedigree shows the inheritance of a very rare birth defect syndrome in a family (short answer): 35. Given the pedigree, which mechanism(s) of inheritance is/are reasonably likely? (You may choose more than one - write down the appropriate letter(s) to indicate your answer.) A. Autosomal recessive B. Autosomal dominant C. X-linked recessive D. X-linked dominant E. Cannot be determined from the information given 36. In alphabetical order, list all the individuals in generations III and IV who MUST be heterozygous for the disease-causing allele in order for the pedigree to make sense. 37. The chance that individual N is a carrier of the disease causing allele is ____. (You may answer as a fraction or a percentage). 38. Assume that males II-E and III-I and III-L in the pedigree are also affected by the birth defect. In that case, which mechanism(s) of inheritance is/are reasonably likely? (You may choose more than one - writedown the appropriate letter(s) to indicate your answer.) A. Autosomal recessive B. Autosomal dominant C. X-linked recessive D. X-linked dominant

* Rare 35. A. Autosomal recessive. * Not dominant, unaffected parents made affected children. * Autosomal recessive: works I: RR x Rr = Rr, RR II: RR x Rr = Rr, RR Rr x RR = Rr, RR RR x RR = RR III: Rr x Rr = rr, RR, Rr, Rr * X-linked recessive: doesn't work I: XRY x XRXr = XRXR, XRXr, XrY, XRY II: XRXr x XRY =XRXR, XRXr, XrY, XRY XRXr x XRY =XRXR, XRXr, XrY, XRY XRY x XRXR = XRXR, XRY III: XRY x XRXr = XRXR, XRXr, XrY, XRY DOESN'T WORK, NO AFFECTED DAUGHTER. Dad would pass his XR, making the daughter XRXr or XRXR, so she's unaffected. A daughter could only be affected if the dad was affected. 36. I,J Only I and J NEED to be heterozygous for the pedigree to make sense. N and P are both dominant, and it is likely that one or both of them is heterozygous, but they don't NEED to be for the pedigree to make sense. 37. (2/3) * We know N is dominant, because it's not affected. Out of the three dominant genotypes created as a result of the Rr x Rr cross, (2/3) were Rr (carrier). 38. C. X-linked recessive RARE: Recessive, carriers did not marry into the family Still has to be recessive, unaffected parents have affected child. * Autosomal recessive: According to the rare rule: I: Rr x Rr = rr, RR, Rr, Rr II: Rr x RR = Rr, RR (DOESN'T WORK, need rr) rr x RR = Rr Rr x RR = RR, Rr (DOESN'T WORK, need rr) III: Rr x Rr = RR, Rr, Rr rr (DOESN'T WORK, needed rr x Rr) * X- linked recessive: works I: XRY x XRXr = XRXR, XRXr, XRY, XrY II: XRXr x XRY = XRXR, XRXr, XRY, XrY XrY x XRXR = XRXr, XRY XRXr x XRY = XRXR, XRXr, XRY, XrY III: XrY x XRXr = XRXr, XrXr, XRY, XrY

Chapter 10: Chromosome Aberrations & Disease *** CONTINUED *** Compare and contrast non-disjunction of the sex chromosomes in meiosis I vs. II. * Know how to assign a gamete a karyotype. - Know the karyotypes of the resultant gametes of Meiosis II non-disjunction of the Y homologous pair. * What are the resulting zygotes if these sperm gametes fertilize a euploid egg? Know how to predict this!

* The karyotypes of the resultant gametes include: - X gamete: Euploid 23, X * 23 total chromosomes, including 1 X sex chromosome. - 0 gamete: Aneuploid 22,0 * 22 total chromosomes, including no sex chromosomes. - YY gamete: Aneuploid 24, YY * 24 total chromosomes, including 2 Y sex chromosomes. If these sperm fertilize a euploid egg, the resulting zygote is.... * Euploid 23,X -> Euploid female 46, XX (normal female) * Aneuploid 22,0 -> Aneuploid female 45,X (one sex chrm) * Aneuploid 24,YY -> Aneuploid male 47,XYY (extra Y chrm)

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN AGAIN AGAIN AGAIN Describe variations from classical Mendelian genetics, including incomplete dominance, co-dominance, multiple alleles/allelic series, and lethal alleles. Lethal alleles: Lethal alleles in animals * Define lethal alleles (what do they cause). - What genotype is needed in order for this to occur? * How are lethal alleles in animals usually detected? What is this due to? * Understand lethal mutation in mice example: - Was the mutation dominant or recessive? * What kind of mutation is this? - What phenotype did the mutation cause? * Know how this compares to the WT phenotype. - Know the designation for the wild-type allele. What does its normal activity lead to in terms of the phenotype? - Know the designation for the mutant allele. What does its activity lead to in terms of the phenotype? * What kind of allele is it? * Is it dominant or recessive?

*** Don't memorize everything, he just said to understand figure, so just know enough from this to understand the figure *** Certain single-gene mutations are so detrimental that they cause death early in life or terminate gestational development. These life-ending mutations affect genes whose products are essential to life. Homozygosity for mutation of these essential genes is lethal, and the mutations are identified as lethal alleles. Lethal alleles in animals are usually detected by a distortion in segregation ratios due to failure to produce the affected category of progeny. The first case of a lethal allele was identified in 1905 by Lucien Cuénot, who studied a lethal mutation in mice carrying a dominant mutation for yellow coat color. In mice, wild-type coat color is a brown color called "agouti" (a-GOO-tee), produced by the presence of yellow and black pigments in each hair shaft. The Agouti gene is one of the pigment-producing genes found in mammals with furry coats. It produces a yellow pigment called pheomelanin that is found in the hairs of mammalian coats. An independently assorting gene produces the black pigment that is also visible in the hair shafts. The wild-type allele for agouti coat color is designated A, and its normal activity leads to the production of a moderate amount of yellow pigment. The mutant allele AY, designated , is a hypermorphic allele. It is a dominant gain-of-function mutation that produces substantially more yellow pigment than does the wild-type allele.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN AGAIN AGAIN AGAIN AGAIN Describe variations from classical Mendelian genetics, including incomplete dominance, co-dominance, multiple alleles/allelic series, and lethal alleles. Lethal alleles: Lethal alleles in animals Understand lethal mutation in mice example: * What genotypes cannot be produced? - What does this mean in terms of the genotype of mutant mice? - What does this mean in terms of the lethal genotype? * At what point of life is this genotype lethal? - What observations about the genetics of the mutant allele can be made? * What ratio will the cross between a WT mouse and a mutant mouse result in among progeny? * What do crosses between two mutant mice produce? - What is the ratio of these crosses? How does this compare to the ratio that is anticipated? * What is the genetic interpretation of this result in terms of allele segregation? - What is the lethal nature of the mutant allele (is it dominant or recessive)? * What does this prevent? How does this affect the ratio?

*** Don't memorize everything, he just said to understand figure, so just know enough from this to understand the figure *** The AY mutation is dominant, but true-breeding yellow mice cannot be produced. From a genetic perspective, this means that all mice with yellow coat color are heterozygous (AAY) and that the AYAY genotype is lethal in embryonic development. From this information, two important observations about the genetics of the yellow allele can be made. First, mating an agouti mouse and a yellow mouse will always result in a 1:1 ratio of agouti and yellow among progeny (Figure 4.9a). Second, crosses between two yellow mice (both of which are necessarily heterozygous) produce evidence of the recessive lethal nature of the AY allele (Figure 4.9b). The outcome of these crosses is a 2:1 ratio of yellow to agouti, rather than the 3:1 ratio that is anticipated when heterozygotes expressing a dominant allele are crossed. The genetic interpretation of this observation is that alleles of heterozygous yellow mice segregate normally in gamete formation and unite at random to produce a 1:2:1 ratio at conception but that AYAY zygotes do not survive gestation. Recessive lethality of AY prevents embryonic development of homozygotes, eliminating that class among progeny and resulting in the 2:1 ratio seen among progeny of heterozygous parents.

Chapter (2.6/3.5) Advanced Pedigree Analysis (Not a learning objective, but should know) What is Huntingtons Disease? Answer this by knowing: * What mechanism of inheritance does it use? * Does it show complete entrance? * Is it rare, or common? * What does the onset look like for this disease? * What does the mutant protein have that the wild-type lacks? Describe it and how it differs from normal proteins. - What is this difference thought to cause? - What kind of mutation is this an example of? Know why.

*** ONLY KNOW THINGS IN BOLD *** Huntington's Disease Symptoms/Phenotype: * Slowly progressive, hereditary brain disease that causes changes in movement, thinking and behavior. * Neuronal Degeneration. * Death. Genetics: * Frequency = 1/10,000 (European origin) = rare. * Autosomal Dominant - Complete Penetrance. * Late Onset: 35-50 years, 10-20 year course. * Mutation caused by trinucleotide repeat expansion probably as a result of replication slippage. * No effective treatment or cure. Wild-Type HUNTINGTIN Protein (new slide, maybe know all info) * In mutant protein, the poly-Q track expands from 23 Q's in wild-type to 100s of Q's in patients. * The extended glutamine tracts are thought to promote the formation of toxic aggregates, leading to cell death. * An example of a gain-of-function mutation: - The mutant allele has gained a disease-causing function (the ability to from toxic aggregates) that the wild-type doesn't have)

Problem Set #2: Pedigree Problem Set Questions 1-4: refer to the following pedigree, which shows the inheritance of a non-lethal birth defect caused by a single-gene disorder in a family. 1. The defect is most likely caused by: A. An autosomal gene for which the defective allele is recessive to the normal allele B. An X-linked gene for which the defective allele is recessive to the normal allele C. An autosomal gene for which the defective allele is dominant to the normal allele D. An X-linked gene for which the defective allele is dominant to the normal allele E. Cannot be determined from the pedigree 2. James and Jane (individuals 6 and 7 in the pedigree) seek genetic counseling because of the history of the birth defect in their families. Based on the pedigree analysis, what is the percentage chance that their first child will be affected by the birth defect? A. 0% B. 25% C. 33.5% D. 50% E. 100%

1. A. An autosomal gene for which the defective allele is recessive to the normal allele Has to be recessive, because unaffected parents have affected child. * Autosomal recessive: works RR x rr = Rr (1,2,5,6) Rr x Rr = Rr, RR, rr (3,4,7,8,9,10) * X-linked recessive: doesn't work. XrY x XRXR = XRXr, XRY - consistent. XRY x XRXr = XRXR, XRXr, XrY, XRY - not consistent (no daughter affected) 2. D. 50% James' side: rr x Rr = Rr, rr OR rr x RR = Rr * Either way, James must be Rr. Jane's side: Rr x Rr = RR, Rr, rr * Jane must be rr (obvious) Rr x rr = Rr (50%), rr (50%)

Problem Set #1: Sex Linkage and Sex Determination For questions 1-2, imagine a world where there are three human sexes, males, females, and bivalves. Bivalveshave two sex chromosomes: X and Z. Bivalves can mate with either males or females, but conceptions with the YZ genotype are early embryonic lethal. The Z chromosome has no genes in common with either X or Y, but (just as Y does) it pairs with X during meiosis 1. Which couple could have kids of any of the three sexes? A. Male-Bivalve B. Female-Bivalve C. Male-Female D. A and B E. Neither A, B or C 2. After many generations of random mating, the population would be expected to consist of: A. An equal number of all three sexes B. More bivalves than any other sex C. More males than any other sex D. More females than any other sex E. Only bivalves

1. A. Male-Bivalve XY-XZ = XY XZ XX 2. D. More females than any other sex - All sexes have X's to give away, adding up the X chromosome count, making females most common.

Chapter 3: Mitosis, Meiosis and Sex Determination Features of X-Linked Recessive Inheritance

1. Many more males than females have the trait due to hemizygosity. 2. A recessive male mated to a homozygous dominant female produces all offspring with the dominant phenotype, and all female offspring are carriers. EX: "Affected male mates with normal female" Punnett square. * We also did "Female carrier mates with normal male" 3. Matings of recessive males with carrier females give half dominant and half recessive offspring of both sexes. EX: Affected male mates with female carrier - Draw this. 4. Matings of homozygous recessive females with dominant males produce all dominant (carrier) female offspring and all recessive male offspring. EX: Affected female mates with normal male - Draw this.

Problem Set #2: Pedigree Problem Set 11-19. The pedigree shows the inheritance of trait in four generations of a family. 11. Based on the pedigree, the mode of inheritance of this trait could be... A. Autosomal recessive B. X-linked recessive C. Autosomal dominant D. X-linked dominant E. A or B 12. Assume #7 & #8's next daughter is unaffected. Based on this new information and the pedigree, the mode of inheritance of this disease could be... A. Autosomal recessive B. X-linked recessive C. Autosomal dominant D. X-linked dominant E. A or B 13. Inbreeding occurred in this family when... A. #1 & #2 mated B. #3 & #4 mated C. #5 & #6 mated D. #7 & #8 mated E. No inbreeding occurred in this family

11. E. A or B * Can't be dominant because two unaffected parents had an affected child. * Autosomal recessive: * rr x R_ = Rr (#4, #5) * Rr x Rr = rr (#7) * Rr x R_ = Rr (#8) * rr x Rr = Rr, rr (#9,10,11,12) * X-linked recessive: * XrY x XRXR = XRXr, XRY Unaffected daughters (#4,5) * XRXr x XRY = XRXR, XRXr, XRY, XrY Half sons affected (#7) * XRXr x XRY = XRXR, XRXr, XRY, XrY No daughters affected (#8) * XrY x XRXR = XRXr, XRY No daughters or sons affected. (This cross isn't possible) OR XrY x XRXr = XRXr, XrXr, XrY, XRY Half daughters affected, half sons affected (This cross is possible) (#9,10,11,12) 12. E. A or B * Autosomal recessive: rr x Rr = Rr, rr (#9,10,11,12) Half affected, half not. (Still works) * X-linked recessive: XrY x XRXr = XRXr, XrXr, XrY, XRY Half daughters affected, half sons affected. (Still works) 13. D. #7 & #8 mated

Problem Set #2: Pedigree Problem Set 11-19. The pedigree shows the inheritance of trait in four generations of a family. For questions 15-16, assume that the pedigree shows the inheritance of the rare x-linked genetic disease icthyosis, which is characterized by flaky skin. 15. As it turns out, #8 knows something about genetics. Prior to mating with #7, #8 estimated her chances of being a carrier for icthyosis at ____. After mating with #7 and observing the phenotype of her kids, she re-evaluated her chances of being a carrier and decided that it was ____. A. 25%, 50% B. 50%, 67% C. 25%, 100% D. 50%, 100% E. 12.5%, 100%

15. D. 50%, 100% * Rare X-linked recessive If the problem "tells" you the disease is rare, then when considering the hypothesis of recessiveness, you can assume that multiple carriers did not marry into the family. - We only had to assume for #2 since the other "married in" individuals were dominant males (XRY). * "#8 estimated her chance of being a carrier" -> She's heterozygous for a recessive allele. X-linked recessive: * XrY x XRXR = XRXr, XRY Unaffected daughters (#4,5) * XRXr x XRY = XRXR, XRXr, XRY, XrY Half sons affected (#7) * XRXr x XRY = XRXR, XRXr, XRY, XrY No daughters affected (#8) * XrY x XRXR = XRXr, XRY No daughters or sons affected. (This cross isn't possible) OR XrY x XRXr = XRXr, XrXr, XrY, XRY Half daughters affected, half sons affected (This cross is possible) (#9,10,11,12) Before having children, she had no way of knowing if she was XRXR or XRXr, since the cross that made her was: * XRXr x XRY = XRXR, XRXr, XRY, XrY There are two genotype options that she could have, with equal possibility. Therefore, she had a 50% probability of being a carrier. After seeing her children's phenotypes, she could see that she is 100% a carrier since if she was XRXR, her cross would've made zero children with the disease.

Problem Set #2: Pedigree Problem Set 11-19. The pedigree shows the inheritance of trait in four generations of a family. 17. Great-Grandma (#2) develops a form of inherited dementia that is autosomal dominant with a very late age of onset. What is the risk that #8 inherited the dominant dementia-causing allele from #2. Assume the inherited dementia is rare. (Note that this dementia is a different disease than the trait/disease shown in the pedigree, but it is running in the same family.) A. 0% B. 100% C. 50% D. 25% E. 12.5% 18. Estimate the risk that #9 inherited the dominant dementia-causing allele from #2. A. 0% B. 100% C. 50% D. 25% E. 12.5% 19. Determine the exact probability that #9 inherited at least one copy of the dominant dementia-causing allele from #2_______ ß- write answer here Note: This is an advanced question that is too hard to be put on a final exam, but you may try it if you wish. You need to calculate the probability of the following outcomes: i) 7 got the allele and 8 did not ii) 8 got the allele and 7 did not iii) both 7 and 8 got the allele

17. D. 25% If the problem "tells" you the disease is rare, then... when considering the hypothesis of dominance, you can assume that the founder is a heterozygote. Late age-of-onset problems: * Individuals with the genotype may not yet be expressing the phenotype. * We assumed great-grandpa wasn't affected because it wasn't given, and the disease is rare. We assumed great-grandma was heterozygous since the disease is rare (rare rule for dominance). Dd x dd = (1/2) Dd, (1/2) dd. Therefore, the probability of #5 being Dd is (1/2). When #5 mates with #6, we assume #6 is dd since the disease is rare. When they mate: Dd x dd = (1/2) dd, (1/2) Dd. Therefore, the possibility of progeny of this cross being Dd is (1/2). The possibility of #5 being Dd and #8 being Dd is (1/2)(1/2)=(1/4) = 25%. 18. D. 25% * Insert . Now, lets look at the other side of the cross... As we said earlier, #1 x #2 = (1/2) Dd, (1/2) dd. Therefore, the probability of #4 being Dd is (1/2). #4 then mates with #3, which is assumed to be dd (rare), giving a (1/2) probability that #7 will be Dd (Dd x dd = (1/2) Dd, (1/2) dd). Then, when #7 Dd mates with #8 Dd, the probability of their offspring (including #9) being Dd is (2/4)=(1/2). When we consider the possibility of #9 being infected, we need to consider both routes: 2->5->8->9 and 2->4->7->9. When we combine these possibilities, we get: (1/2)(1/2)(1/2)x(1/2)(1/2)(1/2)=(1/8)+(1/8)=(2/8)=(1/4). Therefore, the probability of #9 inheriting the dominant dementia-causing allele from #2 is (1/4).

Problem Set #2: Pedigree Problem Set 20-22 The pedigree shows the inheritance of a birth defect in three generations of a family. 20. Based on the pedigree, the mode of inheritance of this disease could be A. Autosomal recessive B. X-linked recessive C. Autosomal dominant D. X-linked dominant E. A or B 21. Assume that #7 was an affected female, instead of an affected male as shown. Given that assumption, then the mode of inheritance of this disease is most likely... A. Autosomal recessive B. X-linked recessive C. Autosomal dominant D. X-linked dominant E. Y -linked dominant 22. Assume that #7 was an affected female. What is the probability that #8 is heterozygous for the disease-causing allele? (short answer)

20. E. A or B * Can't be dominant, unaffected parents made affected child. * Autosomal recessive: works rr x RR = Rr OR rr x Rr = rr, Rr Rr x Rr = rr Rr x Rr = rr Rr x Rr = Rr, RR, rr Rr x Rr = RR, Rr, rr * X-linked recessive: works - XrY x XRXR = XRXr, XRY XRXr x XRY = XRXR, XRXr, XRY, XrY (x2) OR - XrY x XRXr = XRXr, XrXr, XrY, XRY XRXr x XRY = XRXR, XRXr, XRY, XrY 21. A. Autosomal recessive If #7 is an affected female, X-recessive inheritance can't work since the cross between #3 and #4 would give no affected females. 22. (2/3) The cross between #5 & #6 gives RR,Rr,Rr,rr. We know #8 is dominant, because it isn't affected. Of the three dominant phenotypes, there is two heterzygous genotypes. So there's a (2/3) probability that #8 is heterozygous. * If #7 is an affected female, only mechanism is autosomal recessive inheritance. (1,2 cross) rr x RR = Rr OR rr x Rr = rr, Rr (3,4 cross) Rr x Rr = rr, RR, Rr Rr x Rr = rr, RR, Rr (5,6 cross) Rr x Rr = Rr, RR, rr Rr x Rr = RR, Rr, rr

Problem Set #2: Pedigree Problem Set 23-24. The pedigree shows the inheritance of a trait in three generations of a family. 23. Based on the pedigree, the mode of inheritance of this trait could be.. A. Autosomal recessive B. X-linked recessive C. Autosomal dominant D. X-linked dominant E. C or D 24. Assume that #10 marries an unaffected woman, and they have an unaffected daughter. Given that assumption, then the mode of inheritance of this trait is most likely... A. Autosomal recessive B. X-linked recessive C. Autosomal dominant D. X-linked dominant E. Y -linked dominant

23. E. C or D Dominance? * Autosomal dominant: works RR x rr = Rr Rr x Rr = rr, RR, Rr Rr x rr = Rr, rr * X-linked dominant: works XRY x XrXr = XRXr, XrY XRXr x XRY = XRXR, XRXr, XRY, XrY XRXr x XrY = XRXr, XrXr, XRY, XrY Recessive? * X-linked recessive: won't work - affected parents had unaffected child. - XrY x XrXr = XrXr, XrY <- what is expected for X-linked recessive. * Autosomal recessive: won't work. rr x Rr = Rr, rr rr x rr = rr (Autosomal recessive won't work, since when two affected parents crossed in the pedigree, they had an unaffected child. This cross shows what is expected for autosomal recessive). For this same reason, X-linked inheritance won't work. 24. C. Autosomal dominant We know that prior, the cross was only dominant. Let's see what this new information changes. * X-linked dominant: XRY x XrXr = XRXr, XrY (doesn't work). X-linked dominant isn't compatible with this new information. Dad passes his XR to his daughter, making her XRXr. This makes her affected. This cross doesn't make any unaffected daughters. * Autosomal dominant: works RR x rr = Rr OR Rr x rr = Rr, rr Second cross is compatible.

Problem Set #2: Pedigree Problem Set 25-26. The pedigree shows the inheritance of a genetic disease in three generations of a family 25. Based on the pedigree, the mode of inheritance of this disease could be... A. Autosomal recessive B. X-linked recessive C. Autosomal dominant D. X-linked dominant E. A or B 26. Assume that the frequency of disease-causing alleles in the general population, including those (#3 and#6) who married into the family, is very low. Given that assumption, then the mode of inheritance of this disease is most likely... A. Autosomal recessive B. X-linked recessive C. Autosomal dominant D. X-linked dominant E. Y -linked dominant

25. E. A or B Can't be dominant, unaffected parents make affected child. * Autosomal recessive: works rr x RR = Rr Rr x Rr = rr, Rr, RR Rr x Rr = rr, Rr, RR * X-linked recessive: works XrY x XRXR = XRXr, XRY XRXr x XRY = XRXR, XRXr, XRY, XrY (x2) 26. B. X-linked recessive * Autosomal recessive: won't work rr x RR = Rr Rr x RR = RR, Rr (won't work, doesn't give recessive progeny) * X-linked recessive: works XrY x XRXR = XRXr, XRY XRXr x XRY = XRXR, XRXr, XRY, XrY * Rare If the problem "tells" you the disease is rare, then when considering the hypothesis of recessiveness, you can assume that multiple carriers did not marry into the family.

Problem Set #2: Pedigree Problem Set 28-29. The pedigree shows the inheritance of a genetic disease in three generations of a family 28. Based on the pedigree, the mode of inheritance of this disease could be... A. Autosomal recessive B. X-linked recessive C. Autosomal dominant D. X-linked dominant E. C or D 29. Assume the disease is rare. Also, assume that many more generations of this family are followed, and it is observed that affected men never have affected sons, whereas, on average, about half of the sons born to affected women are also affected. Given these assumptions, then the mode of inheritance of this disease is most likely... A. Autosomal recessive B. X-linked recessive C. Autosomal dominant D. X-linked dominant E. Y -linked dominant

28. E. C or D * Can't be recessive, because two affected parents made an unaffected child. * Dominant: * Autosomal dominant: works RR x rr = Rr Rr x Rr = rr Rr x rr = Rr rr * X-linked dominant: works XRY x XrXr = XRXr, XrY XRXr x XRY = XRXR, XRXr, XrY, XRY XRXr x XrY = XRXr, XrXr, XRY, XrY 29. D. X-linked dominant Dominant If the problem "tells" you the disease is rare, then... when considering the hypothesis of dominance, you can assume that the founder is a heterozygote. * Autosomal dominant: doesn't work. * X-linked dominant: works * X-linked dominant: "rare" doesn't change anything for X-linked, since male is the founder (XRY). - "Affected men never have affected sons" means that affected mean are only mating with non-affected females - "Half the sons born to affected women are also affected". XrY x XRXr = XRXr, XrXr, XRY, XrY. * Can't be Y-linked since they observed that affected men never have affected sons. This isn't possible for Y-linked inheritance.

Problem Set #2: Pedigree Problem Set 3. With respect the gene controlling the birth defect, which of the following is a correct statement about the family members shown? A. Individual 1 is a heterozygote B. Individual 2 is a heterozygote C. Individual 3 is a heterozygote D. Individual 4 is a homozygote E. Individual 5 is a homozygote 4. (Short answer) What are the chances that #10 is a carrier of the birth defect ? _____

3. C. Individual 3 is a heterozygote * Using work above, individual 2 might be homozygous or heterozygous, we don't know. 4. (2/3) or 67% Rr x Rr = RR, Rr, Rr, rr -> (2/3). There is a (2/3) chance rather than a (1/2) chance since we know that individual #10 is dominant. So, we see that (2/3) of the dominant phenotypes are Rr, meaning there's a (2/3) chance of #10 being a carrier (heterozygous for recessive allele).

Problem Set #2: Pedigree Problem Set 39-40 The pedigree shows the inheritance of a genetic disease in three generations of a family. 39. Based on the pedigree, the mode of inheritance of this disease could be.. A. Autosomal B. X-linked C. Y-linked D. A or B E. A or C 40. Based on the pedigree, the mode of inheritance of this disease could be... A. Dominant B. Recessive C. A or B D. None of the above are consistent with the pedigree.

39. D. A or B Can't be recessive because affected parents had unaffected child. Dominant: * Autosomal dominant: works I: RR x rr = Rr II: rr x Rr = rr, Rr Rr x Rr = Rr, RR, rr * X-linked: works. I: XRY x XrXr = XRXr, XrY II: XRXr x XrY = XRXr, XrXr, XRY, XrY XRXr x XRY = XRXR, XRXr, XRY, XrY 40. A. Dominant

Chapter 3: Mitosis, Meiosis, and Sex Determination (Not a learning objective, but must know) TEXTBOOK EXPLANATION * During the moment of conception that culminated your birth, what happened? - How is one's chromosomal sex determined? * What is the mechanism of cell division that produces most of your cells? - Describe this process: what happens and what does it produce (specific)? * Describe characteristics of what is produced. - Generally, what does this process produce? What are these things? * Somatic cells: - What do human somatic cells' nuclei contain? * Describe these things: - How many sets? - What do they belong to? - What is the total number of these things called? - How many individual things of these do each of our somatic nuclei contain? In how many pairs? What does this make the "total number"? * What is the characteristic "total number" for animal species in general? - Know what each component of this "total number" represents, and what this means. - In respect to humans, know what our "total number" is, and what that means for the components.

A number of years ago, at the moment of conception that culminated in your birth, two gametes united to form the single fertilized cell—the zygote—from which you developed. Your chromosomal sex was determined in that instant. Your mother's egg carried an X chromosome, and your sex was determined by whether your father's sperm carried an X chromosome, making you female (XX), or a Y chromosome, making you male (XY). Shortly after fertilization, cell division began and over the next few hours increased the tiny zygote to two cells, then four cells, then eight cells, and so on. These cell divisions continued, and genetically controlled processes of cell differentiation and cell specialization began to form the first embryonic organs and structures. These processes eventually determined the structure and function of each cell in your body. Since then, your body has produced thousands of generations of cells. The mechanism of cell division that produced most of them is called mitosis. It is an ongoing process that with each division creates two identical daughter cells. These two cells are exact genetic replicas of one another and of the parental cell from which they are derived. Mitosis produces somatic cells, the structural cells of the body. There are trillions of somatic cells in your body, and nearly all of them contain a nucleus in which the chromosomes are located. Human somatic cells' nuclei contain two sets of chromosomes: Each chromosome belongs to a homologous pair, and the total number of chromosomes is called the diploid number. Your somatic cell nuclei contain 46 chromosomes each, in 23 homologous pairs, so your diploid number is 46. The diploid number varies among species (each species having its characteristic number of pairs), so the characteristic diploid number for animal species in general is described as 2n. The value n represents the haploid number of chromosomes, and it is one-half the diploid number. Humans have a diploid number of 2n=46, so the human haploid number is n=23.

Chapter 10: Chromosome Aberrations & Disease TEXTBOOK EXPLANATION CONTINUED AGAIN Describe non-disjunction. * Define uniparental disomy. - When does it occur? - What are its mechanisms? * Understand the examples provided. * Understand this example in terms of trisomy rescue. * Define trisomy rescue.

A rare abnormality of chromosome content called uniparental disomy has been identified in humans. Uniparental disomy occurs when both copies of a homologous chromosome pair originate from a single parent. Uniparental disomy has two mechanisms of origin. The rarer mechanism requires nondisjunction of the same chromosome in both sperm and egg, with the result that a fertilization occurs in which one gamete contributes two copies of the chromosome and the other does not contribute a copy of the chromosome. The second mechanism is more common. It involves nondisjunction in one parent that results in an aneuploid gamete contributing two copies of a chromosome and the other parent contributing a normal gamete with a single copy of the chromosome. In the case of either Angelman syndrome or Prader-Willi syndrome, the chromosome involved is chromosome 15. Union of the gametes with two copies and one copy, respectively, of chromosome 15—results in trisomy 15 in the fertilized egg. This is a condition that is invariably incompatible with survival. By a process known as trisomy rescue,however, some fertilized eggs that are initially trisomic can survive and lead to the formation of a zygote that can survive. In trisomy rescue, one of the extra copies of the chromosome is ejected in one of the first mitotic divisions following fertilization. Which of the three chromosomes is ejected is apparently random. Thus, one result of trisomy rescue can be a cell with one chromosome from each parent. Zygotes with this result have normal chromosome content. Alternatively, trisomy rescue could result in a zygote that retains two copies of the chromosome from the same parent, and this is uniparental disomy.

Problem Set #2: Pedigree Problem Set Same question as above, but now assume that we know that freckled Fred and Dimpled Doris are homozygous for both the freckling and dimpling genes. What can we say withcertainty about their kids? A. All of their kids will be both freckled and dimpled B. Their kids will either be dimpled and not freckled or freckled and not dimpled C. None of their kids will be both freckled and dimpled D. Whether or not the kids are freckled or dimpled will depend upon the sex of the kids E. None of the above (A-D) can be said with certainty

A. All of their kids will be both freckled and dimpled * Autosomal dominant traits FFdd x ffDD = FfDd

Problem Set #1: Pedigrees Examine the accompanying pedigree and determine the inheritance pattern. A. This trait could only be autosomal recessive B. This trait could only be autosomal dominant C. This trait could only be X-linked recessive D. This trait could only be X-linked dominant E. This trait could be a, b, c, or d

A. This trait could only be autosomal recessive * Has to be recessive since unaffected parents have affected children. * Affected boy: XrY * Affected girl: XrXr * Autosomal recessive: Rr x Rr = rr * X-linked recessive doesn't work: XRY x XRXr The cross gives XRXR, XRXr, XrY, XRY - no recessive girls, so the daughter wouldn't have the disease, only the son.

