Unit 2

A computer randomly generates two nucleotide sequences, each of which is 4000 bases long. At how many positions are they expected to match just by chance, even if they are not related by ancestry?

(1/4)(4000) = 1000

Pigeons exhibit a checkered or plain feather pattern. In a series of controlled matings, the following data were obtained.: F1 A) Check x check checkered: 34 plain: 0 B) Check x plain checkered: 38 plain: 0 C) Plain x plain checkered: 0 plain: 35 F2 (A) check x (C) plain checkered: 34 plain: 0 (B) check x (C) plain checkered: 17 plain: 14 (B) check x (B) plain checkered: 28 plain: 9 (A) check x (B) check checkered: 39 plain: 0 Give the genotype of each pigeon, and explain the inheritance pattern. For simplicity, use A to represent the dominant allele and a to represent the recessive allele.

(A) AA (B) Aa (C) aa AxC - Aa BxC - Aa and aa BxB - AA Aa and aa AxB - AA and Aa

The last Russian tsar had a daughter who was a carrier for hemophilia B, which is caused by a very rare allele. Had the daughter lived to be an adult, what is the probability that her first child would have been a son with the disease if her partner were a male with hemophilia?

25%

A family has three children, none of whom are twins. A) What is the probability that they have two daughters and a son? B) What is the probability that they have at least one son and one daughter? C) Suppose that we know that their oldest child is a girl. Based on this additional information, what is the probability that they have two daughters and a son?

A) 3(0.5*0.5*0.5) = 0.375. The son could be the first-born, the middle child, or the youngest child, so there are three outcomes, each of which has a probability of 1/8. B) P(all sons) = 0.5 * 0.5 * 0.5 = 0.125 P(all daughters) = 0.5 * 0.5 * 0.5 = 0.125 P(at least one son and daughter) = 1 - (.0125 + .0125). In other words, they do not have three sons or three daughters, they must have at least one child of each sex.

Assume that the trait indicated by the filled in circles of the pedigree in Question 41 is very rare. A) What is the most likely mode of inheritance? B) What is the likelihood that child 1 will be affected by the trait?

A) Autosomal Recessive B) 0% because it is assumed the mother will not be carrier of the trait.

7.21 Shown below are three pedigrees for rare genetic traits in humans. Affected individuals are shown in black, while unaffected individuals are shown in white. Although each pedigree has too few members to be absolutely confident about how the trait is inherited in each case, certain modes of inheritance might be likely and some might be impossible. For each of the pedigrees below, which modes of inheritance - that is X linked recessive, X linked dominant, Y linked, autosomal recessive, and autosomal dominant - can be ruled out? Which is most likely? A) Pedigree A B) Pedigree B C) Pedigree C D) For the most likely mode of inheritance in each pedigree, calculate the likelihood that child #1 will be affected by the trait. E) For the most likely mode of inheritance in each pedigree, calculate the likelihood that child #2 will not be affected by the trait.

A) Autosomal dominant is most likely. This trait is not X linked because a son inherits it from the father. This trait could be autosomal recessive if the mother is a carrier; however, if we assume the trait is a rare trait, it is most likely that the mother is not a carrier and the trait is autosomal dominant. B) Autosomal recessive is most likely. The trait is not dominant, as there are affected children of unaffected parents. The trait can not be Y linked because a daughter is affected. The trait cannot be X linked because it is a recessive trait and an affected daughter must have a father with the trait. C) Y-linked is most likely. This trait can not be X linked because it is passed from a father to his sons. It is possible for the trait to be autosomal dominant or autosomal recessive, but the fact that it is passed only from the father to the sons suggests that it is likely to be a Y-linked trait. D) a. 50%. Since the trait is autosomal dominant, there is a 50% chance that the child 1 will inherit the dominant allele and be affected. b. 0%. If the trait is autosomal recessive, then there is a 66% chance that the father of child 1 is a carrier. However, if we assume that the trait is rare and that it is unlikely for the mother of the child 1 to be a carrier, then there will be a 0% chance that the child is affected. C. Assuming that the trait is Y linked, there is a 100% chance of child 1 being affected if he is a boy and a 0% chance of the child being affected is she is a girl. Since there is a 50% chance that the child is a boy, there is a 50% chance of the child being affected. E) a. 100% B. 100%, if we assume that the trait is rare and that it is unlikely for the father of child 1 to be a carrier, then there will be a 100% chance that the child is unaffected. C. 100%

