Genetics to study

Human Sex Chromosomes

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A 3:1 distribution (among progeny) is absolutely typical of a monohybrid cross. You see 3:1, think "hey, the parents were a monohybrid cross"; you are told monohybrid cross, think "hey progeny will be in a 3:1 distribution".

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So what do we know just looking at this pedigree? Actually, we can figure out quite a bit: First, both parents MUST be heterozygotes. That is the ONLY way they can both be phenotypically normal and still have a child with a recessive condition. The son clearly must be homozygous recessive The daughter is either homozygous dominant OR heterozygous but we don't know which. There is a (2/3) chance she is heterozygous and a (1/3) chance she is homozygous dominant Why (2/3) and (1/3)? This is where the idea of "conditional probabilities" comes into play. We know she CANNOT be (aa); if she was she would be an albino. Looking at the punnett square beside the pedigree, if we eliminate the homozygous recessive genotype there are three possible parental gamete unions left: two of those produce a heterozygote and one a homozygote. Let us take this just a little further. Suppose the daughter were to marry an individual KNOWN to be heterozygous for albinism. What is the probability their first child would be albino? To be an albino, the following must happen: The mother to be must be heterozygous. The probability of that, as we determined, is (2/3) The father to be must be heterozygous. The probability of that is (1) - we are told he is heterozygous Even then, there is no guarantee that two heterozygotes will have an albino child. At any given birth, the odds of an albino child are (1/4). As these are all independent events, the odds of ALL three occurring is the product of their individual probabilities: (2/3)*(1)*(1/4) = (1/6). Conditional probabilities can only be employed WHEN YOU KNOW SOMETHING ABOUT AN INDIVIDUAL IN A PEDIGREE. Imagine we go back to the original pedigree and ask ourselves "What is the probability the parents will have a third child that is an albino?". Since we cannot yet see the child, there is a 25% chance of a homozygous dominant child, a 50% chance of a heterozygous child, and a 25% chance of an albino. Once the child is born, if it is albino, we immediately know it is homozygous recessive. If it is phenotypically normal, there is a (2/3) chance of being heterozygous and a (1/3) chance of being homozygous dominant.

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In a mating between individuals with the genotypes IAIB × ii, what percentage of the offspring are expected to have the O blood type?

0%

How many Barr bodies would one expect to see in cells of Turner syndrome females and Klinefelter syndrome males respectively? Enter your answer as two numbers separated by a comma.

0,1

10. In humans, the ability to taste PTC is dominant over non-tasting. Two heterozygous tasters have 3 children. What is the probability that 1 or more of the children's will be non-tasters?

0.578

Indicate the expected number of Barr bodies in interphase cells of individuals with the following karyotypes. 1. 46 XX 2. 47 XXY 3. 47 XYY 4. 48 XXXX

1,1,0,3

The sex of birds, some insects, and other organisms is determined by a ZW chromosomal arrangement in which the males have like sex chromosomes (ZZ) and females are ZW (similar to XY in humans). Assume that a recessive lethal allele on the Z chromosome causes death of an embryo in birds. What sex ratio would result in the offspring if a cross were made between a male heterozygous for the lethal allele and a normal female?

2:1 male to female

1. You perform a trihybrid cross. What is the probability of getting (A_bbCc)?

3/32

2. A couple decides to have (3) children. What is the probability they will have 2 boys and one girl?

3/8

4. Two parents are heterozygous for albinism. If they have 3 children, what is the probability AT LEAST one will be phenotypically normal?

63/64

Aneuploidy

Aneuploidy In humans there are several aneuploid conditions that can exist in a living individual: Sex chromosome aneuploidies Monosomic: Turners syndrome Trisomics: Klinefelters syndrome, Triplo-X syndrome, and Jacobs syndrome Due to (X) chromosome inactivation the phenotypic effects of these sex chromosome aneuploidies is generally mild. Turners', however, is quite lethal to the zygote - it is believed that only aboutr 1% of Turners syndrome zygotes survive to term. Autosomal Aneuploidies. There are only (3) aneuploidies involving autosomes that routinely show up in newborns. All of them are trisomics Downs Syndrome. 47 (total chromosome number) +21 (the extra chromosome is a copy of 21). Trisomy 21. Survivable but with many problems, including heart and immune system. Retardation can be severe or mild. Almost all adult survivors (past age 50) will suffer from Alzheimers; the beta amyloid gene linked to Alzheimers, is on chromosome 21. Overall incidence is 1:700 but increases with increasing maternal age. Pataus Syndrome. 47 + 13, Trisomy 13. Incidence of about 1:10K. Many defects including cardiac and central nervous system. Rarely survive more than a few days. Essentially a lethal. Maternal age increases incidence Edwards Syndrome. 47 + 13, Trisomy 18. Incidence about 1:7K. Maternal age effect. 90% don't survive past 6 months, very small percentage may make it longer. Again, essentially a lethal. The severe effects of autosomal aneuploidies are believed to be due to gene dosage effects; unlike the sex chromosomes, there is no mechanism to inactivate extra autosomes. The incidence of all these autosomal aneuploidies increases as the age of the mother increases. The odds of a sex chromosome aneuploidy also increase but not quite as dramatically. Maternal Age Odds of Trisomy 21 Odds of Trisomy 18 Odds of Trisomy 13 20 - 24 1:1400 1:14000 1:25000 30 - 34 1:700 1:7100 1:14000 40 - 44 1:70 1:700 1:1600 The reason(s) have not been clearly identified. But there are several theories: Meiotic defects. Remember, all of a female's ova begin meiosis I prior to birth and are arrested in prophase 1. This theory states that the older the mother, the longer the oocyte was arrested in prophase 1 and the greater the risk that something happened to the spindle apparatus, ultimately leading to non-disjunction. Maternal selection. This theory states that non-disjunction is equally likely in younger or older women but that there exists a mechanism in place to prevent the implantation of a chromosomally abnormal zygote in the endometrium (a selective mechanism). As a woman ages, this mechanism begins to break down. All in all, aneuploid conditions have been identified FOR ALL HUMAN CHROMOSOMES in miscarried embryos / fetuses. In fact, it seems to be the single most common reason for a miscarriage (when the reason can be identified). Various studies have indicated that chromosomal abnormalities, especially aneuploidies, are present in anywhere from 50% to over 90% of miscarriages. The table below shows some frequencies of aneuploid conditions in miscarried fetuses and live births: Aneuploid Condition Percentage of miscarriages if chromosomal abnormality present Odds of surviving to term Trisomy 13 2% 5% Trisomy 18 3% 5% Trisomy 21 5% 20% Trisomy 16 15% (very common in miscarriages) 0% (never seen in live births) XXX, XXY, XYY 2.5% 75% All other trisomies 26% 0% XO Turners (monosomic) 20% 1% Note that Turners may be the single most common aneuploid condition to occur at fertilization but the zygotes rarely survive to term. The sex chromosome trisomics have minimal effects, thus pretty good survival to term, due to (X) inactivation. Trisomy 16 is surprisingly common but these zygotes never survive to term. Generally speaking, aneuploidy isn't tolerated any better in other animals than it is in humans. Plants may tolerate it - the plant may survive and grow - but the odd chromosome numbers typically render them sterile.

Circle is a female while a square is a male. Open circles or squares represent PHENOTYPICALLY normal individuals, filled in circles or squares indicate individuals affected by what ever condition you are studying. An arrow pointing to an individual indicated the "proband" - the first individual in the pedigree to be identified by whatever condition you are looking at. They might not the be the first person in the pedigree to HAVE the condition, but you noticed it first in that person.

Below is a pedigree reflecting the inheritance of albinism (lack of pigment) in a family. It is a recessive trait (you can see a recessive hedgehog on the right). The circled individuals are particularly diagnostic for this:

Conclusion

Congratulations - you have completed your first module in BIOL 3453. By now you should be able to: Define and explain the meaning of Mendel's rules Apply these rules to predict the outcomes from simple genetic crosses Define terminology fundamental to Mendelian genetics Analyze simple human pedigrees using Mendelian genetics. In our next section, we will be examining some variations on inheritance patterns that Mendel never encountered.

Conclusion 2

Congratulations! Two modules down. By this time you should have a good understanding of: The concept of multiple alleles; what constitutes an allele, and lethal alleles How sex linkage affects the inheritance of traits and how sex can affect the expression autosomal traits Different types of genetic interactions, both with each other and the environment Remember, it is important that you have mastered not only the materials in this module but also in the associated text chapter.

Lethal Alleles

Consider two genes: TYR: produces the enzyme tyrosinase; produces pigment beginning with the amino acid tyrosine. Recessive mutations result in NO tyrosinase being produced. Albinos but clearly you can survive with this enzyme. HEXA: produces an enzyme called hexosaminidase A. Responsible for degradation, in lysosomes, of a complex membrane lipid called ganglioside GM2. Homozygous recessive individuals produce NO hexosaminidase A; deficiency is lethal: Tay-Sachs disease. The TYR gene is NOT essential (you can survive without its product) but the HEXA gene is essential - you MUST have functional product to survive. Sometimes, essential genes that are essential have alleles that result in the death of the individual. Tay-Sachs is considered a recessive lethal. You only suffer (and die) from Tay-Sachs if you inherit two recessive, loss of function alleles. You can see below the disease pathway for Tay-Sachs disease with its pleiotropic effects: Individuals inevitable don't survive past the age of four, with deaf often caused by respiratory failure and /or pneumonia. Lets consider some conditional probabilities involving Tay-Sachs. Look at the following pedigree - it reflects two phenotypically normal parents (clearly heterozygotes) who have a phenotypically normal daughter and had a son who died of Tay-Sachs. The daughter is getting married and she wants to know "What is my chance of having a child with Tay-Sachs?". There are (3) different scenarios Scenario #1. Her fiancé is known to be homozygous dominant (yes, you can be tested and know your genotype for Tay-Sachs). In this case, as Tay-Sachs is a recessive lethal, the is NO chance of an affected child. No matter what mom's genotype may be, a dominant allele will come from dad. Scenario #2. Her fiancé is known to be heterozygous. In this case, both parents will have to be heterozygous to have an affected child. The probability dad is heterozygous is (1), the probability mom is heterozygous is (2/3); she clearly DOESN'T have Tay-Sachs (she is alive) so there is a (2/3) chance of being heterozygous and (1/3) chance of being homozygous dominant. Even if both are heterozygous, there is a (1/4) chance they will have a homozygous recessive, Tay-Sachs child. So, via the product rule, the odds become: (1)*(2/3)*(1/4) = 1/6 Scenario 3. Her fiancé doesn't know his genotype BUT he had a younger sibling die of Tay-Sachs. His probability of being heterozygous now becomes (2/3), mom's chance is still (2/3) and the (1/4) chance of having a Tay-Sachs child hasn't changed. Now, via the product rule, the odds become: (2/3)*(2/3)*(1/4) = 1/9 Unlike the HEXA gene, mutations in the HTT gene are dominant lethals. HTT encodes a protein called huntingtin - defects in this protein (gain of function) lead to Huntingtons disease, or Huntingtons chorea. This is a degenerative neurological condition that typically leads within 10 - 20 years of diagnosis. Huntingtons is also "late onset" in that most patients are diagnosed somewhere between 30 - 50. In this case, of the three genotypes: (HH), (Hh), and (hh), two of them (HH, Hh) will produce Huntingtons. Fortunately, the dominant allele is quite rare and the disease is thus rare - occurring at a frequency of about 4 - 8 individuals per 100,000 population. Because the allele is so rare, almost all people who have Huntingtons are heterozygous - the odds of someone being homozygous dominant are somewhere around 1 in 10-12. The American folk singer Woody Guthrie (This Land Is Your Land) died from Huntingtons in 1967. Several of his children also inherited, and died of, Huntingtons. The last type of lethal I want to talk about is something a little different: it is sometimes called recessive lethal / dominant phenotype. There is a gene, the (A) gene that controls fur color in mice: AY > A. (AY) = yellow allele; (A) = wild type or agouti Image Source, CC BY 3.0 AY is an embryonic lethal when in the homozygous state; when heterozygous it confers the abnormal "yellow" fur phenotype. Yellow mice also tend to obesity, have metabolic disturbances, and are prone to several types of cancer. Consider a cross between two yellow mice: AY (x) AY Note distribution among progeny: 2/3 yellow:1/3 wild type or agouiti. Homozygous (AY) mice never appear thus, among living offspring, the 2/3:1/3 ratio. Called "Recessive Lethal With Dominant Phenotype". Lethality requires TWO (AY) alleles, thus the lethal phenotype is recessive. On the other hand, a single (AY) allele will produce the abnormal phenotype, thus the "yellow" phenotype is dominant. In a sense this is almost like a type of partial dominance with normal mice and an embryonic lethal at opposite ends of the phenotypic spectrum and a "yellow" mouse is the intermediate phenotype. Turns out the most human "dominant" genetic disorders fall into this category - the affected individuals are heterozygous and homozygotes with the abnormal allele never appear in the living offspring.