Problem Set #2: Pedigree Problem Set 11-19. The pedigree shows the inheritance of trait in four generations of a family. For questions 15-16, assume that the pedigree shows the inheritance of the rare x-linked genetic disease icthyosis, which is characterized by flaky skin. 16. Assume that #4, #5 and #8 are found to have patches of their skin that are scaly and patches that are normal, while the skin of #2, #3, #6 and #11 is completely normal. Could these observations be explained by a model involving x-chromosome inactivation in the heterozygous females? A. Yes B. No

A. Yes * Rare X-linked recessive If the problem "tells" you the disease is rare, then when considering the hypothesis of recessiveness, you can assume that multiple carriers did not marry into the family. (only heterozygous females have the patches on them, homozygous females are totally normal).

Problem Set #2: Pedigree Problem Set The pedigree shows the inheritance of the phenotypes cardiac arrhythmia (a) and partial deafness (d) in the W and F families. A study was undertaken to determine if the arrhythmia in these families was likely (1) the autosomal recessive form, Jervell Lang-Neilson syndrome, also characterized by partial deafness, (2) Ward-Romano syndrome, an autosomal dominant form, not associated with deafness, or (3) perhaps a third type, inherited in some other fashion. The problem does not specify that the conditions are rare, but you should assume that they are (except when answering question 64). If arrhythmia without associated deafness was multigenic, then: A. one would have to consider the possibility that the allele segregating in the W family was of a different locus than the one segregating in the F family. B. none of the kids born from the marriage of the W and F families would be arrhythmic C. all of the kids born from the marriage of the W and F families would be arrhythmic D. environmental influences would be a major determinant of the arrhythmic phenotype. E. A and B

A. one would have to consider the possibility that the allele segregating in the W family was of a different locus than the one segregating in the F family. * Arryhythmia without associated deafness: AAdd or Aadd

Chapter 4: Gene Interactions *** CONTINUED *** Describe variations from classical Mendelian genetics, including incomplete dominance, co-dominance, multiple alleles/allelic series, and lethal alleles. * Allelic series. - Understand example.

Allelic Series: There is an order of dominance when multiple alleles are present. (most dom: C -> most rec: c). * PIC: The C gene encodes a tyrosinase enzyme involved in melanin synthesis ----> most dominant = most pigmentation. Brown is dominant to chinchilla, himalaya and albino - (3:1 phenotypic ratio in all crosses). Chinchilla is dominant to himalayan and albino. - (1:2:1 phenotypic ratio) -> elements of codominance. - (3:1 phenotypic ratio) Himalayan is dominant to albino. - (3:1 phenotypic ratio) In a series of dominant relationships, there is diversity of dominance and codominance.

Chapter 10: Chromosome Aberrations & Disease Review homologous vs. non-homologous chromosomes vs. sister chromatids. TEXTBOOK EXPLANATION CONTINUED AGAIN Chromosome Structure and Chromatin Organization of Eukaryotic Chromosomes: * Among eukaryotes, know how the amino acid sequence of histone proteins has evolved. - What does this suggest? - What does this say about the role histones in eukaryotic chromosome organization? * Know the function of histones. * What is the fundamental unit of histone protein organization? - Describe this. - Know the components. - Describe its assembly process. - How is DNA involved with this complex? * Know what DNA does, and what it causes. * Know the presence of histones in eukaryotic cells. - How present are they? - What is the cause of this?

Among eukaryotes, there is very strong evolutionary conservation of the amino acid sequences of histone proteins. This consistency among eukaryotes suggests that there is significant evolutionary pressure to retain the structure and function of each histone protein. A comparison of the amino acid sequences of H4 in cows and pea plants, for example, demonstrates this high degree of evolutionarily retained identity. The comparison tells us that since the time when plants and animals last shared a common ancestor, extraordinarily strong evolutionary pressure has maintained H4 DNA and its amino acid sequence identity in organisms. This example of evolutionary conservation speaks to the importance of histones in eukaryotic chromosome organization. Histones are the principal agents in chromatin packaging, and the fundamental unit of histone protein organization is the nucleosome core particle. The nucleosome core particle is a heterooctameric protein complex that contains two molecules each of four histones—H2A, H2B, H3, and H4 (Foundation Figure 10.24). These proteins are continuously transcribed and translated in eukaryotic cells, and histone genes are one family of genes that are present in multiple copies in eukaryotic genomes. Nucleosome core particles self-assemble. The histone proteins first self-assemble into dimers containing two different histones each: H2A-H2B dimers contain one molecule each of histone 2A and histone 2B, and H3-H4 dimers contain one molecule each of histone 3 and histone 4. Current evidence indicates that nucleosome core particles are formed in steps that begin with two H3-H4 dimers assembling to form a histone tetramer. The tetramer is then joined by two H2A-H2B dimers to form the octameric nucleosome core particle. Nucleosome core particles are flat-ended structures approximately 11 nm in diameter by 5.7 nm thick (see Figure 10.24a). Each nucleosome core particle is wrapped by approximately 146 base pairs of DNA that twist one and two-thirds times around the core particle. This wrapping is the first level of DNA condensation, and it condenses the DNA approximately sevenfold.

Chapter (2.6/3.5) Advanced Pedigree Analysis Calculate the probability that the proband has a disease or disease-causing allele. Genetic Counseling A man who is a carrier for cystic fibrosis marries a woman who also happens to be a carrier for cystic fibrosis. What is the probability that their first kid will be affected? A. (2/3) 67% B. (1/2) 50% C. (1/4) 25% D. (1/6) 17% E. (1/8) 12.5%

Another thing we need to be able to do with pedigrees is to calculate the probability that diseases or alleles are inherited. Genetic Counseling C. (1/4) 25%

Chapter (2.6/3.5) Advanced Pedigree Analysis Calculate the probability that the proband has a disease or disease-causing allele. Genetic Counseling A woman's brother died from Tay Sash's Disease (autosomal recessive, lethal), but she is unaffected. What are the chances that she is a carrier of the disease? A. (2/3) 67% B. (1/2) 50% C. (1/4) 25% D. (1/6) 17% E. (1/8) 12.5%

Another thing we need to be able to do with pedigrees is to calculate the probability that diseases or alleles are inherited. Genetic Counseling Hint: we know both parents must be carriers. B. (1/2) 50%

Chapter 10: Chromosome Aberrations & Disease Review homologous vs. non-homologous chromosomes vs. sister chromatids. TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN AGAIN AGAIN AGAIN AGAIN AGAIN AGAIN AGAIN AGAIN Position Effect Variegation: Effect of Chromatin State on Transcription: * How does this explain the variegation of the mutants and the variability of this variegation pattern? * How do limitations affect the phenotype of the fruit flys? * How do extensions affect the phenotype of the fruit flys? * What have we concluded from this? - How does chromatin structure and the degree of chromatin compaction relate to gene expression in eukaryotic genomes? - What has research on PEV and chromatin state led us to conclude? * How does the state of the chromatin in which a gene is located relate to gene expression? * How can gene expression or silencing be dictated by chromatin structure?

As Figure 10.28b indicates, if centromeric heterochromatin spread is limited (top image) and does not reach the new position of the red-eye-color allele, the allele is expressed in the cell. All cells descending from this initial cell grow in a cluster in the eye, and the cells in such a cluster will have red pigment and form red patches in the variegated eye. If, on the other hand, centromeric heterochromatin spread is more extensive and the relocated red-eye-color allele is covered by reformed heterochromatin (silenced) (bottom image), the allele is not expressed in the cell. All cells descending from this one also grow in a cluster in the eye, and they have no pigment (i.e., they are white). This is the source of white patches in the variegated eye. Because the spread of centromeric heterochromatin can vary from chromosome to chromosome and the development of patches of eye tissue is also variable from eye to eye, there is a great deal of observed variation in the patterns of eye color variegation. Chromatin structure and the degree of chromatin compaction are critical components of gene expression in eukaryotic genomes. Research on PEV and on chromatin state have led to two central conclusions: (1) Gene expression can be controlled by the state of the chromatin in which a gene is located, and (2) gene expression or gene silencing can be dictated by chromatin structure that is transmissible from one cell generation to the next.

Chapter 3: Mitosis, Meiosis and Sex Determination (Not a learning objective, but should know) TEXTBOOK EXPLANATION CONTINUED * Know the eukaryotic cell cycle. * What principal phases is the cell cycle divided into? - What occurs during each of these phases. - What phases are each of these principle phases divided into? * What happens during these smaller phases?

As they approach the end of G1, cells follow one of two alternative paths. Most cells enter the S phase, or synthesis phase, during which DNA replication (DNA synthesis) takes place. On the other hand, a small subset of specialized cells transition from G1 into a nondividing state called G0 ("G zero"), a kind of semiperpetual G1-like state in which cells express their genetic information and carry out normal functions but do not progress through the cell cycle (see Figure 3.1b). Several kinds of cells in your body, including certain cells in your eyes and bones, reach a mature state of differentiation, enter G0, and rarely if ever divide again. Most G0 cells maintain their specialized functions until they enter programmed cell death (apoptosis) and die. Cells only rarely leave G0 and resume the cell cycle. DNA replication takes place during S phase and results in a doubling of the amount of DNA in the nucleus—by creating two identical sister chromatids that are joined to form each chromosome. Prior to S phase, each chromosome is composed of a long DNA double helix. During S phase, the DNA strands separate, and each acts as a template to direct the synthesis of a new daughter strand of DNA. This DNA synthesis forms the sister chromatids that are genetically identical to one another. The completion of S phase brings about the transition to the G2, or Gap 2, phase of the cell cycle, during which cells prepare for division. Interphase ends when cells enter M phase, from which two identical daughter cells emerge. The successive generations of cells produced through mitosis as one cell cycle follows the next are known as cell lines or cell lineages. Each cell line or cell lineage contains identical cells (i.e., clones) that are all descended from a single founder cell. Mitosis ensures that the genetic information in cells is faithfully passed to successive generations of cell lineages.

Chapter 10: Chromosome Aberrations & Disease *** CONTINUED AGAIN *** Describe euploidy, polyploidy, and aneuploidy. * What do autosomal aneuploidies lead to? - What is the leading known cause of this? * What is the most common autosomal aneuploidy? - What is the genetic disorder resulting from this autosomal aneuploidy? * What are other autosomal aneuploidies? Know the diseases caused by these autosomal aneuploidies when relevant. He says "You do not need to memorize these, but you do need to know how to work non-disjunction problems", referring to a list of autosomal and sex-chromosome aneuploidies, the syndromes they cause, their syndrome characteristics, and their frequencies at birth.

Autosomal Aneuploidies Chromosome abnormalities and pregnancy loss: * About (1/3) of human pregnancies are lost spontaneously after implantation. * Chromosome abnormalities are the leading known cause of pregnancy loss. * A minimum of 10-15% of conceptions have a chromosome abnormality. * At least 95% of these conceptions spontaneously abort because the fetus has died. * Trisomy of chromosome 21 = most common autosomal (non-sex chromosome) aneuploidy (1/1500 live births). * Down syndrome = genetic disorder due to trisomy 21. * Approximately 75% of trisomy 21 conceptions are spontaneously aborted. - Those that do make it to live birth have a spectrum of severity symptoms. Down syndrome Varying severity of symptoms: * Mental impairment. * Gastrointestinal tract obstruction. * Congenital heart defects. * Respiratory infections. * 15-20x higher risk of leukemia. * Characteristic appearance. Other Autosomal Aneuploidies Trisomy 13 - Patau Syndrome (1/10,000 live births) Trisomy 18 - Edward Syndrome (1/6,000 live births). * 95% affected fetuses spontaneously aborted. * 90% mortality during first year of life. * All other trisomies (of chromosomes 1,2,3, etc.) and monosomies (of chromosomes 1,2,3, etc.) are embryonic lethals, resulting in spontaneously aborted fetuses.

Quiz: Pedigree #1 A unaffected couple has a child affected with Klansky's disease. Which mode(s) of inheritance can be RULED OUT for Klansky's disease (choose all that are correct): Autosomal Dominant X-linked Dominant Autosomal Recessive X-linked Recessive

Autosomal Dominant X-linked Dominant This is a straight-forward application of Rule 1 from lecture. Rule 1 states that "If unaffected parents have an affected kid, dominance can be ruled out"

Quiz: Pedigree #1 An unaffected couple has a daughter affected with Koopskiʼs disease. Which type(s) of inheritence can be ruled out for Koopskiʼs disease? (Choose all that are correct) Autosomal Dominant X-linked Dominant Autosomal Recessive X-linked Recessive

Autosomal Dominant X-linked Dominant X-linked Recessive This is an example of the "Rule 1 & 4 Combo" from lecture.

Quiz: Pedigree #1 A husband and wife are both affected with Kloxie's disease, but both their children are unaffected. Which mode(s) of inheritance can be RULED OUT for Kloxie's disease (choose all that are correct): Autosomal Dominant X-linked Dominant Autosomal Recessive X-linked Recessive

Autosomal Recessive X-linked Recessive Here we apply Rule 2 from lecture, which states "If two affected parents have an unaffected kid, recessiveness can be ruled out"

Problem Set #2: Pedigree Problem Set Assume, in the previous pedigree, that the condition is very rare, and that couple #1 and #2 had a third daughter who was unaffected. Then the mode of inheritance of this condition is most likely...(again, write down as manyas are correct):

Autosomal dominant * RARE If the problem "tells" you the disease is rare, then when considering the hypothesis of recessiveness, you can assume that multiple carriers did not marry into the family. If the problem "tells" you the disease is rare, then... when considering the hypothesis of dominance, you can assume that the founder is a heterozygote. * Recessive? Autosomal recessive: I: rr x RR = Rr (doesn't work, need rr) X-linked recessive: I: XrY x XRXR = XRXr, XRY (doesn't work, need XrXr) * Dominant: X-linked dominant: I: XRY x XrXr = XRXR, XrY (doesn't work, need XrXr) Autosomal dominant: works I: Dd x dd = Dd, dd II: Dd x dd = Dd, dd Dd x dd = Dd, dd

Problem Set #2: Pedigree Problem Set Assume the condition in the previous pedigree is very rare in the general population. Then the mode ofinheritance of this condition is most likely...(again, write down as many as are correct):

Autosomal dominant, X-linked dominant * RARE If the problem "tells" you the disease is rare, then when considering the hypothesis of recessiveness, you can assume that multiple carriers did not marry into the family. If the problem "tells" you the disease is rare, then... when considering the hypothesis of dominance, you can assume that the founder is a heterozygote. Autosomal recessive: doesn't work I: rr x RR = Rr (need rr) X-linked recessive: doesn't work I: XrY x XRXR = XRXr, XRY (need XrXr for #4,5) Autosomal dominant: works I: Rr x rr = Rr, rr II: Rr x rr = Rr, rr Rr x rr = Rr, rr X-linked dominant: (doesn't change, founder is male)

Problem Set #2: Pedigree Problem Set The pedigree shows the inheritance of a genetic condition in three generations of a family. Based on the pedigree, the mode of inheritance of this condition could be (write down as many as are correct): Autosomal recessive (AR) X-linked recessive (XR) Autosomal dominant (AD) X-linked dominant (XD) Y-linked (Y)

Autosomal recessive, Autosomal dominant, X-linked dominant Recessive? * Autosomal recessive: works I: rr x Rr = Rr, rr II: RR x rr = Rr rr x Rr = rr, Rr * X-linked recessive: doesn't work I: XrY x XRXr = XRXr, XrXr, XRY, XrY II: XrXr x XRY = XRXr, XrY (doesn't work) In cross, we would see Mom pass her Xr to her son, making him affected if this mechanism is used. (#7 not affected) Dominant? * Autosomal dominant: works I: RR x rr = Rr II: Rr x rr = Rr, rr Rr x rr = Rr, rr * X-linked dominant: works I: XRY x XrXr = XRXr, XrY II: XRXr x XrY = XRXr, XrXr, XRY, XrY XRXr x XrY = XRXr, XrXr, XRY, Xry

Chapter 3: Mitosis, Meiosis and Sex Determination (Not a learning objective, but should know) * What are autosomal traits? EX? * What are sex-linked traits? EX?

Autosomal vs. Sex-linked Traits Autosomal traits are caused by genes on autosomes (chromosomes #1-22). EX: we speak of autosomal recessive or autosomal dominant traits or diseases. Sex-linked traits are caused by genes on the sex chromosomes (X or Y). EX: we speak of X-linked recessive or X-linked dominant traits or diseases.

Problem Set #2: Pedigree Problem Set What can be said regarding the of inheritance of the trait or disease shown in the following pedigree? A) It could be autsomal dominant or x-linked dominant. B) It could be autsomal dominant only. C) It could be autsomal recessive or x-linked recessive. D) It could be autsomal recessive only. E) This pedigree cannot be explained by any of the above mechanisms of inheritance

B) It could be autsomal dominant only. * Can't be recessive, two affected parents made unaffected children * Autosomal dominant: works Rr x Rr = Rr, Rr, RR, rr * X-linked dominant: doesn't work XRY x XRXr = XRXR, XRXr, XRY, XrY (Need XrXr, Dad passes XR to daughter, she would be affected)

Problem Set #1: Pedigrees A girl is affected with Sleepy's disease. Both of her parents are unaffected. What is the most likely mode of inheritance of Sleepy's disease? A. Autosomal dominant B. Autosomal recessive C. X-linked dominant D. X-linked recessive E. B or D

B. Autosomal recessive * Has to be recessive since unaffected parents have affected child. * Girl: XrXr * Autosomal recessive: Rr x Rr = rr * X-linked recessive doesn't work: XRY x XRXr ≠ XrXr The dad doesn't have an Xr for the daughter to grab, so she'd have a dominant phenotype, and therefore wouldn't have the disease.

Problem Set #2: Pedigree Problem Set A girl is affected with Sleepy's disease. Both of her parents are unaffected. What is the most likely mode of inheritance of Sleepy's disease? A. Autosomal dominant B. Autosomal recessive C. X-linked dominant D. X-linked recessive E. B or D

B. Autosomal recessive * Has to be recessive since unaffected parents have affected child. * Girl: XrXr * Autosomal recessive: Rr x Rr = rr * X-linked recessive doesn't work: XRY x XRXr ≠ XrXr The dad doesn't have an Xr for the daughter to grab, so she'd have a dominant phenotype, and therefore wouldn't have the disease.

Chapter 4: Gene Interactions Practice Problem: What is this picture an example of? A. Incomplete dominance B. Co-dominance C. Pleiotropy D. Partial penetrance E. Variable expressivity

B. Co-dominance * The kid expresses the phenotype of both alleles (alleles of each parent). - dad = horizontal stripes - mom = vertical stripes - -> son = horizontal and vertical stripes

Chapter 10: Chromosome Aberrations & Disease Practice Problem: A man is born with Double-Y Syndrome: 47,XYY. The most likely explanation for this is...... A. Non-disjunction during meiosis I in his father. B. Non-disjunction during meiosis II in his father. C. Non-disjunction during meiosis I in his mother. D. Non-disjunction during meiosis II in his mother.

B. Non-disjunction during meiosis II in his father. Extra Y in person = extra Y in sperm. -----> has to be from dad (women don't have Y chromosome) -----> must've occurred during meiosis II where nondisjunction at the spermacyte holding the homologous pair of Y chromosomes leads to sperm with two Ys. * Nondisjunction in meiosis I gives sperm with X and Y. This is what told us that nondisjunction occurred during meiosis II.

Problem Set #1: Pedigrees A male affected with Klopski's disease mates with an unaffected female and has an affected son and an unaffected daughter. Which type(s) of inheritence can be ruled out for Klopski's disease? A. Autosomal dominant B. X-linked dominant C. Autosomal recessive D. X-linked recessive E. B, C and D.

B. X-linked dominant Recessive? * Autosomal recessive: works - rr x Rr = Rr, rr Some children affected, some not. * X-linked recessive: works - XrY x XRXr = XRXr, XrXr, XRY, XrY 1/2 daughters affected, 1/2 sons affected Dominant? * Autosomal dominant: works - Rr x rr = Rr, rr Some children affected, some not. * X-linked dominant: doesn't work. - XRY x XrXr = XRXr, XrY Daughters affected, sons not affected.

Problem Set #2: Pedigree Problem Set A male affected with Klopski's disease mates with an unaffected female and has an affected son and an unaffected daughter. Which type(s) of inheritence can be ruled out for Klopski's disease? A. Autosomal dominant B. X-linked dominant C. Autosomal recessive D. X-linked recessive E. B, C and D.

B. X-linked dominant Recessive? * Autosomal recessive: works - rr x Rr = Rr, rr Some children affected, some not. * X-linked recessive: works - XrY x XRXr = XRXr, XrXr, XRY, XrY 1/2 daughters affected, 1/2 sons affected Dominant? * Autosomal dominant: works - Rr x rr = Rr, rr Some children affected, some not. * X-linked dominant: doesn't work. - XRY x XrXr = XRXr, XrY Daughters affected, sons not affected.

Problem Set #2: Pedigree Problem Set Wilson disease is a rare autosomal-recessive copper overload disorder caused by mutations of the Wilson disease gene ATP7BA located on chromosome 13. A man and a woman, neither of whom is affected by Wilson's disease, have a daughter who has Wilson's disease. The woman then gets pregnant with a second child. If the second child is a girl, what are the chances that she will have Wilson's disease?If the second child is a boy, what are the chances that he will have Wilson's disease? A. girl = 50%; boy = 50% B. girl = 25%; boy = 25% C. girl = 25%; boy = 50% D. girl = 67%; boy = 100% E. girl = 50%; boy = 100%

B. girl = 25%; boy = 25% * RARE, autosomal recessive. * Rare If the problem "tells" you the disease is rare, then when considering the hypothesis of recessiveness, you can assume that multiple carriers did not marry into the family. Rr x Rr = rr, RR, Rr, Rr Rr x Rr = (1/4) rr for girl, (1/4) rr for boy. * Autosomal inheritance isn't infected by sex.

Problem Set #2: Pedigree Problem Set The pedigree shows the inheritance of the phenotypes cardiac arrhythmia (a) and partial deafness (d) in the W and F families. A study was undertaken to determine if the arrhythmia in these families was likely (1) the autosomal recessive form, Jervell Lang-Neilson syndrome, also characterized by partial deafness, (2) Ward-Romano syndrome, an autosomal dominant form, not associated with deafness, or (3) perhaps a third type, inherited in some other fashion. Note: this series of questions (except for question 64, which Dr. Bardwell wrote) is taken from an actual MCAT exam. The problem does not specify that the conditions are rare, but you should assume that they are (except when answering question 64). The fact that arrhythmia in the F family is most likely autosomal is shown by: A. female to male transmission B. male to male transmission C. lack of male to male transmission D. male to female transmission E. actually, the arrhythmia is most likely X-linked

B. male to male transmission (Note you already are assuming that it's dominant at this point because of question 6, so this question is just about determining whether it is AD or XLD) * We know it's dominant, determine AD vs. XD: A. female to male supports X-linked inheritance. C. lack of male to male supports X-linked inheritance. D. male to female transmission supports X-linked inheritance. E. can't be X-linked since a father that had the dominant trait did not give it to all his daughters. The problem does not specify that the conditions are rare, but you should assume that they are (except when answering question 64). * RARE If the problem "tells" you the disease is rare, then when considering the hypothesis of recessiveness, you can assume that multiple carriers did not marry into the family. If the problem "tells" you the disease is rare, then... when considering the hypothesis of dominance, you can assume that the founder is a heterozygote. * a means affected by arrhythmia -> A * d means affected by partial deafness. -> D

Problem Set #1: Sex Linkage and Sex Determination People afflicted with Turner syndrome (45, X) appear female, while those afflicted with Klinefelter syndrome(47, XXY) appear male. From this, and our knowledge of the karyotype of normal human males and females,we can tell that sex in humans is determined by: A. the number of X chromosomes B. the presence or absence of the Y chromosome C. by the egg, not the sperm D. any of the above (A-C) are consistent with the observations E. none of the above (A-C) are consistent with the observations

B. the presence or absence of the Y chromosome

Problem Set #2: Pedigree Problem Set Given what you know of the inheritance pattern of Sickle Cell Disease, what can be said of individual II-5? A) He is unaffected, and the mutant allele is passed from II-4 to their son. B) His children with II-4 will all have a 25% chance of being affected (βSβS). C) He is a carrier, and his children with II-4 will have a 50% chance of being affected (βSβS) and a 50% chance of being a carrier (βAβS). D) He is a carrier, and his children with II-4 will have a 50% chance of being affected (βSβS) and a 50% chance of being unaffected (βAβA). E) None of the above are true.

C) He is a carrier, and his children with II-4 will have a 50% chance of being affected (βSβS) and a 50% chance of being a carrier (βAβS). * Mechanism of SCD: Autosomal recessive. I: Rr x Rr = Rr, Rr, RR, rr RR x rr = Rr II: rr x Rr = Rr (50%), rr (50%) * II-5 highlighted - he's unaffected, carrier.

Problem Set #2: Pedigree Problem Set Given what you know of the inheritance pattern of Sickle Cell Disease (SCD), is the following pedigree representative of a typical SCD pedigree? A) Yes; this is a dominant pedigree, and SCD exhibits dominant inheritance. B) Yes; this is a recessive pedigree, and SCD exhibits recessive inheritance. C) No; this is a dominant pedigree, and SCD exhibits recessive inheritance. D) No; this is a recessive pedigree, and SCD exhibits dominant inheritance. E) No, SCD is X-linked, and this pedigree is not typical of X-linked inheritance.

C) No; this is a dominant pedigree, and SCD exhibits recessive inheritance. * Mechanism of SCD: Autosomal recessive Pedigree doesn't work, affected parents make unaffected children, so it can't represent autosomal recessive.

Problem Set #1: Mitosis, Meiosis Consider the following 4 Statements about the events of meiosis and mitosis and then pick the one answer that most accurately and most completely describes these events. Statements: 1. Disjunction of sister chromatids occurs in mitosis 2. Disjunction of sister chromatids occurs in meiosis 3. Disjunction of homologous chromosomes occurs in mitosis 4. Disjunction of homologous chromosomes occurs in meiosis Answer: A. 1 and 3 are TRUE B. 2 and 4 are TRUE C. 1, 2 and 4 are TRUE D. 1, 2, and 3 are TRUE E. 1, 2, 3 and 4 are TRUE

C. 1, 2 and 4 are TRUE * Disjunction of sister chromatids occurs in mitosis * Disjunction of sister chromatids occurs in meiosis * Disjunction of homologous chromosomes occurs in meiosis

Problem Set #2: Pedigree Problem Set Your uncle (your mother's brother) has a rare genetic disease that is inherited in an autosomal recessive manner, but you, your parents and your grandparents are unaffected. What is the approximate chance that you are a carrier (heterozygote) for this disease? A. 1/6, or 17% B. 1/4, or 25% C. 1/3, or 33% D. 1/2, or 50% E. 2/3, or 67%

C. 1/3, or 33% Rare autosomal recessive disease. If the problem "tells" you the disease is rare, then when considering the hypothesis of recessiveness, you can assume that multiple carriers did not marry into the family. * Grandma and grandpa had to be carriers to have uncle affected. This gave a (2/3) probability that mom was a carrier (Rr x Rr cross), since we know she's dominant. We assumed dad wasn't a carrier since the disease is rare, making Rr x RR = (1/2) probability of producing Rr progeny. This means that I have a (2/3)(1/2) = (1/3) = 33% chance of being a carrier.

Chapter (2.6/3.5) Advanced Pedigree Analysis Practice Problem: Your maternal grandfather has been recently diagnosed with Huntington's disease. Your mom is unaffected (so far) and does not want to be tested. What are the chances that you will get the disease? a. 2/3 B. 1/2 C. 1/4 D. 1/8 E. Cannot be determined.

C. 1/4 * Late age of onset problems: - Individuals with the genotype may not yet be expressing the phenotype! - Generally, if a problem does not specify, then assume early onset. - But look for clues about late age-of-onset. Use of the rare disease clue: * D has HD * What's the chance that G will get HD? * What's the chance that H will get HD? From the information in the pedigree, we don't know if D is homozygous or heterozygous for the dominant, disease causing allele. But, because we know the disease is rare, we assume it was a heterozygous. * 1/2 (50% chance) mom got the allele. * 1/2 (50% chance) she gave it to you if she had it. * (1/2) x (1/2) = (1/4) = chance that you inherited the disease-causing allele from Grandpa D.

Problem Set #2: Pedigree Problem Set Your first cousin died of Tay-Sachs disease (autosomal recessive, incidence 0.09%). You are not affected, nor is anyone else in your extended family. What is the chance that you are a carrier? A. 3% B. 5.8% C. 25% D. 50% E. 67%

C. 25% 0.09% = Rare autosomal recessive disease. If the problem "tells" you the disease is rare, then when considering the hypothesis of recessiveness, you can assume that multiple carriers did not marry into the family. IN DARK CHERRY RED: there is an option between picking Rr or RR, both of which will give you Rr progeny when crossed with their mate, which is needed in both cases. In these cases, you need to pick RR since the disease is rare and we can assume that multiple carriers did not marry into the family. You'll need to make this assumption whenever there is a cross between two individuals (in the case of a rare autosomal recessive disease). We see this with grandma and grandpa as well as mom and dad. The reason the last cross (between aunt and uncle) doesn't assume uncle is RR is because Rr is needed as his genotype in order to produce the rr genotype for his son, my sick first cousin.

Problem Set #2: Pedigree Problem Set Sally's maternal grandmother (i.e. Sally's mother's mother) is affected by a rare x-linked recessive condition, but noone else in the family is. What are the chances that Sally is a carrier of this condition? A. 100% B. 67% (2/3) C. 50% (1/2) D. 25% (1/4) E. 0%

C. 50% (1/2) * RARE, X-linked recessive * Rare If the problem "tells" you the disease is rare, then when considering the hypothesis of recessiveness, you can assume that multiple carriers did not marry into the family.

Chapter 4: Gene Interactions Practice Problem: Morning glories are flowers that can be blue, purple, or red, depending on the alleles at two loci, A and B. A_B_ = blue A_bb or aaB_ = purple aabb = red If two AaBb plants are crossed, what is the phenotypic ratio? What is this an example of? A. 9:3:3:1 B. 9:3:4 C. 9:6:1 D. 12:3:1

C. 9:6:1 This is an example of Dominant Gene Interaction.

Problem Set #2: Pedigree Problem Set An unaffected couple has a daughter affected with Koopski's disease. Which type(s) of inheritence can NOT be ruled out for Koopski's disease? A. Autosomal dominant B. X-linked dominant C. Autosomal recessive D. X-linked recessive E. More than one of the above (A-D) cannot be ruled out.

C. Autosomal recessive * Not dominant, since an unaffected couple has a affected child. * Autosomal recessive works: Rr x Rr = rr. * X-linked recessive doesn't work: XRY XRXr = XRXR, XRXr, XRY, XrY No daughters affected.