Define the following terms: A) Centrosome B) Centromere and Kinetochore C) Cohesion D) Synapsis E) SPO11 F) Non-disjunction

A) Centrosome - the structures in an animal cell from which microtubules are grown and organized. B) Centromere and Kinetochore - the protein structure that holds together two sister chromatids (centromere) and the proteins around this structure that provide a place for microtubules to attach (Kinetochore) C) Cohesion - the process by which sister chromatids stay together after replication but before metaphase of mitosis or meiosis II. D) Synapsis - the "zipping up" of two chromosomes that have been paired together. E) SPO11 - an evolutionary conserved enzyme among eukaryotes that creates double-stranded breaks for crossing over. F) Non-disjunction - the failure of chromosomes to segregate properly during meiosis.

A drosophila female that is missing cross veins on the wing is mated to a male with normal wings. The female F1 offspring will have normal wings, but the male F1 offspring have crossveinless wings. A) What two conclusions can you draw so far about the inheritance of crossveinless wings? B) One of these F1 females is mated to a normal male. What are the expected results among the F2 offspring? C) Two F2 females are chosen and each is put into a vial and mated with a male with crossveinless wings. One of these matings produces only wild type flies. The other mating yields both crossveinless flies and normal flies. Explain these results. What are the expected proportions of crossveinless and normal flies in this vial?

A) Crossveinless is X linked and recessive. B)(XA and Y) x ( XA and XA) results in males: 50% missing cross veins, females: 100% normal. C) One of the F2 females is XA XA and the other F2 female is XA Xa. For the heterozygous female, 50% of the offspring (both males and females) will be missing cross veins.

7.26 The Romanov family, the last czars of Russia, were executed during the Russian Revolution, although there have been long-standing rumors that some family members might have escaped. The last czarina was Alexandra, Queen Victoria's granddaughter in pedigree, was married to Czar Nicholas II. As shown in Figure 7-6, they had 4 daughters and a son Alexis who had hemophilia. The relevant part of this pedigree is also shown as Panel A. The likely grave site for the last of three Romanov's including Alexis, was identified an din order to confirm that the bones were from the Romanov children, the two X linked genes for clotting Factors VIII and IX were analyzed. A small amount of DNA was isolated from the femur of the skeleton of the boy, and sequenced. For the Factor VIII gene, no sequence changes were found between the DNA from the bones and from sequences in unaffected people. However, when DNA sequenced from the skeleton of the boy, a single base change was found in the Factor IX gene. This sequence change is illustrated in panel A by the red arrow on the sequencing scan below the pedigree. A) Why were the genes sequenced from the boy rather than one of the sisters? B) The single base change shown above appears to have changed the splicing of the exons in the gene coding factor IX. How would this mutation alter the function of factor IX and lead to hemophilia? C) One of the female skeletons was tentatively identified as Anastasia, based on the age of the girl and the presence of some jewelry. Anastasia died before she could have children. Based on the evidence above, what is the probability that her first child would be a boy with hemophilia?

A) Genes were sequenced from the boy because the boy was known to have hemophilia, an X linked disease, so he must have had a mutant form of one of the genes involved in blood clotting. The daughters have two X chromosomes, and may or may not have been carrying the mutant hemophilia alleles. Therefore, to confirm the identity of the remains, only the boy's DNA needed to be analyzed. B) The single base change appears to cause an alteration in splicing that increases the length of exon 4 by a single base. This frameshift mutation in exon 4 that eventually codes for an early stop codon. This shortening of the protein may cause it to lose its function and lead to reduced blood clotting. C) The probability that her first child would have been a boy is 50%. Since she was a carrier, the probability that her first son would have had hemophilia is 50%. The probability that her first child would have been a boy with hemophilia is therefore (.5*.5) = .25 or 25%

Figure Q5-1 shows a pedigree for a rare trait in humans. Affected individuals are indicated by the filled in symbols. The woman II-2 is pregnant, with II-3 being the father. Her unborn child is represented by the ?. A) What is the relationship between the unborn child and individual I2? B) What is the relationship between the unborn child and individual II4? C) Is this trait inherited as the dominant or recessive trait? What is the best evidence for this answer? D) What is the probability that the unborn child will be a boy and not affected by this trait?