How do mammals, including humans, solve the "dosage problem" caused by the presence of an X and Y chromosome in one sex and two X chromosomes in the other sex?

In females, one of the X chromosomes is condensed and largely inactive so that each sex has one active X chromosome.

Polyploidy - Plants

Polyploidy - Plants Unlike animals, many plants tolerate polyploidy quite well. It has been suggested that 35% of angiosperms may be polyploids. That probably doesn't even include the polyploids specifically developed by plant breeders. Autopolyploid (autoploidy). More than 2N sets of chromosomes. All sets are from the same species thus all chromosomes are homologous. Allopolyploid (alloploid). More than 2N sets of chromosomes. Sets come from different ancestral species. The cultivated banana, Musa paradisiaca, is an example of an autoploid, specifically a auto triploid, derived from the Asian banana Musa acuminata. The Asian banana is a diploid species. Its haploid number, N, is (11) and its diploid number, 2N, is (22). One set of chromosomes from M. acuminata is called the (A) set. Understand what this means: one set is the haploid (N) number, so when we say that one set is called the (A) set we mean: A = one copy of chromosome 1 (a1) + one copy of chromosome 2 (a2) etc. Thus, one (A) set of chromosomes has one copy of every M. acuminata chromosome, or (11). This introduces the concept of the monoploid number when we discuss polyploids. The monoploid number, or (x), is the number of chromosomes in ONE BASIC SET. For M. acuminata, x = 11. The cultivated banana, M. paradisiaca, is a triploid derived from M. acuminata. This can arise through the fusion of one unreduced 2N gamete (it failed to have its chromosome number reduced in meiosis) with a normally reduced, N, gamete: This means that the cultivated banana has 3 sets of chromosomes, each with the number x = 11. Thus, the cultivated banana has 3N = 3x = 33 chromosomes; it is sometimes designated as (AAA), indicating three sets of (A) chromosomes from M. acuminata. This has serious implications for fertility. Have you ever seen seeds in bananas? Believe me, the Asian banana, which is 2N has plenty of them: Image Source, opens in new window CC BY-SA 3.0 opens in new window But the cultivated banana has an odd number of chromosomes, in fact, it is called an odd polyploid as it has an odd number of chromosome sets. Simply put, you can't evenly divide an odd number of chromosomes at meiosis. Gametes will end up with varying numbers of chromosomes resulting in sterility. "Seedless" plants - bananas, watermelons, and some types of apples - are seedless because they are odd polyploids and can't successfully reproduce. You can also have "even" polyploids. Potatoes (Idaho, not sweet) are an example. Even polyploids may arise from the fusion of TWO unreduced gametes: Image Source, opens in new window CC BY-SA 3.0 opens in new window Potatoes are tetraploid with 4N = 4x; x = 12. Even polyploids are often fertile since you CAN divide an even number of chromosomes. Alloploids can arise by the union of gametes from different, but related, species. For example, consider the hypothetical example below: Note that one unreduced gamete from species (A), normally a diploid species (2N=4), unites with a reduced gamete from species (B), also normally diploid (2N=6). This results in a hybrid with TWO sets of chromosomes from species (A) and ONE set from species (B) for a total of 7 chromosomes. This means the hybrid is really an odd polyploid (3N) and sterile. But suppose that an UNREDUCED gamete from the hybrid unites with another reduced gamete from species (B)? Now you have a hybrid with two sets from species (A) and two sets from species (B), a tetraploid with a total of 10 chromosomes. During meiosis, the species (A) chromosomes can pair with each other and the chromosomes from species (B) can pair with each other resulting in a gamete with 2 sets from (A) and 2 sets from (B). Union of these gametes will perpetuate the tetraploid hybrid. Wheat (Tricticum aestivum) is a very old allopolyploid. Hybridization of two ancestral diploid (2N) species, possibly Triticum Urartu and Aegilops speltoides is believed to have given rise to "emmer" wheat, T. turgidum (grown as Durum wheat today) about 10K years ago in Turkey. Subsequent hybridization with possibly A. tauschii added the last sets of chromosomes to yield current T. aestivum, about 8K years ago in what is now Armenia. Understand that the derivation I have depicted is not supported by all botanists and alternative proposed pathways do exist. Based upon this pathway, wheat has two A genomes from T. urartu, 2 B genomes from A. speltoides, and two D genomes from A. tauschii, giving us hexaploid wheat (AABBDD).

Cat breeders are aware that kittens expressing the X-linked calico coat pattern and tortoiseshell pattern are almost invariably females. Which one of the following statements supports this conclusion?

Since females normally have two X chromosomes, random X inactivation during development of a heterozygous female will create a mottled pattern. Normal males are not mottled, because they only have one X chromosome and therefore can only express one of the two potential X-linked alleles.

Although triple-X human females typically have normal offspring, what kinds of gametes, with respect to the X chromosomes, would you expect from such XXX females? Lable meiotic stages that show the gametes that are expected to be produced. Drag the appropriate labels to their respective targets.

The double x's go to // // and the single x goes to / /

Mendel began his work with true-breeding pea plants, one for each version of one trait. True breeding means that, if the plant is self crossed, all offspring will have the same phenotype as the parent:

The image above depicts this true-breeding: we have an individual what has purple flowers. When self crossed, ALL progeny will also have purple flowers.

Genetic Background Effects

The last thing we need to discuss is the concept of "genetic background" and how that might affect the expression of genes and phenotypes. By genetic background, I mean the alleles you possess for all 21k or so of your genes. How can multiple combinations of thousands of different alleles affect phenotype? Can environment factors come in to play? The answer is yes. We have already discussed pleiotropy - the multiple phenotypic effects of a single gene. Consider now the idea of expressivity - the idea that not all individuals with the same genetic condition will express the same phenotypic effects. Marfan syndrome is a great example. Phenotypic effects can include skeletal abnormalities, heart valve problems, myopia, respiratory system dysfunction, dislocation of the lens of the eye, and a tendency to aortic aneurysms. Understand, however, that NOT ALL MARFAN PATIENTS HAVE ALL THESE SYMPTOMS. For example, only about 70% have the respiratory dysfunction while 30% have myopia. When different individuals express different subsets of a range of symptoms, this is called expressivity (sometimes called variable expressivity). A pedigree below shows the concept of expressivity - all the offspring show some symptoms but not all look the same: Image Source, CC BY-SA 4.0 Another variation is called penetrance and really refers to dominant characteristics. Sometimes, even though a person carries a dominant allele, THEY MAY NOT EXPRESS IT. Traits like this are said to show incomplete penetrance. Polydactyly (extra digits) is such a trait. Image Source, CC BY-SA 3.0 Look at the pedigree below, especially the individual indicated with an arrow, that shows incomplete penetrance: © Robert Wiggers A quick look might lead us to think that polydactyly is recessive: there are two unaffected parents with an affected son BUT, we would be wrong. Polydactyly shows incomplete penetrance. The affected son HAD to have gotten the dominant allele from his unaffected mother (yes, it COULD have been the father but polydactyly is rare so the mother had to have the allele from HER father) but she did exhibit the phenotype. Penetrance can be calculated by using the formula (# affected / # who carry the allele, affected or not). In our case, there are 3 affected individuals (obviously they carry the allele) and there are 4 total individuals who carry the allele - the 3 affected plus the one unaffected female who nonetheless must be carrying the allele. Our penetrance value is thus 75%. Penetrance and expressivity come in various combinations: Incomplete penetrance only: not all who carry the allele show phenotypic effects but those who do all exhibit the same phenotypic effects. 100% penetrance AND variable expressivity: ALL those affected will show at least some symptoms Incomplete penetrance AND variable expressivity: Not all those who carry the allele will show symptoms AND those who do exhibit different subsets of symptoms. Below is an image that shows this: on the left we can see that not all the offspring are affected but that those who are show the same phenotype (incomplete penetrance only); in the middle we see that all offspring are affected but exhibit different phenotypic spectrums (100% penetrance AND variable expressivity); on the right we see both variable expressivity AND incomplete penetrance. These are not the only genetic background effects that exist. As you read through section 4.13 in your book, you will learn about others that exist.

Probabilities The product rule The sum rule

The product rule: The probability of multiple independent events occurring is the product of their individual probabilities In carnations, red flower color is dominant to white and long stems are dominant to short. If a dihybrid is self crossed, what is the probability of getting an offspring that has red flowers and short stems? The sum rule: the probability of mutually exclusive events occurring is the sum of their individual probabilities An individual with the genotype ( AaBbCCDd ) is testcrossed. What is the probability of getting an offspring with the genotype The binomial: for rapidly figuring out multiple combinations Two parents are heterozygous for albinism. They plan on having (4) children. What is the probability that (3) will be phenotypically normal? N = # of events x = number of one possible outcome p = probability of outcome (x) In our case: N = 4 (four children) X = # of one possible outcome (3) p = ¾

Which of the following statements about Barr bodies is true?

They ensure that males and females have an equal dosage of the genes on the X chromosome

Gregor Mendel worked with garden peas. He examined the inheritance of seven different characteristics (e.g. flower color, seed shape or form, plant size). Each characteristic had two alternative forms (e.g. purple vs. white flowers or yellow vs. green seeds):

This physical appearance is called phenotype.