Chapter 10: Chromosome Aberrations & Disease Which of these chromosome rearrangements is most likely to result in lethality? A. Reciprocal Translocation B. Inversion C. Deletion D. Invitation

C. Deletion Lethality can result from deletion if a big chunk of the chromosome is lost.

Chapter 4: Gene Interaction Which of the following is FALSE? A. Epistasis does not alter genotypic ratios. B. Epistasis alters phenotypic ratios, e.g. a dihybrid cross will produce a modified 9:3:3:1 ratio of phenotypes. C. Epistasis can occur when only a single gene controls a phenotype. D. You can detect epistasis using a chi-squared test. E. There are many types of epistatic interactions.

C. Epistasis can occur when only a single gene controls a phenotype. A: true, only alters phenotypic ratios. B.: true. C.: false. D.: true, the chi-squared value would be really big. Your expected would be 9:3:3:1, so you can use that and compare it with your observed value. A really big chi-squared value/really small P-value can hint to you that epistasis has occurred. E.: true

Problem Set #2: Pedigree Problem Set The following statements refer to Duchenne Muscular Dystrophy (DMD), a single-gene, x-linked recessive disorder. Which of the following statements (A-D) is FALSE (pick E if you think A-D are all true). A. If a female carrier of DMD mates with a normal man, half of her sons will be affected, on average. B. If a female carrier of DMD mates with a normal man, half of her daughters will be carriers, on average. C. If a female carrier of DMD mates with a male affected with DMD, all of their sons will be affected. D. If a male affected with DMD mates with a normal woman, all of his daughters will be carriers. E. None of the statements are false.

C. If a female carrier of DMD mates with a male affected with DMD, all of their sons will be affected. * X-linked recessive A. true XRXr x XRY = XRXR, XRXr, XRY, XrY B. true XRXr x XRY = XRXR, XRXr, XRY, XrY C. false XRXr x XrY = XRXr, XrXr, XRY, XrY D. true XrY x XRXR = XRXr, XRY

Problem Set #1: Sex Linkage and Sex Determination The following statements refer to Duchenne Muscular Dystrophy (DMD), a single-gene, x-linked recessive disorder. Which of the following statements (A-D) is FALSE (pick E if you think A-D are all true). A. If a female carrier of DMD mates with a normal man, half of her sons will be affected, on average. B. If a female carrier of DMD mates with a normal man, half of her daughters will be carriers, on average. C. If a female carrier of DMD mates with a male affected with DMD, all of their sons will be affected. D. If a male affected with DMD mates with a normal woman, all of his daughters will be carriers. E. None of the statements are false.

C. If a female carrier of DMD mates with a male affected with DMD, all of their sons will be affected. * X-linked recessive inheritance - Females: XrXr, Males: XrY - Female carrier: XRXr - Normal female: XRXR, Normal Male: XRY A. XRXr x XRY = XRY, XrY, XRXR, XRXr - true. B. XRXr x XRY = XRY, XrY, XRXR, XRXr - true C. XRXr x XrY = XRY, XrY, XRXr, XrXr - false D. XrY x XRXR = XRXr, XRY - true

Problem Set #2: Pedigree Problem Set The pedigree shows the inheritance of the phenotypes cardiac arrhythmia (a) and partial deafness (d) in the W and F families. A study was undertaken to determine if the arrhythmia in these families was likely (1) the autosomal recessive form, Jervell Lang-Neilson syndrome, also characterized by partial deafness, (2) Ward-Romano syndrome, an autosomal dominant form, not associated with deafness, or (3) perhaps a third type, inherited in some other fashion. The problem does not specify that the conditions are rare, but you should assume that they are (except when answering question 64). The deafness and the arrhythmia in the W family are the result of alleles at different loci because: A. non-disjunction occurs among the progeny of the first couple in generation II B. Multifactorial inheritence is suggested from comparing the concordance of the twins to their siblings C. independent assortment of these traits can be observed in generation III D. they exhibit epistasis E. actually, the deafness and arrhythmia are associated with the sam locus, or with two tighly-linked loci.

C. independent assortment of these traits can be observed in generation III The problem does not specify that the conditions are rare, but you should assume that they are (except when answering question 64). * RARE If the problem "tells" you the disease is rare, then when considering the hypothesis of recessiveness, you can assume that multiple carriers did not marry into the family. If the problem "tells" you the disease is rare, then... when considering the hypothesis of dominance, you can assume that the founder is a heterozygote. * a means affected by arrhythmia -> A * d means affected by partial deafness. -> D

Problem Set #2: Pedigree Problem Set The pedigree shows the inheritance of the phenotypes cardiac arrhythmia (a) and partial deafness (d) in the W and F families. A study was undertaken to determine if the arrhythmia in these families was likely (1) the autosomal recessive form, Jervell Lang-Neilson syndrome, also characterized by partial deafness, (2) Ward-Romano syndrome, an autosomal dominant form, not associated with deafness, or (3) perhaps a third type, inherited in some other fashion. Note: this series of questions (except for question 64, which Dr. Bardwell wrote) is taken from an actual MCAT exam. The problem does not specify that the conditions are rare, but you should assume that they are (except when answering question 64). The arrhythmia in the W family is most likely dominant, because it: A. affects both monozygotic twins in the third generation B. is sometimes associated with deafness C. is passed from generation to generation without skipping D. is transmitted by both sexes E. actually, it is most likely recessive

C. is passed from generation to generation without skipping The problem does not specify that the conditions are rare, but you should assume that they are (except when answering question 64). * RARE If the problem "tells" you the disease is rare, then when considering the hypothesis of recessiveness, you can assume that multiple carriers did not marry into the family. If the problem "tells" you the disease is rare, then... when considering the hypothesis of dominance, you can assume that the founder is a heterozygote. * a means affected by arrhythmia -> A * d means affected by partial deafness. -> D Autosomal dominant: works I: aa x Aa = Aa, aa Aa x aa = Aa, aa II: Aa x aa = Aa, aa Aa x aa - Aa, aa Aa x Aa = Aa, Aa, AA, aa Aa x aa = Aa, aa aa x aa = aa Autosomal recessive: doesn't work I: AA x aa = Aa (none have it) (DOESN'T WORK)

Problem Set #2: Pedigree Problem Set The pedigree shows the inheritance of the phenotypes cardiac arrhythmia (a) and partial deafness (d) in the W and F families. A study was undertaken to determine if the arrhythmia in these families was likely (1) the autosomal recessive form, Jervell Lang-Neilson syndrome, also characterized by partial deafness, (2) Ward-Romano syndrome, an autosomal dominant form, not associated with deafness, or (3) perhaps a third type, inherited in some other fashion. Note: this series of questions (except for question 64, which Dr. Bardwell wrote) is taken from an actual MCAT exam. The problem does not specify that the conditions are rare, but you should assume that they are (except when answering question 64). The fact that the deafness in the W family is most likely autosomal is shown by: A. male to male transmission B. lack of male to male transmission C. two affected males have daughters whose hearing is normal D. an affected female has a daughter whose hearing is normal E. actually, the deafness is most likely X-linked

C. two affected males have daughters whose hearing is normal (Note you have figured out it's dominant at this point, so this question is just about determining whether it is AD or XLD) - if inheritance was X-linked, the daughters would've had affected hearing. Recessive? * Autosomal recessive (rare): I: dd x DD = Dd (Doesn't work, would give only non-affected). * X-linked recessive: I: XDY x XdXd = XDXd, XdY (Doesn't work, only affected boys expected) We know it's dominant: determine AD vs. XD: A. no male-to-male transmission present. B. lack of male-to-male transmission supports X-linked inheritance. D. Doesn't reject X-linked. (XDXd x XdY = XDXd, XdXd, XDY, XdY) E. Not X-linked, dominant dad didn't affect his daughter. The problem does not specify that the conditions are rare, but you should assume that they are (except when answering question 64). * RARE If the problem "tells" you the disease is rare, then when considering the hypothesis of recessiveness, you can assume that multiple carriers did not marry into the family. If the problem "tells" you the disease is rare, then... when considering the hypothesis of dominance, you can assume that the founder is a heterozygote. * a means affected by arrhythmia -> A * d means affected by partial deafness. -> D

Chapter 10: Chromosome Aberrations & Disease Review homologous vs. non-homologous chromosomes vs. sister chromatids. TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN AGAIN AGAIN AGAIN AGAIN AGAIN AGAIN Position Effect Variegation: Effect of Chromatin State on Transcription: * What role does chromatin state play in the transcription of genes in eukaryotes? * In what regions of the chromosome are most expressed genes located? - What are characteristics about this region? * In what regions of the chromosome are few expressed genes located? - What are characteristics about this region? * How does the level of DNA compaction by chromatin relate to gene transcription? - Define PEV. * Understand the example given. - When a biologist used X-rays to mutate fruit flys, giving them white variegated (patchy) eye color instead of the wild-type red eye color, what was puzzling about the product of mutation? * What did he find when looking at the chromosomes of mutant flies with variegated eye color? How did this occur mechanistically? * How did this affect the gene causing the puzzling abnormality?

Chromatin state plays a critical role in the transcription of genes in eukaryotes. Most expressed genes are located in euchromatic regions of chromosomes, where DNA is not as tightly affiliated with histones. In contrast, relatively few expressed genes are found in heterochromatic regions, where histones and other proteins tightly bind DNA. The experimental evidence that first suggested a direct link between gene transcription and the level of DNA compaction by chromatin came from research into a phenomenon called position effect variegation (PEV). Position effect variegation was discovered in connection with a red and white variegated eye color seen in certain Drosophila mutants. Recall that the wild-type X-linked allele produces red eye color in the fruit flies. In the 1920s and 1930s, Hermann Muller created mutations in fruit flies using X-rays. In one experiment, he irradiated flies with wild-type red eye color and generated flies with mutant variegated eye color. He noticed that the pattern of variegation differed from fly to fly and that the two eyes of a single fly also had different variegation patterns. At the time Muller studied this trait, the production of the white eye color was known to be caused by a mutation, but the variability of variegation pattern was puzzling. Looking at the chromosomes of mutant flies with variegated eye color, Muller discovered that the X chromosome had undergone an inversion. Exposure to X-rays had broken the X chromosome, and the broken ends reattached to form a paracentric inversion. Muller examined the banding patterns in the inverted X chromosome and noticed that variegated flies had a particular kind of paracentric inversion. Their inversions had moved the w gene from its normal location near the telomere of the X chromosome to a new location very near the chromosome centromere (Figure 10.28).

Chapter 10: Chromosome Aberrations & Disease Know the four major types of chromosome rearrangements. * Know what is happening in each rearrangement. - Know what causes this rearrangement. - Know what each rearrangement leads to. - For each type of rearrangement, be able to classify it as a balanced rearrangement, or a unbalanced rearrangement. - For each type of rearrangement, know if it results in a loss or gain of a lot of chromosomal material, and if so, know where this loss/gain of material occurs. * Of the rearrangement(s) that results in a loss or gain of a lot of chromosomal material, know which one is the most severe. - When relevant, know: * Benefit of the rearrangement. * How frequent the rearrangement is. * Where the rearrangement occurs with respect to the chromosome.

Chromosome Rearrangements: 1. Reciprocal Translocation: * Interchange of genetic info between non-homologous chromosomes. * Present in at least 1/500 people. * Probably result from a mistake by the recombination machinery. PIC: Reciprocal Translocation between p arms of 1 and 9 * The chromosomes switch p arms. There is not a big gain/loss of genetic info, this is only possible at the break point. 2. Inversion: Genetic rearrangement in which the order of genes is reversed in a chromosome segment. PIC: A gene or two is messed up at the breakpoints. Summary of Translocation and Inversion * Balanced rearrangements. * Do not result in a loss or gain of much chromosomal material, only the breakpoints are the issue where this happens. 3. Chromosomal Deletions * Deletions = missing chromosomal segment. * Large deletions often lethal (even in heterozygote). 4. Gene Duplications * Chromosome segment present in multiple copies. * Provide material for evolution. * Tandem duplications are frequent. - Repeated segments are adjacent. - Often results from unequal crossing-over. Summary of Deletion and Duplication * Unbalanced rearrangements. * Result in a loss (deletion) or gain (duplication) of a chunk of the chromosome. * Usually severe consequences for affected patients, with deletions more severe than duplications.

Chapter 10: Chromosome Aberrations & Disease TEXTBOOK EXPLANATION AGAIN Know the four major types of chromosome rearrangements. "Skim": * Define chromosome breakage. * What does breakage not followed by reattachment of the broken segments lead to? * Define chromosome inversion. - What does it cause? * Define chromosome translocation. * How might individuals that have a chromosome inversion or translocation not experience any phenotypic abnormalities? * Define reciprocal balanced translocation. - How are the members involved in this process altered? * Describe the relationships among the members involved.

Chromosome breakage involves double-strand DNA breaks that sever a chromosome. Breakage that is not followed by reattachment of the broken segment leads to partial chromosome deletion—but what happens if the broken chromosome reassembles with the broken segment reattached in the wrong orientation or if the broken segment reattaches to a nonhomologous chromosome? The answers are that reattachment in the wrong orientation produces a chromosome inversion, whereas attachment to a nonhomologous chromosome results in chromosome translocation. A repeating theme that emerges is that as long as no critical genes or regulatory regions are mutated by chromosome breakage, and as long as dosage-sensitive genes are retained in their proper balance, individuals that have a chromosome inversion or a chromosome translocation might not experience any phenotypic abnormalities. However, complications during meiosis may affect the efficiency of chromosome segregation, and fertility may be affected in those individuals. Chromosome inversions occur as a result of chromosome breaks followed by reattachment of the free segment in the reverse orientation. Chromosome inversion causes a difference in linear order of genes on homologous chromosomes by a 180-degree reorientation of the inverted segment. Chromosome translocation takes place when chromosome breakage is followed by the reattachment of a broken segment to a nonhomologous chromosome. Reciprocal balanced translocation is produced when breaks occur on two nonhomologous chromosomes and the resulting fragments switch places when they are reattached. In reciprocal balanced translocation, one member of each homologous pair is altered by translocation and none of the four chromosomes has a fully homologous partner. Instead, the translocated chromosome segments homologous to the normal member of each pair are dispersed on two other chromosomes.

Chapter 10: Chromosome Aberrations & Disease Review homologous vs. non-homologous chromosomes vs. sister chromatids. TEXTBOOK EXPLANATION Know characteristics of chromosomes. * What is the p arm? * What is the q arm? * In general, how do less condensed chromosome regions compare to more heavily compacted regions in terms of the activity of gene transcription? * Define euchromatin. * Define heterochromatin. * In what regions are few expressed genes found? * In what regions are most expressed genes found? * What DNA (heterochromatic or euchromatic) is more likely to contain repetitive DNA sequences that may be located in multiple regions of the genome?

Chromosomes are divided by their centromere into segments known as chromosome arms that are almost always of unequal lengths. One chromosome arm, called the short arm, also known as the p arm, is shorter than the other arm that is known as the long arm, or the q arm. In general, the chromosome regions populated by actively transcribed genes are relatively less condensed than chromosome regions with few transcribed genes, which are more heavily compacted. Regions of lesser chromatin compaction are identified as euchromatin, or as euchromatic regions. Most expressed genes are located in euchromatic regions, where condensation is variable during the cell cycle. Conversely, chromosome regions in which chromatin is tightly condensed are said to contain heterochromatin and are called heterochromatic regions. Heterochromatic regions contain many fewer expressed genes than do euchromatic regions. With fewer expressed gene sequences, heterochromatic DNA is more likely than euchromatic DNA to contain repetitive DNA sequences that may be located in multiple regions of the genome.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED Describe variations from classical Mendelian genetics, including incomplete dominance, co-dominance, multiple alleles/allelic series, and lethal alleles. Codominance: * What does codominance lead to? * What is codominance characterized by? * When is codominance most clearly identified? How would one identify this? - Compare and contrast codominance to incomplete dominance. How are they similar, and how are they different? Allelic series: * What do diploid genomes contain? * How many alleles, at most, can each individual organism possess at a locus? How does this compare to the population? * Define multiple alleles. - What do many multiallelec genes display? - What often emerges among the alleles? * What is this based on? * What does this form? - What are characteristics of alleles in this? What can they display?

Codominance, like incomplete dominance, leads to a heterozygous phenotype different from the phenotype of either homozygous parent. Unlike incomplete dominance, however, codominance is characterized by the detectable expression of both alleles in heterozygotes. Codominance is most clearly identified when the protein products of both alleles are detectable in heterozygous organisms, typically by means of some sort of molecular analysis or a biochemical assay that can distinguish between the different proteins. An example of codominance is presented in the following discussion of ABO blood type. Diploid genomes contain pairs of homologous chromosomes; thus, each individual organism can possess at most two alleles at a locus. In populations, however, the number of alleles is theoretically unlimited, and some genes have scores of alleles. At the population level, a locus possessing three or more alleles is said to have multiple alleles; and like the ABO gene, many multiallelic genes display a variety of dominance relationships among the alleles. Commonly, an order of dominance emerges among the alleles, based on the activity of each allele's protein product, forming a sequential series known as an allelic series. Alleles in an allelic series can be completely dominant or completely recessive, or they can display various forms of incomplete dominance or codominance.

Chapter 4: Gene Interactions Interpret the results of complementation analysis crosses. * When studying a specific biological process, what will biologists sometimes do? Why? * What does complementation analysis do? * When organisms have the same mutant phenotype, what do genetecists ask? - How can they answer these questions? * Describe this process. * What will happen if complementation occurs? What does this mean? * What will happen if the mutations fail to complement? What does this mean? * Understand picture attached, which explains complementation analysis. - If the wild-type flower is purple, what are the possible scenarios that can occur (mutants are white flowers). - What does it mean for mutants to "complement"? What can we conclude from this? What does the result imply? - What does it means for mutants to "fail to complement"? What can we conclude about this?

Compelementation Analysis When studying a specific biological process, sometimes biologists will conduct a "screen" to find many mutants with phenotypes related to that process. Complementation Analysis Distinguishes Mutations in the Same Gene from Mutations in Different Genes * When organisms have the same mutant phenotype, geneticists ask: - Do these organisms have mutations in the same or in different genes? - How many genes are responsible for the phenotypes observed? * They can answer the questions through complementation testing. - Two pure-breeding strains with similar mutant phenotypes are mated. - If complementation occurs, wild type offspring are obtained -> the mutations affect two different genes. - If the mutations fail to complement, the offspring have the mutant phenotype -> the mutations affect the same gene. PIC: Wild-type is purple. Scenario 1: Mutant phenotype can result from mutations in either of two genes. We say that these mutants "complement" and can conclude they occurred in two different genes. (And these two genes are epistatic - the result implies "complementary gene interaction" epistasis). Scenario 2: Mutant phenotype is caused by mutations in the same gene. We say that these mutants "fail to complement" and can conclude they occurred in the same gene.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN Define epistasis, compare and contrast complementary gene interaction epistasis and duplicate gene action epistasis. Complementary Gene Interaction * What phenotypes do wild-type sweet peas have? - When two pure-breeding mutants that had white flowers were crossed, what was expected, and what resulted? - When these F1 progeny were crossed to produce the F2 generation, what was the phenotypic ratio that resulted? * How could these results possibly be explained? - What does this mean for the genotypes of the parental line (the mutant flowers)? What must their genotypes be? - What does this make the mutants if this is their genotypes (what would happen if these parental mutants selfed)? What is this a result of? - What does a cross of these two lines of pure-breeding white parents produce (in terms of phenotype and genotype)? * How do we know the phenotype? What causes the phenotype?

Complementary Gene Interaction (9:7 Ratio) Wild-type sweet peas have purple flowers, and the experiments began with crossing two pure-breeding mutant plants that had white flowers. Bateson and Punnett expected these mutants to produce mutant (white-flowered) progeny, but to their surprise, the F1 generation all had purple flowers. When Bateson and Punnett crossed F1 plants, the F2 produced a ratio of (9/16) purple-flowered plants to (7/16) white-flowered plants. Bateson and Punnett recognized that their results could be explained if two genes interacted with one another to produce sweet pea flower color. Assuming two genes are responsible for a single pigment that gives the sweet pea flower its purple color, each parental line—represented by the genotypes ccPP and CCpp—is pure-breeding for white flowers as a result of homozygosity for recessive alleles at one of the genes. The cross of these two lines of pure-breeding white parents produces dihybrid purple-flowered F1 plants — genotype CcPp—because the dominant allele at each locus enables completion of each step of the pathway leading to the synthesis of purple pigment.

Chapter 4: Gene Interactions (Not learning objective, but should know) * What are complex traits? What all plays a role? - Give an example of a complex trait.

Complex traits: when many different genes each contribute a little bit to a phenotype.. and the environment (nutrition, etc.) also plays a role. - Example: human height.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN Describe variations from classical Mendelian genetics, including incomplete dominance, co-dominance, multiple alleles/allelic series, and lethal alleles. Allelic series: * What do crosses between animals with different genotypes at the C gene indicate? - How would complete dominance of an allele over other alleles in the series be demonstrated? * What allele shows complete dominance over other alleles in the series? - How is the dominance order of alleles in a series revealed? - Know what alleles are completely/partially dominant over eachother. - Know and understand the dominance relationship within the given allelic series locus.

Crosses between animals with different genotypes at the C gene indicate the dominance relations of the alleles. For example, in Crosses A, B, and C in Figure 4.6, complete dominance of C over other alleles in the series is demonstrated by the finding that all of the progeny of an animal with the genotype CC have full color, regardless of the genotype of the mate. The dominance order of alleles in the series is revealed by the pattern of 3:1 ratios obtained from crosses of various heterozygous genotypes shown in Figure 4.6. Cross D shows that chinchilla is completely dominant over albino. Himalayan, too, is completely dominant over albino (Cross E). Cross F shows that the chinchilla allele (c^ch) is partially dominant over the Himalayan allele (c^h). Note the F2 of this cross have a 1:2:1 ratio of phenotypes, with the heterozygous F2 displaying Himalayan markings and dilute coat color over the rest of the body that are both somewhat lighter than in their homozygous counterparts. The dominance relationships within this allelic series locus can be expressed as C > c^(ch) > c^h > c.

Problem Set #2: Pedigree Problem Set 47. Your sister suffers from a rare genetic disorder syndrome (incidence 1/10,000), but you and your parents are unaffected. What is the approximate chance that you are a carrier? A. 1/100, or 1% B. 1/4, or 25% C. 1/2, or 50% D. 2/3, or 67% E. 100%

D. (2/3) or 67% * RARE If the problem "tells" you the disease is rare, then when considering the hypothesis of recessiveness, you can assume that multiple carriers did not marry into the family. If the problem "tells" you the disease is rare, then... when considering the hypothesis of dominance, you can assume that the founder is a heterozygote. * Can't be dominant, unaffected parents make affected child. Recessive: Autosomal recessive: Rr x Rr = RR, Rr, Rr, rr X-linked recessive: doesn't work I: XRY x XRXr = XRXR, XRXr, XRY, XrY (need XrXr) We know I'm dominant, since I'm not affected, so given the three dominant genotypes available from producing the Rr x Rr cross, I have a (2/3) chance of being a carrier.

Problem Set #2: Pedigree Problem Set A female carrier of hemophilia A mates with a man who carries one wild-type and one disease-causing allele of Huntingtin (the gene that when mutated can cause Huntington's disease). What percentage of their sons will have both hemophilia and Huntington's disease? (In order to answer this question, you need to have memorized the mechanism of inheritance of hemophilia A and Huntington's disease - both are typically discussed in class) A. 100% B. 75% C. 50% D. 25% E. 0%

D. 25% * Huntington: autosomal dominant * Hemophilia A: X-linked recessive XRXr x XRY = XRXR, XRXr, XRY, XrY dd x Dd = Dd, dd * (1/2 sons have hemophilia) x (1/2 children have huntington) = (1/4 songs will have both)

Problem Set #2: Pedigree Problem Set Judy's paternal grandmother (that is, her father's mother) and great grandmother have breast cancer. Both have tested positive for a cancer-predisposing allele of the BRCA2 gene. Judy's father does not want to get tested, and will not let Judy (who is only 17) get tested. Based on the information available, what is the probability that Judy has inherited the cancer-predisposing allele? Note: BRCA2 is on chromosome 13, and cancer-predisposing alleles of BRCA2 are dominant (and rare) A. 100% B. 67% (2/3) C. 50% (1/2) D. 25% (1/4) E. 0%

D. 25% (1/4) * Autosomal dominant: - BRCA2 is on chromosome 13 -> autosomal - Cancer-predisposing alleles of BRCA2 are dominant (and rare) -> dominant If the problem "tells" you the disease is rare, then... when considering the hypothesis of dominance, you can assume that the founder is a heterozygote. Judy's dad has a half chance of inheriting the cancer-predisposing allele. If he has it, that gives Judy a half chance of inheriting the cancer predisposing allele. * (1/2) x (1/2) = (1/4) or 25%

Problem Set #2: Pedigree Problem Set The pedigree shows the inheritance of the phenotypes cardiac arrhythmia (a) and partial deafness (d) in the W and F families. A study was undertaken to determine if the arrhythmia in these families was likely (1) the autosomal recessive form, Jervell Lang-Neilson syndrome, also characterized by partial deafness, (2) Ward-Romano syndrome, an autosomal dominant form, not associated with deafness, or (3) perhaps a third type, inherited in some other fashion. Note: this series of questions (except for question 64, which Dr. Bardwell wrote) is taken from an actual MCAT exam. The problem does not specify that the conditions are rare, but you should assume that they are (except when answering question 64). Strictly speaking, the only mode(s) of inheritance that can absolutely be ruled out for the transmission of the deafness characteristic is (for this problem do not assume the deafness is rare): A. Autosomal recessive B. X-linked dominant C. X-linked recessive D. B and C E. A and B and C

D. B and C Recessive? * Autosomal recessive: works I: Rr x rr = Rr, rr II: Rr x rr = Rr, rr rr x Rr = Rr, rr rr x Rr = Rr, rr * X-linked recessive: doesn't work XRY x XrXr = XRXr, XrY (this says no daughters are expected to be affected) Dominant? * Autosomal dominant: works I: rr x RR = Rr II: Rr x rr = Rr, dd Rr x rr = Rr, rr Rr x rr = Rr, rr * X-linked dominant: doesn't work I: XrY x XRXR = XRXr, XRY II: XRXr x XrY = XRXr, XrXr, XRY, XrY XRY x XrXr = XRXr, XrY (all daughters should be affected) XRY x XrXr = XRXr, XrY (all daughters should be affected)

Problem Set #2: Pedigree Problem Set A male with hypophosphatemic rickets marries a normal woman. They have 8 children (4 boys and 4 girls). All the boys are normal but all the females are affected. What type of inheritance does this suggest (that is, which is the most likely given the observations)? A. Autosomal recessive B. X-linked recessive C. Autosomal dominant D. X-linked dominant E. Y -linked dominant

D. X-linked dominant XRY x XrXr = XRXr, XrY * Normal: not a carrier Hint: Dad passes his XR to his daughters, making them affected. Mom is normal, so she's recessive. Since she gives her Xr to her sons, none of them will be affected.

Problem Set #1: Pedigrees A female affected with Klutzki's disease mates with an unaffected male and has an affected daughter. Which type(s) of inheritence can be ruled out for Klutzki's disease? A. Autosomal dominant B. X-linked dominant C. Autosomal recessive D. X-linked recessive E. A, B and D

D. X-linked recessive Recessive? * Autosomal recessive: works. - rr x Rr = Rr, rr Half affected, half not. * X-linked recessive: doesn't work. XrXr x XRY = XRXr, XrY Girls not affected, boys affected. Dominant? * Autosomal dominant: works - Rr x rr = Rr, rr OR RR x rr = Rr Half affected, half not, or all affected. * X-linked dominant: XRXr x XrY = XRXr, XrXr, XRY, XrY Half girls have it, half don't. Half boys have it, half don't. XRXR x XrY = XRXr, XRY All children have it.

Problem Set #2: Pedigree Problem Set A female affected with Klutzki's disease mates with an unaffected male and has an affected daughter. Which type(s) of inheritence can be ruled out for Klutzki's disease? A. Autosomal dominant B. X-linked dominant C. Autosomal recessive D. X-linked recessive E. A, B and D

D. X-linked recessive Recessive? * Autosomal recessive: works. - rr x Rr = Rr, rr Half affected, half not. * X-linked recessive: doesn't work. XrXr x XRY = XRXr, XrY Girls not affected, boys affected. Dominant? * Autosomal dominant: works - Rr x rr = Rr, rr OR RR x rr = Rr Half affected, half not, or all affected. * X-linked dominant: XRXr x XrY = XRXr, XrXr, XRY, XrY Half girls have it, half don't. Half boys have it, half don't. XRXR x XrY = XRXr, XRY All children have it.

Problem Set #2: Pedigree Problem Set For this pedigree, if you cannot assume that the disease or trait is rare, can you rule out any of the 4 standard mechanisms of inheritance? A. Yes, you can rule out autosomal recessive B. Yes, you can rule out x-linked recessive C. Yes, you can rule out autosomal dominant D. Yes, you can rule out x-linked dominant E. No you cannot rule out any

D. Yes, you can rule out x-linked dominant Recessive? Autosomal recessive: works I: rr x Rr = Rr, rr X-linked recessive: works I: XrY x XRXr = XRXr, XrXr, XRY, XrY Dominant? Autosomal dominant: works I: Rr x rr = Rr, rr X-linked dominant: doesn't work. I: XRY x XrXr = XRXr, XrY - means all daughters and no sons should be affected.

Chapter 4: Gene Interactions (Not a learning objective, but should know) * How is ABO blood type identified? - What does this identification process involve? - Know the components of the process and their functions. - When does a positive reaction occur? What happens during this reaction? What does this result indicate to the identifier? What does "no result" indicate? * What would blood from a person with blood type A show? * What would blood from a person with blood type B show? * What would blood from a person with blood type AB show? * What would blood from a person with blood type O show? - What is the antigen-antibody relationship in humans? * What antibodies would people with blood type A carry? * What antibodies would people with blood type B carry? * What antibodies would people with blood type AB carry? * What antibodies would people with blood type O carry?

Determining ABO Blood Type ABO blood type is identified by an antigen-antibody reaction on a microscope slide. The test involves placing a drop of blood into a drop of anti-A antiserum in one well of a microscope slide and placing another drop of blood into anti-B antiserum in the other well of the slide. The two antisera contain antibodies, molecules produced by the immune system that bind to a specific antigen (for each kind of antibody there is a specific antigen). Each antigen in the case of ABO blood type is a carbohydrate group (sugar) embedded on the surface of red blood cells. A positive reaction occurs when an antibody detects its antigen target. The antibody binds the antigen and also attaches to other antigen-bound antibodies, causing red blood cells to form visible clumps. Clumping indicates that the antibody has detected its antigen target, whereas an absence of clumping indicates that the blood does not contain the antigen target of the antibody. Blood from a person with blood type A shows clumping with anti-A antiserum but not with anti-B (Figure 4.3). Conversely, blood type B is identified when clumping occurs with anti-B but not with anti-A. If clumping occurs with both antisera, the blood type is AB. Clumping with neither antiserum identifies blood type O. The antibodies anti-A and anti-B develop in humans from birth, but people do not carry an antibody if they also carry the corresponding antigen. Thus people with blood type A, who have the A antigen, also carry the anti-B antibody. People with blood type B have the B antigen and the anti-A antibody. Those with blood type AB have both antigens and neither anti-A nor anti-B antibody. Finally, people with blood type O have neither A nor B antigen and have both anti-A and anti-B antibodies.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN AGAIN AGAIN Describe the different classes of gain-of-function and loss-of-function mutations and relate these classes to Mendel's definitions of dominant and recessive alleles. * Know dominant loss-of-function mutations. - What is their occurrence? - What do they produce? - How do these mutations do this? What do they alter? * Know the components of what they alter. * How are these things subject to dominant negative mutation? * How would this "thing" containing an abnormal component suffer? - Why are mutations of this kind dominant? - How are these mutations characterized? Why?