A) I2 is the maternal grandmother of the unborn child B) II4 is the aunt of the unborn child C) Dominant. The two affected individuals in generation I have only two children, but in each case one of the two is affected. If the trait is rare, then the affected individuals are expected to have an affected parent. D) II2 is aa and II3 is Aa, so half of their children are expected to be affected by this trait and half will be unaffected. The probability that the child will be unaffected and a boy is then 0.5 x 0.5 or 1/4.

Define the following terms an discuss how they are used in this chapter: A) Locus B) Diploid C) Recessive and dominant

A) Locus: Refers to the location of a gene on the chromosome, either from a physical or genetic map. Often used as a general term for a gene. B) Diploid: Means having two copies of every chromosome (one inherited from each parent) C) Recessive and Dominant: A dominant allele masks the phenotype of a recessive allele. The dominant allele is the one whose phenotype appears in a heterozygote.

In c elegans, the genes Lon-2 and unc-18 are both X linked and are very close to each other on the X chromosome. Mutations in the Lon-2 gene result in worms that are unusually long while mutations in the unc-18 gene result in worms that are uncoordinated. A) A wild type male is mated to a hermaphrodite that is both lon and unc. The F1 hermaphrodites are wild type, while the F1 males are both Lon and unc. One of the F1 hermaphrodite is mated to a wild type male. What are the expected male progeny from this cross, and in what proportions will they arise? B) A Lon male is mated to a hermaphrodite that is Unc. The F1 hermaphrodites are wild-type while the males are unc. One of the F1 hermaphrodites is mated to a wild type male. What are the expected male progeny from this cross, and in what proportions will they arise? C) While nearly all of the F1 males from the cross in Part B have the expected phenotypes, a few of the males are completely wild-type. Based on what your learned about meiosis in chapter 6, postulate the origin of these few wild-type males.

A) Lon-2 and Unc-18 are linked genes that are inherited together on the chromosome. Therefore, the F2 males will be approximately 50% wild type and 50% Lon and Unc - they will get one of the X chromosomes from their mother but not the other. Crossovers between the two genes may cause a few males to be Lon but not unc and a few to be unc but not Lon. We cannot predict how many of these males there will be without knowing how frequently the crossovers can happen between the two genes. However, since the question states that the genes are very close together, it is unlikely that the crossover on the X chromosome will occur between them. B) The hermaphrodite's X chromosomes have the genotypes Lon-2/wild type and wild type/unc-18. The F2 males could inherit either X chromosome, so the male offspring will be 50% Lon and 50% Unc. A few may be Lon and Unc and a few may be wild type due to crossover between Lon-2 and Unc-18 genes on the X chromosome. C) These wild-type males arise due to crossover between the Lon-2 and Unc-18 genes on the hermaphrodite's X chromosomes. As a result of the crossover, some of the hermaphrodite's gametes will have X chromosomes carrying the wild type versions of both Lon-2 and the Unc-18 genes, and some will have X chromosomes carrying both of the Lon-2 and Unc-18 alleles.

During the Korean War, soldiers (males) and nurses (females) were given antimalarial drugs such as chloroquine. Some soldiers developed a severe and life-threatening hemolytic anemia in which their blood cells lysed; other soldiers were completely unaffected. Subsequent genetic analysis showed that the hemolytic response was due to a particular allele of the X linked gene encoding the enzyme G6PD. A) Explain why the response of males to this drug was biphasic - some with a severe reaction and other with no reaction, with no intermediate phenotypes. B) Surprisingly, some female nurses also showed this severe hemolytic response, although the allele is rare enough that no nurse was expected to be homozygous. Postulate an explanation for this response.

A) Males are hemizygous for all alleles on the X chromosome. Therefore they cannot have a heterozygous phenotype—they either have one or another allele for the enzyme G6PD. The males with one allele would have a strong reaction, while the males with the other allele would have no reaction. B) When X inactivation occurs in the embryo, either X chromosome can be inactivated. Perhaps the hematopoietic system (which gives rise to the red blood cells) is derived from only a few cells at the time when inactivation occurred. By chance then, some females will have the same X chromosome inactivated in all of their hematopoietic cells. By knowing either when X inactivation occurs or how many cells give rise to the hematopoietic system in the embryo at this time, it is possible to estimate the other parameter from the fraction of women who show the hemolytic response. If the hematopoietic system is derived from only a single cell when X inactivation occurs, then half of women should show this hemolytic anemia; if it is derived from two cells that inactivate their X chromosomes independently, then about 25% of women should show the hemolytic responses, and so on.