Mapping 3 Genes - Trihybrid Allele Arrangement

To begin mapping three genes we will need to analyze the data set and determine two things: The arrangement of alleles on the trihybrid's chromosomes: are all three dominant alleles on one and the recessives on the other? Does each chromosome have some dominant and recessive alleles? The linear order of the three genes. By this I mean which gene is "in the middle", or between the other two. Anytime you have three genes, there are three possible linear orders: (A B C), (B A C), or (B C A); the underline is used to represent the chromosome. You may wonder, what about (A C B), for example. Well, that is really just a mirror image of (B C A). In other words, so long as we properly determine which gene is in the middle, how we write the two flanking genes doesn't matter. Our cross is once again a trihybrid testcross. We are using the same cinnabar and vestigial genes but instead of (e), we will be using: (b); the black body gene. It has two alleles (b+) the dominant wild type that produces a wild type body color and (b), a recessive allele that, when the fly is homozygous recessive, has a dark black body. It is very similar in phenotype to the ebony gene. Our cross is therefore: cn+cn b+b vg+vg (x) cn cn b b vg vg Our shorthand notation is: + + + (X) cn b vg Below is a data table indicating the phenotype of progeny, the gamete (with 3 recessive alleles) that came from the testcross parent, the gamete that HAD to come from the trihybrid parent to give the stipulated phenotype, and how many of each progeny we saw: Progeny Phenotype Testcross gamete tri-hybrid gamete Observed + + + cn b vg + + + 22 cn b + cn b vg cn b + 179 + b vg cn b vg + b vg 46 + + vg cn b vg + + vg 173 cn + vg cn b vg cn + vg 4 cn + + cn b vg cn + + 52 cn b vg cn b vg cn b vg 22 + b + cn b vg + b + 2 Make sure you understand how we derived the trihybrid gamete. As an example, if progeny show the dominant (wild type) phenotype for all three traits - (+ + +) - they MUST have gotten a gamete from the trihybrid parent that had all three wild type alleles as the testcross parent had to contribute all recessive alleles. First thing to note - the progeny ARE NOT OCCURRING in equal frequencies. I don't even need a chi-square test to tell me that. Clearly our genes are linked. Next thing we need to do is classify each gamete as a SCO, DCO, or NCO. Remember, NCO's will be the two most frequent and the DCO's will be the two least frequent. We can identify those - we will have to simply label the remaining four as SCO's. Below is the table above (I removed the "testcross gamete") with each category labeled as NCO, DCO, or SCO: trihybrid gamete Obs gamete type + + + 22 CO cn b + 179 NCO + b vg 46 CO + + vg 173 NCO cn + vg 4 DCO cn + + 52 CO cn b vg 22 CO + b + 2 DCO I have highlighted the NCO's in yellow and the DCO's in blue - they are important. First, we determine the allele arrangement on the trihybrid chromosomes. To do this, look at the NCO gametes, remembering they are produced in the ABSENCE of crossing over - they reflect the trihybrid's chromosomes unaltered: (cn b +) (+ + vg) In other words, one chromosome had the recessive cinnabar, recessive black body, and the wild type vestigial allele while the other chromosome had the wild type alleles for cinnabar, body color, but the recessive allele for vestigial: While we now the arrangement of alleles on the trihybrid's chromosomes, WE DON'T KNOW IF THIS REFLECTS THE CORRECT LINEAR ORDER - we don't know for sure if (b) is in the middle. Next, I will show how to determine correct gene order.

A typical XX human female has one Barr body per cell.

true

Generally: Dominant characteristics DON'T skip generations, if an offspring has it, they had to get the allele from a parent; being dominant, the parent will also be affected. This means that affected individuals will have at least one affected parent. Two affected parents MAY have a phenotypically normal child - they could both be heterozygous and thus have a homozygous recessive, unaffected child.

......

In mice, there is a set of multiple alleles of a gene for coat color. Four of those alleles are as follows:C = full color (wild type)cch = chinchillacd = dilutionc = albinoGiven that the gene locus is not sex-linked and that each allele is dominant to those lower in the list, give the phenotypic ratios expected from the following cross.chinchilla (heterozygous for albino) × albino

1 chinchilla :1 albino

8. Suppose the couple described in the previous question is expecting another child. What is the probability it will be heterozygous for the albino allele?

1/2

Choose the offspring phenotypes you would expect from the following cross: R 1 R 2 ×× R 2 R 2

1/2 pink, 1/2 white

7. Albinism is recessive to normal color. A phenotypically normal couple have a daughter who is albino and a normal son. What is the probability their son is homozygous dominant?

1/3

5. In fruit flies, straight wings and red eyes are dominant to curved wings and brown eyes. A homozygous straight winged, red eyed male is crossed to a curved winged, brown eyed female. The resulting F1 flies are crossed to produce an F2. What is the probability of producing a straight winged brown eyed fly OR a curved winged, red eyed fly in the F2 generation?

3/8

Klinefelter and Turner syndromes have how many chromosomes, respectively?

47, 45

5. Red eyes and straight wings are dominant in Drosophila, while white eyes and curved wings are recessive. Suppose two dihybrid flies are crossed. What is the probability of getting a fly with red eyes and curved wings or white eyes and straight wings?

6/16

1. A Dihybrid (AaBb) is self crossed. What is the probability of getting a progeny of the genotype (A_bb) or (AaBb)

7/16

P1: yellow, smooth (YYSS) (x) green wrinkled (yyss) The principle of segregation says that the yellow, smooth parent will produce gametes with one (Y) and one (S) allele while the green wrinkled parent will produce gametes with one (y) and one (s) allele. When they unite they will produce an F1 heterozygous for both genes. Since yellow and smooth are dominant, the F1 will, phenotypically speaking, be yellow and smooth

Allow the F1s to cross amongst themselves. Principle of segregation says the alleles for the color gene (yellow vs green alleles) will segregate one to a gamete and the same will happen for the alleles for the seed shape (or form) gene (smooth vs wrinkled alleles). The principle of independent assortment says the alleles for the color gene will assort independently of the alleles for the shape gene. This means that each parent can produce (4) different gametes: Note that in this cross, the parents are both heterozygous for TWO genes - this is a dihybrid cross (right - monohybrid meant heterozygous for one gene, thus dihybrid means heterozygous for two genes).

The image above depicts this true-breeding: we have an individual what has purple flowers. When self crossed, ALL progeny will also have purple flowers.

Allow true breeding plants, each exhibiting one of the two forms of the trait, to cross and you will always see the F1 generation look like ONLY one of the parents Allow these F1 offspring to cross amongst themselves and the version of the trait that disappeared in the F1 would reappear in the F2 AND always at a 3:1 ratio. For example, if a true breeding tall plant was crossed with a true breeding short plant (P1), all the F1's were tall. In the F2's short made a reappearance in a 3 tall:1 short ratio.

Alleles

Alleles of a gene are simply variants of the DNA sequence that encodes the gene product - thus different alleles produce different variants of the protein. The "normal" or "wild-type" allele of the gene produces the normal, functional gene product. Any other sequence variant is called a "mutant" allele. Mutant alleles may be: A null allele: the alteration of DNA sequence leads to an allele that either produces NO product at all OR a product that doesn't function. Also called a "loss of function" allele. These are usually recessive to the wild-type allele. Generally, having 50% the normal level of a gene product is sufficient for a normal phenotype. A gain of function allele: in this case, the altered DNA sequence is producing a product that has a new, novel function. These alleles are often (but not always) dominant; even though you have 50% normal gene product, the remaining 50% now has a new a novel function that affects the phenotype. Below are two examples of DNA sequence changes that lead to alleles of a gene: Hexosaminidase A, an enzyme: Normal allele: ....CGTATATCCTATGCCCCTGAC... Null allele: ...CGTATATCTATCCTATGCCCCTGAC... Note that the null allele has (4) additional bases (in red) versus the normal allele. Beta globin, a component of hemoglobin Normal allele: ...CTT... (also called the bA allele) Mutant allele: ...CAT... (also called the bS allele; a single base change) Mutant alleles, be they dominant or recessive, can affect the phenotype in many ways. Consider the three genotypes involving b A and b S: b Ab A: 100% normal beta-globin, normal phenotype bAbS : 50% normal beta-globin:50% abnormal beta-globin. Normal phenotype bSbS : 100% abormal beta-globin, results in "sickle cell anemia" Patients with sickle anemia (homozygous for the recessive bS allele) suffer from anemia, intensive joint and body pain, fevers, organ damage, kidney failure, impaired mental function, and often die relatively (<50) young. These multiple effects traced to a defect in a single gene is termed pleiotropy. Pleiotropy is the rule rather than the exception - all conditions caused by mutant alleles of a single gene have multiple effects. The term was coined decades ago when it was thought that most genes would only influence one aspect of the phenotype. As you read chapter 4, sections 4.1 - 4.6, you will learn a lot more about alleles. As you study, keep the following points in mind: (1). Previously, we only considered there to two alleles for a gene, one dominant and one recessive. There may, however, be many alleles that exist for a gene. Yes, an individual can only inherit two, but there may be many that exist. Think about how this would effect inheritance patterns, especially as dominance patterns may change depending upon which two alleles are being compared. For example, in dogs, white patterning seems to be controlled by one gene with four alleles: (s): solid allele - no white at all; color determined by other genes (si): irish pattern - white on chest, front legs, and maybe paws (sp): piebald - extensive white on body (sw): extreme - entire dog may be white or have only SMALL patches of color. In this case, the dominance relationships can be indicated as: s>si>sp>sw. This means that (s) is dominant to all other alleles, the irish allele is recessive to solid but dominant to piebald and extreme, the piebald allele is recessive to all but the extreme allele, and the extreme allele is recessive to all others - the only way to get an extreme white dog is for it to be homozygous for the extreme allele. (2). In some cases, there may not be a clear cut dominant or recessive allele; alleles may exhibit partial (or incomplete) dominance or even co-dominance. How will this change phenotypic distributions from crosses? (3). Lethal alleles exist. These alleles result in the death of the individual. Some are dominant (rare) while others are recessive. In what type of genes do you find lethal alleles and how can they affect phenotypic distributions in progeny? Your book does a very good describing and defining these but I do want to dwell just a little bit more on the concept of a lethal allele.

The following coat colors are known to be determined by alleles at one locus in horses:palomino = golden coat with lighter mane and tailcremello = almost whitechestnut = brownThe following table gives ratios obtained in matings of the above varieties: CrossParentsOffspring 1cremello x cremelloall cremello 2chestnut x chestnutall chestnut 3cremello x chestnutall palomino 4palomino x palomino1/4 = chestnut ​1/2 = palomino ​1/4 cremello

C1C1 = cremello, C2C2 = chestnut, C1C2 = palomino

Pedigree Analysis

Doing crosses and examining large numbers of offspring to determine distributions (3:1, 1:1, 9:3:3:1, etc) is great if you are working with an organism that produces large numbers of offspring. Other organisms, i.e. humans, require a different analytical technique - pedigree analysis. Pedigrees are "family trees" that show relationships across multiple generations. Below is just a sample of the symbols used in pedigree construction:

Normally in humans, all the sons of a male showing a sex-linked phenotype will inherit the trait.

False

If a (Y) chromosome is present than the SRY is present. Contained within this region is a single gene, tdf, or testes determining factor. In its presence: The bi-potential gonads develop into testes The testes begin to produce small amounts of testosterone which triggers the development of the Wolffian ducts into the male reproductive tract The testes produce MIS, Mullerian Inhibiting Substance, which triggers the degeneration of the Mullerian ducts Small amounts of testosterone are converted into DHT, which triggers the development of the male external genitalia. In the absence of a (Y) chromosome, there is no SRY and no tdf gene: The bipotential gonads develop into ovaries Under the influence of estrogen (maybe other signals too?), the Mullerian ducts develop into the female reproductive tract and external genitalia The Wolffian ducts degenerate. The (XY) system used by humans and other mammals is FAR from the only method of sex determination that exists. The next section will give you an idea of how diverse sex determination is.