Dominant loss-of-function mutations are also known to occur. Some of these produce dominant mutant phenotypes through alterations in the function of a multimeric protein of which the mutant polypeptide forms a part (Figure 4.1d). Multimeric proteins, composed of two or more polypeptides that join together to form a functional protein, are particularly subject to dominant negative mutations as a consequence of some change that prevents the polypeptides from interacting normally to produce a functional protein. A multimeric protein that contains an abnormal polypeptide may suffer a reduction or total loss of functional capacity. Mutations of this kind are dominant due to the substantial loss of function of the multimeric protein (as illustrated in the following paragraph). These mutations are characterized as "negative" due to the spoiler effect of the abnormal polypeptide on the multimeric protein. An example of dominant negative mutation is seen in the human hereditary disorder osteogenesis imperfecta (OMIM 116200, 116210, and 116220), which is caused by defects in the bone protein collagen and has multiple forms with different severity. Collagen protein is composed of three interwoven polypeptide strands—two polypeptides from the COL1A1 gene and one polypeptide from the COL1A2 gene. The trimeric collagen protein is subject to dominant negative mutation as a consequence of COL1A1 mutations that produce a defective polypeptide. The trimeric structure of collagen and the 2:1 ratio of incorporation of COL1A1 polypeptide over COL1A2 polypeptide means that in individuals who are homozygous wild type for COL1A2 and heterozygous for COL1A1 mutation, most collagen protein contains one or two mutant COL1A1 proteins. As a result, most collagen protein is defective, and osteogenesis imperfecta develops.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN AGAIN AGAIN Duplicate Gene Action * What constitutes a redundant genetic system? * What does this mean in terms of genotype/phenotype? - What alleles are required to produce the dominant phenotype? - What alleles are required to produce the recessive phenotype? * What kind of action are the genes in a redundant system said to have? - What does this mean? * Understand Mendel's bean experiment and what it identified. - When Mendel crossed pure-breeding purple-flowered bean plants to pure-breeding white-flowered bean plants, what F1 progeny phenotypes were produced? * Of the F2 progeny, what phenotypes were produced? - What does this say about their genotypes? * What would occur if at least one copy of a dominant allele were present at either loci (in terms of the protein product and its function)? * What would occur if homozygous recessive alleles were present at both loci ( ). * What do white flowers result from (know ratio, genotype)?

Duplicate Gene Action (15:1 Ratio) Two genes that duplicate one another's activity constitute a redundant genetic system in which any genotype possessing at least one copy of a dominant allele at either locus will produce the dominant phenotype. Only when recessive homozygosity is present at both loci does the recessive phenotype appear. The genes in a redundant system are said to have duplicate gene action; either they encode the same gene product or they encode gene products that have the same effect in a single pathway or compensatory pathways. Figure 4.21 ➋: illustration and explanation of duplicate gene action identified inadvertently by Mendel in an experiment involving flower color in bean plants. Mendel described an experiment with beans that began with the cross of a pure-breeding purple-flowered bean plant to a pure-breeding white-flowered bean plant. The F1 plants all had purple flowers, and Mendel probably assumed that flower-color determination in beans would follow the same pattern as in peas. Among the 32 F2 plants Mendel produced, however, 31 had purple flowers and only 1 had white flowers. Among the F2 plants, (15/16) have a genotype containing at least one copy of either P or R, and only (1/16) have the genotype pprr and the white-flowered phenotype. Figure 4.21 ➋ shows that the protein product of the dominant allele of either gene is capable of catalyzing the conversion of a precursor to anthocyanin and producing the dominant phenotype. Conversely, if homozygous recessive alleles are present at both loci, no functional gene product is produced, and the synthesis pathway is not completed. White flowers result from the absence of pigment in the (1/16) of the F2 progeny that are homozygous recessive for alleles of both genes.

Problem Set #1: Mitosis, Meiosis Certain kinds of cells (e.g., some cells in the eyes and bones) mature and differentiate into a state in which they have a specialized function but do not divide or progress through the cell cycle. These cells are "stuck" inwhich stage? A) M phase B) G1 C) G2 D) S E) G0

E) G0

Problem Set #2: Pedigree Problem Set A man affected with the autosomal recessive disorder cystic fibrosis (CF) mates with a normal woman. Which of the following statements about the transmission of CF to his kids is TRUE: A. All of his sons will be affected B. All of his sons will be carriers C. All of his daughters will be affected D. All of his daughters will be carriers E. B and D

E. B and D rr x RR = Rr * Normal: not a carrier, (RR) in this situation

Problem Set #2: Pedigree Problem Set A girl is born with Yutski's disease, but her parents are unaffected. Which type(s) of inheritance can be ruled out for Yutski's disease. A. Autsomal recessive B. Autosomal dominant C. X-linked recessive D. X-linked dominant E. B, C and D

E. B, C and D. * Has to be recessive since unaffected parents have affected child. * Girl is XrXr or rr. * Autosomal recessive: works Rr x Rr = rr * X-linked recessive: doesn't work XRXr x XRY = XRXR, XRXr, XRY, XrY Unaffected daughters, half affected sons. Dad has to be affected in order to pass his "bad X" onto his daughter.

Problem Set #2: Pedigree Problem Set A boy is born with an autosomal recessive genetic disease that is determined by a single-gene defect. Both his parents are unaffected. What can we say about the parents genotypes with respect to the relevant locus?: A. One parent is homozygous B. Both parents are homozygous C. Only one of the two parents is a carrier D. Mom is a heterozygote, and dad is hemizygous wild-type E. Both his parents must be heterozygotes

E. Both his parents must be heterozygotes Only way to go from two dominant genotypes and get dd as a possible progeny is to cross: Dd x Dd = dd * Autosomal recessive

Problem Set #2: Pedigree Problem Set In humans, both freckling (that is, having freckles) and dimpling (that is, having dimples are autosomal dominant traits. If a freckled man without dimples (Fred) mates with adimpled woman with no freckles (Doris), what can we say with certainty about their kids? A. All of their kids will be both freckled and dimpled B. Their kids will either be dimpled and not freckled or freckled and not dimpled C. None of their kids will be both freckled and dimpled D. Whether or not the kids are freckled or dimpled will depend upon the sex of the kids E. None of the above (A-D) can be said with certainty

E. None of the above (A-D) can be said with certainty * Autosomal dominant traits FFdd x ffDD = FfDd OR Ffdd x ffDd = FfDd, Ffdd, ffDd, ffdd OR FFdd x ffDd = Ffdd, FfDd OR Ffdd x ffDD = FfDd, ffDd

Problem Set #2: Pedigree Problem Set In humans, both freckling (that is, having freckles) and dimpling (that is, having dimples) are autosomal dominant traits. If a woman with freckles and dimples (Felicia) mates with a non-dimpled, non-freckled man (David), what can we say with certainty about their kids? A. All of their kids will be both freckled and dimpled B. We would expect some of their kids to be dimpled and some not to be C. None of their kids will be both freckled and dimpled D. Whether or not the kids are freckled or dimpled will depend upon the sex of the kids E. None of the above (A-D) can be said with certainty

E. None of the above (A-D) can be said with certainty FFDD x ffdd = FfDd OR FfDd x ffdd = FfDd, ffDd, Ffdd, ffdd OR FFDd x ffdd = FfDd, Ffdd OR FfDD x ffdd = FfDd, ffDd

Problem Set #2: Pedigree Problem Set Which of the following statements are true? (assume the person in the question has mated with a normal partner) A. In x-linked dominant inheritance, it is possible for an unaffected woman to have an affected son B. In x-linked dominant inheritance, it is possible for an affected man to have an unaffected daughter C. In x-linked recessive inheritance, it is possible for an affected woman to have an unaffected son D. In x-linked recessive inheritance, it is possible for an unaffected man to have an affected daughter E. None of the above are true

E. None of the above are true. A. Not true, an unaffected women would be XrXr, and she'd give her Xr to her sons, making her sons unaffected. B. Not true, an affected man would be XRY, and he'd give his XR to his daughters, making them affected. C. Not true, an affected women would be XrXr, and she'd give her Xr to her sons, making them affected. D. Not true, an unaffected man would be XRY, and he'd give his XR to his daughters, making them unaffected.

Chapter 4: Gene Interactions (Not a learning objective, but should know) * What can be added to modify the H antigen? By what? * If the IA allele is present in the genotype, what gene product does it produce? What is the function of this gene product? * If the IB allele is present in the genotype, what gene product does it produce? What is the function of this gene product? * How do A and B alleles differ? What does this cause in relation to the gene products? What does this lead to? * What is the i allele due to? What kind of allele is it? What does it produce? * At the cellular level, what does anti-A recognize? * What does anti-B recognize? * What recognizes the H antigen? In the people of what blood type is the H antigen found? * What alleles in a genotype is sufficient to produce an ABO antigen detectable by the corresponding antibody? * What alleles are dominant/recessive to what alleles? Why? * When the IAIB genotype is present, what gene products are produced? How does this alter the H antigen? In the case of this, what do the H-antigens that the red blood cells carry look like? What does this say about the phenotype? What does this conclude?

Either of two alternative sugars can be added to the H antigen by the respective gene products of the IA or IB allele. If the IA allele is present in the genotype, it produces the gene product alpha-3-N-acetyl-d-galactosaminyltransferase, or simply, "A-transferase." A-transferase catalyzes the addition of the sugar N-acetylgalactosamine to the H antigen, producing a six-sugar oligosaccharide known as the A antigen. The IB allele, on the other hand, produces alpha-3-d-galactosaminyltransferase , commonly called "B-transferase," which catalyzes the addition of a different sugar, galactose, and produces a six-sugar oligosaccharide known as the B antigen. Molecular analysis reveals that the A and B alleles differ in several nucleotides, causing four amino acids of the resulting transferase enzymes to differ and leading to differences in enzymatic activity. In contrast, the i allele is due to a single base-pair deletion and is a null allele that does not produce a functional gene product capable of adding a sixth sugar to the H antigen. At the cellular level, anti-A antibody recognizes the N-acetylgalactosamine addition mediated by IA, and anti-B antibody identifies the galactose addition produced by the action of IB. Neither of these antibodies has any reactivity with the unmodified H antigen, so unmodified H antigen, present in individuals with blood type O, is not recognized by either antibody. One copy of the IA or the IB allele in a genotype is sufficient to produce an ABO antigen detectable by the corresponding antibody; and both IA and IB are dominant to i, since IA and IB produce enzymes that modify the H antigen but i does not. When the IAIB genotype is present, on the other hand, both A-transferase and B-transferase are produced, resulting in the addition of N-acetylgalactosamine to some H antigens and the addition of galactose to other H antigens. In this case, all red blood cells carry both types of H-antigen modifications; about one-half of the red cell surface antigens are A antigens, and the rest are B antigens. As a result, the action of both alleles is detected in the phenotype, leading to the conclusion that IA and IB are codominant to one another.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED Provide an example of gene-environment interaction. * In some cases, the expression of a given gene is entirely dependent on the presence of what? - How is this relevant in the medical management world? * What is PKU? - What is its mechanism of inheritance? - What is it caused by? What does this lead to? - What is the benefit of newborn screening? - What is the key management feature shared by all of the hereditary diseases screened by newborn genetic testing? * What might this environmental modification consist of? * What does this environmental modification look like specifically for PKU patients? - What are the effects of making this modification and sticking to it? - What does this environmental modification prevent? * In the case of homozygous recessive PKU patients that have made this environmental modification, what does their phenotype look like?

Environmental Modification to Prevent Hereditary Disease In some cases, the expression of a given gene is entirely dependent on the presence of certain environmental conditions. * An example of the manipulation of this relationship to achieve a desired outcome is found in an element of the medical management that prevents development of the human autosomal recessive condition known as phenylketonuria (PKU) PKU is caused by the absence of the enzyme phenylalanine hydroxylase (PAH), which catalyzes the first step of the pathway that breaks down the amino acid phenylalanine, a common component of dietary protein. Infants with PKU are normal at birth, but over the first several months of life the body's inability to carry out the normal breakdown of phenylalanine leads to the buildup of a compound that is toxic to developing neurons. Tests of newborns identify the disease before the disease has had a chance to manifest itself and begin to damage the body.The key feature shared by all of the hereditary diseases screened by newborn genetic testing is that the disease symptoms can be prevented or substantially reduced in severity by strict and consistent dietary management. Dietary control either prevents individuals from consuming compounds that allow the disease to develop or it provides the essential compound missing in those with the disease. The key dietary control for management of PKU is elimination of the amino acid phenylalanine from the diet as soon as PKU is diagnosed. An infant who is started on the phenylalanine-free diet soon after birth and kept on it through adolescence avoids the complications of PKU and will develop and function normally despite having PKU. Thousands of people with PKU are living fully normal and productive lives today, thanks to this simple environmental modification that prevents the expression of the devastating PKU phenotype. In this case, people who are homozygous recessive for the mutant PKU allele do not express the trait if they are raised in a largely phenylalanine-free environment.

Chapter 4: Gene Interactions Provide an example of gene-environment interaction. * Know environmental impact. * What all is responsible for the variation seen between organims? * Understand gene-environmental interactions: - Know what they are. - Know what they do. - Know an example. * Know the characteristics of this example. - What mode of inheritance is it? - Is it rare? - What is it caused by? Know the components of this. - What are symptoms? - What is treatment? - What is expected for people diagnosed early that maintain a strict treatment plan? - What does screening look like for this?

Environmental impact: genetically identical individuals show different phenotypes as a result of environmental factors. Gene-Environment Interactions Genes alone are not responsible for all the variation seen between organisms. Gene-environment interaction is the result of the influence of the environment on the expression of genes and on the phenotype of the organism. One example: PKU (phenylketonuria) - diet can greatly modify the severity of the disease. * The diet is the environment. PKU (phenylketonuria): * Autosomal recessive. * Very rare - incidence about 1 in 25,000 births. * Caused by the absence of an enzyme involved in phenylalanine breakdown. Enzyme = phenylalanine hydroxylase; Gene = PAH. * Symptoms: infants with PKU are normal at birth, but over time, the inability to break down phenylalanine is toxic to developing neurons. * Treatment is with a diet low in foods that contain phenylalanine, plus special supplements. People who are diagnosed early and maintain a strict diet can have normal health and life spans. * One of the hereditary disorders infants are routinely screened for. * Blood test from baby's heel done 1-2 days after birth can detect PKU.

Chapter 4: Gene Interactions *** CONTINUED *** Define epistasis, compare and contrast complementary gene interaction epistasis and duplicate gene action epistasis. * Understand the pathways epistasis uses to give its products. - Pathway #2: What is it called? * What kinds of alleles are the recessive alleles? * Define duplicate gene action. - What does it allow? - What are the requirements for having a mutant phenotype? * Know the _________ of relevant pathways, and what is involved in each one. (for the ones that PIC says, "you don't need to know these in detail", only know the phenotype ratios expected). - genotype ratios - phenotype ratios - gene interactions * Describe the characteristics of biosynthetic pathways. - What kinds of steps do biosynthetic pathways consist of? - What does completion of one step generate? - What is necessary for production of the end product of the pathway? - Know the function of wild-types in terms of biosynthetic pathways, and why this is. - Know the function of mutants in terms of biosynthetic pathways, and why this is.

Epistasis 2 Parallel Pathway: If you're missing E1, you can still make Anth from precursor I if you have E2. If you're missing E2, you can still make Anth from precursor I if you have E1. If you don't have either, then you can't make Anth. If you have both, then you can use both to make Anth. The recessive allele p is a loss-of-function allele of E1. The recessive allele r is a loss-of-function allele of E2. 2. Duplicate gene action: Duplicate gene action allows dominant alleles of either duplicate gene to produce the wild-type phenotype. Only organisms with homozygous mutations of both genes have a mutant phenotype. (Pic shows the phenotype ratio expected on dihybrid cross with this type of epistasis present) There are many types of epistasis: you don't need to memorize them all!!! Textbook: (1) Biosynthetic pathways consist of sequential steps, (2) completion of one step generates the substrate for the next step in the pathway, and (3) completion of every step is necessary for production of the end product of the pathway. These assumptions support the conclusions that wild-type strains are able to complete each pathway step and that mutant strains are unable to complete a pathway because one or more pathway steps are blocked by mutation.

Chapter 4: Gene Interactions Define epistasis, compare and contrast complementary gene interaction epistasis and duplicate gene action epistasis. * What does epistasis require? * When does it occur? - Give an example. * What does a dihybrid cross look like for epistatic genes? What phenotype ratio is produced? Talk about the ratio. * What have previous encounters with dihybrid crosses involved (how many phenotypes)? * Know what no epistasis looks like on a cross. What ratio does it give, and what phenotypes do you see? * Understand the pathways epistasis uses to give its products. - Pathway #1: What is it called? * What kinds of alleles are the recessive alleles? * Define complementary gene interaction. - What is required for this to occur? - What does mutation cause? * What is another way that epistasis occurs using this pathway?

Epistasis: a gene at one locus alters the phenotypic expression of a gene at a second locus. * Epistasis requires two (or more) different gene loci that control the same trait. * Epistasis occurs when there is phenotypic interaction between these two different genes: e.g. the alleles of one gene modify, allow or prevent expression of alleles for the other gene. * For epistatic genes, a dihybrid cross produces a modified 9:3:3:1 ratio of phenotypes (e.g., 9:7, 15:1). - Genotype ratios do not change, just the phenotypes produced. *Our previous encounters with dihybrid crosses (9:3:3:1 rations, etc.) have involved two different phenotypes, such as seed color and seed shape. * Epistasis requires two (or more) different genes that control the same trait. No Epistasis Here two genes control the 'same' trait, but there's no epistasis, giving a 9:3:3:1 ratio of phenotypes in the F2. Epistasis 1: Sequential pathway: the absence of either E1 or E2 will lead to no product being formed (no pigment). The fxn of both enzymes are required to produce the WT phenotype. - anthocyanin is a pigment that makes flowers purple. A plant that can't make anthocyanin has white flowers. * C and c are alleles of E1, c is a recessive loss-of-fxn allele. * P and p are alleles of E2, p is a recessive loss-of-fxn allele. Complementary gene interaction: Complementary gene interaction occurs when genes must act in tandem to produce a phenotype. The WT action from both genes is required to produce the WT phenotype. Mutation of one or both genes produces a mutant phenotype. * You get the same type of epistasis (sequential pathway) when two proteins combine to form an active enzyme = product only made when gene A and B are combined and functional. If one of the genes is missing, the active site won't be formed, (enzyme won't be active).

Chapter 10: Chromosome Aberrations & Disease Describe euploidy, polyploidy, and aneuploidy. * Define euploid. - Are human somatic cells euploid? Be able to explain why or why not. - Are human gametes euploid? Be able to explain why or why not. * Define aneuploid. - What are characteristics of aneuploid cells? * What are common types of aneuploidy? * Know these types and be able to define them. * What is the most frequent cause of aneuploidy? * Define polyploidy. - How common is polyploidy in flowering plants? - What is a characteristic of triploid plants? - For a polyploidy organism, what is disrupted in meiosis? - In humans, what does polyploidy result in?

Euploid, aneuploid A cell that contains a whole number multiple of the haploid set of chromosomes is called euploid. {you ploid // you = good} * Euploid set is a good set. Thus, human somatic cells (2 haploid sets, or 2n) and human gametes (1n) are both euploid. Cells that do not contain a multiple of the haploid set are called aneuploid. Aneuploid cells contain missing or additional chromosomes (or chunks). Common types of aneuploidy * Monosomy - only 1 copy of a given chromosome in an otherwise diploid cell. * Trisomy - 3 copies of a given chromosome in an otherwise diploid cell. The most frequent cause of aneuploidy is chromosomal nondisjunction. Polyploidy * Euploid, but > 2n (EX: 3N, 4N, 5N etc). * About 1/3 of flowering plants are polyploid. * Triploid plants often are infertile. - Pairing and segregation is disrupted in meiosis. - Commercially beneficial (e.g. seedless fruit: banana). * In humans, polyploidy is lethal before or shortly after birth.

Chapter (2.6/3.5) Advanced Pedigree Analysis TEXTBOOK EXPLANATION CONTINUED AGAIN Analyze pedigrees and determine the mechanism of inheritance of the disease or trait being followed. * By analyzing a pedigree, be able to determine the genotypes of individuals in each generation. - Know what conditional probability is and how to apply it.

Figure 2.19a shows a pedigree in which both parents (I-1 and I-2) have the dominant phenotype. The parental genotypes for this trait are initially unknown. They have had four children: Three of the children also have the dominant phenotype (II-1, II-3, and II-4), but one child (II-2) has an autosomal recessive condition. Through retrospective genetic analysis based on the family pedigree, we can obtain some of the missing genotypic information. Using allele symbols D and d for the dominant and recessive alleles of the gene, and knowing that II-2 has the recessive trait, we can assign her the genotype dd. This means that she must have received a recessive allele from each of her parents, who must each be heterozygous carriers of the condition. Therefore, both I-1 and I-2 have the genotype Dd. The three other children have the dominant phenotype but their genotypes are unknown. Figure 2.19b shows a Punnett square with the possible outcomes for the Dd x Dd parental cross. In the pedigree, the three children with the dominant phenotype can be assigned the D- genotype to indicate that they have at least one copy of the dominant allele. Their second allele is unknown without additional information. To make estimates of the two possibilities, recall the earlier discussion of conditional probability in Section 2.4. It tells us to focus only on the children with the dominant phenotype as the group of interest. Within this group, as the Punnett square shows, each child has a chance of (1/3) being DD and a (2/3) chance of being Dd.

Chapter 3: Mitosis, Meiosis and Sex Determination TEXTBOOK EXPLANATION CONTINUED AGAIN Compare and contrast chromosomal behavior in mitosis and meiosis. * Know what a nucleus contains at each step of the cell cycle, and how it changes as meiosis progresses. - Explain this by knowing the steps of the cell cycle and what is occurring.

Figure 3.14a shows the profile of the content of a nucleus that begins with 2 ng of DNA and 46 chromosomes composed of one chromatid each. As we discussed for somatic cell nuclei, the amount of DNA and the number of duplexes double during S phase. These values are maintained until homologous chromosomes are separatedin anaphase I. The end of meiosis I leaves the nucleus with one-half the DNA, chromosomes, and chromatids it contained at the end of S phase. Anaphase II brings the separation of sister chromatids and a further reduction by one-half in DNA amount and in the numbers of chromosomes and chromatids. The products of meiosis II, containing 1 ng of DNA and 23 chromosomes composed of one chromatid each, are gametes. The union of a sperm and an egg with this nuclear profile produces a fertilized egg with 2 ng of DNA and 46 chromosomes (Figure 3.14b). This is the profile of a cell ready to initiate its first somatic cell cycle.

Chapter 4: Gene Interactions (Not a learning objective, but should know) Understand how a dominant lethal allele can efficiently be eliminated by the action of natural selection. - Knowing this, how is it possible that there are numerous dominant lethal hereditary conditions ("how do these mutations persist in the population", "how are these mutations able to sidestep natural selection")? * Define this. * What is an example of a human hereditary disorder displaying this? - What is this disease caused by? - How does the mutant allele persist in the population? * Functionally, why does this happen?

From an evolutionary perspective, it is easy to understand that a dominant lethal allele can be efficiently eliminated by the action of natural selection when it is expressed during gestation or very early in life (everyone would just die off, and the lethal allele would "go extinct"). Even so, there are numerous examples of dominant lethal hereditary conditions, and a pertinent evolutionary genetic question concerns how these mutations persist in populations. One reason, in the case of a small number of dominant lethal alleles, is that they sidestep natural selection by having a delayed age of onset; the abnormalities they produce do not appear until after affected organisms have had an opportunity to reproduce and transmit the mutation to the next generation. One well-characterized human hereditary disorder displaying delayed age of onset of a dominant lethal allele is the condition called Huntington disease (HD). This progressive neuromuscular disorder, usually fatal within 10 to 15 years of diagnosis, is caused by mutation of a gene near one end of chromosome 4. The HD mutant allele persists in the population because symptoms do not begin in about half of all cases until the person's late thirties or early forties, well after most people have begun having children. Functionally, the onset of symptoms of HD is delayed because the symptoms are due to neuron death, which usually takes place over an extended period of time that often stretches over several decades.

Chapter 4: Gene Interactions Describe the different classes of gain-of-function and loss-of-function mutations and relate these classes to Mendel's definitions of dominant and recessive alleles. * Know the different categories of the functional types of mutations/alleles. - Know what alleles are included in each category, and the protein function of these alleles. Compare these mutations/alleles to the normal function expected. * Know which mutations/alleles cause a gain of function, and which cause a loss of function, and be able to define/give examples of these mutations/allels.

Functional Types of Mutations/Alleles 1. Wild Type - Protein function: normal, wild-type (the way it's supposed to be). 2. Loss-of-function * Null or amorphic - Protein function: dead protein, zero function/zero protein made. * Hypomorphic - Protein product is weak, less is made. - Protein function - leaky, diminished, reduced function but not zero. * ̶D̶o̶m̶i̶n̶a̶n̶t̶ ̶n̶e̶g̶a̶t̶i̶v̶e̶ 3. Gain-of-function * Hypermorphic - Protein function: increased, more than there should be. * Neomorphic - Protein function: new, novel - something completely different from what the wild-type protein does. Protein product does something completely different. Loss of function Null: * deletion of entire coding sequence. * a nonsense mutation that creates a STOP codon at codon 3 of a 400-codon ORF. Hypomorph: single-amino substitution that reduces enzyme activity. Gain of function Hypermorph: mutation in promoter that causes higher-than-normal expression of the gene product. Neomorph: mutation that causes the gene product to aggregate - and these aggregates kill cells.

Chapter 3: Mitosis, Meiosis, and Sex Determination (Not a learning objective, but must know) TEXTBOOK EXPLANATION CONTINUED * What are gametes? - Know examples in humans and in plants. - What are they produced from? * How do these things divide? Describe this process and what it does. * In respect to humans, what is the number of chromosomes in each of our gametes? How many representatives do each of our chromosome pairs have, and where are they found? * What does the union of our gametes' nuclei produce at fertilization? How many chromosomes does it have in its nucleus? * What does human reproduction, like that of other sexually reproducing organisms ensure?

Gametes, produced from germ-line cells, are the germinal, or reproductive, cells: sperm and egg in animals or pollen and egg in plants. Germ-line cells divide by meiosis. Meiotic cell division reduces the number of chromosomes in the nucleus of each daughter cell by one-half to the haploid number. In humans, the number of chromosomes in each egg and sperm nucleus is 23. Each of the 23 human chromosome pairs has one representative in each sperm or egg. The union of the sperm and egg nuclei at fertilization produces the fertilized egg with 46 chromosomes in its nucleus. Thus human reproduction, like that of other sexually reproducing organisms ensures that exactly one-half of the genetic information in an offspring comes from each parent.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED Define epistasis, compare and contrast complementary gene interaction epistasis and duplicate gene action epistasis. * Know how genes contribute to the end product or outcome of biosynthetic pathways. - What is required to achieve the wild-type (normal) outcome? What does this mean? - What are mutant phenotypes a result of? Explain. - Define gene interaction. - Define epistasis. * How does the phenotypic ratio resulting from epistatic interactions differ from the expected phenotypic ratio of F2 generations of dihybrid crosses? - What is this alteration due to? - What causes the distinct patterns of epistatic interactions? - What is true for all of the distinct patterns of epistatic interactions? - What do these epistatic ratios result from? What is this a consequence of? - What is the expected phenotypic ratio of a dihybrid cross illustrating "no interaction"? * What law is responsible for this ratio? How does this explain "no interaction"?

Genes contributing to different steps of a multistep pathway work together to produce the pathway end product or outcome. Each gene is required to produce its normal product to achieve the wild-type (normal) outcome; thus, a mutation of any gene in the pathway can result in a failure of the pathway to be complete. Mutant phenotypes are the result of these pathway breakdowns. All the genes but one involved in a pathway can be normal, but the one mutant gene results in a mutation. In this context, gene interaction is the result of one gene influencing whether and how other pathway genes are expressed or how they function. The discussion that follows describes certain alterations of the 9:3:3:1 phenotype ratios that may be seen in F2 generations of dihybrid crosses when the mutant alleles belong to one or more multistep pathways. These patterns of altered phenotype ratios result from gene interaction phenomena known collectively as epistasis. We describe and illustrate six distinct patterns of epistatic interactions that result from different ways gene products may interact in pathways. Figure 4.19 gives an overview of these patterns, showing the modification of dihybrid ratios that characterizes each form of epistasis. In these six examples, each of the two interacting genes has a dominant and a recessive allele. As we describe the patterns, notice that the epistatic ratios result from the merging of two or more of the F2 phenotype categories as a consequence of the epistatic gene interaction. First, however, we describe a dihybrid cross in which there is no interaction between the two genes in question, genes that both contribute to feather color in budgerigar parakeets (popularly known as "budgies"). The result is the 9:3:3:1 phenotypic ratio expected for the independent assortment of alleles of two genes.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION Provide an example of gene-environment interaction. * Understand that genes alone are not responsible for all the variation seen between organisms. - Know the other essential contributor to observable variation between organisms. * Define this. * Define gene-environment interactions. * Understand the examples of gene-environment interactions. - Understand how an ideal vs. not ideal growth environment among identical organisms can influence phenotypes. - Understand how seasonal changes can influence the phenotypes of organisms. * How did these capacities to make seasonal changes evolve?

Genes control innumerable differences between species. The genome of an organism lays out the body plan and biochemical pathways of the organism, and it controls the progress of development from conception to death. But genes alone are not responsible for all the variation seen between organisms. The environment—the myriad of physical substances, events, and conditions an organism encounters at different stages of life—is the other essential contributor to observable variation between organisms. Gene-environment interaction is the term describing the influence of environmental factors (i.e., nongenetic factors) on the expression of genes and on the phenotypes of organisms. As an example, consider the tall and short pure-breeding lines of pea plants studied by Mendel. Inherited genetic variation dictates that one line will produce tall plants and the other line will produce short plants, but the environment in which the individual plants are grown also has a significant influence on plant height. Environmental factors such as variations in water, light, soil nutrients, and temperature each influence plant growth. It is not hard to imagine that genetically identical plants of a type adapted to temperate zones might grow to different heights if one plant has an ideal growth environment while the other faces a hot, arid environment with poor soil. Phenotypic expression of genotypes can also depend on the interaction of genetically controlled developmental programs and external factors operating on organisms. For example, the seasonal change in coat color observed in arctic mammals that are nearly white in winter but have darker coats in spring and summer results from an interaction between numerous genes and external environmental cues such as day length and temperature. Similarly, environmental cues that induce plants to bloom in the spring trigger changes in gene expression that stimulate the growth and development of multiple plant structures, including flowers and reproductive structures. Such capacities to make seasonal changes evolved by aiding the survival of these organisms, and they suggest that gene-environment interaction is pivotal in understanding and interpreting phenotypic variation.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED AGAIN Allelic series: * What does genetic analysis of coat color in mammals reveal? * What gene is responsible for coat color in mammals? * What enzyme does the C gene produce? - What is the fxn of this enzyme? - What is this enzyme responsible for? * What do the C-gene alleles form? How is this revealed? * Is allele C dom. or rec. to other alleles of the gene? What does this mean for genotypes with a copy of C? What phenotype is produced? * How does allele C compare to the other alleles in terms of activity of the enzyme produced? What do the other alleles form with allele C? * What phenotype does the allele c^(ch) in homozygotes produce? Compare this phenotype to the WT. - What kind of allele is this? What does it generate in terms of phenotype, comparing it to the WT (red. or incr. coat color)? What is this a result of? * What phenotype does the c^h allele in homozygotes produce? - What is it characterized by? * What does the c allele produce? Know enzyme activity/fxn. - Is the c allele dom. or rec.? What kind of allele is it? - What phenotype does homozygosity for this allele produce?