A male mouse with a pale yellow coat called cream is mated to a female mouse with a wild-type gray brown coat. All of the F1 mice, both males, and females have the wild-type gray brown color. One of these F1 female mice is mated to a cream colored male. A) What are the expected results if cream is X linked and recessive? B) What are the expected results if cream is autosomal and recessive? C) In fact, cream is X linked. When the female mouse is examined more closely, it is apparent that she has patches of fur that are cream colored rather than gray brown. Explain this result.

A) Males: 50% cream and 50% wild-type gray brown Females: 50% cream and 50% wild-type gray brown B) 50% cream and 50% wild-type gray brown C) This can be explained by X inactivation. Different X chromosomes form a Barr body in different cells of the mouse. In some cells, the chromosome containing the dominant allele for gray brown was inactivated, resulting in the recessive allele for cream fur being expressed, resulting in cream colored fur.

C elegans has two sexes, males and hermaphrodites. Hermaphrodites are essentially females what make sperm for a few hours, and then shut off spermatogenesis for the remainder of their life. Thus, a hermaphrodite can either reproduce by self-fertilization of its own sperm with its own ova, or by cross-fertilization with a male. Hermaphrodites have five pairs of autosomes and a pair of X chromosomes (XX). As described in Chapter 7, males have five pairs of autosomes and a single X chromosome (X0); there is no Y chromosome in nematodes, so males have 11 chromosomes rather than 12. This mode of reproduction and sex determination has made meiotic mutants particularly easy to identify in c elegans. A) most of the offspring of self-fertilization by a hermaphrodite are also hermaphrodites. However, approximately one in 500 offspring of self-fertilization is a male. Explain these results. B) Mutations that result in nondisjunction in the c elegans hermaphrodite are referred to as HIM mutations. HIM is the phenotypic designation for High Incidence of Male progeny. Rather than having 1 in 500 male offspring, HIM mutants have 3% or more male offspring, with the exact percentage depending on the gene and the mutant allele. Explain why mutations that affect meiosis have a HIM phenotype. C) What are some of the biological processes that might be encoded by a HIM gene? D) Although HIM mutants have a high incidence of male progeny, most of them also lay many eggs that do not hatch. Explain why most HIM mutations have so many non viable offspring?

A) Nondisjunction of the X chromosomes during meiosis caused one egg to not have an X chromosome. When this egg joins with a sperm that has an X chromosome, the fertilized egg will only have one X chromosome, the fertilized egg will only have one X chromosome, making the fertilized egg male. B) Normal segregation of the chromosomes is the outcome after all of the steps of meiosis, so mutations that affect meiosis are likely to cause nondisjunction. If the chromosomes are not being segregated properly this is likely to create gametes that are missing an X chromosome, which will produce an X0 male offspring if fertilized. C) Possible answers could be synapsis, pairing, crossing over, structures of synaptonemal complex, checkpoints, or any of the steps in meiosis. D) Not only do these mutants have gametes that are missing an X chromosome, many of them will have nondisjunction of the other chromosomes as well. Monosomy for the X chromosome affects sex determination, while monosomy and possibly trisomy for other chromosomes is likely to be lethal.

The nematode C elegans can reproduce by self-fertilization or cross-fertilization. Hermaphrodites are capable of self-fertilization or cross-fertilization by mating with a male. A wild-type male with normal movement is mated to a paralyzed hermaphrodite. All of the F1 offspring are capable of movement. One of the F1 offspring is allowed to self-fertilize to produce an F2 generation. A) Which allele is dominant and which is recessive? B) Among the F2 generation worms, what fraction of them do you predict will be paralyzed? C) When an observant student did the cross, she noticed some of the worms in the F2 generation moved normally, some were paralyzed, and some moved sluggishly, although they could move. She decided to count the number or worms in each category. How might you explain these results? What do these results suggest about the F1 worms that are sluggish but capable of movement?

A) Normal movement is dominant and paralysis is recessive. B) The F1 offspring must be heterozygous since the parent genotypes are MM and mm. 25% of the F2 worms will be paralyzed. C) Perhaps the sluggish worms are the heterozygotes. The concept of incomplete or partial dominance could explain these results. In incomplete dominance the heterozygous phenotype is in between the dominant and recessive phenotypes.