Dosage Compensation Females have 2 (X) chromosomes while males have only 1. Potential problem - gene dosage. For genes on the (X) chromosome, females would be expected to have twice the gene product as males as evidenced by the hypothetical frequency plot shown below: Image Source, opens in new window CC BY-SA 3.0 opens in new window Many genes are dosage sensitive while others are not: PMP22 gene: produces a myelin protein. TOO MUCH leads to a neurological condition called Charcot-Marie-Tooth disease while TOO little leads to a different neurological condition called Hereditary Neuropathy With Liability To Pressure Palsies. Adenosine deaminase (enzyme). Levels can vary 5000x across individuals with no ill effects. What about dosage sensitive genes on (X) chromosome. How can females cope with twice the gene product (or males with only half the gene product)? Consider a normal (XX) female. Very early in embryogenesis, ONE (X) chromosome is inactivated via condensation. The process is called Lyonization while the condensed, inactive (X) is called a Barr body. You can see a Barr body (indicated by arrow) in the image of an (XX) female. The Barr body appears darkly stained near the edge of the nucleus: Image Source, opens in new window CC BY-SA 3.0 opens in new window Initial inactivation is random (some cells inactivate maternal X while others inactivate paternal X). Once pattern is established, ALL DAUGHTER cells will inactivate the same (X). If a female is heterozygous for any genes on her (X) chromosomes, she will become a tissue mosaic - some cells will be expressing one allele while other cells express the other allele. The distributions of these cells and tissues is completely random Source (right), Medline Plus courtesy of NLM, public domain opens in new window As you can see in the image above, initial (X) inactivation is completely random - some cells will inactivate the paternal (XP) while other inactivate the maternal (XM). Once established, this pattern is continued by daughter cells. On the right you can see the concept of a mosaic female: some tissue patches (purple) are expressing the (XP) alleles, having inactivated the maternally inherited (X) while other areas (peach) are expressing the (XM) alleles, having inactivated the paternally inherited (X) Tissue distribution is random. This process of (X) inactivation is a way of ensuring females and males now have the same amount of gene product from the (X) chromosomes. Interesting twist: in individuals with additional (X) chromosomes - Triplo-X, Klinefelters - inactivate ALL BUT ONE (X), while Turners patients don't inactivate their single (X): XX female: 1 Barr body XY male: 0 Barr bodies XXX female: 2 Barr bodies XXY male: 1 Barr body XO female: 0 Barr bodies How the cells "count" (X) chromosomes is unknown. You may be asking why then do sex chromosome variants (Turners, Klinefelters, etc.) have ANY phenotypic effects (while typically mild, there some effects). Turns out that a number of genes (15 - 20%) on the (X) are active even from the so-called "inactive X". Thus, imbalances in these genes probably account the mild phenotypic effects seen in people with abnormal sex chromosome complements.

Based upon these results, Mendel formulated, initially, three fundamental laws governing inheritance:

For each trait a unit factor exists. We now call this a gene. In other words, a gene exists that controls flower color, another for seed form, etc. Alternative forms of these unit factors exist. For example, for the unit factor that controls flower color, there must be a form that confers purple flowers and another form that will confer white flowers. Today we call the unit factors genes and the alternative forms alleles. So, a gene for flower color with one allele for purple and another for white. Furthermore, Mendel postulated that these alleles are paired. In other words, each individual inherits one allele from each parent. Alleles may be dominant or recessive. When an individual inherits two different alleles for a gene, only one will determine the phenotype. This allele is said to be dominant. The allele which is "hidden" or masked is called recessive. Principle of segregation. During gamete formation, alleles segregate from another one to a gamete. If an individual possesses two identical alleles, all gametes will get the same allele. If an individual possesses two different alleles, 50% of the gametes will get one allele and 50% will get the other.

The Dihybrid Cross

From study of inheritance of TWO traits, Mendel formulated last fundamental principle of genetics: Principle of Independent Assortment: alleles of DIFFERENT genes assort themselves into gametes independently of one another. In other words, what allele a gamete gets for one gene (dominant or recessive) has NO INFLUENCE on which allele it gets for another gene. Consider the P1 cross in which two true-breeding peas are crossed - one true breeding for yellow seeds and a round (or smooth) shape and one true-breeding for green seeds and a wrinkled shape. Yellow and round are both dominant to green and wrinkled. As true breeding means homozygous, one is homozygous dominant and one is homozygous recessive:

They are both PHENOTYPICALLY normal BUT have children that are albinos. Clearly the children got the condition from parents, thus the parents HAD to have the albino allele BUT they weren't expressing it, the hallmark of a recessive allele.. In other words, albinism is caused by being homozygous recessive - the parents were both heterozygotes; unaffected heterozygotes such as these are often called carriers - they carry the allele but it is hidden by the presence of a dominant allele.

Generally: Recessive traits MAY skip generations; they can be hidden in heterozygotes only to reappear if two heterozygotes have children If you find two affected individuals AND they have children, 100% of children will be affected. If the parents are affected it means they are homozygous recessive and all children will also be heterozygous recessive. Below is a pedigree reflecting a condition, dentinogenesis imperfecta (literally, bad teeth), that is dominant:

Goals And Objectives

Goals: When you complete this exercise you will understand the effects of epistasis (gene interaction) on the distribution of phenotypes among progeny of different crosses Objectives: When you complete this exercise, you will be able to Based upon a given phenotypic distribution, recognize whether or not epistasis is occurring; Based upon Chi-square data testing, determine which phenotypic distribution is supported by the data Based upon given genetic interactions, genes, and alleles, determine genotypes that, when crossed, will produce specific phenotypic distributions.

2. Dominant phenotype is heterozygous: Dd (x) dd Punnett Square predicts 1:1 ratio of heterozygotes to homozygous recessive offspring and thus a 1:1 ratio of dominant and recessive phenotypes

Just as a 3:1 ratio is typical of a monohybrid cross, a 1:1 ratio is typical of a heterozygote that has been testcrossed.

The presence of chromosomes linked to sex determination has been known since 1891. It has been known for decades that humans (indeed all mammals although the monotremes are a bit different) utilize the (X/Y) system, in which females are (XX) and males are (XY). In this system:

Males are heterogametic: they will produce two kinds of gametes with respect to sex chromosomes an (X) bearing sperm and a (Y) bearing sperm Females are homogametic: all of their gametes will possess an (X) chromosome although the alleles may be different.

Mapping 3 Genes - Introduction

Lets consider three genes in Drosophila: (cn): cinnabar; two alleles - wild type (cn+) and the recessive (cn). When homozygous recessive, flies display a bright red eye color instead of the reddish brown wild type. (vg): vestigial; two alleles (vg+) and the recessive (vg). When homozygous recessive, flies have poorly developed vestigial wings instead of wild type, normal functional wings. (e): ebony; two alleles - wild type (e+) and the recessive (e). When homozygous recessive, flies have a very dark body as opposed to the light yellow wild type. We will take a trihybrid (cn+cn vg+vg e+e) that has the wild type phenotype for all three genes - reddish brown eyes, normal wings, normal body - and testcross it to an individual that is homozygous recessive (cncn vgvg ee) - cinnabar eyes, vestigial wings, and ebony body. This cross can be written in a shorthand that will be useful for mapping exercises: + + + (x) cn vg e We are simply saying that our trihybrid has the wild-type phenotype for all three genes and the testcross parent has the recessive phenotype for all three genes. Think about the trihybrid and understand that there are EIGHT different types of gametes that could be produced: For ease of demonstration, I have assumed that one chromosome carries all three wild type alleles and the other carries all three recessive - that may not always be the case. NCO's: in the absence of any recombinational event there will be two NCO gametes produced (+++) and (cn vg e). As always, if genes are linked, the two NCO gametes will be the most frequent. CO gametes IF A SINGLE RECOMBINATIONAL EVENT BETWEEN (cn) and (vg). These type of gametes are often called single cross over, or SCO's. This will produce two what are called SINGLE CO's: (+ vg e) and (cn ++). CO gametes IF A SINGLE RECOMBINATIONAL EVENT BETWEEN (vg) and (e). This will produce another set of SCO's: (++e) and (cn vg +) DCO gametes - DOUBLE CROSSOVER GAMETES - this will be produced IF THERE ARE TWO RECOMBINATIONAL EVENTS: one between (cn) and (vg) and a second between (vg) and (e). These gametes will be (+ vg + ) and (cn + e). While NCO gametes will always be the two most common, we can't say just how frequently the single CO gametes will be. We can say, however, that no matter what, the DCO gametes will be THE LEAST FREQUENT. Why? Imagine there are (20 cM) between (cn) and (vg). Means that SCO gametes produced by this single recombinational event should account for 20% of the total gametes Imagine there are (10 cM) between (vg) and (e). Means that SCO gametes produced by this single recombinational event should account for 10% of the total gametes What about DCO's? Probability of a SCO between (cn) and (vg) is 20% or (0.20); probability of a SCO between (vg) and (e) is 10% or (0.10). Thus the probability of a crossover between (cn) / (vg) AND (vg) / (e) is the product of (0.20) * (0.10), or 0.02 - 2%. No matter what the map distances between the gene pairs are, the odds of two crossover events is always less than the odds of a single cross over between either pair. Thus, DCO gametes ARE ALWAYS THE LEAST FREQUENT. So, if you are given data from such a cross - trihybrid testcrossed - and assuming the genes are linked to one another, you should be able to identify the NCO gametes (two most frequent) and the DCO gametes (two least frequent) and the other four gametes, well all we can say is that they represent the two categories of SCO gametes. Note I said "assuming the genes are linked". How could we tell? Chi-square! If you have a trihybrid testcross AND the genes are unlinked, you would expect (8) phenotypic categories in the progeny, all in equal frequencies. Let us look at that trihybrid testcross and, using independent assortment, ask what gametes we will get from the trihybrid: cn+ cn vg+ vg e+ e (x) cn cn vg vg e e ½ chance of (cn+) vs (cn); ½ chance of (vg+) vs (vg); ½ chance of (e+) vs (e). Odds of (cn+ vg+ e+) becomes 1/8; odds of (cn vg+ e+) becomes 1/8. You get the picture: any combination of alleles from the trihybrid is 1/8, thus 8 different gametes, all in equal frequencies, all meeting with a (cn vg e) gamete from the testcross parent. We would therefore expect the following phenotypic categories, all at equal frequency among the offspring: (+ + +): wild type (cn vg e): cinnabar eyes, vestigial wings, ebony body (cn + +): cinnabar eyes, wild type wings and body (+ vg e): wild type eyes, vestigial wings, ebony body (cn + e): cinnabar eyes, wild type wings, ebony body (+ vg +): wild type eyes, vestigial wings, ebony body (+ + e): wild type eyes, wings, but ebony body (cn vg +): cinnabar eyes, vestigial wings, wild type body In other words, if you do a trihybrid testcross and all progeny phenotypes ARE NOT IN EQUAL FREQUENCIES, the genes are linked. Up next, we will see how to apply this in an actual trihybrid testcross.

Historical Mapping In Humans

Mapping as detailed so far involves looking at many progeny. Difficult in humans as we don't produce enough offspring to be statistically significant. LOD scores (Logarithm Of Odds) help assess whether two traits are inherited together due to linkage or independent assortment. Developed in 1947, involves several calculations: What is the probability the traits were inherited together because genes are linked What is probability the traits are inherited together by chance, i.e. independent assortment is occurring Take ratio (odds of linkage / odds of independent assortment) then convert to Log10 LOD scores of 3 (or greater) indicate linkage is favored by 1000:1. Why? The logarithm of 1000 = 3. If the ratio of probabilities has a Log10 of 3, that means it was equivalent to the numerator being 1000 and the denominator being 1 - in other words the odds favoring linkage were 1000 times greater than the odds favoring independent assortment. A LOD score of 4 would mean odds favored linkage by 10,000 :1. On the other hand a LOD score of 1 would mean that the odds favor linkage by only 10:1. Historically, if two genes had a LOD score of 3 or greater, they were deemed linked. If the LOD score was (-)2 or less, they were deemed independently assorting. Between (-)2 and 3, the results were inconclusive. An improvement on LOD scores came about with the development of cell hybridization and synteny testing - be sure to read about this in your text. Today, mapping is instead done using DNA sequence analysis, something that will be discussed later in the course.