Genetic analysis of coat color in mammals reveals that many genes are required to produce and distribute pigment to the hair follicles or skin cells, where they are displayed as coat color or skin color. Although various interactions among these genes can modify color expression, we focus here on just one gene, the C (color) gene that is responsible for coat color in mammals such as cats, rabbits, and mice. This gene has dozens of alleles that have been identified over nearly a century of genetic analysis, but we limit our discussion to just four alleles that form an allelic series. The C gene produces the enzyme tyrosinase, which is active in the first two steps of a multistep biochemical pathway that synthesizes the pigment melanin, which imparts coat color in furred mammals and skin color in humans. In the initial melanin pathway steps, tyrosinase is responsible for the breakdown (catabolism) of the amino acid tyrosine. The C-gene alleles form an allelic series that is revealed by the phenotypes of offspring of various matings. Allele C is dominant to all other alleles of the gene, and any genotype with at least one copy of C produces wild-type coat color. These genotypes are written as to indicate that regardless of the second allele in the genotype, the phenotype is dominant. Three other alleles, producing tyrosinase enzymes with reduced or no tyrosinase activity, form an allelic series with C(Figure 4.5). The allele c^(ch) in homozygotes produces a phenotype called chinchilla, a diluted coat color. This allele is hypomorphic and generates reduced coat color as a result of the reduced level of activity of the gene product. The c^h allele in homozygotes produces the Himalayan phenotype, characterized by fully pigmented extremities (paws, tail, nose, and ears) but virtually absent pigmentation on other parts of the body. This allele is temperature sensitive, as we describe momentarily. Finally, the c allele produces a protein product with no enzymatic activity. This is a fully recessive null (amorphic) allele that does not produce a functional gene product. Homozygosity for this allele produces an albino phenotype.

Chapter 3: Mitosis, Meiosis and Sex Determination Compare and contrast homologous vs. non-homologous chromosomes vs. sister chromatids. * What is a chromosome? What does it contain? * What is a chromatid? * Do homologous chromosomes have the same genes or different genes? What about alleles? - Where are homologous chromosomes found? - What do homologous chromosomes do during meiosis? - On average, how do homologous chromosomes differ from each other in terms of bps? * What are sister chromatids? - What are they a product of? - Are they likely to be fraternal or identical? - What do they contain? - What do sister chromatid structures look like?

Haploid and Diploid * Homologous chromosomes have same genes. * They may have different alleles of those genes. * On average, they differ from each other about 1 per 1000 bps, just as any two people do. Duplicated Chromosomes and Sister Chromatids * Sister chromatids are the two double strands that result from one round of semiconservative DNA replication. * They are highly likely to be 100% identical to each other. - The structure we see of sister chromatids is what chromosomes look like after full condensation. Chrosomes, Chromatids, ETC. Review * A chromosome is a DNA-containing structure containing a centromere. * A chromatid is a double-stranded DNA molecule. * Sister chromatids are two copies of the same double-stranded DNA molecule, joined by a centromere. * Homologous chromosomes are found in diploid cells. They pair during meiosis and contain the same genes, but possibly different alleles. * "Chromosome" can mean one double stranded DNA molecule on its own with nothing bound at its centromere, or two double stranded DNA molecules joined at a centromere. Textbook: The meaning and usage of the terms chromosome, chromatid, and sister chromatid sometimes cause confusion, and this is a good time to provide functional definitions. The term chromosome is used throughout the cell cycle to identify each DNA-containing structure that has a centromere. At the end of G1, a chromosome consists of a single DNA duplex (double helix) with associated proteins. After the completion of S phase, a chromosome consists of two replicated DNA duplexes with associated proteins. The two DNA molecules making up this chromosome are identical. Individually, these DNA molecules are identified as chromatids, and together they are identified as the sister chromatids.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED AGAIN Describe the different classes of gain-of-function and loss-of-function mutations and relate these classes to Mendel's definitions of dominant and recessive alleles. * Illustrate the basis' of dominance and recessiveness. - Know what the wild-type allele produces, and what the mutant allele produces. Understand the illustrations. - Know what completion of a step means, and what failure of a step means. - Know what makes an allele dominant over another (what is dominance based on)? - Know what makes an allele recessive to another (what is recessiveness based on)? - Define haploinsufficient, and know what alleles are identified as haplosufficient.

Haploinsufficient Wild-Type Allele Is Recessive The second example involves gene T, for which the wild-type allele is recessive to a mutant allele. Gene T produces an enzyme required to catalyze a critical reaction step that produces a wild-type phenotype if it is completed. The inability to complete the reaction step results in a mutant phenotype. For the reaction step in question, 18 units of enzyme activity are required. The wild-type allele T1 produces 10 units of activity. A mutant allele, T2, generates 5 units of enzyme activity. Homozygous T1T1 organisms generate 20 units of catalytic enzyme activity, enough to catalyze the critical reaction step and produce the wild-type phenotype. Heterozygous organisms, on the other hand, produce only 15 units of enzymatic activity and have the mutant phenotype because they fall short of the 18 units required to catalyze the reaction step. Similarly, homozygous T2T2 organisms, which produce 10 units of enzyme activity, also have a mutant phenotype. In this case, the mutant allele T2 is dominant over the wild-type allele T1 since both the heterozygous (T1T2) and homozygous (T2T2) organisms have a mutant phenotype. In cases like this, the wild-type allele (recessive) is identified as haploinsufficient because a single copy is not sufficient to produce the wild-type phenotype in the heterozygous genotype.

Chapter 3: Mitosis, Meiosis and Sex Determination (Not a learning objective, but should know) What is Hemophilia A? * What are the symptoms? * What are the stats on who it affects? * What type of pattern of inheritance is it? * What is it caused by? * What do disease-causing alleles encode? - What makes a disease severe vs. mild? - Are the alleles dominant or recessive? * What do you see in severe forms of this disease? - What is used for treatment? * What is the timeline of progression of Hemophilia A treatment? * What are some other human diseases/traits with this type of pattern of inheritance?

Hemophilia A Symptoms: Excessive bleeding (including internally) and easy bruising. Affects 1/5000 to 1/10000 males worldwide. X-linked recessive pattern of inheritance. Caused by mutation in the gene encoding Factor VIII, which is required for blood clotting. Disease-causing alleles encode non-functional factor VIII (severe disease) or reduced function of Factor VIII (milder disease) -> recessive. Severe forms often fatal by age 20 in absence of treatment with purified or recombinant Factor VIII. * Fatality often due to excessive internal bleeding of organs. Hx of Tx of Hemophilia A * Up until mid 1960's: no treatment (often fatal by age 20) * Mid 60's: Factor VIII purified from donor plasma (and injected into hemophiliacs). * 1978-1985: half of hemophiliacs treated with donor plasma get HIV. * 1984: Factor VIII gene cloned by Genentech. * 1994: Recombinant factor VIII available. Some X-linked recessive human diseases/traits * Hemophilia A * Hemophilia B * Duchenne muscular dystrophy * Retinitis pigmentosum (one of many loci) * Lesch-Nyhan Syndrome * Red-green color blindness * Many others

Chapter 3: Mitosis, Meiosis and Sex Determination TEXTBOOK EXPLANATION (Not a learning objective, but should know) What is Hemophilia A? * What is it caused by? * What does it cause? Name? * In what manner is it transmitted? - Most often, who is it transmitted by? * Who do they pass it to? - What is the ratio of this? * What is something about Hemophilia A that is common among other conditions of this same pattern of inheritance? - Explain how this works. * What is responsible of the appearance of hemophilia in some families?

Hemophilia A, a serious blood-clotting disorder, is caused by mutation of an X-linked gene called factor VIII (F8) that produces a blood-clotting protein called factor VIII protein. Hemophilia A is transmitted in an X-linked recessive manner, most often by a carrier mother who passes the mutant allele to an affected son. In typical X-linked recessive fashion, approximately one-half of the sons of carrier mothers have the disease. Also as is common for X-linked recessive conditions, hemophilia often appears to "skip" a generation because the mutant allele is passed from affected father to carrier daughter and on to an affected grandson. In some families, a de novo (newly occurring) mutation of the F8 gene is responsible for the appearance of hemophilia. An example occurred in the royal families of England and Europe: An apparent de novo mutation of the F8 gene affected Queen Victoria of England (Figure 3.24). Victoria had five sons, one of whom had hemophilia, along with four daughters, two of whom were known carriers. Victoria's carrier daughters had normal blood clotting but introduced the mutation to the royal families of Russia, Germany, and Spain through intermarriage. These daughters passed the mutation to their sons, who had hemophilia, and to their daughters, who were carriers like their mothers.

Chapter 10: Chromosome Aberrations & Disease Review homologous vs. non-homologous chromosomes vs. sister chromatids. TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN AGAIN AGAIN AGAIN Nucleosome Disassembly, Synthesis, and Reassembly during Replication: * Describe chromatin organization of newly synthesized DNA. - When the amount of DNA doubles, does the number of nucleosomes also double to organize the newly synthesized DNA? * If so, how are the new nucleosomes constructed? - Are old nucleosomes recycled during replication or are new nucleosomes composed entirely of newly produced proteins? * When does this doubling of the DNA occur? - What is assembly of nucleosome core particles in connection with replication driven by? * Describe this process step-by-step. - Are the histones old or new? What do the combinations of histones look like? * What is the function of these new nucleosome core particles?

Histones being a basic organizing element of chromatin and the occurrence of DNA replication in the nucleus invite a question about the chromatin organization of newly synthesized DNA. Specifically, when the amount of DNA doubles during S phase, does the number of nucleosomes also double to organize the newly synthesized DNA? If so, how are the new nucleosomes constructed? It would be of interest to know whether old nucleosomes are recycled during replication or whether the new nucleosomes are composed entirely of newly produced proteins. The assembly of nucleosome core particles in connection with replication is driven by the partial denaturing of old nucleosome octamers. These are partially broken down into dimers and tetramers that are then randomly joined with other dimers and tetramers—some old and some new—to form the nucleosome core particles that organize newly synthesized DNA.

Chapter 3: Mitosis, Meiosis, and Sex Determination (Not a learning objective, but must know) * Are human somatic cells haploid or diploid? * How many chromosomes do human somatic cells have? - How many are autosomes? Which ones? - How many are sex chromosomes? Which ones? * Are human gametes haploid or diploid? * How many chromosomes do gametes have? - How many are autosomes? Which ones? - How many are sex chromosomes? Which ones? - What is a nickname for this group of chromosomes? * What type of chromosomes do eggs have (maternal or paternal)? - Is an egg haploid or diploid? What does this mean? - In an actual human egg, how many chromosomes does it have (what does n = ?)? * What types of chromosomes does sperm have (maternal or paternal)? - Is sperm haploid or diploid? What does this mean? - In an actual human spec, how many chromosomes does it have (what does n = ?)? * Know chromosome numbering, be able to tell which one would be labeled #1 vs. #3. * In a human zygote? 2n = ? * In all somatic cells of embryo/child/adult, 2n = ? * On average, how do any two people differ in terms of basepairs?

Human somatic cells are diploid: * 46 chromosomes. - 22 pairs of autosomes (1,2, ... 22) - 1 pair of sex chromosomes (XX or XY). Human gametes are haploid: * 23 chromosomes. - 1 each of the 22 autosomes. - X or Y. - = "haploid set" or "monoploid set". Review: Haploid and Diploid * Egg: - Maternal chromosomes. - Haploid = 1 copy of each. - n = 3 in this example (n = chromosomes). - (In an actual human egg, n = 23). * Sperm - Paternal chromosomes. - Haploid = 1 copy of each. - n = 3 in this example (n = chromosomes). - (In an actual human sperm, n = 23). Chromosome numbering: the longest chromosome is numbered #1, and then in order of longest->shortest, the rest are #2, #3, etc. * 2n = 6 in this example. - (In an actual human zygote, and in all somatic cells of the embryo/child/adult, 2n = 46). * There are two copies of each #1, #2, #3 chromosome: homologous chromosomes, one from mom one from dad. We're all different: on average, any two people differ at 1 in every 1000 basepairs.

Chapter 10: Chromosome Aberrations & Disease TEXTBOOK EXPLANATION CONTINUED Describe non-disjunction. * What are humans enormously sensitive to changes in? * What is true for almost all human aneuploidies? * In humans, how many trisomies and monosomies are potentially possible? - Understand this. * Which chromosomal abnormalities are seen with measurable frequency in newborn infants? * What happened to the other trisomies and monosomies expected? - Why is this? * Define mosaicism. - What is an example of this? - What can it develop as a consequence of? * Understand the example provided.

Humans are enormously sensitive to changes in gene dosage, and almost all human aneuploidies are incompatible with life. Theoretically, there are potentially 24 different kinds of trisomy in humans—one for each autosome, and one each for the X and Y chromosomes—and an equal number of potential monosomies. Yet only autosomal trisomies of chromosomes 13, 18, and 21, and no autosomal monosomies, are seen with any measurable frequency in newborn human infants. Multiple forms of sex-chromosome trisomy are detected with some frequency at birth, however, as is one type of sex-chromosome monosomy. Each of the aneuploidy conditions, along with the other chromosome abnormalities that occur, result in significant phenotypic abnormalities. Human biologists know that other trisomies and monosomies also occur at conception, but the resulting zygotes almost never survive to be born alive. The explanation for this outcome is that the abnormalities of development that are produced are so severe that implantation in the uterine wall does not occur, early zygotic mitotic division is so disrupted that the zygote dies, or fetal development comes to a halt and the fetus spontaneously aborts. Random X-inactivation is an example of naturally occurring mosaicism, in which different cells of the organism contain differently functioning X chromosomes. Mosaicism is the condition of being composed of two or more cell types having different genetic or chromosomal makeup. In addition to the random X-inactivation process, mosaicism can also develop as a consequence of mitotic nondisjunction early in embryogenesis. Mosaicism derived in this way is one of the many kinds of chromosome abnormalities that occur in newborn infants. For example, 25-30% of cases of Turner syndrome, the X-chromosome monosomy (XO), occur in females exhibiting mosaicism in which some cells are 45, XO and others are 46, XX. Some individuals with mosaic Turner syndrome carry 47, XXX cells as well. This kind of mosaicism is usually derived from mitotic nondisjunction in a 46, XX zygote (Figure 10.9).

Chapter 10: Chromosome Aberrations & Disease *** CONTINUED AGAIN *** Compare and contrast non-disjunction of the sex chromosomes in meiosis I vs. II. * Predict the gametes that would result from Meiosis I non-disjunction in a human female. * What are the resulting zygotes if these gametes get fertilized by normal euploid sperm? - Determine the karyotype of the gametes, and then the resulting zygote. * Predict the gametes that would result from Meiosis II non-disjunction in a human female. * What are the resulting zygotes if these gametes get fertilized by normal euploid sperm? - Determine the karyotype of the gametes, and then the resulting zygote.

If these eggs get fertilized by normal euploid sperm, the resulting zygote is: .... * Aneuploid 24,XX. -------fertilization------------> Aneuploid male 47,XXY OR Aneuploid female 47,XXX Normal euploid sperm karyotype: Euploid 23,Y OR Euploid 23,X

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION: DON'T MEMORIZE, JUST USE TO HELP YOU UNDERSTAND ***CONTINUED*** Interpret the results of complementation analysis crosses. * Cross 1: What does the production of F1 progeny with the wild-type phenotype by the cross of two different pure-breeding mutants tell you? (What is the genetic interpretation?) - What do genotypes given for each mutant indicate? * Cross 2&3: What does the production of F1 progeny with the mutant phenotype by the cross of two different pure-breeding mutants tell you? - Cross 2: What does cross two illustrate in terms the genotype of the parental mutants? - Cross 3: What does cross three illustrate in terms of the genotype of the parental mutants? * What can genetic complementation analyses using numerous crosses of different pure-breeding mutants determine? - What does complementation analysis of this type initially focus on? Why is this? * What does it mean for mutations to mutually "fail to complement" one another?

In discussing complementary gene interaction, we described production of the purple-colored pigment anthocyanin as requiring the action of dominant alleles of the C gene and the P gene. Figure 4.22 shows three crosses involving four pure-breeding white-flower mutants. Cross 1, between mutant A and mutant B, produces F1 progeny that have wild-type purple flowers. The genetic interpretation of this result is that genetic complementation is observed. Genotypes given for each mutant indicate homozygosity for recessive alleles on different genes in the parents and a dihybrid genotype in the F1. In contrast to this result, Cross 2 and Cross 3 are also made using pure-breeding white-flower parentals. In both crosses, however, the F1 have the mutant phenotype. This indicates that there is no genetic complementation and that the mutant parents in the respective crosses carry mutations on the same gene. Cross 2 illustrates mutant parental plants that are homozygous for the C gene (ccPP), and Cross 3 illustrates mutant parental plants for gene P (CCpp). Genetic complementation analyses using numerous crosses of different pure-breeding mutants can determine which mutants represent mutations of a certain gene, which represent mutations of certain other genes, and how many different mutant genes are represented in a group of mutants. Complementation analysis of this type initially focuses on crosses that indicate no complementation, as this is a sign of mutations that are in the same gene. Mutations that mutually fail to complement one another are identified as a complementation group, which can consist of one or more mutant alleles of a single gene. All members of a complementation group will fail to complement other members of the group, but they will complement members of other complementation groups that represent mutations of other genes. In the genetic context, a "complementation group" is synonymous with a "gene" because the mutant alleles of each complementation group all affect the same phenotypic characteristic. Thus, in genetic complementation analysis, the number of complementation groups equals the number of genes.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN Describe the different classes of gain-of-function and loss-of-function mutations and relate these classes to Mendel's definitions of dominant and recessive alleles. * What would be the phenotype of an organism with two copies of the wild-type allele? * What would be the phenotype of an organism with a single copy of a fully dominant wild-type allele? * What do we use to categorize mutant alleles into loss-of-function and gain-of-function mutations? * What does a loss-of-function mutation result in? - Are they usually dominant or recessive? * What determines whether the mutant allele is dominant or recessive? * What do gain-of-function mutations identify? - Are they usually dominant or recessive? - What does this mean for the mutant phenotype? - What is a consequence of their newly acquired functions?

In the study of mutations, a central question concerns the mechanism through which the mutation disrupts normal (wild-type) gene function and leads to the mutant phenotype. From a functional perspective, organisms with two copies of the wild-type allele have the wild-type phenotype (Figure4.1a). The same would be true if an organism had a single copy of a fully dominant wild-type allele. Using the level of activity of the protein products of the wild-type allele as the basis for comparison, mutant alleles can often be placed into either a loss-of-function or a gain-of-function category. A loss-of-function mutation results in a significant decrease or in the complete loss of the functional activity of a gene product. This common mutational category includes mutations like those described in the R-gene and T-gene examples. Loss-of-function mutant alleles are usually recessive, but under certain circumstances they may be dominant, depending on whether the wild-type allele is haplosufficient or haploinsufficient. Gain-of-function mutations identify alleles that have acquired a new function or have their expression altered in a way that gives them substantially more activity than the wild-type allele. Gain-of-function mutations are almost always dominant and usually produce dominant mutant phenotypes in heterozygous organisms. As a consequence of their newly acquired functions, certain gain-of-function mutations are lethal in a homozygous state.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED Define penetrance, expressivity and pleiotropy. Incomplete Penetrance: * Define penetrant. - Give an example of a penetrant organism. * Define nonpenetrant. * Define fully penetrant. * Define incomplete penetrance. * What is the human condition, Polydactyl? - What is its method of inheritance? - What does it display in terms of penetrance? - Is the allele dominant or recessive? Is it penetrant or nonpenetrant? - What abnormality do most individuals with polydactyl have? - What do some people with the mutant allele have? What do they express? - Understand the pedigree of polydactyl. * Nine individuals carry a copy of the polydactyl allele, but how many are pentrant? How many are nonpenetrant? Understand this. * When nonpenetrant individuals have a child or grandchild with polydactyl, what does this say about their genotype? * Understand how to calculate the penetrance value of polydactyl for this pedigree.

Incomplete Penetrance When the phenotype of an organism is consistent with the organism's genotype, the organism is said to be penetrant for the trait. In such a case, if the organism carries a dominant allele for the trait in question, the dominant phenotype is displayed. Sometimes an organism with a particular genotype fails to produce the corresponding phenotype, in which case the organism is nonpenetrant for the trait. Traits for which a genotype is always expressed in the phenotype are identified as fully penetrant. In contrast, traits that are nonpenetrant in some individuals are characterized as displaying incomplete penetrance. The human condition known as polydactyly ("many digits") is an autosomal dominant condition that displays incomplete penetrance. Individuals with polydactyly have more than five fingers and toes—the most common alternative number is six (Figure 4.13). Polydactyly occurs in hundreds of families around the world, and in these families the dominant allele is nonpenetrant in about 25-30% of individuals who carry it. Most people who carry the dominant mutant polydactyly allele have extra digits; but at least one in four people with the mutant allele do not have extra digits and instead express the normal five digits. Figure 4.14 shows a family in which polydactyly segregates as a dominant mutation. Nine individuals in the family carry a copy of the polydactyly allele. Six of them are penetrant for the phenotype (meaning that they express the phenotype), but at least three family members—II-6, II-10, and III-10—are nonpenetrant. Each of these individuals has a child or grandchild with polydactyly; thus, each carries the dominant allele for polydactyly but is nonpenetrant for the condition. When nonpenetrant individuals are relatively common, the magnitude of frequency of penetrance can be quantified. Penetrance values vary between families, but for the family shown in Figure 4.14, the penetrance of polydactyly is (6/9), or 66.7%, which is about the average seen worldwide among hundreds of families with polydactyly.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN AGAIN Complementary Gene Interaction * What genotypic classes does independent assortment of alleles result in? - What is the ratio of these genotypic classes? Is this what is expected from a dihybrid cross? * How does the phenotypic ratio of the F2 progeny of this cross differ from the phenotypic ratio expected from a dihybrid cross? - Understand this ratio and what causes it (what is required to produce the WT phenotype, what is required to produce the mutant phenotype). * Understand the mutant progeny. Why are they unable to synthesize pigment? Do all mutants have the same phenotype? * What phenotypic ratio results from complementary gene interaction? * What does complementary gene interaction require? * Understand the flower example: - What is required for purple flower color to be produced (or pigment anthocyanin to be deposited in petals)? - What genotypes at gene loci cause blockage of pathways, and in turn, production of white flowers containing no pigment? * Define genetic complementation. What does it indicate?

Independent assortment of alleles results in four genotypic classes, C_P_ , ccP_ , C_pp, and ccpp, produced in the 9:3:3:1 ratio that is expected from a dihybrid cross. Among the F2, however, only the (9/16) carry the genotype that confers the ability to produce purple pigment. The remaining (7/16) of the F2 are homozygous either for one of the recessive alleles c and p or for both sets of alleles. None of these plants is able to synthesize pigment, due to the absence of functional gene products from one or both loci, and they all have the same mutant phenotype. A 9:7 phenotypic ratio results from complementary gene interaction that requires genes to work in tandem to produce a single product. Figure 4.21 ➊ shows that at the molecular level, purple flower color in sweet peas is produced when the pigment anthocyanin is deposited in petals. Since anthocyanin production requires the action of the product of C as well as the product of P, both steps must be successfully completed for anthocyanin production and deposition in flower petals. The presence of the homozygous recessive genotype at the C locus (cc), the P locus (pp), or both results in blockage of the pathway, and production of white flowers containing no pigment. The ability of two mutants with the same mutant phenotype to produce progeny with the wild-type phenotype is called genetic complementation, and it indicates that more than one gene is involved in determining the phenotype.

Chapter 3: Mitosis, Meiosis and Sex Determination (Not a learning objective, but should know) TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN * FOCUS ON PIC FOR WHAT YOU NEED TO KNOW, INFO PROVIDED IS "SKIM" * Know the eukaryotic cell cycle. * What principal phases is the cell cycle divided into? - What occurs during each of these phases. - What phases are each of these principle phases divided into? * What happens during these smaller phases?

KNOW ONLY PROPHASE, METAPHASE, ANAPHASE, TELOPHASE, METAPHASE PLATE Anaphase is the part of M phase during which sister chromatids separate and begin moving to opposite poles in the cell. Anaphase includes two distinct events tied to microtubule action: anaphase A, characterized by the separation of sister chromatids, and anaphase B, characterized by the elongation of the cell into an oblong shape. Anaphase A begins abruptly with two simultaneous events. First, the enzyme separase initiates cleavage of polypeptides in cohesin, thus breaking down the connection between sister chromatids. Second, kinetochore microtubules begin to depolymerize at their ends to initiate chromosome movement toward the centrioles. The separation of sister chromatids in anaphase A is called chromosome disjunction. As anaphase progresses, sister chromatids complete their disjunction and eventually congregate around the centrosomes at the cell poles. The next part of anaphase, anaphase B, is characterized by the polymerization of polar microtubules that extends their length and causes the cell to take on an oblong shape. The oblong shape facilitates cytokinesis at the end of telophase, which leads to the formation of two daughter cells. In telophase, nuclear membranes begin to reassemble around the chromosomes gathered at each pole, eventually enclosing the chromosomes in nuclear envelopes. Chromosome decondensation begins and ultimately returns chromosomes to their diffuse interphase state. At the same time, microtubules disassemble. As telophase comes to an end, two identical nuclei are observed within a single elongated cell that is about to be divided into two daughter cells by the process of cytokinesis.

Chapter 3: Mitosis, Meiosis and Sex Determination (Not a learning objective, but should know) TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN AGAIN * FOCUS ON PIC FOR WHAT YOU NEED TO KNOW, INFO PROVIDED IS "SKIM" * Know the eukaryotic cell cycle. * What principal phases is the cell cycle divided into? - What occurs during each of these phases. - What phases are each of these principle phases divided into? * What happens during these smaller phases? * Keep track of DNA amount, chromosomes and the number of duplexes in each stage of the cell cycle. Be able to explain changes at each stage.

KNOW ONLY PROPHASE, METAPHASE, ANAPHASE, TELOPHASE, METAPHASE PLATE In animal cells, a contractile ring composed of actin microfilaments creates a cleavage furrow around the circumference of the cell; the contractile ring pinches the cell in two (Figure 3.5). In plant cells, cytokinesis entails the construction of new cell walls near the cellular midline. In both plant and animal cells, cytokinesis divides the cytoplasmic fluid and organelles. The human nucleus has approximately 2 nanograms (ng) of DNA in G1, with 46 chromosomes, each composed of one DNA duplex. DNA amount and the number of duplexes double (forming sister chromatids) with the completion of S phase, and the separation of sister chromatids into separate daughter cell nuclei in anaphase reduces the amount of DNA by one-half. At the end of mitotic M phase, the nucleus again contains 2 ng of DNA and 46 chromosomes composed of one duplex each, at which point the cell is ready to enter stage of the following cell cycle. Notice that despite changes in the amount of DNA and chromatid number, the chromosome number remains at 46 throughout the cell cycle.

Chapter 3: Mitosis, Meiosis and Sex Determination (Not a learning objective, but should know) TEXTBOOK EXPLANATION CONTINUED AGAIN * FOCUS ON PIC FOR WHAT YOU NEED TO KNOW, INFO PROVIDED IS "SKIM" * Know the eukaryotic cell cycle. * What principal phases is the cell cycle divided into? - What occurs during each of these phases. - What phases are each of these principle phases divided into? * What happens during these smaller phases?

KNOW ONLY PROPHASE, METAPHASE, ANAPHASE, TELOPHASE, METAPHASE PLATE M phase follows interphase and is divided into five substages—prophase, prometaphase, metaphase, anaphase, and telophase—whose principal features are described in Figure 3.2. These five substages accomplish two important functions of cell division—(1) the equal partitioning of the chromosomal material into the nuclei of the two daughter cells, a process called karyokinesis, and (2) the partitioning of the cytoplasmic contents of the parental cell into the daughter cells, a process known as cytokinesis. During interphase chromosomes are diffuse and cannot be clearly seen by light microscopy. Chromosome condensation, a process that progressively condenses chromosomes into more compact structures, begins in early prophase. Chromosomes become visible in mid-prophase, and the process continues until chromosomes reach their maximum level of condensation in metaphase. Nuclear envelope breakdown also occurs in prophase, and chromosome centromeres become visible as do the sister chromatids of each chromosome. The centromere is a specialized DNA sequence on each chromosome, and its location is identified as a constriction where the sister chromatids are joined together. Centromeric DNA sequence binds a specialized protein complex called the kinetochore that facilitates chromosome movement and division later in M phase. Metaphase chromosomes have condensed more than 10,000-fold in comparison with their form at the beginning of prophase. This makes them easily visible under the microscope and allows them to be easily moved within the cell. Because they are tethered to kinetochore microtubules from opposite centrosomes, the sister chromatids experience opposing forces that are critical to the positioning of chromosomes along an imaginary midline at the equator of the cell. This imaginary line is called the metaphase plate.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED Describe the different classes of gain-of-function and loss-of-function mutations and relate these classes to Mendel's definitions of dominant and recessive alleles. * Illustrate the basis' of dominance and recessiveness. - Define wild-type traits. - Know what the wild-type allele produces, and what the mutant allele produces. Understand the illustrations. * Know what completion of a step means, and what failure of a step means. - Know what makes an allele dominant over another (what is dominance based on)? - Know what makes an allele recessive to another (what is recessiveness based on)? - Define haplosufficient, and know what alleles are identified as haplosufficient.

Let's compare two examples to illustrate the molecular basis of dominance and recessiveness. In both examples, a wild-type allele produces an enzyme with full activity and a mutant allele produces either very little enzyme activity or none at all. In the first example the mutant allele is recessive, but in the second example the mutant allele is dominant. The Wild-type trait or allele is the most common allele in a natural (wild) population. Haplosufficient Wild-Type Allele Is Dominant In the first example, gene R has a dominant wild-type allele R+ and a recessive mutant allele r. Gene R produces an enzyme that must generate 40 or more units of catalytic activity to drive a critical reaction step. Successful completion of this step produces the wild-type phenotype, whereas failure to complete the step generates a mutant phenotype. Each copy of allele R+ produces 50 units of enzyme activity. The mutant allele r produces no functional enzyme and leads to 0 units of activity. Homozygous R+R+ organisms produce 100 units of enzyme activity (50 units from each copy of R+), far exceeding the minimum required to achieve the wild-type phenotype. Heterozygous organisms (R+r) produce a total of 50 units of enzyme activity, which is sufficient to produce the wild-type phenotype. Homozygous rr organisms produce no enzymatic action, however, and display the mutant phenotype. Based on its ability to catalyze the critical reaction step and produce the wild-type phenotype in either a homozygous (R+R+) or heterozygous (R+r) genotype, R+ is dominant over r. Dominant wild-type alleles of this kind are identified as haplosufficient since one (haplo) copy is sufficient to produce the wild-type phenotype in the heterozygous genotype.