Mitosis and Meiosis are both processes of eukaryotic cell division, with the same names used to describe different stages. A) What are the significant cellular and chromosomal events that occur during each stage of mitosis? B) What are the similarities and differences between the events that occur during each stage of mitosis and the corresponding stage of meiosis I? C) What are the similarities and differences between the events that occur during each stage of mitosis and the corresponding stage of meiosis II?

A) Prophase: Chromosomes condense and spindles begin to form. Metaphase: Microtubules attached to the Kinetochore begin to pull at the chromosomes creating tension that lines up the chromosomes along the center of the cell. Anaphase: Cohesin releases and the sister chromatids are pulled towards opposite poles of the cell. Telophase/Cytokinesis: The cell cleaves down the middle to form two new cells as the nuclear envelope starts to reform in each cell and the chromosomes decondense. B) Prophase I includes bouquet formation and synapsis of homologues, unlike mitosis prophase. The synapsed chromosomes then cross over and recombine at the end of prophase I. During metaphase I, the synapsed chromosomes line up along the metaphase plate, with the tension from the spindle poles holding them in place. The main difference between metaphase in meiosis I and metaphase in mitosis is that homologous pairs line up along the center of the cell, rather than sister chromatids. In anaphase, the homologous chromosomes split apart from one another, instead of the sister chromatids splitting apart in mitosis. The separate cells that are formed during telophase I will be haploid, but will have two sister chromatids for each homologue—unlike a cell resulting from mitosis which will be diploid and have one copy of each chromatid. C) Meiosis II is very similar to mitosis, except that the meiosis II occurs with half of the chromosomes, resulting in cells that are haploid instead of diploid.

Figure Q5-2 shows a pedigree for a rare trait in humans. Affected individuals are indicated by filled in symbols. The woman II2 is pregnant, with II3 being the father. Her unborn child is represented by the ?. A) Is this trait inherited as a dominant or recessive trait? What is the best evidence for your answer? B) What is the probability that both II2 and II3 are heterozygous for the trait? C) what is the overall probability that the unborn child will be unaffected by the trait and a girl?

A) Recessive, Individual II1 is evidence because the trait must be recessive for two unaffected individuals to give birth to an affected individual. B) II3 must be heterozygous (Aa) since I3 is affected (aa). Since I1 and I2 are both carriers there is a 67% chance that II2 is heterozygous. It is not one half since we know that II2 is not affected and has the genotype A_. P(both are heterozygous) = 1 *0.67 = 0.67. C) There are numerous combinations that would result in an unaffected child so it is simpler to calculate the probability that the child will be affected. The child can only be affected if both parents are heterozygotes, and we found in part b that the probability of that is 0.67. Even if both parents are heterozygotes, the probability that the child will be affected is one fourth so that the total probability that the child will be affected is .67 x .25 x 1/6. So that the probability that the child will not be affected is 5/6. The probability that the unborn child is a girl is one-half. So the probability of the unborn child will be unaffected by the trait ΝΔ girl will be 5/6 x 1/2 or 5/12.

The Rh blood factor is a single gene trait with Rh negative recessive to Rh positive. Two Rh positive parents have an Rh negative child. A) What must be their genotypes? B) What is the probability that their next child will be Rh positive?

A) Rh+/Rh- and Rh+/Rh- B) 0.75

In many mammals, including rabbits, there is a hair texture known as angora, in which the hair is long and soft. Angora rabbits are highly prized for the quality of their fur. A breeder has four rabbits: angora male, angora female, a short-haired male, and a short-haired female. He made the following crosses with the outcomes shown. Angora female x Angora male = all Angora Angora female x short male = all short Short female x Angora male = 4 short 5 angora A) Which phenotype is dominant? B) What are the genotypes of each of the parents? C) If the short female and the short male were mated to each other, what fraction of the offspring is expected to have short hair? D) Even for experts, the sex of rabbits is very difficult to determine but the breeder wants to keep the males and females separate as much as possible. When the mating in part C is performed, a litter of 5 rabbits is born. What is the probability that at least one of the five is female?