The Testcross

Mendel ultimately devised a simpler way of determining if a plant with DOMINANT phenotype was homozygous dominant or heterozygous - easier than allowing everything to self cross then see what phenotypic distribution was produced in offspring. It is called the test-cross: an individual with dominant phenotype but unknown genotype (is it homozygous dominant or heterozygous?) is crossed to an individual with the recessive phenotype and therefore the homozygous genotype. There are two possibilities: Dominant phenotype has homozygous dominant genotype: DD (x) dd

A Punnett Square (16 positions this time) reveals all possible gamete combinations and thus phenotypes and genotypes amongst the F2 progeny:

Note the distribution of progeny: 9/16 yellow AND smooth 3/16 yellow AND wrinkled 3/16 green AND smooth 1/16 green AND wrinkled

Conditional Probabilities

Pedigrees, and their analysis can be used to make predictions regarding inheritance. Sometimes, we can use what a pedigree tells us about individuals to refine our predictions. Consider two parents, both phenotypically normal, that have two children: a son who is an albino and a daughter who is phenotypically normal. The pedigree and a punnett square are shown below:

Polyploid plants

Polyploid Plants - Summary A surprising number of agronomically important crop plants are polyploid. The table below shows you some of them. Note, for some, there are multiple levels of polyploidy in cultivated varieties: coffee has diploid, tetraploid, hexaploidy, and octaploid varieties. species (x) (Total chromosomes) potato (Solanum tuberosum) 12 48 coffee (Coffea arabica) 11 22, 44, 66, 88 banana (Musa paradisiaca) 11 33 alfalfa (Medicago sativa) 8 32 peanut (Arachis hypogaea) 10 40 sweet potato (Ipomoea batata) 15 90 tobacco (Nicotiana tabacum) 12 48 cotton (Gossypium hirsutum) 13 52 wheat (Triticum aestivum) 7 42 oats (Avena sativa) 7 42 sugar-cane (Saccharum officinarum) 10 80 plum (Prunus domesticus) 8 16, 24, 32, 48 strawberry (Fragaria grandiflora) 7 56 apple (Malus sylvestris) 17 34, 51 pear (Pyrus communis) 17 34, 51 Polyploid plants usually grow larger and yield more than their diploid progenitors. Below is an image of a diploid wild strawberry on the left and an octaploid cultivated variety on the right - note the size difference. Image Source, opens in new window CC BY-SA 3.0 opens in new window For years this was a mystery but modern molecular biology seems to have solved this puzzle. Turns out that polyploid cells spend more time in G1 of the cell cycle, due to repressed levels of the G1 cyclins, than their diploid brethren. As this is the primary period of cellular growth, the polyploids simply get bigger all the way round.

Mendel observed ALL short plants were true-breeding. Observed that 2/3 of Tall plants produced 3:1 ration amongst offspring. Why 2/3?

Short plants are (dd). If you consider ONLY tall plants, 1/3 are (DD) and true breeding while 2/3 are heterozygous and will thus produce progeny in a 3:1 ratio.

Sex - Linkage In Humans

Sex-linked (or X-linked) genes exist in humans. As with other genes they have mutant alleles. Most are recessive but a few are dominant. Below is an image showing possible inheritance patterns for a sex-linked recessive allele depending upon the parental genotypes: Image Source, opens in new window CC BY-SA 3.0 opens in new window (1). A unaffected father. Can be written as (AY) or (XAY); either denotes his single (X) chromosome has a normal, dominant allele. A mother who is affected (XaXa) denoting she is homozygous recessive. All sons will be affected as they all receive an (X) chromosome from their mother. All daughters will be heterozygous (XAXa) but phenotypically normal. (2). An unaffected father (XAY) and a heterozygous, phenotypically normal mother. In this case 50% of sons will be affected and 50% normal, depending on which (X) they inherit from their mother. All daughters will be phenotypically normal although 50% will be heterozygous carriers and the other half homozygous dominant. (3). An affected father and a heterozygous, carrier mother. In this case half of sons and half of daughters will be affected while the other half will be phenotypically normal, although daughters may be heterozygous. Note that we finally see an affected girl. In fact, for females to be affected by a sex-linked recessive condition, their fathers MUST be affected - they need two recessive alleles; mom can be heterozygous but the other allele has to come from dad, meaning he will be affected. Below is a pedigree reflecting the inheritance of a sex linked recessive condition. There are some things that can help you distinguish it from autosomal inheritance: First understand that all the "rules" we talked about regarding autosomal recessive inheritance apply here with the following modifications: There will be more affected males than females. Males need inherit only ONE recessive allele but females need two. Sex-linked recessive conditions tend to be uncommon - the odds of seeing a male with the one allele is FAR greater than a female inheriting two alleles. In fact, one more common sex-linked recessive condition, colorblindness, occurs in less than 1% of women but affects almost 10% of males. Affected females MUST have had an affected father. If there is an affected female, and she has any sons, 100% will be affected. Sons get their (X) chromosome from mom, so all receive a recessive allele. Sex-linked dominant conditions also exist but they are far less common than sex-linked recessive ones. Some of them are embryonic lethals in males (only the single X chromosome) and affected females are typically heterozygous and less severely affected than males (when the males survive that is). Is an image showing the inheritance of a sex-linked dominant condition under two scenario's: Image Source, opens in new window CC BY-SA 3.0 opens in new window (1). Left: Normal male (XaY) and a heterozygous, affected female (XAXa). For both boys and girls there as a 50% chance they will be affected, depending upon which (X) chromosome they inherit from the mother (2). Right: Affected male (XAY) and unaffected mother (XaXa) . ALL DAUGHTERS will be affected due to inheritance of the dominant allele from the father. NO SONS will be affected. As with sex-linked recessive conditions, there are some things that can help you distinguish sex-linked dominant traits from autosomal inheritance and sex-linked recessive inheritance. First understand that all the "rules" we talked about regarding autosomal dominant inheritance apply here with the following modifications: All affected sons MUST have an affected mother - sons get their (X) chromosome from mom - if the son is affected he must have received the dominant allele from his affected mother Affected fathers will have 100% affected daughters. All daughters get one (X) from dad - if he is affected, daughters will get at least that dominant allele (and be affected) from him. To try your had at identifying modes of inheritance, below are four pedigrees. One of them is autosomal recessive (AR), one is autosomal dominant (AD), one is sex-linked recessive (SLR), and one is sex-linked dominant (SLD). See if you can figure out which is which. Hint: sometimes you may not be able to say for certain what a pedigree is and instead will have to use the process of elimination. Once you have given it a try, link you can check your answers here opens in new window and see the reasoning behind those answers.

Gene Interactions

So far we have thought about genes acting alone or perhaps in pairs but controlling different phenotypic characteristics. In many cases, however, multiple genes may control the same phenotypic trait. In a situation such as this, interactions between the genes and their products may occur. Epistasis is just such one interaction. In epistasis, the effects of one gene will mask or hide the effects of another gene. This sort of interaction can occur when multiple genes all influence the same phenotype. Consider coat color in Labrador retrievers: yellow, black, or brown. Image Source, CC BY-SA 1.0 Collectively, these three colors are controlled by two genes: (B) gene: (B__); this genotype will lead to the production of a black pigment while (bb) leads to the production of a brown pigment. Whether or not this pigment is actually deposited in the fur depends on a second gene (E) gene: produces a protein called the melanocortin receptor. It controls the deposition of pigment in hair. (E__) and pigment will be deposited in the hair while (ee) means no pigment, even if produced, will be deposited in hair. Dogs that are (ee) will be yellow no matter what alleles they may possess for pigment production. Thus, in order to be either black or brown, a Labrador must FIRST be at least (E__); any dog that is (ee) will be yellow no matter their alleles for the (B) gene: Image Source, opens in new window CC BY-SA 3.0 opens in new window Thus, the presence of the homozygous (ee) genotype will mask or hide any alleles that may be present for the (B) gene. Consider the following cross: EeBb (x) EeBb; both dogs are black in color This is a dihybrid cross and, while epistasis may affect phenotypes, it won't affect the genotypic distribution; we will get our anticipated 9:3:3:1 genotypic distribution: 9 E__B__: Black 3 E__bb: Brown 3 eeB__: yellow 1 eebb: yellow Note that our phenotypic distribution becomes 9:3:4 as both genotypes that are (ee) are yellow regardless of their alleles for the second (B) gene. This is classic epistasis - it will change phenotypic distributions away from what we might anticipate based on a simple dihybrid (or other) cross. Epistasis is not the only type of interaction that genes may exhibit. Sometimes, when two genes influence the same characteristic, the production of novel phenotypes may be seen. Novel phenotypes are those that are not apparent in the P1 or F1 generation but appear among F2's. For example, consider two genes (bw; brown) and (st; scarlet) in Drosophila. Flies homozygous for brown (bw bw) have brown eyes while flies with at least one dominant allele (bw+__) have wild type (brownish red) eyes. Flies homozygous recessive for scarlet (st st) have bright red eyes while those with at least one dominant allele (st+__) have, once again, wild type eyes. bw bw st+ st+ (P1: true breeding, homozygous recessive for brown, homozygous dominant, or wild type, for scarlet) phenotype is brown eyes. bw+bw+ st st (P1; true breeding, homozygous recessive for scarlet, homozygous dominant, or wild type, for brown) phenotype is scarlet eyes. Consider the cross P1 cross: bw bw st+ st+ (x) bw+bw+ st st The F1 will be, genotypically bw+bw st+ st, a dihybrid and will have wild type eyes. If we allow the F1s to produce an F2, as per Mendel's rules, we will get: 9/16: bw+__ st+ __; wild type eyes 3/16: bw+__ st st; scarlet eyes 3/16: bw bw st+ __: brown eyes 1/16: bw bw st st: white eyes; this is the novel phenotype The novel phenotype appears in the F2 because there is a genotype (homozygous recessive for both genes) that has not appeared before. In this particular case, since both genes affect eye color, the double recessive genotype has eye color effects not seen before. As you read sections 4.7 - 4.9 in your text, you will learn about some more types of genetic interaction. Always consider: No matter how genes interact, Mendels basic principles of segregation and independent assortment still apply. You will always be able to apply those principles to determine the frequencies of genotypes among progeny - it is the phenotypes that may change. Be sure to note variations on standard phenotypic distributions. How have they changes and why? What sort of variations are possible and can you account for those. The addition of factors such as co-dominance, multiple alleles, and lethal alleles may complicate genetic interactions and further change phenotypic distributions. Be sure to understand how these influence distributions. Always ask yourself - how will genotypes be grouped into phenotypes.