Chapter 4: Gene Interactions *** CONTINUED AGAIN*** Describe variations from classical Mendelian genetics, including incomplete dominance, co-dominance, multiple alleles/allelic series, and lethal alleles. * Lethal Alleles - Understand example.

Lethal Alleles. * Sometimes the homozygous recessive genotype is embryonic lethal. * If one counts only the live births, this will distort the genotype and phenotype ratios, e.g. for Aa x Aa (PIC). - For example, a 3:1 phenotypic ratio will appear as a 3:0 ratio since the homozygous recessive genotype died. * 1:2:1 genotypic ratio will appear as a 1:2 ratio.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN AGAIN Describe the different classes of gain-of-function and loss-of-function mutations and relate these classes to Mendel's definitions of dominant and recessive alleles. * How do mutations resulting in loss of function vary? * Define null mutation. - What's another name for it? - What do these mutant alleles produce? - What is notable about their homozygous genotype? - Know examples of null mutations. - Is it usually dominant or recessive? * Define leaky mutation. - What's another name for it? - How does this mutant allele compare to the wild-type? - What does the severity of the phenotypic abnormality depend on? - What does a greater percentage of activity from a leaky allele result in? - Is it usually dominant or recessive? - What is notable about their homozygous genotype?

Loss-of-Function Mutations As the previous discussion suggests, mutations resulting in a loss of function vary in the extent of loss of normal activity of the gene product. A loss-of-function mutation that results in a complete loss of gene function in comparison with the wild-type gene product is identified as a null mutation, also known as an amorphic mutation (Figure 4.1b). The word null means "zero" or "nothing," and the word amorphic means "without form." These mutant alleles produce no functional gene product and are often lethal in a homozygous genotype. The elimination of functional gene products can result from various types of mutational events, including those that block transcription, produce a gene product that lacks activity, or result in deletion of all or part of the gene. Alternatively, a mutation resulting in partial loss of gene function may be identified as a leaky mutation, also known as a hypomorphic mutation (Figure 4.1c). Hypomorphic means "reduced form"; like the term leaky, it implies that a small percentage of normal functional capability is retained by the mutant allele but at a lower level than is found for the wild-type allele. The severity of the phenotypic abnormality depends on the residual level of activity from the leaky mutant allele. A greater percentage of activity from a leaky allele results in a less severely affected phenotype than when the mutation incurs a more substantial loss of function. Both null and hypomorphic loss-of-function mutations are often recessive and homozygous lethal.

Chapter 4: Gene Interactions *** CONTINUED *** Describe the different classes of gain-of-function and loss-of-function mutations and relate these classes to Mendel's definitions of dominant and recessive alleles. * Know characteristics of loss-of-function alleles. - Are they usually dominant or recessive? What does this mean? * Is this always the case? What does this mean? - Define haplo-sufficient and haplo-insufficient. * What does haplo-insufficient mean in terms of the state of the loss-of-function allele? * Know characteristics of gain-of-function alleles. - Are they usually dominant or recessive? What does this mean? * Understand picture that explains this.

Loss-of-funciton alleles are usually recessive. * In most cases, having ≥ 1/2 the normal functional amount of a given protein is enough. - That is, +/- genotype -> normal phenotype (no disease). * (+) = wild-type allele, (-) = LOF allele. - In other words, the loss-of-function allele is recessive. * LOF allele is recessive to WT. * You either have no protein made (null) or reduced amount made/reduced function in protein (hypomorphic) Loss-of-function alleles are usually, BUT NOT ALWAYS, recessive. * When the +/- genotype -> normal phenotype (no disease), we say that the locus is haplo-sufficient (half is enough). * When the +/- genotype -> disease, we say the locus is haplo-insufficient (half is NOT enough). - In such cases, a loss-of-function allele is dominant. * Because heterozygotes get the disease. (EX: hosoprolencephaly). Gain-of-function alleles are usually dominant. * Gain of function: neomorphic mutation - heterozygous -> forms toxic aggregates -> kills neurons. PIC: Mutation makes protein fold differently, leading to formation of toxic aggregates that kill neurons. The WT is unable to prevent this since GOF alleles are dominant.

Chapter 3: Mitosis, Meiosis and Sex Determination Describe the chromosomal basis of human sex determination. * What chromosomes do Females have? * What chromosomes do males have? * What is the relevance of the Y chromosome? What does its presence/absence mean? * Are X and Y chromosomes identical or fraternal? - Do they share any genes? * What do X and Y chromosomes do at Meiosis I? * How many base pairs are X and Y chromosomes? * What are Y-specific genes involved in? * What do most X-specific genes do? * What is the male-to-female ratio? * Where does a male get his X-chromosome from? Where does the male transmit this X chromosome to? Compare male and female sex chromosomes of the same species. * How are they similar/different? * What does this lead to?

Mammalian Sex Determination * Females have two X chromosomes (XX genotype). * Males = have XY genotype. * Y chromosome: key genes that initiate the male developmental program. * No Y -> Female. The X and Y Chromosomes * Non-identical but share a small number of genes. * Nevertheless, pair and segregate at meiosis I. * X = 160 million base pairs (Mb), Y = 70 Mb. * Y-specific genes are involved in male sexual differentiation. * Most X-specific genes encode functions essential to both males and females. PIC: the male-to-female ratio is one-to-one (average). - A male gets his X chromosome from his mother and transmits it only to his daughters. Textbook: Sex chromosomes differ between males and females of a species and have very few DNA sequences in common outside the pseudoautosomal regions. This means that the number of copies of sex-linked genes usually varies between males and females, and it leads to patterns of inheritance of sex-linked genes that differ from those seen for autosomal genes.

Chapter 3: Mitosis, Meiosis and Sex Determination Compare and contrast chromosomal behavior in mitosis and meiosis. MEIOSIS * Know what a homologous pair is. * Know the process, and what happens in each step. * Are the stages of interphase and actions and functions of subcellular structures the same among somatic and germ-line cells? - Do germ-line cells experience mitosis? Or is mitosis exclusive to somatic cells? Explain why or why not. * What is the purpose of germ-line cells undertaking meiosis?

Meiosis * Homologous pair: two copies of chromosome 1, one from Mom (M) and one from Pa (P). Textbook: Meiosis Features Two Cell Divisions Interphase of the germ-line cell cycle contains stages G1, S, and G2 that are indistinguishable from those in somatic cells. Similarly, the actions and functions of subcellular structures such as centrosomes and the microtubules they produce are the same in all cells. Nor is mitosis exclusive to somatic cells. Germ-line cells of plants and animals are created and maintained by mitotic division. These cells undertake meiosis solely for the purpose of producing gametes.

Chapter 3: Mitosis, Meiosis and Sex Determination TEXTBOOK EXPLANATION Interpret independent assortment in terms of meiotic chromosome behavior. * How can we understand segregation in terms of meiotic chromosomes? - What type of chromosomes would we need to follow? - In an organism of what genotype would these chromosomes need to be found? - Know the steps in which things occur, and what it shows. - Know the end result and what it explains.

Meiosis Generates Mendelian Ratios The separation of homologous chromosomes and sister chromatids in meiosis constitutes the mechanical basis of Mendel's laws of segregation and independent assortment. The connection between meiosis and Mendelian hereditary principles was first suggested, independently, by Walter Sutton and Theodor Boveri in 1903. Based on microscopic observations of chromosomes during meiosis, Sutton and Boveri proposed two important ideas. First, meiosis was the process generating Mendel's rules of heredity; and second, genes were located on chromosomes. Over the next two decades, work on numerous species proved these hypotheses to be correct. We can understand segregation by following a pair of homologous chromosomes through meiosis in a heterozygous organism. Figure 3.15 illustrates meiosis in a pea plant with the heterozygous Gg genotype. Recall that Mendel's law of segregation predicts that one-half (50%) of the gametes produced by a heterozygote will contain G and the remaining one-half will contain g. How does meiosis generate this outcome? DNA replication in S phase creates identical sister chromatids for each chromosome. At metaphase I, the homologs align on opposite sides of the metaphase plate; and at anaphase I, the homologs separate from one another. This movement segregates the chromosome composed of two G-bearing chromatids from the chromosome bearing the two g-containing chromatids. Following these cells through to the separation of sister chromatids in meiosis II, we find that among the four gametes are two containing the G allele and two containing g. This outcome explains the 1:1 ratio of alleles that the law of segregation predicts for gametes of a heterozygous organism.

Chapter 3: Mitosis, Meiosis and Sex Determination TEXTBOOK EXPLANATION CONTINUED Compare and contrast chromosomal behavior in mitosis and meiosis. * Know Meiosis II. - Know each step and what is occurring in each step. - Know the end result. - Compare and contrast relevant steps to mitosis.

Meiosis II The second meiotic division divides each haploid product of meiosis I by separating sister chromatids from one another in a process that is reminiscent of mitosis, except that the number of chromosomes in each cell is one-half the number observed in mitosis. The products of meiosis II mature to form the gametes that contain a haploid set of chromosomes. The four stages of meiosis II—prophase II, metaphase II, anaphase II, and telophase II. Meiosis II bears a general resemblance to mitosis in that kinetochore microtubules from opposite centrosomes attach to the kinetochores of sister chromatids. Also, as in mitosis, in meiosis II the chromosomes align randomly along the metaphase plate. Furthermore, sister chromatid separation is accompanied by cohesin breakdown, the action of motor proteins, and depolymerization of microtubules. Cytokinesis takes place at the end of telophase II. There are, however, only a haploid number of chromosomes present in each cell during meiosis II. Four genetically distinct haploid cells, each carrying one chromosome that represents each homologous pair, are the products of meiosis II.

Chapter 10: Chromosome Aberrations & Disease Compare and contrast non-disjunction of the sex chromosomes in meiosis I vs. II.

Meiosis of sex chromosomes in Human Male * Sex chromosomes pair at meiosis I (X pair and Y pair). They pair and segregate as if they were a homologous pair (they act as a homologous pair in meiosis). Meiosis I non-disjunction: Meiosis I The homologous pairs of chromosomes fail to disjoin properly. Both pairs end up in one descendant of cell division that characterizes meiosis I (the pairs end up in the same spermacyte I cell). The other spermacyte cell has no sex chromosomes. Meiosis II, Disjunction II Assuming meiosis II precedes properly (sister chromatids disjoin properly), we end up with these gametes (Two XY gametes, and 2 gametes without sex chromosomes). Meiosis II non-disjunction Meiosis I The homologous pairs (X pair and Y pair) disjoin, resulting in the X homologous pair and the Y homologous pair being pulled into separate Spermacyte I cells. Meiosis II, Disjunction II Disjunction doesn't work properly in meiosis II: * X homologous pair fails to disjoin, leading the sister chromatids of this pair to end up in the same sperm. The other sperm has no sex chromosomes. * Disjunction occurs properly for the Y homologous pair, so we have Y in each sperm. Meiosis II non-disjunction Meiosis I * The homologous pairs (X pair and Y pair) disjoin, resulting in the X homologous pair and the Y homologous pair being pulled into separate Spermacyte I cells. Meiosis II, Disjunction II * Y homologous pair fails to disjoin in meiosis II, leading to one sperm containing 2 Y sex chromosomes, and the other sperm containing no sex chromosomes. Disjunction occurs properly for the X homologous pair, so we have X in each sperm.

Chapter 10: Chromosome Aberrations & Disease Review mitosis vs. meiosis. Meiosis review * How do sex chromosomes act during meiosis? What do they do? * Know what occurs during meiosis I, meiosis II, and disjunction II. - Know the steps, and what they result in. * Are gametes genetically identical to each other? - How are they the same, and how are they different? * What does each gamete contain?

Meiosis of sex chromosomes in Human Male * Sex chromosomes pair at meiosis I (X pair and Y pair). They pair and segregate as if they were a homologous pair (they act as a homologous pair in meiosis). Meiosis I: The homologous pairs (X pair and Y pair) disjoin, resulting in the X homologous pair and the Y homologous pair being pulled into different Spermacyte I cells. Meiosis II, Disjuction II: Sister chromatids of each pair disjoin, producing 4 gametes: 2 X gametes and 2 Y gametes. * The result is two X-carrying sperm and two Y-carrying sperm from the germline stem cell in a male human. Gametes are not genetically identical to each other; they differ at about 1 per 1000 base pairs. * They have the same genes, but different alleles of the genes. Each of these gametes contain 23 total chromosomes. This slide just shows the X and Y. * In each gamete (2 X and 2 Y), there will be 22 autosomes included as well in a single copy. - X gamete includes: X sex chromosome and 22 autosomes. - Y gamete includes: Y sex chromosome and 22 autosomes. Textbook: (don't memorize, just understand) Chromosomes contain genetic material. Their movement and separation during meiosis, and their union at fertilization mirror the separation and transmission of genes.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION Describe the different classes of gain-of-function and loss-of-function mutations and relate these classes to Mendel's definitions of dominant and recessive alleles. * What did Mendel wisely choose to examine? - Define complete dominance. * Know how it relates to heterozygous dominant, homozoygous dominant, and homozygous recessive organisms, and how it affects their phenotypes. - What does the complete dominance of one allele result in for the F1 and F2 generations? - What are the phenotypes of the traits Mendel studied controlled by? - Define dominant alleles in relation to function of the gene. - Define recessive alleles in relation to function of the gene. - Define dominant characters in relation to phenotype and genotype. - Define recessive characters in relation to phenotype and genotype. - What kinds of basis' does dominance and recessiveness have? - What are phenotypes a consequence of? - What is the dominance of one allele over another determined by?

Mendel wisely chose to examine traits presenting in one of two easily distinguishable forms. One form of each trait he studied displayed complete dominance over the other form. Complete dominance makes the phenotype of a heterozygous organism indistinguishable from that of an organism homozygous for the dominant allele; thus, only organisms homozygous for the recessive allele display the recessive phenotype. The complete dominance of one allele also results in the exclusive expression of the dominant phenotype among the heterozygous progeny of a cross between pure-breeding homozygous parents, while the progeny display a 3:1 ratio of dominant to recessive phenotypes. We now know that the phenotypes of the seven traits that Mendel studied are controlled by two alternative alleles at seven different genes. For the four traits of Mendel that have been described at the molecular level (see Section 2.6), the dominant alleles produce full function of the gene, while the recessive alleles encode gene products with reduced or no functional activity. A character is called dominant if the same phenotype is seen in organisms with the homozygous and heterozygous genotypes. The correlative character is called recessive if it is observed only in a single homozygous genotype. In this sense, dominance and recessiveness have a phenotypic basis. The phenotypes are, however, a consequence of the characteristics of proteins produced by the alleles of a gene. In this sense, dominance and recessiveness also have a molecular basis. The dominance of one allele over another is determined by the protein products of the allele—by the manner in which the protein products of alleles work to produce the phenotype.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION Describe variations from classical Mendelian genetics, including incomplete dominance, co-dominance, multiple alleles/allelic series, and lethal alleles. * Compare classical Mendelian genetics to incomplete dominance. - Which occurs more frequently in nature? * Define incomplete dominance. - When incomplete dominance exists among alleles, what do we observe about the phenotype of heterozygous organisms in comparison with the homozygotes of different phenotypes? - When traits display incomplete dominance, two pure-breeding parents with different phenotypes produce F1 heterozygotes having what phenotype? - Understand the flowering example. * How would we determine which allele is partially dominant to the other?

Mendel's description of inheritance of traits controlled by single genes having a dominant and a recessive allele is a simple hereditary process that is relatively rare in nature. More commonly with single-gene traits, the dominance of one allele over another is not complete but instead is described as incomplete dominance, also known as partial dominance. When incomplete dominance exists among alleles, the phenotype of the heterozygous organism is distinctive; it falls somewhere on a phenotypic continuum between the phenotypes of the homozygotes and is typically more similar to one homozygous phenotype than the other. When traits display incomplete dominance, two pure-breeding parents with different phenotypes produce F1 heterozygotes having a phenotype different from that of either parent. One of the many traits displaying incomplete dominance is the trait described as flowering time in Mendel's pea plants (Pisum sativum). In peas, the first appearance of flowers is under the genetic control of a gene that we will call T, for flowering time. The earliest-flowering strain of pea plants has the homozygous genotype T1T1; the flowering time of this strain is described as day 0.0. The latest-flowering strain is homozygous T2T2, and it flowers 5.2 days later on average than T1T1 plants. A cross of pure-breeding early-flowering and late-flowering strains produces T1T2 heterozygous progeny that begin to flower 3.7 days later on average than the earliest-flowering strain (Figure 4.2a). Genetic crosses show that flowering time is controlled by a single locus. Self-fertilization of T1T2 plants produces a 1:2:1 ratio of early-, intermediate-, and late-flowering progeny (Figure 4.2b). We say the T2 allele is partially dominant, but not completely dominant, to T1 because the heterozygous phenotype is distinct from either homozygous phenotype but more closely resembles the late-flowering strain.

Chapter (2.6/3.5) Advanced Pedigree Analysis Give examples of Y-linked and mitochondrial patterns of inheritance. * What does a mitochondria's genome contain? What is it made of? (humans) * Where does a zygote's mitochondria come from? Where does it not come from? * What are patterns of mitochondrial inheritance. * What is Y-linked inheritance? * What are patterns of Y-linked inheritance? * Where is the Y chromosome found? What does this mean in terms of transmission? * What is a likely role of the genes found on the Y chromosome? * How do the genes on the X chromosome relate to the genes on the Y chromosome? * Do females carry Y chromosomes? * What does the evolution of the Y chromosome tell you about its components and their roles?

Mitochondrial Inheritance Mitochondria have their own DNA genome, which contains 37 genes (in humans). A zygote's mitochondria comes from the mom's egg, not from the dad's sperm. * MI is typically partially penetrant. * Inheritance only through maternal line, affected males do not pass on the genes. You do not need to worry about M.I when working pedigree problems for this class. Textbook: Whereas females receive one copy of X-linked alleles from each parent, males receive their X-linked alleles from their mother and their Y chromosome from their father. This means that Y-linked inheritance, the inheritance of genes on the Y chromosome, is an exclusively patrilineal (father to son) pattern of hereditary transmission. The key to Y-linked inheritance is that the Y chromosome is found only in males. This means Y-linked genes are transmitted in a male-to-male pattern. In mammals, fewer than 50 genes are found on the Y chromosome; and like SRY, those genes are likely to play a role in male sex determination or development. The genes on the human Y chromosome do not have counterparts on the X chromosome, although the DNA sequences in the pseudoautosomal regions are shared by the X and Y chromosomes to facilitate synapsis of the chromosomes during meiosis. There is crossing over between the pseudoautosomal regions, but this does not involve expressed genes. Females never carry Y chromosomes, so from an evolutionary perspective it makes sense that the genes carried on a Y chromosome should be male-specific, having either to do with male sex determination or reproduction. Indeed, the most recent genomic evidence suggests that the mammalian Y chromosome has rapidly evolved over the past 300 million to 350 million years, undergoing multiple changes in structure but preserving a handful of genes that are essential to male fertility and survival.

Chapter 3: Mitosis, Meiosis and Sex Determination Compare and contrast chromosomal behavior in mitosis and meiosis. MITOSIS * Know what a chromosome is. * Know what a homologous chromosome is. - Know how two homologous chromosomes differ from each other, on average, in terms of base pairs. * Know what a homologous pair is. * Are sister chromatids likely to be identical or fraternal? * What is the result of metaphase (what happens, and what is the result of this happening)? - What is important to note about the very end result?

Mitosis (diploid w/ 2n=4) * A chromosome is a long double-stranded DNA molecule. * Homologous chromosomes have same genes; they differ from each other an average of 1 per 1000 bps, just as any two people do. * Homologous pair: that is, 2 copies of chromosome 2, one from Mom (M) and one from Pa (P). * Sister chromatids: likely to be 100% identical. - Disjunction and segregation of sister chromatids leads to: genetically identical daughter cells (although one of them might have a new mutation)

Chapter 3: Mitosis, Meiosis and Sex Determination Compare and contrast chromosomal behavior in mitosis and meiosis. * Compare mitosis and meiosis (processes). * Know Meiosis I and Meiosis II. - Know what is occurring in each, and the end result of each.

Mitosis vs. Meiosis Mitosis: - A diploid somatic cell replicates its DNA once and divides once to form 2 diploid, genetically-identical daughter cells. Meiosis: - A diploid germline stem cell replicates its DNA once and divides twice to form 4 haploid cells that are not genetically identical. Textbook: Meiosis is distinguished from mitosis by having two successive cell divisions during M phase, by distinctive movement of homologous chromosomes and sister chromatids, and by the production of four haploid gametes. Meiotic interphase is followed by two successive cell-division stages known as meiosis I and meiosis II. There is no DNA replication between these meiotic cell divisions, so the result of meiosis is the production of four haploid daughter cells (Figure 3.9). In meiosis I, homologous chromosomes separate from one another, reducing the diploid number of chromosomes (2n) to the haploid number (n). In meiosis II, sister chromatids separate to produce four haploid gametes, each with one chromosome of every diploid pair. Following the completion of meiosis, each gamete contains a single nucleus holding a haploid chromosome set. The gametes of the two sexes are often dramatically different in size and morphology, however. Female gametes are generally much larger than male gametes and have a haploid nucleus, a large amount of cytoplasm, and a full array of organelles. In contrast, male gametes contain a haploid nucleus but very little cytoplasm and virtually no organelles. As the fertilized ovum begins mitotic division, the organelles and cytoplasmic structures provided by the maternal gamete support its early zygotic growth.

Chapter 3: Mitosis, Meiosis and Sex Determination (Not a learning objective, but should know) What is mitosis (what happens)? - Know each phase and what occurs during each phase. * Know the mitotic cell cycle, and what happens during each phase.

Mitosis: A diploid somatic cell replicates its DNA once and divides once to form 2 diploid, genetically-identical daughter cells. - G1 phase: This cell contains two pairs of homologous chromosomes with the genotype AaBb. - S phase: DNA replication creates identical sister chromatids for each chromosome. - Metaphase: Chromosomes align randomly along the metaphase plate with the aid of the mitotic spindle. - Telophase: Two daughter cells are produced by mitosis. Each AaBb following sister chromatid separation to form daughter chromosomes. The Mitotic Cell Cycle G1: Active gene expression and cell activity; preparation for DNA synthesis. G0: Terminal differentiation and arrest of cell division. -> Cell remains specialized but does not divide. -> Eventual cell death (apoptosis). S phase: DNA replication and chromosome duplication. G2: Preparation for cell division. M phase: Cell division. - Mitosis (somatic cells) - Meiosis (germ-line cells)

Chapter 4: Gene Interactions (Not a learning objective, but should know) * Codominance: ABO Blood Type - Where does codominance and recessiveness occur in the alleles of the gene determining human ABO blood type? - What do our blood types result from? - What are the alleles of the ABO gene? - What are the phenotypes? What are they produced by? - What is the relationship between the alleles of the ABO gene? * How is this determined?

More than one pattern of dominance between the alleles of a gene can occur under certain circumstances. Here we examine the codominance of two alleles and the recessiveness of a third allele of the gene determining human ABO blood type. All of us have one of the four common blood types—type O, type A, type B, or type AB—that result from our genotype at the ABO blood group gene located on chromosome 9 (OMIM 110300). The three alleles of the ABO gene are identified as IA, IB and i, and the four blood groups are phenotypes produced by six genotypes. On the basis of genotype-phenotype (i.e., blood type) correlation, geneticists have concluded that IA and IB have complete dominance over i, and that IA and IB are codominant to one another. The complete dominance of IA and IB to i is indicated by the identification of blood type A in individuals whose genotype is IAIA or IAi, and of blood type B in individuals whose genotype is IBIB or IBi. The completely recessive nature of the i allele is confirmed by the observation that only ii homozygotes have blood type O. Lastly, codominance of IA and IB to one another is confirmed by the observation that blood type AB occurs only in individuals who have the heterozygous genotype IAIB.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN AGAIN AGAIN AGAIN Describe the different classes of gain-of-function and loss-of-function mutations and relate these classes to Mendel's definitions of dominant and recessive alleles. Gain-of-function mutations: * What determines whether a mutation is hypermorphic or neomorphic? * Define hypermorphic mutations (how do they differ from the wild-type)? - Are they usually dominant or recessive? - How does the gene product of a hypermorphic allele compare to the wild-type allele? * What does this lead to? - What are some examples of hypermorphic mutations? - How does the phenotypic effect compare between mutation in homozygotes and heterozygotes. - What is notable about mutant homozygotes? * Define neomorphic mutations. - Are they usually dominant or recessive? - Describe the gene products of neomorphic mutants. How do they compare to the wild-type? * What do these differences lead to? - How do homozygotes for neomorphic alleles differ from heterozygotes?

Mutations resulting in a gain of function fall into two categories that depend on the functional behavior of the new mutation. Hypermorphic ("greater than wild-type form") mutations produce more gene activity per allele than the wild type (Figure 4.1e) and are usually dominant. The gene product of a hypermorphic allele is indistinguishable from that of the wild-type allele, but it is present in a greater amount and thus induces a higher level of activity. The excess concentration is the functional equivalent of overdrive, pushing processes forward more rapidly, at the wrong time, in the wrong place, or for a longer time than normal. Hypermorphic mutants often result from regulatory mutations that increase gene transcription, block the normal response to regulatory signals that silence transcription, or increase the number of gene copies by gene duplication. The phenotypic effect may be more severe in mutation homozygotes than in heterozygotes, but often, particularly in humans, mutant homozygotes are not seen because homozygosity is lethal. Gain-of-function mutations resulting from neomorphic ("new form") mutations acquire novel gene activities not found in the wild type (Figure 4.1f) and are usually dominant. The gene products of neomorphic mutants are functional but have structures that differ from the wild-type gene product. The altered structures lead the mutant protein to function differently than the wild-type protein. Homozygotes for a neomorphic allele may exhibit a more severely affected phenotype than do heterozygotes.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED AGAIN Define epistasis, compare and contrast complementary gene interaction epistasis and duplicate gene action epistasis. * How is epistasis most easily identified? * What does the "expected" F2 ratio result from? * How can specific types of epistasis be identified? - Understand the example of "no interaction".

No Interaction (9:3:3:1 Ratio) Epistasis is most easily identified through specific deviations from the expected 9:3:3:1 ratio among the F2 progeny of a dihybrid cross involving dominant and recessive alleles. This "expected" F2 ratio results from the action of two independently assorting genes in the absence of epistasis. Specific types of epistasis can be identified by the characteristic change in phenotypic ratio each produces. As an example of no interaction between two genes, consider the feather color of the budgie. Two genes, B and Y, contribute to separate pigment-producing biosynthetic pathways that produce a blue pigment and a yellow pigment. Wild-type budgies have feathers that are green, a mixture of blue and yellow. Budgies are also found to have blue feathers (due to the absence of yellow pigment), yellow feathers (due to the lack of blue pigment), and white feathers (the absence of both pigments). Consider the mating of a pure-breeding blue budgie (BByy) to a pure-breeding yellow budgie (bbYY). The F1 progeny have wild-type green feather color and are dihybrid (BbYy), and they are shown at the left in Figure 4.20, across from the F2 progeny that have all four feather-color phenotypes. As predicted by independent assortment, green feather color (wild type) is observed (9/16) in of the progeny, blue feathers and yellow feathers are each seen in (3/16) of the F2, and the white-feather phenotype appears in (1/16) of the F2 progeny.

Chapter (2.6/3.5) Advanced Pedigree Analysis ** CONTINUED AGAIN ** Know the rules that allow particular mechanisms of inheritance to be ruled out, and understand the logic behind these rules. PROBLEM: The pedigree given illustrates an inheritance pattern typical of a dominant trait. * Could it be X-linked dominant? - If yes, explain why, specifically. - If no, explain what rules rule this possibility out. - When answering, point out specific family groups. * Could it be autosomal recessive? - If yes, explain why, specifically. - If no, explain what rules rule this possibility out. - When answering, point out specific family groups.

No, this could not be X-linked dominant. Parent-child trios/groups in this pedigree allow us to rule out X-linked dominance. * Look at the I-1 I-2, II-6 trio: - How did the son get the disease? Mom isn't affected. * Look at the II-6 II-7, III-13 III-15 group: - How did the sons get the disease? Mom isn't affected. * Looks at the II-6 II-7, III-14 trio: - The daughter would have to be infected since her father is infected. Dad gives XA to daughter, why isn't she infected? Yes, it could be autosomal recessive because we can't apply rule #2 to rule out recessiveness. * #2: If two affected parents have an unaffected kid, recessiveness can be ruled out. But, if this is a recessive pedigree, I-1, II-1, II-4 and II-7 would all need to be carriers that married into the family.

Chapter 10: Chromosome Aberrations & Disease TEXTBOOK EXPLANATION Compare and contrast non-disjunction of the sex chromosomes in meiosis I vs. II. * What does nondisjunction in germ-line cells produce? - Define this. * What do these lead to? * Where can meiotic nondisjunction occur? - What does meiotic nondisjunction most often affect? * Define the cells involved. * Define meiosis I nondisjunction. - What does it result in? - What does each cell contain (generally and numerically, assuming only one chromosome pair is affected)? * How does meiosis II proceed when meiosis I is aberrant? How does its completion affect the sister chromatids? * If nondisjunction occurs in meiosis I, what resulting gametes are produced? - Union of these gametes with normal haploid gametes at fertilization results in fertilized eggs with what number of chromosomes (with respect to n)? * Define the possibilities (including the definition in terms of n).

Nondisjunction in germ-line cells produces aneuploid gametes—reproductive cells that have one or more extra or missing chromosomes. These errors lead to the production of aneuploidy of fertilized eggs. Meiotic nondisjunction can occur in either meiosis I or II and most often affects just a single homologous pair or a single pair of sister chromatids in a gametocyte (gametocytes are the cells that undergo meiosis to produce gametes). Meiosis I nondisjunction is the failure of homologous chromosomes to separate. It results in both homologs moving to a single pole. One of the gametocytes produced in meiosis I contains both chromosomes, and the other contains neither chromosome (Figure 10.6). These gametocytes contain aneuploid chromosome numbers of n+1 and n-1 (assuming only one chromosome pair is affected). Meiosis II usually proceeds normally even when meiosis I is aberrant, and its completion sends the sister chromatids to different gametes. If nondisjunction occurs in meiosis I, each of the four resulting gametes are aneuploid—either n+1 or n-1. The union of an aneuploid gamete with a normal haploid gamete at fertilization results in a fertilized egg with an aneuploid number of chromosomes that will be either trisomic (2n+1) having three of one of the chromosomes rather than a homologous pair, or monosomic (2n-1), having just a single copy of one of the chromosomes rather than a homologous pair. PIC: Homologous chromosomes fail to disjoin in meiosis I, and all resulting gametes are aneuploid. Fertilization by a normal haploid gamete produces fertilized eggs that are trisomic (2n+1) or monosomic (2n-1).