A) Short hair - note that when the angora and the short are mated, the offspring have short hair. B) Angora female x Angora male : ss and ss , Angora female x short male : ss and SS , short female x angora male : Ss and ss C) Short female (Ss) and short male (SS) results in 100% offspring with short hair. D) The easiest way to determine the probability that at least one is a female is to calculate the probability that all the rabbits are males and then subtract from 1. (.05)^5 = .03125 (1-.03215) * 100 = 96.875%

Factor VIII and factor IX are two X linked genes in humans involved in blood clotting, and mutations in these genes result in hemophilia a or hemophilia b. Originally it was not known which form affected the royal families but the inheritance pattern is the same in either case. Use figure 7-6 to answer the following questions. A) Queen Victoria's daughter Alice had 2 daughters, Irene and Alexandra. Both of these daughters were carriers for hemophilia, as seen from their children. What is the probability that the two daughters born to Alice would both be carriers of the disease? B) Note that Irene married her first cousin Henry of Prussia. Normally, the marriage of first cousins results in a higher occurrence of recessive disorders. However, that is irrelevant to the explanation of the occurrence of hemophilia in this family. Why is this information on the family relationship not important? C) Irene's cousin Alice had a hemophiliac son, Viscount Tremation, as well a daughter Mary, and a son who died as a newborn. What is the probability that this son was hemophiliac?

A) Since Alice is a carrier, there is a 50% chance that her daughter will inherit the mutant allele for hemophilia. Since this is true of each daughter, the probability of both daughters being carriers is (0.5 * 0.5) = 0.25. B) Normally, marriage between cousins increases the chances of recessive disorders showing up in the offspring because being related increases the chance that both will be carriers for the same recessive genetic diseases. This increases the chance that their children will be homozygous for a recessive disease. However, men who have the X linked gene for hemophilia will have the disease. This means that a healthy male does not have a recessive copy of the gene. Inheritance of the disorder is completely dependent on the mother. It doesn't matter how closely related the male is, just whether or not he has hemophilia. C) 50%

In tomatoes, the shape of the leaf called potato is recessive to a leaf shape called cut. A true-breeding variety called Mortgage Lifter with cut leaves is crossed to a true-breeding variety called Hillbilly with potato leaves, and seeds are collected. The F1 plants are grown. A) What is the genotype of these F1 plants and what is the phenotype of their leaves? B) The F1 plants are self-fertilized and their seeds are collected These seeds are planted to create and F2 generation. What fraction of the F2 plants are expected to have potato leaves? C) Approximately how many seeds have to be planted to be confident (at a 95% level) that at least one seedling has a potato leaf?

A) Since each plant is true-breeding the genotypes of the parents can be inferred to be CC (Mortgage Lifter - Cut) and cc (Hillbilly - Potato). A Punnett square will show that all of the F1 plants are heterozygous (Cc) and thus have cut leaves, the dominant phenotype. B) 25% of the F2 plants are expected to have potato leaves (cc). A Punnett square will show 75% cut leaves ( 25% homozygous dominant and 50% heterozygous dominant). C) If one seed is planted there is a 25% chance that it will have a potato leaf and 75% chance that it will be a non-potato leaf. If two seeds are planted, there is .75 x .75 that neither of them will have a potato leaf. So the general formula is the probability of not having a potato leaf raised to the power of the number of seeds. (.75)^11 = .0422; If 11 seeds are planted then there is a less than 5% chance (4.22% to be precise) that all of the seedlings have cut leaves. This means that there is a 95.78% chance that at least one seedling has a potato leaf. A very crude rule of thumb used by geneticists in the lab is to figure out the probability of something not happening, and look at three times as many offspring to be sure that you find one.

A gene known as Lon-2 is X linked in c elegans; mutations in Lon-2 result in worms that are unusually long. A Lon-2 mutant male was mated to a wild type hermaphrodite. A) What will be the cross progeny of this mating if Lon-2 is recessive? B) What will be the cross progeny of this mating if Lon-2 is dominant? C) In fact, Lon-2 is recessive. One of the F1 hermaphrodites is mated with wild type males. What are the expected progeny of the offspring of this progeny among the offspring of this mating, and in what proportions they will be found.

A) The hermaphrodite would be Xl 0 and the male would be XL XL therefore 100% of the progeny (male and female) will be normal length. B) The hermaphrodite would be XL 0 and the male would be Xl Xl therefore 100% of the females will be long and 100% of the males will be normal length. C) The hermaphrodite would be XL 0 and the male would be XL Xl therefore 100% of the females will be normal length and 50% of the males will be long.