Sex Linkage

Technically, any gene on a sex chromosome is termed sex - linked. Practically speaking, in humans, when we say sex-linked, we are referring to genes on an (X) chromosome. This is because both males and females can have an (X) chromosome; only males have genes on the (Y) so we would never have to consider females with respect to these genes. Because males have only one (X) and females have two, patterns of inheritance differ between males and females for these x-linked, or sex-linked, genes. Sex linkage was first identified by T.H. Morgan and Calvin Bridges working with Drosophila melanogaster. Like humans, Drosophila use an (X/Y) sex determination system: Females are (XX) and males are (XY). I should point out this is where the similarities to humans STOPS but more on that later. Image Source, CC BY-SA 4.0 Morgan and Bridges followed the inheritance of a recessive eye color, white (different from the novel phenotype that was white), that is located on the (X) chromosome. All images depicting the following described crosses are from Wikimedia Commons, CC BY-SA 3.0 In the first cross they did (left), a truebreeding wild type female (w+ w+; homozygous dominant) was crossed to a white eyed male. As males have only one (X), they can't really be called homozygous - instead they are termed hemizygous. In this case hemizygous recessive (w Y). The females had wild type eyes and the males had white eyes. In terms of gametes, 100% of the female ova will contain a single (X) with a wild type, or dominant allele (w+). 50% of male gametes will contain an (X) with a recessive white allele (w) while the remaining 50% will carry a (Y) chromosome with no allele for this gene. This will produce an F1 generation where 100% of females are wild type BUT heterozygous for the gene (w+ w) and 100% of the males will be wild type and be hemizygous dominant (w+ Y): When considering F2 flies produced by the F1s: Female F1 gametes: 50% carry wild type allele (w+), 50% carry recessive (w) allele Male F1 gametes: 50% carry (w+) allele and 50% carry (Y). F2 flies are thus 50% female, either homozygous wild type or heterozygous; thus all females have wild type eyes 50% male, either hemizygous wild type or hemizygous recessive; thus half of males have white eyes and half have wild type eyes Note the overall ratio is 3:1 wild type to white BUT, the only flies with white eyes are males. This is a consequence of how (X) chromosomes are inherited: females get an (X) from both parents while males can only get an (X) from the female parent. Next up, Morgan and Bridges did another P1 cross (at right), this one was a reciprocal cross: now the female was white eyes (homozygous recessive) and the male was hemizygous wild type (w+Y). Reciprocal simply means the phenotypes of the parents were flipped. You can follow through the phenotypes and genotypes all the way to the F2: Note the differences: P1 gametes: female gametes all contain the recessive (w) allele; male gametes are 50% (w+), 50% 9Y) F1 flies: all males have white eyes (hemizygous recessive) while all females are wildtype (heterozygous) F1 gametes: female gametes are 50% (w+) and 50% (w); male gametes are 50% (w) and 50% (Y) F2 flies: 50% females, evenly split between wild type (heterozygous) and white (homozygous recessive); 50% males evenly split between white and wild type (hemizygous recessive or hemizygous wild type, respectively). Overall, when comparing the two P1 crosses: In the first P1 cross, all F1 flies had wild type eyes. In the first P1 cross there was an overall 3:1 distribution in the F2 BUT only males could be white eyes. In this F2, 100% of females had wild type eyes while the ratio of wild type to white was 1:1 in males. In the second P1 cross, there was a 1:1 distribution in F1 flies with all females being wild type and all males white eyed In the second P1 cross, there was an overall 1:1 distribution of phenotypes in the F2 with the same 1:1 distribution present in BOTH males and females. The above differences are because of how (X) chromosomes are inherited in the offspring. Sex linked traits will often show differing distributions of phenotypes when the sexes are compared.

In humans, the (Y) chromosome is critical for male development. You can see the (Y) chromosome below:

There are no essential genes on the (Y); since females lack a (Y), there can't be. There probably are not more than 150 total genes on the (Y). Most are involved in male development and fertility although new research hints at other roles too. Note the PAR regions. Stands for "pseudoautosomal"; more about that later. The bulk of the (Y) is termed the MSY - male specific, Y region. Especially important is the SRY - sex determining region, Y. This gene is enough to trigger the development of testes in an early embryo. Very early in development, the embryo is sexually indifferent with "bi-potential" gonads (can become either testes or ovaries) and two side by side internal duct systems. The Wolffian ducts, which can develop into the male reproductive system, and the Mullerian ducts, which can develop into the female reproductive system:

This 9:3:3:1 ratio is typical of a dihybrid cross. If you see 9:3:3:1, you should think "hey, the parents were dihybrids"; if you are told that you are doing a diybrid cross, you should think "hey, the offspring will be in a 9:3:3:1 ratio.

To make things easier when working crosses (lets face it, a 16 place Punnett Square is unwieldy) be sure you read about the "forked line" method for working crosses (no matter how many genes are involved) in the text.

The Degree Of Linkage

To begin calculating the degree of linkage, we first have to decide if the linkage is complete (no recombination occurred between the genes) or partial, meaning there was some. We know we have a dihybrid. There are two potential configurations for that dihybrid: The two horizontal lines indicate the chromosomes. The configuration on the left, in which one member of the chromosome pair carries both dominant alleles while the other carries both recessive alleles is called a coupled (aka coupling or cis) configuration. The configuration on the right, where one member of each pair carries one dominant and one recessive allele is termed the repulsion, or trans, configuration. If the dihybrid is in the cis configuration AND linkage is complete, only two gametes would be produced: (AB) and (ab). This would have given progeny with only the (AB) and (ab) phenotypes in a 1:1 distribution. You can see this in the figure below on the left. If the dihybrid is in the trans configuration AND linkage is complete, only two gametes would have been produced: (Ab) and (aB). This would have given progeny with only the (Ab) and (aB) phenotypes in a 1:1 distribution. You can see this in the figure below on the right. We saw four categories of progeny phenotypes, that means that linkage WAS NOT COMPLETE and that there must have been some recombination going on. In order to calculate the degree of linkage (cM or m.u.) we need to identify the NCO vs the CO gametes from the dihybrid parent. Remember, when genes are linked, the NCO's will always account for more than 50% of the gametes. In other words, the NCO gametes will be the TWO MOST FREQUENT. CO gametes will always account for less than 50% of the gametes and will thus be the TWO LEAST FREQUENT: Phenotype Genotype Dihybrid gamete Gamete Type A B (64) Aa Bb A B nco A b (11) Aa bb A b co a B (12) aa Bb a B co a b (68) aa bb a b nco Note, 64 progeny exhibited the (AB) phenotype so those 64 must have gotten a gamete with both dominant alleles from the dihybrid. The same logic applies to the rest of the table entries. Notice that the two MOST frequently occurring gametes are the NCO's, while the TWO least frequently occurring gametes are the CO's. You can see the gametes below; again, the lines indicate a chromosome. Out of 155 total gametes (155 progeny who each got ONE gamete from the dihybrid parent), only 23 of them were the result of recombination between the (A) and (B) gene. We can formalize our last two conclusions: Conclusion #3: the dihybrid parent was in the coupled configuration. Note that the two NCO gametes, produced when crossing over DID NOT OCCUR contain both dominant OR both recessive alleles. The only way this can happen is if one member of the chromosome pair has both dominant alleles and the other has both recessive alleles. Conlcusion #4: 14.8% of our dihybrid's gametes were the result of crossing over (23 out of 155). Remember, one cM (or m.u.) was equivalent to 1% recombination. Thus, with 14.8% recombination, there are 14.8 cM between our two genes. This is our degree of llinkage. This is how you can analyze a data set with respect to linkage.

These individuals demonstrated that a (Y) chromosome was necessary for a male to develop, not that two (X) chromosomes led to a female. There are two other sex chromosome abnormalities that also occur:

Triplo-X Syndrome: 46 XXX. Female, with few, if any fertility issues "Super Males" or Jacobs Syndrome: 47 XYY. Male with an extra (Y). In the 1960's a poorly done study seemingly linked this condition to a propensity to violence and incarceration. Further studies found this is untrue. No fertility issues None of these conditions produce significantly deleterious phenotypes other than the fertility issues for Turners and Klinefelters. In fact, most cases of Triplo-X and Jacobs syndromes are probably never detected. They seem to be surprisingly common, occurring once in every several thousand births. Some statisticians believe that 5 - 6 girls with the (XXX) sex chromosome complement are born every day.

The study of humans with sex chromosome abnormalities led to the conclusion that the presence of a (Y) chromosome, as opposed to the presence of only one (X) chromosome was necessary for males to develop as males:

Turners Syndrome: 45 X. Forty five total chromosomes but only a single sex chromosome, an (X). Female but sterile due to ovarian dysgenesis. Most cases arise by fertilization of an ova by an "empty" sperm - one missing a sex chromosome. Klinefelters Syndrome: 47 XXY. An additional (X) chromosome. Male but with fertility issues due to disturbed estrogen:testosterone balance.

Map Distance - A Final Word

When you hear "map distance" I'm sure you are thinking distance. If two genes have a relatively large map distance between them, they must be physically further apart than two genes with a lower map distance. Maybe, maybe not. Map distance is determined by RECOMBINATION. Anything that affects recombination can affect map distances. Yes, generally speaking, map distance is related to actual physical distance. In humans, 1 m.u. (1 cM) is equal to roughly 1 million base pairs of DNA. Thus, genes with 20 cM between them will be about 20 million base pairs apart while genes with 10 cM between them will be only 10 million base pairs apart. Look at the image below and ask yourself "Which pair of genes (A) and (B), labeled as gene pair 1, or (C) and (D), labeled gene pair 2, would have the largest map distance between them?". The answer is actually gene pair 1. Notice that between (C) and (D) lies the centromere. Centromeres generally suppress crossing over. Thus, while (C) and (D) are physically further apart, because the centromere resides between them, they will have fewer map units between them than (A) and (B). So, while map distance does often correlate with physical distance, there are times when it won't.

In Drosophila, sex is determined by a balance between the number of haploid sets of autosomes and the number of ________.

X chromosomes

Working With Probabilities - The Sum Rule

Working With Probabilities - The Sum Rule "The sum rule". This principle states that the probability of either of two mutually exclusive events occurring (each of which is independent and has its own probability of occurring) is the sum of their individual probabilities. To illustrate this rule, consider: Suppose a trihybid (AaBbCc) is self-crossed. What is the probability of getting offspring of the genotype (A_Bbcc) or (AABbCc) o Probability of (A_Bbcc) = (3/4)*(1/2)*(1/4) = 3/32; figured using the product rule o Probability of (AABbCc) = (1/4)*(1/2)*(1/2) = 1/16; also figured using the product rule These offspring represent mutually exclusive events - if you get the first stipulated genotype in an offspring, then the second is precluded (a single offspring can only have one genotype!). Conversely, if the second stipulated genotype occurs, then the first is likewise precluded. Thus, getting an offspring of either one of the stipulated genotypes is mutually exclusive of the other; the probabilities must be added together - this gives the probability of EITHER the first OR second genotype in an offspring: Probability of one or the other = 3/32 + 1/16 = 5/32; figured using the sum rule Let us try one that is just a little more complex: Assume two parents are heterozygous for albinism AND galactosemia, both of which are recessive conditions. What is the probability their first child will be a boy with albinism ONLY or a girl with both albinism and galactosemia. To do this we must use the product rule to see what the odds are of the albino boy: o Probability of (boy, albino, no galactosemia): (1/2)*(1/4)*(3/4) = 3/32 o Probability of (girl, albino, galactosemia): (1/2)*(1/4)*(1/4) = 1/ 32 Obviously these are mutually exclusive events - if the boy is born OR the girl is born - so we must add the individual probabilities together via the sum rule: o Probability of (boy, albino, no galactosemia) OR (girl, albino, galactosemia): 4 /32 Essentially, the sum rule allows you to make predictions where only one out of several outcomes is possible. Sometimes, when there are a great many potential outcomes, and we desire only one, the binomial theorem can make our lives easier. To illustrate that, let's consider tossing two coins simultaneously. How could we calculate the probability of getting a combination of one head and one tail? This is a little more complex that it first appears; there are actually two ways to get the desired result. To illustrate, let's consider one coin to fall on the left and one to fall on the right. Perhaps the left coin lands heads and the right one lands tails: The probability o f this occurring is ¼ (this was calculated using the product rule) But there is one more way to get one head and one tail - perhaps the left coin lands tails and the right one lands heads: The probability of this occurring is also ¼ Clearly, these are mutually exclusive events: if the first combination occurs, then second is impossible and vice versa. This is where the "sum rule" comes into play and therefore the probability of getting one head and one tails when two coins are tosse d simultaneously becomes: (probability of one head and one tail) = (¼) + (¼) = (½) Now consider this: what is the probability of getting five heads and two tails if I toss seven coins simultaneously? One way to do this would be to try and come up wit h all the different ways 7 coins could land to give five heads and two tails, figure a probability for each and then sum them altogether: Etc, etc, etc.,... I think you can see where this is going. This type of solution would be time consuming and probably wouldn't work. Think about it, I did just three possibilities. What are the chances I would think of ALL the possible combinations (there are 128) and get the right answer? A better way would to be to use the binomial; the binomial will tell us the probability distribution for eve nts in which there multiple outcomes or combinations, such as simultaneous coin tosses. The binomial takes the following form: Probability of desired combination = Where (N) = the total sa mple size Where (x) = the number of one possible outcome W h ere (N - x) = the number of the alternative outcome Where (p) = the probability of outcome (x) Where (1 - p) = the probability of the alternative outcome When one considers the binomial, each part of the equation tells us something different: tells us how many ways (combinations) will give us what we want tells us the probability of one of those combinations. Putting the two together gives us our overall probability. In our example involving the tossing of seven coins: (N) = 7 (x) = 5 heads; we could just as easily have termed (x) = 2 for the number of tails (p) = (½); the probability of getting outcome (x), a head Now we can plug all our numbers into the binomial and get an answer: probability = = (21/128) As you can see from the binomial, there are 21 ways to get 5 heads and 2 tails and the probability of any one of them is (1/128). Suppose we toss the seven coins and want to know the probability of no heads and 7 tails? In this case: (N) = 7 (x) = 0 heads (p) = (1/2); even though we don't wa nt any heads, we still have a (½) chance of getting one for any given coin Probability == (1/128) You could also use the binomial to figure a probability distribution - the probabilities for ALL combinations of heads and tails: Note that the sum of the probability distribution is (1); obviously, if you toss the seven coins, some combination of heads and tails MUST occur - you just don't know which