Chapter 10: Chromosome Aberrations & Disease TEXTBOOK EXPLANATION AGAIN Compare and contrast non-disjunction of the sex chromosomes in meiosis I vs. II. * What does nondisjunction occurring in meiosis II typically follow? * If nondisjunction occurs in meiosis II, how many secondary gametocytes will be affected? Understand why this is. * Among the four resulting gametes, how many are normal? Understand why this is. - What is the name for any non-normal gametes? * How many chromosomes (in terms of n) do these non-normal gametes contain? * What kinds of fertilized eggs are produced when one of these non-normal gametes unites with a normal gamete at fertilization?

Nondisjunction occurring in meiosis II typically follows a normal meiosis I that produced normal secondary gametocytes, both containing the haploid (n) number of chromosomes (Figure 10.7). Since these gametocytes are separate cells, they independently divide during meiosis II; thus, if nondisjunction occurs, only one of the secondary gametocytes will be affected. Among the four resulting gametes, two are normal because a normal disjunction took place during each meiotic division. The other two gametes are aneuploid: One contains n+1 chromosomes and the other n-1 chromosomes. Trisomic or monosomic fertilized eggs are produced when one of these aneuploid gametes unites with a normal gamete at fertilization. PIC: Sister chromatid disjunction fails in meiosis II. Normal fertilization of the resulting gametes generates trisomy, monosomy, or normal diploidy at fertilization.

Chapter 10: Chromosome Aberrations & Disease Review homologous vs. non-homologous chromosomes vs. sister chromatids. TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN AGAIN AGAIN AGAIN AGAIN AGAIN Nucleosome Displacement and Positional Restoration During Transcription: * How do nucleosomes, and the chromatin of which they are a part of, relate to eukaryotic gene transcription? - Understand how this occurs. * Understand how the structure of chromatin plays a role in transcr. - What are mechanisms of this, and what are the effects on transcription? * How do the following components associated with transcription interact with nucleosomes? Know where they interact with nucleosomes. - transcription factors - RNA polymerase II * Know the mechanism of nucleosome displacement. - Know all components involved, and their role in the mechanism.

Nucleosomes, and the chromatin of which they are a part, are foundational regulators of eukaryotic gene transcription. Broadly speaking, transcriptional control occurs by remodeling chromatin, which converts it from a closed structure that inhibits transcription to a more open structure that permits transcription. Numerous mechanisms by which the access to DNA is controlled by chromatin structure have been described. These processes have varying effects on transcription. Some shut off all transcriptional access to certain genes and genome regions, some periodically open and close chromatin in some genome regions in response to the transcriptional needs of the organism, and others permit relatively free and continuous transcription of certain genes by maintaining an open chromatin structure. RNA polymerase II and the transcription factors associated with transcription interact with nucleosomes at sites of transcription by locally separating DNA from nucleosomes, displacing nucleosomes to allow RNA pol II to carry out transcription, and then restoring nucleosome positioning after the transcription machinery passes. Research examining the interactions between the hundreds of transcription factors that inhabit eukaryotic cells reveals some of the mechanisms by which nucleosome displacement takes place. Transcription factors and elongation factors appear to peel DNA away from nucleosomes in advance of the passage of RNA polymerase. Several dozen base pairs of DNA are separated from nucleosome association as the transcriptional machinery approaches. The DNA-strand separation exposes nucleotides of the template strand, then the strands re-close after RNA polymerase passes through the gene region. The displaced nucleosomes remain intact and are moved back to their original locations after RNA polymerase passes.

Problem Set #2: Pedigree Problem Set Next two questions: Fred is a 30 year old man who is about to get married. Fred has a brother who has cystic fibrosis and a sister and a father who have red-green color blindness. Fred and his mother are unaffected by either condition. Part one: What is the chance that Fred is a carrier of a disease-causing allele for cystic fibrosis? A. 100% B. 67% (2/3) C. 50% (1/2) D. 25% (1/4) E. 0% Part two: What is the chance that Fred is a carrier of a disease-causing allele for red-green color blindness? A. 100% B. 67% (2/3) C. 50% (1/2) D. 25% (1/4) E. 0%

Part one: B. 67% (2/3) * CF: autosomal recessive * CF carrier probability: (2/3) Part two: E. 0% R/G color blindness: X-linked recessive * Fred has to be XRY since he's unaffected.

Chapter (2.6/3.5) Advanced Pedigree Analysis Analyze pedigrees and determine the mechanism of inheritance of the disease or trait being followed. Dominant inheritance of disease/trait: * What does a heterozygous genotype give you in terms of phenotype? * Compare the mutant, disease-causing allele with the wild-type allele, is it dominant or recessive over the WT? * If D is the disease allele and d is normal, then what genotypes are disease free? * What are typical patterns of dominant traits/diseases when looking at a pedigree? * Can an affected kid be born to unaffected parents? Explain why or why not. Autosomal Dominant Inheritance * What are typical patterns of dominant traits/diseases when looking at a pedigree? * Can an affected kid be born to unaffected parents? Explain why or why not. * Can affected parents have an unaffected kid? Explain why or why not. X-linked Dominant Inheritance * Where is the gene containing the allele(s) in question? - What is the dominant disease causing allele? - What is the recessive WT allele? * What is the genotype of affected males? Who will they pass the disease onto? Who will they not pass it onto? * Gtype of affected F?

Pedigree Symbols * Note that many pedigree problems will not show you who the carriers are. You will need to figure this out as part of the pedigree analysis. Dominant Inheritance * Heterozygous genotype -> disease phenotype. * The mutant, disease-causing allele is DOMINANT over the wild-type allele. * If D is the disease allele and d is normal, then only dd genotypes are disease free. * Dominant trait/disease typically found in every generation. * Affected kid NEVER born to unaffected parents. ASSUMING 100% PENETRANCE & NO NEW MUTATION Common Characteristics of Autosomal Dominant Inheritance * Dominant trait/disease typically found in every generation. * Affected kid NEVER born to unaffected parents. * Two affected parents CAN have an unaffected kid. X-linked Dominant Inheritance * Gene in question is on the X-chromosome. - D is dominant disease causing allele. - d is recessive wild-type allele. * Affected males are XDY, and will pass the disease to all of their daughters and none of their sons. * Affected Females are XDXD or XDXd.

Chapter (2.6/3.5) Advanced Pedigree Analysis Understand the "100% penetrance" and "no new mutation" assumptions. * What is penetrance? - What is partial penetrance? - What is 100% penetrance? * What is the disease that experiences irregular penetrance? Describe it. - What type of penetrance does it experience? - What trait is it due to, dominant or recessive? * When looking at a pedigree, how would we distinguish dominant inheritance from recessive inheritance if there is partial penetrance? * What is a "new mutation"? - What are other names for this? - Define it, and know in what scenarios it occurs.

Penetrance * Polydactyly - Dominant trait - partially penetrant pedigree. - partial penetrance: genotype is there, but phenotype doesn't necessarily show. - Penetrance: probability that gene manifests as phenotype. * 100% penetrance: genotype shows phenotype. It's hard/impossible to distinguish a dominant inheritance from recessive inheritance if there is partial penetrance. New Mutation * AKA "de novo mutation, germline mutation" A genetic alteration that is present for the first time in a new child as a result of a mutation in a germ cell (egg or sperm) of one of the parents, or a mutation that arises in the embryo itself during the first few embryonic cell divisions. *** For pedigree questions for this course: Assume 100% penetrance, and no new mutation (unless otherwise stated). ***

Chapter 4: Gene Interactions Define penetrance, expressivity and pleiotropy. * Know examples when relevant. * To interpret phenotype ratios and to identify the distribution of genotypes among phenotypic classes, geneticists make what assumption? - When is this assumption valid? * Define complete penetrance. - When this does not occur, what are the usual reasons for this? * How are incomplete penetrance and variable expressivity related? - What are these circumstances due to?

Penetrance and Expressivity: a mutation does not affect every individual or may cause phenotypes that differ in severity Pleiotropy: one gene affects > one phenotypic character. * A mutant allele has multiple distinct effects in different cells/tissues. * SCD: The same point mutation causes several different consequences/symptoms in different tissues. Penetrance: the probability that a mutation will manifest its phenotype. EX: Polydactyly: dominant trait, partially penetrant pedigree. * Circled individuals have the mutant genotype but do not express the phenotype (have normal hands). Expressivity: the same mutant allele produces different phenotypes in different individuals. EX: Waardenburg syndrome: Dominant trait, variable expressivity. Multiple symptoms, only some manifest in each individual. * Different individuals express a different subset of phenotypes of the overall disease. Textbook: To interpret phenotype ratios and identify the distribution of genotypes among phenotypic classes, geneticists make the assumption that phenotypes differ because their underlying genotypes differ. This assumption is valid only to the extent that a particular genotype always produces the same phenotype. If the correspondence between genotype and phenotype holds true in every case, the trait is identified as having complete penetrance. When the correspondence between genotype and phenotype does not consistently hold true—if instead the same genotype can produce different phenotypes—the usual reasons are gene-environment interaction or interactions with alleles of other genes in the genome. Phenomena in which a certain genotype does not always produce the same phenotype: Incomplete penetrance and variable expressivity, are circumstances in which phenotypic variation among organisms with the same genotype is due to some sort of unknown genetic/environmental interaction.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN Define penetrance, expressivity and pleiotropy. Pleiotropy: * Define it. * Compare variable expressivity and pleiotropy. * How do most mutations display pleiotropy? How do they do this? * Understand the examples given. - Know the example in humans. * What does this example really show us? * What is its mechanism of inheritance? * What is it caused by? What does this lead to?

Pleiotropy is a phenomenon describing the alteration of multiple features of the phenotype by the presence of one mutation. It is distinguished from variable expressivity by the fact that variable expressivity affects one trait, whereas pleiotropy alters several aspects of the phenotype. Most mutations displaying pleiotropy do so either by altering the development of phenotypic features through the direct action of the mutant protein or as a secondary result of a cascade of problems stemming from the mutation. Pleiotropy through the direct action of a mutant protein product is frequently encountered in studies of development. One example is the activity of the Drosophila hormone called juvenile hormone (JH), which is active throughout the Drosophila life cycle and influences numerous attributes of development and reproduction. Increased production or increased activity of JH has been shown to prolong developmental time, decrease adult body size, promote early sexual maturity, raise fecundity (the ability to produce offspring), and decrease life span. An evolutionary tradeoff is associated with changes in JH level or activity. On the one hand, producing more JH can lead to production of more offspring through earlier sexual maturity and higher fecundity. On the other hand, body size decreases and life span is shortened because of increased JH activity. Pleiotropy in the human hereditary condition sickle cell disease (SCD) is an example of the phenotypically diverse secondary effects that can occur due to a mutant allele. SCD (OMIM 603903) is an autosomal recessive condition caused by mutation of the gene that, in turn, affects the structure and function of hemoglobin, the main oxygen-carrying molecule in red blood cells. Many of the red blood cells of people with SCD take on a sickle shape and cause numerous physical problems and complications.

Chapter (2.6/3.5) Advanced Pedigree Analysis Know the rules that allow particular mechanisms of inheritance to be ruled out, and understand the logic behind these rules. * When shown a pedigree that's given to illustrate typical X-linked recessive inheritance, how would you determine if it is also illustrating something else? - i.e. Could it be.... Autosomal recessive? Autosomal dominant? X-linked dominant? Know how to rule these possibilities out, or how to confirm them. Define: * Affected * Unaffected * Normal * Wild-Type

Professor Bardwell's Rules for Pedigree Analysis To determine the mechanism of inheritance in a pedigree (recessive or dominant, autosomal or x-linked), rule out as many possibilities as you can, ideally until only one possibility remains. All of these rules assume: * 100% penetrance * no new mutations Terminology: * Affected: has the disease * Unaffected: does not have the disease (but may be a carrier if the disease is recessive). * Normal: does not have the disease, and is not a carrier. * Wild-type: same meaning as "normal". #1: If unaffected parents have an affected kid, dominance can be ruled out (both autosomal and X-linked). #2: If two affected parents have an unaffected kid, recessiveness can be ruled out (both autosomal and X-linked). Sometimes we can't rule out dominant inheritance by #1, but we can still rule out X-linked dominant inheritance... #3: If an unaffected woman has an affected son, OR an affected man has an unaffected daughter, X-linked dominant can be ruled out. Sometimes we can't rule out recessive inheritance by #2, but we can still rule out X-linked recessive inheritance... #4: If affected woman has unaffected son, OR unaffected man has affected daughter, X-linked recessive can be ruled out. #1 & #4: If unaffected parents have an affected daughter, dominance AND X-linked recessive can be ruled out. So, it must be autosomal recessive inheritance in this case. #2 & #3: If affected parents have an unaffected daughter, recessiveness AND X-linked dominant can be ruled out. So, it must be autosomal dominant inheritance in this case.

Chapter 10: Chromosome Aberrations & Disease (Not a learning objective, but should know) Be familiar with recombination mistakes. * What is recombination responsible for? - Besides this, when is recombination used in other situations? * What can recombination mistakes lead to? * What process(es) of rearrangement would occur between similar sequences on non-homologous chromosomes? * What process(es) of rearrangement would occur between similar sequences within a chromosome? * What process(es) of rearrangement would occur between tandemly repeated sequences on homologous chromosomes?

Recombination Mistakes * Recombination is the process responsible for crossing-over in meiosis. * Recombination is also used in some forms of DNA repair. * Recombination mistakes can lead to chromosome rearrangement: - Between similar sequences on non-homologous chromosomes -> translocation. - Between similar sequences within a chromosome -> inversion or deletion. - Between tandemly repeated sequences on homologous chromosomes -> duplication or deletion. Chromosome rearrangement is caused by mistakes in the same process that causes crossing over in meiosis.

Chapter 3: Mitosis, Meiosis and Sex Determination (Not a learning objective, but should know) TEXTBOOK EXPLANATION Know the categories that reproduction can be divided into. * Know what occurs during each type of reproduction. * Know the difference between them. Know the difference between reproduction in bacteria and archea, and reproduction in eukaryotes. * For eukaryotes, know different modes of reproduction. * Know differences between bacteria, archea, and eukaryotes (single-celled and multicellular): which is haploid/diploid?

Reproduction can be divided into two broad categories: (1) asexual reproduction, in which organisms reproduce without mating, giving rise to progeny that are genetically identical to their parent; and (2) sexual reproduction, in which cells called reproductive cells or gametes are produced by cell division and unite during fertilization. Bacteria and Archaea reproduce exclusively by asexual reproduction. These organisms are haploid; they usually have just a single chromosome. Cell division follows shortly after the completion of chromosome replication; each cell produces two genetically identical daughter cells. Single-celled eukaryotes, such as yeast, have multiple chromosomes and may be either haploid or diploid, and these organisms can reproduce either sexually or asexually. Asexual reproduction in yeast is similar to cell division in bacteria. A haploid yeast cell undergoes DNA replication and distributes a copy of each chromosome to identical daughter cells. Although yeast spend most of their life cycle in a haploid state and actively reproduce as haploids, it is also common for two haploid yeast cells to fuse and form a diploid cell that produces haploid spores by meiosis.

Chapter 10: Chromosome Aberrations & Disease *** CONTINUED *** Describe euploidy, polyploidy, and aneuploidy. * Know the karyotypes of Sex Chromosome Aneuploidies. - Know the nicknames of/diseases caused by these karyotypes. * For diseases, know the symptoms related - Be able to determine, when given a karyotype, the sex of the individual. * How do extra/missing sex chromosomes compare to extra/missing autosomes? * What karyotype is never seen or reported among individuals? - Why is this? * How does this karyotype compare to the karyotypes in the attached picture? He says "You do not need to memorize these, but you do need to know how to work non-disjunction problems", referring to a list of autosomal and sex-chromosome aneuploidies, the syndromes they cause, their syndrome characteristics, and their frequencies at birth.

Sex Chromosome Aneuploidies * Trisomy-X = 47,XXX (female). - Extra X. - 0.1% of female births. * Double-Y = 47,XYY (male) - Extra Y - 0.1% of male births. Klinefelter syndrome = 47,XXY (male, sterile) - Extra X. - 0.1% of male births. - Faces sterility problems, experiences less body hair, and breast enlargement. Turner Syndrome = 45,X (female, sterile) - Only one X. - 0.5% of female births. - Faces sterility problems. An extra or missing X or Y chromosome usually has a relatively mild effect in comparison to an extra or missing autosome. - An extra/missing autosome is much more severe. PIC: The reason you don't see the YY karyotype in this picture is because having only Y and no X sex chromosomes is embryonic lethal. This is due to the X chromosome containing essential genes, meaning the absence of essential genes in the YY karyotype. Therefore, the picture includes only the karyotypes that make it to live birth.

Chapter 3: Mitosis, Meiosis and Sex Determination What are these thingies? A. Sister chromatids B. Homologous chromosomes C. Chaismata D. Single strands of DNA E. Purple blobules

Sister chromatids

Problem Set #1: Pedigrees What are the standard assumptions we need to make for all questions? - Terminology assumptions: * Affected * Unaffected * Normal * Filled in symbol on pedigree * Blank symbol on pedigree - Penetrance Assumptions - Mutation Assumptions - Disease Type Assumptions

Standard assumptions for all questions: Affected = has the disease. Unaffected = does not have the disease (but may be a carrier if the disease is recessive). - If someone is a heterozygous carrier of a recessive disease, they will be depicted as unaffected. Normal = does not have the disease and is not a carrier.• For all questions, assume any trait or disease mentioned has 100% penetrance, unless it is stated otherwise. For inheritance questions, assume no new (de novo) mutations, unless it is stated otherwise. All diseases mentioned are genetic diseases, unless it is stated otherwise. Pedigrees will only show you if someone is affected (black symbol) or unaffected (white symbol).

Chapter 10: Chromosome Aberrations & Disease Compare and contrast balanced vs. unbalanced rearrangements. * Which rearrangements are balanced? * Which rearrangements are unbalanced? * What do balanced rearrangements result in? - Compare them in terms of their severity/lack thereof. * What do unbalanced rearrangements result in? - Compare them in terms of their severity/lack thereof.

Summary of Translocation and Inversion * Balanced rearrangements. * Do not result in a loss or gain of much chromosomal material, only the breakpoints are the issue where this happens. Summary of Deletion and Duplication * Unbalanced rearrangements. * Result in a loss (deletion) or gain (duplication) of a chunk of the chromosome. * Usually severe consequences for affected patients, with deletions more severe than duplications.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION: DON'T MEMORIZE, JUST USE TO HELP YOU UNDERSTAND. Interpret the results of complementation analysis crosses. * When geneticists encounter organisms with the same mutant phenotype, what are their initial questions? * What can mutations of different genes produce? - Explain this with an example. * Define genetic complementation analysis. - What is it designed to test? - What can the results determine? - How is genetic complementation testing done? (What is crossed? What does the phenotype of the F1 progeny tell you?) * The wild-type phenotype tells you: ________. - Has genetic complementation occurred? * The mutant phenotype tells you:___________. - Has genetic complementation occurred?

TEXTBOOK EXPLANATION: DON'T MEMORIZE, JUST USE TO HELP YOU UNDERSTAND (SKIM) When geneticists encounter organisms with the same mutant phenotype, two initial questions are (1) do these organisms have mutations on the same gene or on different genes, and (2) how many genes are responsible for the mutations observed? Mutations of different genes can produce the same, or very similar, abnormal phenotypes. For example, in a multistep pathway whose end point is the production of a pigment that colors flower petals, it is possible that a mutation of any of the genes in the pathway could block production of the pigment and produce mutant flower color. In this section, we discuss genetic complementation analysis, an experimental analysis of crosses designed to test alternative genetic explanations of an abnormal phenotype. The results of genetic complementation analysis can determine whether mutant organisms carry mutations of different genes that produce the abnormal phenotype or if the abnormal phenotype occurs due to allelic mutations on the same gene. Genetic complementation testing is done by crossing pure-breeding mutants for a recessive mutation and observing the phenotype of F1 progeny. If the F1 progeny have the wild-type phenotype, genetic complementation has occurred, and the conclusion is that the mutant alleles are of different genes. On the other hand, if the mutant alleles are of the same gene, the progeny of two pure-breeding mutants will have a mutant phenotype. This result indicates that no genetic complementation has taken place.

Chapter 4: Gene Interaction Duplicate Gene Action * Define gene interaction. * Define the one gene-one enzyme hypothesis. - What do single-gene mutations cause? - What are mutant phenotypes attributable to? - How is an enzyme defect inherited? - What does the one gene-one enzyme hypothesis identify? - What modifications has the one gene-one enzyme concept undergone (what is now taken into account)?

Textbook: Gene interaction, the collaboration of multiple genes in the production of a single phenotypic character or a group of related characteristics. One gene-one enzyme hypothesis: each gene produces an enzyme and each enzyme has a specific functional role in a biosynthetic pathway. Single-gene mutations block the completion of biosynthetic pathways and lead to the production of mutants that are deficient in their ability to produce the end product of the pathway. Their hypothesis proposed that each mutant phenotype was attributable to the loss or defective function of a specific enzyme. The consequences of these enzyme losses or defects were the blockage of a biosynthetic pathway and the absence of the end product of the pathway. Since each enzyme defect was inherited as a single-gene defect, the one gene-one enzyme hypothesis identifies the direct connection between genes, proteins, and phenotypes. The one gene-one enzyme concept has undergone modifications since it was first proposed. These changes take account of three observations: (1) Many protein-producing genes do not produce enzymes, but produce transport proteins, structural proteins, regulatory proteins, or other nonenzyme proteins; (2) some genes produce RNAs rather than proteins; and (3) some proteins (e.g., β-globin) must join with other proteins to acquire a function. Despite these modifications, the fundamental conclusion linking each gene to a particular product is valid and forms the basis for understanding gene function.

Chapter 10: Chromosome Aberrations & Disease TEXTBOOK EXPLANATION Describe non-disjunction. * Compare disjunction and nondisjunction. * What consequences does aneuploidy have (what does it affect)? - What do these abnormalities associated with aneuploidy result from? * Define this. - In a diploid organism where there are two copies of a gene on a homologous pair of chromosomes, what % is generated, and for what? - In a monosomic mutant that has just one gene copy, what % is generated, and for what? - In a trisomic mutant that has three copies, what % is generated, and for what? * What do changes in this lead to?

Textbook: We've previously discussed the connection between Mendel's two laws of heredity and the disjunction of homologous chromosomes and sister chromatids during meiosis. In the discussion that now follows, we focus on nondisjunction, the failure of chromosomes and sister chromatids to properly disjoin during cell division. As we describe, nondisjunction is the cause of abnormalities of chromosome number in cells. The changes in chromosome number exert their effects primarily by addition or removal of one or more chromosomes of the normal complement in a nucleus. Such changes are mutations that add or remove large numbers of genes. These abnormalities almost always alter the phenotype and can have an effect on the development and reduce fertility and viability of the affected organism There are phenotypic consequences of aneuploidy. Aneuploidy profoundly affects the phenotype and development of nearly all animal species. The phenotypic and developmental abnormalities associated with aneuploidy result from changes in gene dosage, the number of copies of a gene in the genome. Aneuploidy changes the dosage of all the genes on the affected chromosome. In a diploid organism where there are two copies of a gene on a homologous pair of chromosomes, 100% of gene dosage is generated for each gene on the chromosome, a monosomic mutant has just one gene copy and just 50% of normal gene dosage for each gene on the chromosome. In contrast, a trisomic mutant has three copies and 150% of normal gene dosage for each of the genes on the chromosome. Changes in gene dosage lead to an imbalance of gene products from the affected chromosome relative to unaffected chromosomes, and this imbalance is at the heart of alterations of normal development and the production of abnormal phenotypes.

Chapter 10: Chromosome Aberrations & Disease Review homologous vs. non-homologous chromosomes vs. sister chromatids. TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN Chromosome Structure and Chromatin Organization of Eukaryotic Chromosomes: * Define core DNA. * Define nucleosome. - Know what level of organization it is. * Define the 10-nm fiber. - Know what level of organization it is. - What is a nickname for this level of organization? * Understand this nickname and what the components represent. - Describe this state of chromatin. What does this mean (where is the 10-nm fiber found)? * Define linker DNA. * What molecular interactions are important to the types of chromatin structure present in different regions of eukaryotic chromosomes? * Define the 30-nm fiber. - Know what level of organization it is. - Where is this found? - Describe its assembly, and what components take part in this process. * Describe its structure (what assembly produces). - What is another name for this?

The 146 bp of DNA wrapped around a nucleosome core particle is called core DNA, and the combination of a nucleosome core particle wrapped with core DNA is identified as a nucleosome. Electron micrographs of chromatin fibers in a highly decondensed state show a regular series of circular structures strung together by connecting filaments (see Figure 10.24b). This form of chromatin is identified as the "beads on a string" morphology of chromatin. The "beads" are nucleosomes that are a little more than 11 nm in diameter, and the "string" is called linker DNA. Linker DNA is the DNA between regions of core DNA. This beads-on-a-string form of chromatin is identified as the 10-nm fiber, since the diameter of nucleosomes is approximately 10 nm. Additional molecular interactions between the N-terminal (amino-terminal) tails of histone proteins and core and linker DNA are critically important to the types of chromatin structure present in different regions of eukaryotic chromosomes. The 10-nm fiber is an unnatural state for chromatin, not achieved in cells. Under in vitro conditions, chromatin forms the 30-nm fiber, although it is not certain this structure forms in vivo (in cells). Electron micrographs and molecular modeling help us visualize how the 30-nm fiber is assembled. It is produced by coalescence of the 10-nm fiber into a cylindrical filament of coiled nucleosomes that is hollow in the middle. Due to its coiled structure and open middle, the 30-nm fiber is often also called the solenoid structure (like the coil of wire in the starter of a car). Each turn of the solenoid structure contains six to eight nucleosomes. The diameter of the solenoid is approximately 34 nm.

Chapter 4: Gene Interactions (Not a learning objective, but should know) * What is the general rule for safe blood transfusion? - What occurs if these rules aren't followed? - What blood type is a "universal donor"? Explain why. - What blood type is a "universal recipient"? Explain why. * What are the two ABO blood group antigens on the surfaces of red blood cells? - How do they differ from one another? - Describe their structure/components of their structure. Know where each component is found. - Know what the components are composed of. - What is the H antigen? What does it result from? Where is it present? * How can the H antigen be modified? What do modifications depend on?

The ABO system is one of several blood types that must be tested before blood transfusion to ensure the safety of the procedure. The general rule for safe blood transfusion is that the recipient blood must not contain an antibody that reacts with an antigen in the donated blood. When such a reaction occurs, blood clots produced by clumping blood cells form at the site of transfusion. These adverse reactions can potentially cause life-threatening complications. Notice that people with blood type O are "universal donors" who can donate to people of any blood type. This is because type O contains neither A nor B antigens. Notice also that people with blood type AB are "universal recipients" who can receive blood from any blood type. This is because their blood contains neither anti-A nor anti-B antibodies. The Molecular Basis of Dominance and Codominance of ABO Alleles The two ABO blood group antigens on the surfaces of red blood cells each have a slightly different molecular structure. The antigens are glycolipids that contain a lipid component and an oligosaccharide component. The lipid portion of the antigen is anchored in the red blood cell membrane, and the segment protruding outside the cell contains the oligosaccharide. Initially, the oligosaccharide is composed of five sugar molecules and is called the H antigen. It results from the activity of an enzyme produced by the H gene (Figure 4.4). The H antigen is present on the surfaces of all red blood cells, but it can be further modified, in two alternative ways, by the addition of a sixth sugar, or it can be left unmodified. The final modification of the H antigen depends on the enzymatic activity of the protein product of the ABO blood group locus.

Chapter 10: Chromosome Aberrations & Disease Review homologous vs. non-homologous chromosomes vs. sister chromatids. TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN AGAIN AGAIN AGAIN AGAIN AGAIN AGAIN AGAIN Position Effect Variegation: Effect of Chromatin State on Transcription: * What kind of region is the chromosome region immediately surrounding the centromere? - How many expressed genes are found here? * What happens to nucleosomes during the S phase? - What happens to nucleosomes directly after this? * What does this lead to? - How does this vary? * Why is this variablility permissible? * Specifically, how does this vary for X-chromosome centromeric heterochromatin? - How does a greater extent of this variation affect the DNA sequence included in the heterochromatic region? * What does this mean in terms of the amount of genes expressed? Does this cause an increase or decrease? * How does this explain the variegation of the mutants and the variability of this variegation pattern? * How do limitations affect the pheno

The chromosome region immediately surrounding the centromere is a heterochromatic region that in Drosophila and most eukaryotes contains very few expressed genes. During S phase there is a temporary dissociation of nucleosomes as DNA replicates. The reassociation of nucleosomes after DNA replication leads to the reformation of heterochromatin around the region of the centromere, but the distance to which the reformed heterochromatin extends can vary from chromosome to chromosome (the heterochromatic region spreads like a virus and infects previously euchromatic regions, silencing previously expressed genes). This variability is permissible because the centromeric region normally contains few if any expressed genes (normally no genes at risk). Specifically, on some X chromosomes the centromeric heterochromatin spreads a greater distance outward from the centromere than on other X chromosomes. A greater extent of heterochromatin spread leads to more DNA sequence being included in the heterochromatic region. This provides an explanation for both the patches of red and white eye color and the variability of the variegation pattern. The X-chromosome-to-X-chromosome variability of centromeric heterochromatin spread following DNA replication has a very specific consequence for the expression of the red-eye-color alleles that are close to the centromere as a result of paracentric inversion (red-eye-color allele close to centromere = more likely to become "infected", or a part of the heterochromatic region, meaning it is more likely to be silenced, causing an absence of red-eye-color) The wild-type red-eye-color allele is expressed in inverted X chromosomes as long as the centromeric heterochromatin does not spread to cover the gene. If the spread of centromeric heterochromatin covers the new gene location in inverted X chromosomes, the allele is silenced (no red eye color). (Can't reach/limited spread = expressed) (Can reach = silenced)

Chapter 10: Chromosome Aberrations & Disease Review homologous vs. non-homologous chromosomes vs. sister chromatids. TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN AGAIN AGAIN Chromosome Structure and Chromatin Organization of Eukaryotic Chromosomes: * Define the chromosome scaffold. Answer this by: - Know what level of organization it is. - Describe its structure. - Know the function of the chromosome scaffold. - What is the relationship between chromatin and the scaffold? - Stripped of chromatin, what is the chromosome scaffold reminiscent of (be specific)? - Stripped of chromatin, what is the chromosome scaffold composed of? * What is the role of these components? - At what sites does this occur? - What do contemporary models of chromatin organization predict about the structure of the chromosome scaffold over time? * What components play a part in this? - Know characteristics of the chromosomes found in metaphase. * What is achieved? What components are involved?

The chromosome scaffold is a filamentous framework made up of a large number of distinct nonhistone scaffold proteins. The chromosome scaffold gives a chromosome its shape. The scaffold is in some ways like the steel infrastructure that provides the shape, strength, and support for a building. In the case of chromosomes, the chromatin is "hung" on the scaffold. By comparing a fully condensed chromosome at metaphase to the protein scaffold of a metaphase chromosome after being stripped of DNA, we can see the shape of the chromosome scaffold is clearly reminiscent of the metaphase chromosome structure, consisting of sister chromatids joined at the centromere. Stripped of chromatin, the chromosome scaffold is composed of nonhistone proteins that form an infrastructure that anchors DNA loops and gives the chromosome its shape. Chromatin loops containing 20,000 to 100,000 bp of DNA are anchored to the chromosome scaffold by other nonhistone proteins at sites called matrix attachment regions (MARs). Contemporary models of chromatin organization predict that the chromatin loops progressively consolidate and are further compressed by nonhistone proteins. Ultimately, the compaction of chromatin achieved by metaphase is approximately a 250-fold compaction of the 300-nm fiber, which already represents significant compaction.