A couple has a child with trisomy-21, that is, with three copies of chromosome 21 rather than two. Geneticists analyzed chromosome 21 of both parents and the child in order to understand the origin of this nondisjunction event. Some of the results are shown in Figure QB-4. A) The gel at the top of the figure shows a polymorphic locus on chromosome 21 referred to as a copy number variation (CNV). At this locus, different individuals have a different number of copies of repeated sequence; a higher copy number results in a larger band (closer to the top of the gel) than a low copy number. The locus itself has nothing to do with meiosis; it is simply a marker to track the parents' and child's chromosomes. Based on this gel, which parent experienced the non disjunction and how can you tell? B) The bottom part of the figure shows two different SNPs at different loci on chromosome 21 in each parent and the child. The line is a chromosome from each parent, with the dashed line indicating that the two loci are far apart on chromosome 21. The chromosomes are arranged in the same order as on the helm so that the top chromosome shown on the gel (in the mother and the child) has the A at the first locus and the C at the second locus. What does this sequence information tell you about the "cause" of the nondisjunction event?

A) The mother of this child experienced nondisjunction during meiosis because two of the three bands on the child's gel match up with the mother and only one matches with the father. B) Crossing over did not occur between the mother's chromosomes.

The ability to taste bitterness in certain foods such as Brussel sprouts depends on the T allele, which is dominant to non-tasting (tt). A couple has a daughter and a son. Both parents can taste bitter but their daughter cannot taste bitter. A) What is the probability that their son can taste bitter foods? B) How does the phenotype of the daughter help you answer this question? C) When he is tested, their son can taste bitter foods. He marries a woman who cannot taste bitter foods, and they have a son. What is the probability that their son will be able to taste bitter foods?

A) The parents' genotype are both Tt. Therefore, the probability that their son can taste bitter foods is 0.75. B) The fact that their daughter is a non-taster (tt) means that each parent must be a carrier of the non-taster allele (t). Since the parents are both taster their genotypes must each be Tt. C) The son's wife's genotype is tt. The son could have the genotype Tt or TT. There is a 67% chance that he has the genotype Tt and a 33% chance that he has the genotype TT. If the son's genotype is TT then there is a 100% Probability that his son will be able to taste bitter foods. If the son's genotype is Tt then there is a 50% chance that his son will be a taster. We can combine this to find that (0.33*1)+(.67*.5) = 0.67. The reason that the son has a 67% probability of being Tt is that we know he is not tt; of those who can taste, two-thirds are heterozygotes.

The woman II2 shown in the pedigree of Q6-5 Panel A is pregnant with her first child. Both her family and her husband have a sister who is affected by an autosomal recessive trait as shown. A) What is the probability that their first child will not be affected by this trait?

A) The probability is 8/9. It is simpler to work out the probability that the child will be affected. That can only occur if both parents are heterozygotes. There is a 2 in 3 chance that either parent is heterozygous for the trait as both parents have parents that are heterozygous for the trait but are not themselves affected. The chance that both parents are heterozygous for the trait is 4 in 9. If they are both heterozygous, there is a 1 in 4 chance that their child will be homozygous recessive, which makes the probability that both parents are heterozygous and their child is homozygous recessive 4/36 or 1/9. Therefore, the chance that the child does not have the trait is 8/9.

Some parakeets and other birds of the parrot family have a phenotype known as lutino. Lutino is a much prized beautiful coloration pattern. A) A lutino male parakeet is mated with a normal green female. The female offspring are all lutino, while the male offspring are all green. Explain this result. B) One of the male offspring in part A is mated to a lutino female. What are the expected outcome of this mating?

A) The trait lutino is Z linked recessive while green is dominant. (Males are ZZ and females are ZW in birds.) Male parents were Zl/Zl and female parents were ZL/W. B) In this cross, the male was ZL/Zl and the female was Zl/W. In this cross, 50% of all offspring will be lutino and 50% will be green.

7.11 The pedigree below is an example of an X linked recessive trait. A) Could this inheritance pattern occur if the trait were autosomal? If so, what additional assumptions would need to be made to explain this inheritance pattern? B) Could this inheritance pattern occur if the trait were X linked but dominant? C) For each of the following individuals, calculate the probability that their first son will be affected by the trait. I. III3 II. III5 III. III9 IV. III10

A) This pattern could occur if the trait were autosomal recessive but the trait would have to be assumed to be a common one, since II1, II3, and II7, all of whom married into the family, would have to be carriers. Therefore, X linked recessive inheritance is a much better explanation of this pedigree. B) No, I2, II4, II7, and III2 are all carriers of the trait. If the trait were dominant then it would be expressed in all of those individuals. C) I. 50% II. 25%, Her mother II4 is heterozygous and her father is unaffected so she could have gotten either of her mother's X chromosomes. There is a 50% probability that she inherited the trait and a 50% probability that she will pass it on to her son. III. 0%, assuming that the trait is rare (so that II6 is not a carrier) then the chance that III9's son will be affected is 0%. IV. 0%, assuming that the trait is rare there is a 0% chance that III10's son will be affected (remember that males pass on their Y chromosome, not their X chromosome to their male offspring)