Dosage compensation is accomplished in humans by inactivation of the Y chromosome.

false

Predict the potential effect of the Lyon hypothesis on the retina of a human female heterozygous for the XX-linked red-green color blindness trait.

females will display mosaic retinas with patches of defective color perception, and surrounding areas with normal color perception.

Which mutations are generally dominant since one copy in a diploid organism is sufficient to alter the normal phenotype?

gain of function

A mutation in a gene often results in a reduction of the product of that gene. The term for this type of mutation is ________.

loss of function or null (in the case of complete loss)

For an individual with the XXY chromosomal composition, the expected number of Barr bodies in interphase cells is ________.

one

Assume that a cross is made between two organisms that are both heterozygous for a gene that shows incomplete dominance. What phenotypic and genotypic ratios are expected in the offspring?

phenotypic 1:2:1; genotypic 1:2:1

You can see karyotypes below. Top left - Turners Syndrome with single (X), and no partner, circled. Top Right - Klinefelteers with (XX) and (Y) chromosomes circled. Bottom left - Jacobs Syndrome with the (XXY) complement circled. Bottom right - Triplo-X with the 3 (X) chromosomes circled.

picture printed out

An individual with Klinefelter syndrome generally has one Barr body

true

Normally in humans, all the sons of a female homozygous for a sex-linked recessive gene will inherit that trait.

true

Two forms of hemophilia are determined by genes on the X chromosome in humans. Assume that a phenotypically normal woman whose father had hemophilia is married to a normal man. What is the probability that their first daughter will have hemophilia?

zero

Another test of Mendel's Principle of Segregation: allow F2 plants described above to "self cross" - each plant serves as both male and female parent:

¼ DD: all progeny will be DD and thus tall - plant is homozygous thus true breeding ¼ dd: all progeny will be dd and thus short - plant is homozygous thus true breeding ½ Dd: will produce progeny in 3:1 distribution - this is really just a monohybrid cross.

3. What is the probability that a trihybrid cross (AaBbCc) x (AaBbCc) will produce an individual that is homozygous recessive for all 3 genes?

1/64

2. What is the probability that a trihybrid (AaBbCc) will produce a gamete that is the genotype (A b C)?

1/8

4. What is the probability that a testcross involving one of the above trihybrids will produce an individual homozygous recessive for all 3 genes?

1/8

Introduction

In this module, we will be looking at some variations on the rather simple inheritance schemes observed by Mendel. We will start by defining exactly what an allele is and then move on to some (slightly) more complicated inheritance patterns than the rather simple traits in peas. This will include looking at how genes interact with one another through physiology and environment, how alleles on sex chromosomes follow parental specific inheritance patterns, variations on simply dominance, and the concept of multiple alleles. As you work through this material, always remember: Mendel's rules regarding segregation and independent assortment still (and always will) apply. Always look at how various combinations of alleles can affect phenotype and how genetic interactions may change some of the phenotypic distributions we saw in the previous module. In the case of sex-linkage, be sure to understand how the parental origin of sex chromosomes (and their alleles) influences phenotypic distribution amongst offspring.

Introduction To Linkage

Mendel's principle of independent assortment: alleles of different genes will assort independently of each other into gametes Result: in gametes, ALL combinations of alleles are equally Consider the dihybrid below (heterozygous for two genes, genes are located on DIFFERENT chromosome pairs): As chromosomes segregate independently at metaphase I, and then centromeres divide and sister chromatids separate at anaphase II, four total gametes are produced; each contains ONE formerly sister chromatid from each chromosome pair. Probability of chromatid with (A) - 50%; probability of chromatid with (B) - 50%. Thus, probability of gamete with (A) AND (B) - 25%. Note all allele combinations equally at 25% each - independent assortment. What if genes are on SAME CHROMOSOME: Note that now NOT ALL GAMETE COMBINATIONS ARE EQUALLY LIKELY. If a gamete gets (A) IT MUST ALSO GET (B). Only two gametes possible: (A) AND (B); (a) and (b). Independent assortment no longer holds - genes are said to be linked. In this case, the linkage is said to be complete. Sometimes, all four combinations of alleles can be restored IF crossing over occurs during prophase I: Note that two gametes will get chromatids that were not involved in crossing over: (A) / (B) and (a) / (b). These are called non-crossover, or NCO, gametes. The other two gametes got chromatids that were involved in the cross over event. They got allele combinations of (A) / (b) and (a) / (B). These are called cross-over gametes, or CO, since it took crossing over to produce them. The frequency of CO gametes depends upon how frequently crossing over between the two genes occurs in meiosis I. When this occurs, the linkage is said to be partial. To be linked: Genes must first of all reside on the same chromosome; Additionally, the distribution of alleles in gametes will deviate from that predicted by independent assortment. This means progeny phenotype distributions will also deviate from what independent assortment would predict. When genes are linked, NCO gametes WILL ALWAYS ACCOUNT FOR MORE THAN 50% of gametes and CO gametes WILL ALWAYS ACCOUNT FOR LESS THAN 50% of gametes. What is so magic about this 50%? Imagine two genes, on the same chromosome, but so far apart from one another that recombination between them OCCURS EVERY SINGLE MEIOSIS (i.e. 100% of the time): Note that two of the four chromatids have undergone recombination while two have not. This means that two our of four gametes will be NCO while the other two out of four will be CO. This means that each NCO will account for 25% of total gametes: (A) / (B) and (a) / (b) while each CO will also account for 25% of the total gametes: (A) / (b) and (a) / (B). This is exactly the same ratio that independent assortment predicts. In other words, if genes are far enough apart, even on the same chromosome, recombination occurs frequently enough that they appear to assort independently. So, if two genes are indeed linked, it means that recombination occurs at some frequency less than 100% and thus CO gametes will account for less than 50% of the total while NCO gametes will account for more than 50% of the total. The degree to which two genes are linked can be measured; the units are sometimes called map units (m.u.) or centimorgans (cM). Essentially, one cM = 1% recombination between two genes. Thus, if recombination were found occur 22% of the time, the genes would be said to be 22 m.u. or 22 cM apart. Up next, we will apply some of these principles

Polyploidy - Animals

Polyploidy - Animals There are only two polyploid conditions that occur in humans, triploid and tetraploid, and neither of them is survivable. Most are miscarried, die shortly after birth, or, rarely, may survive for a period of time. Triploid. 3 sets of chromosomes (3N). May occur in 1% of all conceptions. Present in about 15% of miscarriages. Most cases result of dispermy: simultaneous fertilization of one ova (N) with two spermatozoa (N each) to give the 3N chromosome number. See karyotype below, note 3 copies of each chromosome. Image Source, opens in new window CC BY-SA 3.0 opens in new window Tetraploid. 4 sets of chromosomes (4N). 6% of miscarriages. Only a handful of live births known. Believed to arise from failure of first cleavage in a 2N embryo - the chromosomes double BUT the zygote fails to undergo cytokinesis. So, 4 sets of chromosomes but only one cell. Subsequent mitotic divisions maintain the 4N chromosome number. Polyploidy naturally occur in some other animals - fish, reptiles, amphibians, and fish. Some sturgeon are 12N with 360 chromosomes. No identified polyploid mammals. Several years ago it was believed that a rodent living in the Andes was tetraploid but that is not supported, for the most part, by rigorous genetic analyses.

Sex Effects On Autosomal Traits

Sometimes the expression of an autosomal trait may be affected by the sex of the individual, presumably through different hormones in males vs. females. Sex-influenced. Autosomal traits that occur at different frequencies in males and females. The classic example is pattern baldness: it is a dominant trait in males BUT recessive in females. To understand this, consider the (B) gene with two alleles : (B1) and (B2). Below are the genotypes: B1B1: Pattern bald male AND female B1B2: Pattern bald male BUT normal female B2B2: Normal female AND male Note: B1 is dominant males - it takes only ONE B1 allele to produce pattern baldness. B1 is recessive in females - it requires TWO B1 alleles to produce pattern baldness. Even when women are pattern bald, the hair loss is usually less severe and extensive than in males. Any condition that exhibits difference incidence rates between the sexes can be considered sex-influenced. For example, 90% of Lupus patients are women, 67% of infants born with a club foot are males, and 75% of osteoporosis sufferers are female. Sex-limited. These are autosomal traits that are only ever expressed in one sex, even though both may carry the alleles. For example, a dominant allele for the BRCA1 gene greatly increases a woman's chance of breast cancer. While males can get breast cancer (it is uncommon), this dominant allele DOES NOT INCREASE HIS ODDS. Other examples include bright plumage restricted to some male birds (think cardinals), male restricted bright colors in some tropical fish, antler growth restricted to male deer (reindeer being the exception), and the ability to lactate restricted to females.

Mapping 3 Genes - Gene Order

To determine gene order, we need to focus on the DCO's - they are easily identifiable as the two least frequent. Below is the data table indicating the NCO and DCO gametes: trihybrid gamete Obs gamete type + + + 22 CO cn b + 179 NCO + b vg 46 CO + + vg 173 NCO cn + vg 4 DCO cn + + 52 CO cn b vg 22 CO + b + 2 DCO There are three potential gene orders: (cn b vg), (b cn vg), or (cn vg b). To determine the correct gene order we need to do the following: Write out our trihybrid, maintaining the proper allele arrangement, 3 times; each should reflect ONE of the possible gene orders Examine each possibility and ask ourselves: " If this gene order is correct, would it give me the allele combinations I am seeing in the DCO gametes". If it is, then that reflects the correct linear gene order. Below are the three possible versions of our trihybrid parent and the double crossover events for each Assume that (b) is in the middle. If so, then the indicated DCO would produce gametes that are (+ b vg) and (cn + +) Assume that (cn) is in the middle. If so, then the indicated DCO would produce gametes that are (+ cn vg) and (b + +) Assume that (vg) is in the middle. If so, then the indicated DCO would produce gametes that are (+ + +) and (cn vg b) Now look at the actual DCO's from the table: (cn + vg) and (+ b +). Only one gene order is consistent with this: (b cn vg) - the cinnabar gene must be in the middle. We can now have our tribybrid parent with the correct linear order AND the correct arrangement of alleles on the chromosomes: We now only have the actual map distances and a few related values to calculate.

