Genetics Exam #2

Ace your homework & exams now with Quizwiz!

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 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).

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.

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 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 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.

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?

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

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.

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.

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 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 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 (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: 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

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 (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 #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: 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).

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 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.

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 #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

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.

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 #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 #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 #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

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 #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 (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 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 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 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 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 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 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 (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 (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 (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 (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 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 (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 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 (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 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 (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: 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 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.

Genetics Exam #2

Related study sets

Audit Chapter 8

Chapter 15 Database Basics

OB - Chapter 18: Nursing Management of the Newborn

Organizational Management Ch. 5

CTC MATH-1342 Final Exam

Chapter 4

SIE 2.2 Understanding Product and their Risk

5 AP Psychology Unit 4 Vocab

SE370 quiz 1

Fin 201 Test 2

Patho Exam 2

Chapter 7: By-Products of Meat Animals

MKTG Chapter 12

Internet Technologies

Chap 9 Terminology

CH 19 Review Questions

CS111 Quiz 2

CHAPTER 6

CE SHOP GA PRACTICE QUESTIONS

Real World Examples IB ECONOMICS