Chapter 10: Chromosome Aberrations & Disease Review homologous vs. non-homologous chromosomes vs. sister chromatids. TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN AGAIN AGAIN AGAIN AGAIN Nucleosome Disassembly, Synthesis, and Reassembly during Replication: * What is assembly of nucleosome core particles in connection with replication driven by? Describe this process step-by-step. - Are the histones old or new? What do the combinations of histones look like? * How do the differences in histones affect what they bind to? - H3/H4 - H2A/H2B - old/new if applicable * What is the function of these new nucleosome core particles? - What do current models propose happens as the replication fork passes? * Know _____________, and the mechanism by which this occurs. - What histones attach or reattach to. - Where histones attach or reattach. - How much synthesis of new histone proteins occurs? * How does the end result amount of nucleosomes compare to the initial amount?

The current models propose that as the replication fork passes, nucleosomes break down into H3-H4 tetramers (each tetramer contains two molecules of H3 and two molecules of H4) and H2A-H2B dimers (one molecule of each histone in a dimer). The H3-H4 tetramers reattach at random to one of the sister chromatid products of replication. Meanwhile, H2A-H2B dimers dissociate from the chromosome, and they may disassemble into individual histone molecules. Quickly, however, disassembled H2A and H2B histones reform into dimers or are joined by newly synthesized H2A and H2B histones to form dimers. New H3 and H4 molecules are also synthesized, and they form tetramers that attach to H2A-H2B dimers and to sister chromatids. H2A-H2B dimers also join H3-H4 tetramers already attached to sister chromatids. Enough new synthesis of all four histone proteins takes place to double the number of nucleosomes. All combinations of old and new histone components aggregate in assembling new nucleosomes after DNA replication. Following the passage of the replication fork, "old" H3-H4 tetramers are randomly assigned to daughter strands, and newly synthesized H3-H4 tetramers inhabit strands not bound by old tetramers. Old and new H2A-H2B dimers join the tetramers to form complete nucleosomes. Collectively, these activities provide the histone octamers needed to organize newly synthesized DNA.

Chapter 10: Chromosome Aberrations & Disease Review homologous vs. non-homologous chromosomes vs. sister chromatids. TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN AGAIN Chromosome Structure and Chromatin Organization of Eukaryotic Chromosomes: 30-nm fiber (continued): * What is the relationship between H1 and the 30-nm fiber? - Describe this. * What do we see in chromatin from which H1 has been removed? * When does chromatin exist in a 30-nm fiber state? - During this time, what other states could the chromatin exist in? * Define the 300-nm fiber. - Know what level of organization it is. - Beyond the 30-nm stage, what is integral to the structure of chromosomes and the process of chromosome condensation? * When does this chromosome condensation initiate? - What are the roles of nonhistone proteins? - Which chromosome structure from the cell cycle results from the formation of this fiber? - Describe the structure of this fiber in detail. * What does it look like? What are the components? * What is the foundation of chromosome shape? - With continued condensation, what would the 300-nm fiber form?

The histone protein H1 plays a key role in stabilizing the solenoid structure. The long N-terminal and C-terminal ends of the H1 protein attach to adjacent nucleosome core particles. H1 protein pulls the nucleosomes into an orderly solenoid array and lines the inside of the structure. Experimental analysis shows that chromatin from which H1 has been removed can form 10-nm fibers but not 30-nm fibers. Chromatin exists in a 30-nm-fiber state or a more condensed state during interphase. Beyond the 30-nm stage, chromatin compaction and the presence of nonhistone proteins are integral to the structure of chromosomes and the process of chromosome condensation that initiates with the onset of prophase in the M phase of the cell cycle. Nonhistone proteins perform multiple roles in influencing chromosome structure and in facilitating M phase chromosome condensation. Interphase chromosome structure results from the formation of looped domains of chromatin similar to supercoiled bacterial DNA (see Figure 10.24d). The loops are variable in size, containing from tens to hundreds of kilobase pairs and consisting of 30-nm-fiber DNA looped on a category of nonhistone proteins that are the foundation of chromosome shape. The diameter of looped chromatin is approximately 300 nm, so looped chromatin is called the 300-nm fiber. With continued condensation, the chromatin loops form the sister chromatids. In metaphase, chromosome condensation reaches its zenith, resulting in chromosomes that are easily visualized by microscopy (see Figure 10.24e).

Chapter 4: Gene Interactions Genetic Analysis 4.1 Problem: The MN blood group in humans is an autosomal codominant system with two alleles, M and N. Its three blood group phenotypes, M, MN, and N, correspond to the genotypes MM, MN, and NN. The ABO blood group assorts independently of the MN blood group. A male with blood type O and blood type MN has a female partner with blood type AB and blood type N. Identify the blood types that might be found in their children, and state the proportion for each type.

The male has blood types O and MN. Type O results from homozygosity for the recessive i allele, whereas MN is produced in heterozygotes carrying both alleles. The male genotype is ii MN. The female has blood groups AB and N. The AB blood type is found in heterozygotes, and blood type N in homozygotes. The female blood group genotype is IAIB NN. Independent assortment predicts two gamete genotypes for the male: All gametes contain i, half carry M, and half carry N. Independent assortment predicts two gamete genotypes for the female: All gametes contain N, half contain IA, and half contain IB. Blood types A and B are each expected in 50% of the offspring of this cross, as are blood types MN and N. Four different blood group phenotypes, each with an expected frequency of 25% are predicted.

Chapter 10: Chromosome Aberrations & Disease (Not learning objective, but should know) * Explain why maternal non-disjunction during meiosis II can produce XXY offspring. * Explain why paternal non-disjunction during meiosis II cannot produce XXY offspring.

The reason maternal non-disjunction during meiosis II can produce XXY offspring. In this case, meiosis I occurs as normal so that after the first division the replicated homologous chromosomes, each consisting of two chromatids connected at a centromere, have separated. If a non-disjunction occurs at this point, haploid cells will not be formed. Instead, the sister chromatids of a replicated chromosome will not be separated and a gamete containing two X chromosomes will be formed. When this gamete combines with a sperm containing a Y chromosome, the offspring has the XXY genotype. The reason paternal non-disjunction during meisosis II cannot produce XXY offspring. This is similar to the maternal case except that after meiosis I, in one cell will be two YY chromosomes while the other cell will be two XX chromosomes. When non-disjunction occurs (assuming it occurs in both divisions), these chromosome arrangements remain the same and two cells are formed. One with two XX chromosomes and one with two YY chromosomes. If either of these gametes combines with a female X gamete, either XYY or XXX will be formed but never XXY.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN AGAIN AGAIN Allelic series: Molecular Basis of the C-gene allelic series: * Know the characteristics of the tyrosinase enzyme produced by the c^h allele. - What kind of allele is the c^h allele?. - What phenotype does this allele produce? * What are all the genotypes that produce this phenotype? * Define temperature-sensitive allele. * Understand the siamese cat coat example. - What occurs at cooler extremities of the body in terms of tyrosinase activity of the c^h allele? * What phenotype does this produce in these areas? - What occurs at the warmer portions of the body in terms of tyrosinase activity of the c^h allele? * What phenotype does this produce in these areas? * Know the characteristics of the tyrosinase enzyme produced by the c allele. - What kind of allele is the c allele? - What phenotype does this allele produce? - Describe homozygotes for this allele: * Biochemical processes. * Phenotype.

The tyrosinase enzyme produced by the hypomorphic c^h (Himalayan) allele is unstable and is inactivated at a temperature very near the normal body temperature of most mammals. This type of gene product is an example of a temperature-sensitive allele. Cats with the Siamese coat-color pattern are familiar examples of the action of this temperature-sensitive allele. The parts of cats that are farthest away from the core of the body (the paws, ears, tail, and tip of the nose) at most times tend to be slightly cooler than the trunk. At these cooler extremities, the temperature-sensitive tyrosinase produced by the c^h allele remains active, producing pigment in the hairs there. However, in the warmer central portion of the body, the slightly higher temperature is enough to cause the tyrosinase produced by the c^h allele to denature, or unravel. This inactivates the enzyme and leads to an absence of pigment in the central portion of the body. Animals that are c^hc^h or c^hc have the Himalayan phenotype. The final allele in the series, c, is a null allele that does not produce functional tyrosinase. Homozygotes for this allele are unable to initiate the catabolism of tyrosine. This leads to an absence of melanin and produces the condition known as albinism.

Chapter 3: Mitosis, Meiosis and Sex Determination Genetic Analysis 3.3 Problem Hemophilia A is an X-linked recessive blood-clotting disorder caused by mutation of the factor VIII gene. Suppose a heterozygous woman with normal blood clotting has children with a man who also has normal blood clotting. Determine the probability of each of the following outcomes: a. The probability of a son having hemophilia A. b. The probability of a child of either sex having normal blood clotting. c. The probability of having three children, each of whom has hemophilia A. d. The probability3 of having four children, two of whom have hemophilia A and two of whom have normal blood clotting.

The woman is described as being heterozygous and so her genotype is XHXh, where the uppercase and lowercase superscripts represent the dominant and recessive alleles, respectively. The man has normal blood clotting, so he is hemizygous for the wild-type allele. His genotype is XHY. The Punnett square predicts four different genotypes among the possible children of this couple. A) Taking sex into account, we find that approximately one-half the offspring are male and one-half are female. The Punnett square shows two possible male genotypes, one healthy and one a hemizygous male with hemophilia A. The probability that a son will have hemophilia A is therefore one-half, or 50%. B) The Punnett square shows that three of the four possible offspring genotypes would produce normal blood clotting. The probability that a child of this couple has normal blood clotting is 0.75, or 75%. C) The risk that each child will have hemophilia A is 25%. For three children with hemophilia A, the probability is (.25)(.25)(.25) = 0.0156, or (1/4)(1/4)(1/4)=(1/64). D) The chance the couple has four children, two of whom have hemophilia A and two of whom are healthy, is predicted by the binomial expansion. There are six different ways (birth orders) in which to produce two healthy and two affected children. The probabilities are (3/4) for a healthy child and (1/4) for a child with hemophilia A, so the requested probability is 6[(3/4)(3/4)(1/4)(1/4)] = (54/256), or 0.2109. Six different birth ways because each of the probabilities can be rearranged in 6 different ways. If unaffected is U and affected is A, you can have UUAA, UAUA, UAAU, AUUA, AUAU, or AAUU.

Chapter 10: Chromosome Aberrations & Disease Review homologous vs. non-homologous chromosomes vs. sister chromatids. TEXTBOOK EXPLANATION CONTINUED Chromosome Structure and Chromatin Organization of Eukaryotic Chromosomes: * Understand why the molecular organization of chromatin is necessary. * Why is chromosome compaction by chromatin important? - Without compaction, what would happen? - What important process would be impossible without condensation? - Think about your chromosomes and what they contain. Think about all that is needed to fit into the nucleus, and still allow room for essential processes. * What is the eukaryotic chromosome made up of? What are the proportions of these components? - Divide these component proportions into smaller groups and know what makes up these proportions of the chromosome. * Define histones. - Know their characteristics and what they do. - Know the types of histone proteins, and how they compare. * Define nonhistone proteins. * Compare and contrast histone and nonhistone proteins.

This molecular organization is essential to the normal function and distribution of chromosomes in cell division, and it plays a pivotal role in the regulation of gene expression that typifies all kinds of eukaryotic cells. Why is chromosome compaction by chromatin important? Simply stated, eukaryotic chromosomes would not fit into the nucleus without compaction, and chromosome segregation during cell division would be impossible without chromosome condensation. Each one of your chromosomes contains one long DNA double helix that is incorporated with large amounts of protein into the complex known as chromatin. Each of your somatic cell nuclei contains more than 6 billion base pairs of DNA divided among 46 chromosomes, and all that DNA fits in the nucleus and still allows space for DNA replication, transcription, and mRNA processing, thanks to a remarkable feat of biomolecular engineering brought about by chromatin. By weight, each eukaryotic chromosome is approximately half DNA and half proteins, and about one-half of the protein content of chromatin is histone protein. The histones are five small, basic proteins that are positively charged and bind tightly to negatively charged DNA. Equally abundant in the chromatin, but more diverse, is an array of hundreds of types of other DNA-binding proteins named, by default, nonhistone proteins. This large array of proteins performs a variety of tasks in the nucleus, not all of which are defined. The five types of histone proteins in chromatin are designated H1, H2A, H2B, H3, and H4 (Table 10.4). H1 is the largest and most variable histone protein. The other four histones are considerably smaller and more uniform in size.

Chapter (2.6/3.5) Advanced Pedigree Analysis TEXTBOOK EXPLANATION CONTINUED Analyze pedigrees and determine the mechanism of inheritance of the disease or trait being followed. * What is an autosomal dominant trait? Requirements? * What is an autosomal recessive trait? Requirements? * What are prospective predictions? Be able to give examples. * What does retrospective mean? Be able to give examples.

To be classified as autosomal dominant, a trait must appear both in individuals who have a heterozygous genotype (e.g., Aa) and in those with a certain homozygous genotype (e.g., AA). There are several common characteristics of autosomal dominant traits that can be evident in pedigrees. In the autosomal recessive pattern of heredity, the recessive phenotype appears only in those individuals who have the genotype that is homozygous for the recessive allele (e.g., aa). The major common characteristics of autosomal recessive inheritance in pedigrees differ in several ways from those seen for autosomal dominant traits. In the context of testing his hereditary laws, Mendel made prospective predictions about the outcomes of certain crosses. In other words, when setting up specific crosses between pea plants, he would make a prediction beforehand about the percentages of dominant and recessive phenotypes he expected to see among the cross progeny. That kind of prospective prediction occurs in the field of human genetics. If, for example, a man and a woman know that each is heterozygous for an autosomal recessive disease, they can ask the question, "What is the chance a child of ours will have the recessive condition?" In this case, the genetic cross is Aa x Aa, and there is a (1/4) chance that any offspring will have the homozygous genotype aa. The study of heredity can also be retrospective. One feature making the study of inheritance in humans different from that in other organisms is that human heredity is often examined after reproduction has taken place, when questions may arise about the genotypes of individuals even though their phenotypes are known. For example, it is usually only after an adverse hereditary outcome has been detected in a family that the inheritance of the unusual trait becomes a subject of attention by medical genetic professionals. Construction of a pedigree may show the family to have a history of the hereditary condition; alternatively, it may show the hereditary condition to have previously been unknown in the family. In either case, an adverse reproductive outcome is the trigger for medical genetic investigation of the family.

Chapter 3: Mitosis, Meiosis and Sex Determination TEXTBOOK EXPLANATION (Not a learning objective, but should know) Explain X-linked dominant trait transmission * Know the characteristics of transmission of traits that are controlled by X-linked dominant alleles. - Heterozygous female (Hh) x Recessive male (Yh) = ? - Hemizygous male dominant (HY) x Recessive female (hh) = ?

Transmission of traits that are controlled by X-linked dominant alleles has two distinctive characteristics, one indicating transmission from a female and one indicating transmission from a male. When the transmitting parent is a heterozygous female with the dominant trait (Hh) and her mate is a male with the recessive trait (Yh), about half the progeny of each sex have the dominant condition. When the transmitting parent is a hemizygous male with the dominant trait (HY) and his mate is a female with the recessive trait (hh), we see a hallmark that distinguishes autosomal dominant transmission from X-linked dominant transmission. In these matings, the dominant trait appears in all daughters, who are Hh, and in no sons,who are hY.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED AGAIN AGAIN AGAIN Allelic series: Molecular Basis of the C-gene allelic series * What is the difference between tyrosinase enzymes produced by different C-gene alleles? - What is this difference the basis for? * Know characteristics of the C allele. - Is it dominant or recessive? - How does it compare to the WT? - What does enzymes activity look like for this allele? * What explains the order of the allelic series? * Know the activity of the enzyme produced by the c^(ch) allele, compared to the WT. - What kind of allele is this? - What phenotypes are seen in ________ genotype and how does this compare to the WT? What biochemical process explains this phenotype difference? What does this process lead to in terms of phenotype? * Homozygous c^(ch)c^(ch) * Heterozygous c^(ch)c^h * Heterozygous c^(ch)c - How do heterozygous genotypes compare to homozygous genotypes in terms of the alleles present, and what is the effect of the absence/presence of these alleles on the phenotype?

Tyrosinase enzymes produced by different C-gene alleles have distinctive levels of catabolic activity that are the basis for the dominance relationships between the alleles. The allele C is a dominant wild-type allele producing fully active tyrosinase that is defined as 100% activity. The percentage of wild-type tyrosinase activity produced by each allele explains the order observed for the allelic series. Biochemical examination reveals that the enzyme produced by the c^(ch) hypomorphic allele has much less activity than the wild-type enzyme. In the homozygous c^(ch)c^(ch) genotype or heterozygous c^(ch)c^h genotypes or c^(ch)c, only a small amount of melanin is synthesized. This leads to a decreased amount of pigment, and it has the effect of muting the coat color, more so in heterozygous genotypes, where just one c^(ch) allele is present, than in the homozygous c^(ch)c^(ch) genotype, where two alleles are present.

Chapter 4: Gene Interactions TEXTBOOK EXPLANATION CONTINUED AGAIN Define penetrance, expressivity and pleiotropy. Variable Expressivity: * Define variable expressivity. * Know Waardenburg syndrome. - What is its method of inheritance? - Know the principal features of the syndrome and how individuals with the syndrome vary. - Understand the pedigree of Waardenburg syndrome. - What does molecular genetic analysis tell us about Waardenburg syndrome in terms of the alleles individuals with the syndrome carry, and the patterns of their phenotypes? Variable Expressivity and Incomplete Penetrance: * What interactions may be the cause of incomplete penetrance or variable expressivity? * What is the characterization of a trait as having incomplete penetrance or variable expression acknowledging?

Variable Expressivity Sometimes the discrepancy between genotype and phenotype is a matter of the degree or specific manifestation of expression of a trait rather than presence or absence of the trait altogether. In the phenomenon of variable expressivity, the same genotype produces phenotypes that vary in the degree or form of expression of the allele of interest. Waardenburg syndrome is a human autosomal dominant disorder displaying variable expressivity. Individuals with Waardenburg syndrome may have any or all of four principal features of the syndrome: (1) hearing loss, (2) different-colored eyes, (3) a white forelock of hair, and (4) premature graying of hair. In the pedigree shown in Figure 4.15, notice that the circles and squares representing family members with Waardenburg syndrome may be entirely or only partly colored. Each quadrant of the symbols represents one of the principal features of the syndrome. The diversity of symbol darkening demonstrates the variation in expressivity of Waardenburg syndrome in this family. Molecular genetic analysis tells us that each family member with Waardenburg syndrome carries exactly the same dominant allele, yet among the six affected members of the family, there are five different patterns of phenotypic expression. Pinpointing the cause of incomplete penetrance or variable expressivity is a challenging task. Three kinds of interactions may be responsible: (1) other genes that act in ways that modify the expression of the mutant allele, (2) environmental or developmental (i.e., nongenetic) factors that interact with the mutant allele to modify its expression, and (3) some combination of other genes and environmental factors interacting to modify expression of the mutation. Indeed, the characterization of a trait as having incomplete penetrance or variable expressivity is an acknowledgment that an as-yet-unknown factor is interacting with gene expression to produce variability in expressivity or to reduce penetrance.

Chapter 4: Gene Interactions Describe variations from classical Mendelian genetics, including incomplete dominance, co-dominance, multiple alleles/allelic series, and lethal alleles. * Know all variations from classical Mendelian Genetics, define each. - Know relevant examples.

Variations from Classical Mendelian Genetics Incomplete dominance: heterozygotes have a phenotype intermediate to the two alleles. PIC: Example of incomplete dominance in humans: Hypercholesterolemia - too much cholesterol in blood. * HH: homozygous for ability to make LDL receptors (normal cell). * Hh: heterozygous for the ability to make LDL receptors, makes less (mild disease). * hh (homozygous for inability to make LDL receptors, can't make any (severe disease). Co-dominance: the phenotype of both alleles is fully expressed in heterozygotes. * PIC: One way in which human blood groups are classified is the MN system. Two alleles are present in the population M and N, and an individual may have one or both of them - i.e. LM/LM or LM/LN or LN/LN. When both alleles are present in a heterozygote, both phenotypes can be detected. * Classic example - ABO Blood types in humans. 3 alleles: A, B, and O. The O allele is recessive to A and B. A and B are codominant to each other. Multiple alleles: more than two alleles affect a phenotype. A given gene can have many alleles (in the population). A single individual can only have two alleles of any given gene. PIC: The Brca1 gene has many different alleles. Mutation in Brca1 causes disease, meaning that mutation in any of the several different alleles of Brca1 causes disease. - Allelic Series: There is an order of dominance when multiple alleles are present. Penetrance and Expressivity: a mutation does not affect every individual or may cause phenotypes that differ in severity. Pleiotropy: one gene affects > one phenotypic character. Environmental impact: genetically identical individuals show different phenotypes as a result of environmental factors. Epistasis: a gene at one locus alters the phenotypic expression of a gene at a second locus.

Chapter 10: Chromosome Aberrations & Disease TEXTBOOK EXPLANATION Describe euploidy, polyploidy, and aneuploidy. * How does the number of chromosomes in males and females of a species differ? * What is the number of chromosomes in nuclei of normal cells? - Define this. * What is the total chromosome number in nearly all animal species? - Define this. * What is the total chromosome number in nearly all plant species? - Define this * Define euploid. * Define aneuploidy. - What causes aneuploidy? "Skim" * Define polyploidy. * What can polyploidy result from? * What mechanisms are most commonly the cause of polyploidy? - Know and understand the steps of each mechanism, and their results.

With a few unusual exceptions, the number of chromosomes is the same for males and females of a species, and the number of chromosomes in nuclei of normal cells is a multiple of the haploid number (n), the number in a single set of chromosomes. In nearly all animal species, the total chromosome number is 2n (diploid), but in plants, 3n (triploid) or higher multiples of n are relatively common. Chromosome numbers that are a multiple of the haploid number are identified as euploid. In contrast, the addition or removal of a chromosome alters the euploid number and generates a chromosome count known as aneuploidy (i.e., "not euploid"). Chromosome nondisjunction is the cause of aneuploidy. "Skim", don't memorize, just understand: Polyploidy is the presence of three or more sets of chromosomes in the nucleus of an organism. Polyploidy is common, particularly in plant species, and can result either from the duplication of full sets of chromosomes or from the combining of chromosome sets from different species. Many types of polyploidy are possible—triploids (3n), tetraploids (4n), pentaploids (5n), hexaploids (6n), octaploids (8n), and so on. Two mechanisms are most commonly the cause of polyploidy. The first mechanism derives from meiotic nondisjunction. In these cases, one or both gametes have an extra set of chromosomes that is contributed at fertilization. For example, nondisjunction during oogenesis can produce an egg that is 2n. When fertilized by pollen that is n, the resulting plant will be triploid (3n). Similarly, both egg and pollen could be 2n, resulting in a plant that is 4n. The second mechanism is mitotic nondisjunction that results in a doubling of chromosome number. For example, a 2n cell experiencing mitotic nondisjunction can become 4n. These two mechanisms can also combine to increase polyploidy. For example, the autotriploid plant that results from the union of a 2n egg and an n pollen could become 6n (autohexaploid) by a doubling of the chromosomes through mitotic nondisjunction.

Chapter (2.6/3.5) Advanced Pedigree Analysis TEXTBOOK EXPLANATION Analyze pedigrees and determine the mechanism of inheritance of the disease or trait being followed. * What is autosomal inheritance? - What are humans, diploid or haploid? * Know what it means to be diploid/haploid. - How many pairs of chromosomes do humans have? Categorize them. - Compare and contrast the chromosomes found in females and males. * What is a pedigree? What does it do? - Know how to represent: Males Females Phenotype of interest is present Person is deceased Parents Progeny Generation Specific organism

With the benefit of well over a century of research, geneticists now understand that the patterns of hereditary transmission Mendel described are those of autosomal inheritance. This term refers to the transmission of genes that are carried on the paired chromosomes known as autosomes. In diploid organisms, like humans, one chromosome of each autosomal pair of chromosomes is inherited from the father and the other copy from the mother. Humans have 22 pairs of autosomal chromosomes (a total of 44 autosomes) and these are commonly identified by the numbers 1 through 22. The other two human chromosomes are the sex chromosomes, designated X and Y. Thus, humans have 46 chromosomes: 44 are autosomes and two are sex chromosomes, with two X chromosomes found in females and an X and a Y chromosome found in males The study of hereditary transmission in humans and numerous other species is assisted graphically by the construction of pedigrees, or family trees. A pedigree is drawn using a kind of symbolic shorthand designed to trace the inheritance of traits. In standard pedigree notation, males are represented by squares and females by circles (Figure 2.16). A filled circle or square indicates that the phenotype of interest is present. A line through a symbol indicates the person is deceased. Parents are connected to each other by a horizontal line from which a vertical line descends to their progeny. Individuals in a pedigree are numbered by a Roman numeral (I, II, III, etc.) to indicate their generation combined with an Arabic numeral (1, 2, 3, etc.) that identifies each organism in a generation. Identifying an individual by a Roman numeral followed by an Arabic numeral, as in I-2 or III-6, is an efficient way to ensure clarity in referring to particular organisms and, in the case of humans, allows protection of privacy by not requiring the use of names.

Chapter 3: Mitosis, Meiosis and Sex Determination Explain the inheritance of X-linked traits. * Are males haploid or diploid for most X-linked genes? - Are men more frequently affected by diseases caused by recessive alleles of X-linked genes, or dominant alleles of X-linked genes? * What does hemizygous mean? * What is X-linked gene? * What is an X-linked disease? * Know how to analyze a X-linked Punnett square. * Know how to analyze a X-linked Pedigree. - Compare X-linked inheritance pedigrees to autosomal inheritance pedigrees. What difference do we see?

X-linked Genes (and X-linked Diseases) * Males: Haploid (= hemizygous) for most X-linked genes. ----> Men are more frequently affected by diseases caused by recessive alleles of X-linked genes. *** Hemizygous refers to the condition of a gene where only one set of chromosomes from a chromosomal pair is observed. EX: Female carrier of an X-linked recessive disease or trait mates with normal male. F(xNxA) x M(xNY) xN = normal X chromosome. - Contains wild-type allele of the X-linked gene of interest. - In this example, since we're talking about X-linked recessive traits and diseases, the wild-type allele is dominant. xA = "affected" X chromosome. - Contains recessive allele of the X-linked gene of interest. X-linked gene = A gene that is on the X chromosome. X-linked disease = a genetic disease that results from inheriting disease-causing allele(s) of an X-linked gene. PIC: Female carrier mates with normal male. - half her daughters will be carriers. (on average) - half her sons will be affected. (on average) PIC: Affected male mates with normal female. - All his daughters will be carriers. - None of his sons will be affected. Pedigree Analysis: PIC Pedigree Showing X-linked Recessive Inheritance: PIC * A difference we see between X-linked and autosomal inheritance is that mostly males are affected in X-linked inheritance.

Chapter 3: Mitosis, Meiosis and Sex Determination TEXTBOOK EXPLANATION (Not learning objective, but should know) Describe the expression of X-linked recessive traits and how this works. * When are they expressed in males? When in females? * What is the pattern we see in males in terms of X-linked recessive traits? * What does the transmission of the recessive allele from grandfather to daughter to grandson appear as? * Recessive male (cY) x Homozygous dominant female (CC) = ? - Know genotype and phenotype of progeny. * Recessive male (cY) x Carrier female (Cc) = ? * Recessive female (cc) x Hemizygous dominant male (CY) = ? - Know genotype and phenotype of progeny.

X-linked recessive traits are expressed in hemizygous males who carry the recessive allele and in females who are homozygous for the recessive allele. Because hemizygous males express the single copy of a recessive X-linked allele in their phenotype, one of the hallmarks of X-linked recessive inheritance is the observation that many more males than females express the traits. 1. As a result of male hemizygosity, more males than females have the recessive phenotype. 2. Often, the transmission of the recessive allele from grandfather to daughter to grandson gives the appearance of generation skipping. (grandfather has recessive phenotype, daughter has dominant phenotype (carrier), son has recessive phenotype). 3. If a recessive male (cY) mates with a homozygous dominant female (CC), all progeny have the dominant phenotype. All female offspring are heterozygous carriers (Cc), and all male offspring are hemizygous for the dominant allele (CY). 4. Matings of recessive males (cY) and carrier females (Cc) can produce the recessive phenotype in females. About one-half of the offspring of these matings have the dominant trait and one-half have the recessive trait. 5. Mating of a homozygous recessive female (cc) and a hemizygous dominant male (CY) produces male progeny with the recessive trait (cY) and female offspring who have the dominant trait who are heterozygous carriers of the recessive allele (Cc).

Problem Set #1: Sex Linkage and Sex Determination A male patient with red-green color blindness comes to your genetic counseling clinic. If both his parents have normal color vision, which of his grandparents is most likely to be red-green colorblind? a. maternal grandmother b. maternal grandfather c. paternal grandmother d. paternal grandfather e. either grandfather is equally likely

b. maternal grandfather * X-linked recessive inheritance * Male patient: XrY * Mom and dad: XRXr XRY - In order for male patient to be affected, mom had to be a carrier since the X chromosome comes from mom. In order for mom to get that chromosome, she had to get it either from her mom or dad. Parents If both parents have normal vision, the mother of the affected male must be heterozygous for the X-linked, recessive alleles for red-green color blindness. The father of the affected male could not have been the source of the red-green color blind allele since fathers can only pass X-linked traits to their daughters, and Y chromosomes to their sons. Grandparents The two possible pedigrees for inheritance from a maternal grandparent are shown in the pedigree charts labeled A and B. Which of the mother's parents, the maternal grandmother (pedigree chart B) or maternal grandfather (pedigree chart A), is more likely to both be: 1. red-green color blind, 2. the source of the allele inherited by their grandson? Males with only a single X chromosome are more commonly affected by X-linked, recessive traits than are females with two X chromosomes. Why? A male need only inherit the recessive allele from a heterozygous female carrier. A female would need to inherit the recessive allele from both parents, an affected father and a carrier (or affected) mother.

Problem Set #1: Mitosis, Meiosis A trihybrid has the genotype AaBbCc. Genes A and B are on different chromosomes and gene C is on the same chromosome as gene B. The diagram depicts the chromosome composition of different cells from the animal. Which of the cells could represent a mature gamete? a. cell 1 b. cell 2 c. cell 3 d. cell 4 e. cell 5

d. cell 4 This is the only option that follows the "A and B on different chromosomes, C and B on same chromosome" rule, as well as the gamete rule. Animal gametes only have one allele per gene.

Quiz: Pedigree #2 Refer to the pedigree shown in Question 1. What are some examples of parent-child trios that allow us to rule out x-linked recessive inheritance? (For this question, do NOT assume the disease is rare) Choose all answers that are correct . parents I-1 and I-2 and daughter II-2 parents I-1 and I-2 and son II-6 parents II-1 and II-2 and daughter III-1 parents II-1 and II-2 and son III-3 parents II-6 and II-7 and sons III-13 and III-15 x-linked recessive inheritance cannot be ruled out

parents II-1 and II-2 and daughter III-1 parents II-1 and II-2 and son III-3