9. Again, with respect to question 7, suppose the son marries a woman, who is phenotypically normal but has a brother who is an albino. What is the probability their first child will be an albino?

1/9

6. In a family of 5, what is the chance of having 2 boys and 3 girls?

10/32

With incomplete dominance, a likely ratio resulting from a monohybrid cross would be ________.

1:2:1

A human with the karyotype 49, XXXYY forms _______ Barr bodies.

2

In mice, there is a set of multiple alleles of a gene for coat color. Four of those alleles are as follows:C = full color (wild)cch = chinchillacd= dilutionc = albinoGiven that the gene locus is not sex-linked and that each allele is dominant to those lower in the list, diagram the crosses indicated below and give the phenotypic ratios expected from each.a) wild (heterozygous for dilution) × chinchilla (heterozygous for albino)b) chinchilla (heterozygous for albino) × albino

2 full color: 1 chinchilla : 1 dilution

3. Two parents are heterozygous for albinism. If they have 3 children, what is the probability they will have 2 that are phenotypically normal and one albino?

27/64

introduction 2

Introduction Numerical chromosomal variation is either aneuploid or polyploid. Consider the image below: Image Source, opens in new window CC BY-SA 3.0 opens in new window As we have learned, one set of chromosomes, the haploid, or (N), number is the number of chromosomes in a gamete. When two gametes unite, the zygote receives two sets of chromosomes and is thus 2N, or diploid. What would happen, however, if an individual received THREE sets of chromosomes? Or four? These individuals would then be considered polyploid, specifically triploid and tetraploid. Polyploidy is the state in which entire extra sets of chromosomes are present. Such individuals are labeled as 3N, 4N, 5N, 6N, etc. based upon how many sets of chromosomes are present. Aneuploidy involves numerical variation that is LESS THAN A FULL set. In other words, an individual has one or a few TO MANY or one are a few TO FEW: Aneuploidy typically arises via fusion of gametes that have suffered non-disjunction, or a failure of chromosomes fail to segregate properly during meiosis: Image Source, opens in new window CC BY-SA 3.0 opens in new window The failure can occur during the first meiotic division (primary non-disjunction) or during the second meiotic division (secondary non-disjunction). Failure in the first division results in all gametes either (N + 1) or (N - 1) while failure at the second division results in half of all gametes with the correct chromosome number and the other half (N + 1) and (N - 1).

Mapping In Reverse

Our previous example involved analyzing a data set to determine if linkage was present and, if so, the map units and coupled vs repulsion in the dihybrid parent. Here is an example of "mapping in reverse" - you are given the map units and parental configuration and asked to determine the probabilities of specific phenotypes in the progeny: A dihybrid parent (Aa Bb), cis (or coupled) configuration, with 22 cM separating the two genes, is testcrossed. Probability of getting progeny with genotype (Aa Bb) in progeny? Probability of getting progeny with genotype (Aa bb) in progeny? Probability of getting progeny with genotype (aa bb) OR (aa Bb) in progeny? Below is the cross: Note the dihybrid parent is the cis configuration: one member of the chromosome pair has both dominant alleles and the other has both recessives. The testcross parent is homozygous recessive. 22 map units, or cM, means that recombination will occur 22% of the time, meaning that 22% of the dihybrid's gametes will be CO while the remaining 78% will be NCO. 78% of gametes will thus be (AB) or (ab) with (AB) accounting for 39% and (ab) accounting for 39% 22% of the gametes will be CO and be either (Ab) or (aB); each will account for 11% of the total Whether or not crossing over occurs in the testcross parent doesn't matter - all gametes will be (ab). 39% of progeny will be formed from gametes (AB) and (ab) giving genotype (Aa Bb) 39% of progeny will be formed from gametes (ab) and (ab) giving genotype (aa bb) 11% of progeny will be formed from gametes (Ab) and (ab) giving genotype (Aa bb) 11% of progeny will be formed from gametes (aB) and (ab) giving genotype (aa Bb) 39% + 11% will be either (aa bb) OR (aa Bb)

The Monohybrid Cross

P1 cross between (Tall) and (short) pea plants. Note that (Tall) has two (D; dominant) alleles while (short) has two (d; recessive alleles). When individual has two identical alleles they are homozygous. Principle of segregation says that (Tall) parent can produce gametes with only (D) alleles while (short) parent can produce gametes with only (d) alleles. Imagine that (tall) or (short) were to be self crossed - all progeny would look like parent. Genetic definition of "true-breeding" is that an individual is homozygous. Homozygous for dominant or recessive alleles doesn't matter but homozygotes can pass only ONE TYPE of allele to offspring - true-breeding. Lets look at F1: all Tall (Dd). F1 is heterozygous; has two different alleles. As (D) is dominant, phenotype is (tall). F1 is said to be monohybrid, that is heterozygous for ONE GENE. Allow F1's to cross amongst themselves, called a monohybrid cross. Principle of segregation says each parent will produce gametes that carry either (D) or (d) allele. 50% chance (D) dominant allele and 50% chance (d) recessive allele. Punnett Square can be used to predict outcomes of gamete union. Note offspring distribution is what Mendel observed: 3 (Tall):1 (short). Specifically: ¼: DD; homozygous dominant ½: Dd; heterozygous ¼: DD; homozygous recessive Description of genetic constitution (homozygous vs heterozygous) is termed the genotype. Note the phenotypic distribution becomes: ¾: D__; in other words, 75% of the progeny have AT LEAST ONE dominant allele and will be tall ¼: dd; 25% are homozygous recessive and will be short ¾ D__: ¼ dd 3 D__: 1 dd

Mapping 3 Genes - Map Distances

We have three map distances to calculate: The distance between (b) and (cn) The distance between (cn) and (vg) The distance between (b) and (vg). Fortunately, map distances are additive, once we have the distances between the middle gene and the two outside genes, we simply sum them to get the distance between the two outermost genes. To begin, let us take our trihybrid parent, with correct gene order AND correct arrangement of alleles on the chromosomes, and call the distance between (cn) and one outside gene Region 1 and the other distance between (cn) and the other outside gene as region 2: Now we can ask the following questions: If there is a crossing over event in region 1, what gametes will be produced? The answer is (+ cn +) and (b + vg) If there is a crossing over event in region 2, what gametes will be produced? The answer is (+ + +) and (b cn vg) Below is our data table with SCO progeny now properly identified as region 1 or region 2: trihybrid gamete Obs gamete type + + + 22 CO: 2 cn b + 179 NCO + b vg 46 CO: 1 + + vg 173 NCO cn + vg 4 DCO cn + + 52 CO: 1 cn b vg 22 CO: 2 + b + 2 DCO Notice that 98 of our gametes were the result of a SCO event in region 1 (46 + 52) while 44 of our gametes were the result of a SCO event in region 2 (22 + 22). Remember to calculate map distances, you need to calculate the percentage of recombination, so: Region 1: {(46 + 52 + 4) / 500} x 100 = 20.8% recombination or 20.8 m.u. We added in the (4) as the total number of DCO progeny. DCO's represent a crossing over event IN REGION 1. Yes, they also represent a crossing over event in region 2 as well BUT WE MUST COUNT THEM WHEN FIGURING REGION 1 RECOMBINATION. Clearly, we also have to count them as we calculate region 2 recombination. Region 2: {(22 + 22 + 4)/500} x 100 = 10% recombination or 10 mu. The distance between the (b) and (vg) genes: 10 m.u + 20.8 m.u. = 30.8 m.u. Congratulations, you have just mapped three genes! There are only two more values to calculate: Coefficient of Coincidence (COC): (Observed DCO's) / (Expected DCO's) Interference: (1 - COC) Based upon our map distances, we would have expected, out of our 500 progeny, 10.4 DCO's. Since you can't have "part of a DCO" we round DOWN to 10. How did we get this: Probability of a SCO in region 1 is 20.8%, or (0.208) Probability of a SCO in region 2 is 10%, or (0.10) Thus, the probability of a DCO (a crossover in region 1 AND region 2) is therefore (0.208)*(0.10) = 0.0208 or 2.08% So, out of 500 progeny, we would expect 2.08% to be DCO's: (500)(0.0208) = 10.4, or 10 Notice that we actually saw 6 DCO's. Our COC is thus (6) / (10) = 0.60 or 60%; we only saw 60% of what we expected. The flipside is our interference value: 1 - COC = 0.40 or 40%; 40% of the time SOMETHING interfered with a DCO occurring. This "something" is the physical structure that accompanies a cross over event - the chiasmata. In some cases, the formation of a chiasmata, marking a crossing over event will physically interfere with another, nearby, crossing over event. It is not uncommon at all to observe fewer DCO's than one would expect based on the map distances.

Determining If Two Genes Are Linked

You have two genes, (A) and (B). You have an individual, of unknown genotype, that displays the DOMINANT phenotype for both (A) and (B). You testcross this individual and get the following phenotypic categories in the offspring: Phenotype # A B 64 A b 11 a B 12 a b 68 In other words, 64 progeny exhibited the DOMINANT phenotype for both genes, 11 exhibited the DOMINANT phenotype for (A) but the recessive phenotype for (B), and so forth. Let us figure out everything we can about these genes. Conclusion #1: The unknown genotype was a dihybrid. By definition, a testcross involves an unknown genotype crossed by a homozygous recessive genotype. Thus, the testcross parent would have contributed recessive alleles for both (A) and (B) gene. If progeny exhibit the dominant phenotype for both genes, it means they got dominant alleles for both from the unknown parent; if they exhibit the recessive phenotype for both genes, the unknown parent would have to contributed two recessive alleles. You can see this in the table below: Phenotype Testcross parent gamete Unknown parent gamete A B ab AB A b ab Ab a B ab aB a b ab ab Note that the parent on unknown genotype is capable of contributing both dominant and recessive alleles for the (A) and (B) genes. Our cross is thus a dihybrid testcrossed: Aa Bb (x) aa bb Conclusion #2: These genes are linked. How do we know this? So we know our cross: Aa Bb (x) aa bb If we assume independent assortment, and you should ALWAYS begin by assuming genes will assort independently, we would expect the following in the progeny: ¼: Aa Bb: phenotypically (AB) ¼: Aa bb: phenotypically (Ab) ¼: aa Bb: phenotypically (aB) ¼: aa bb (phenotypically (ab) We observed 155 progeny in total, so we would have expected 39 (25%) of each category: Phenotype # Expected E A B 64 39 16 A b 11 39 20 a B 12 39 18.7 a b 68 39 21.6 Totals 155 156 76.3 Our Chi-square value is 76.3 and we have 3 degrees of freedom. Our p-value is significantly less than 0.05. Thus, our null hypothesis - independent assortment as that was what we assumed to calculate our expected values - is not supported. Independent assortment is NOT occurring. If independent assortment IS NOT occurring, the genes must be linked. Now that we have determined the nature of the cross, a dihybrid testcrossed, AND that the genes are linked, we can move onto calculating the degree of linkage.