A mutant cucumber plant has flowers that fail to open when mature. A cross between the closed flower plant and a wild type open flower plant resulted in F1 progeny with open flowers. A cross between F1 flowers resulted in F2 plants, 145 of which were open and 59 closed. A cross between F1 plant and a closed plant resulted in 81 open and 77 closed progeny. Draw the Punnett squares. How is the closed trait inherited?

Autosomal Recessive

In order for the lengths of the branches on a tree to represent the passage of time between ancestor and descendant, A) bootstrap values must be >95% B) the substitution rate must be known C) highly conserved genes must be used D) all of the above

B. Branch length can correspond to the passage of time

Which statement regarding crossovers is FALSE? A) They're required for proper segregation of homologous at meiosis I. B) They're initiated by random double-stranded breaks. C) They occur between 2 of the 4 chromatids in a tetrad. D) They're important for genetic diversity.

B. Cross over is highly regulated and only occurs in specific spots.

Consider the two possible ways of aligning the two sequences below. If a scoring matrix that takes the differing probabilities of transitions and transgressions into account is used, which alignment will get the higher score. A. ACCTTATACC B. ACCTTATACC ACCTTC. ACC ACCTT CACC

B. G and A are purines. C and T are pyrimidines. Therefore A is a transversion and B is a translation which will be a less negative number since it is more common.

When sequences have been evolving long enough that some sites have accumulated more than one substitution, how does this affect the estimate of when two sequences last shared a common ancestor? A) they cause divergence time to be overestimated B) they cause sequences to appear to be more distantly related than they actually are C) they cause divergence time to be underestimated D) Estimates of divergence times are independence of substitution rates, so substitutions have no effect on them

C, lecture 4

If the numbers next to the nodes on the tree shown below are bootstrap values, what can be assumed about the node with a value of 87? A) The cluster connected by that point last shared a common ancestor ~87 mya B) 87 is the number of substitutions per site C) 87% of trees built from pseudoreplicates of the data set show this same cluster D) The sequences in that cluster are on average 87% identical

C, see lecture 4 slide 35

The rate of evolution of a species refers to... A) the rate at which de novo mutations occur B) the rate of progress from simpler to more complex features C) the rate at which mutations are fixed in somatic cells D) the rate at which substitutions accumulate in the germline

D. The germ line is successive generations aka gametes

True or false: In a cross between 2 heterozygotes, the ratio of genotypes is the same as the ratio of phenotypes.

False, Punnett square.

True or false: Red-green color blindness is more likely to occur in males than females because the opsin genes that encode light receptors are on the Y chromosome in humans.

False, X-linked recessive disorders

As shown with calico cats, female mammals that are heterozygous for an X linked gene have some cells that express one allele and some that express the other allele. However, this difference between two expression patterns does not make a difference in the phenotypes for most X-linked traits. In other words, calico cats have a very striking phenotype, but this pattern of cell differences is not seen for most other X linked traits. What are some of the possible explanations for the lack of difference for most X linked traits?

For some traits, the gene products are exported from the cell. For example, in a carrier for hemophilia, as long as the blood clotting protein is expressed in some cells, the organism will not show the disease. In addition, skin cells that produce color ( the melanocytes ) do not migrate or mix very much during development after they form so that the cells with the same inactivated X chromosomes are clustered near each other in the skin. In some cases, the functions of genes on the X chromosome are duplicated in other genes on the chromosome.

Define the difference between heterogametic and homogametic sexes. Give an example of a species in which the male is heterogametic and one in which the male is homogametic.

Heterogametic sexes have two different sex chromosomes while homogametic sexes have duplicates of the same sex chromosome. In humans males are heterogametic (XY) and females are homogametic (XX). In birds, males are homogametic (ZZ), while females are heterogametic (ZW).

Why is inheritance of X-linked recessive traits sometimes called crisscross inheritance?

If a female that expresses an X-linked recessive trait is mated to a male that does not express an X-linked recessive trait then the male offspring will express the trait and the females will not, as shown in a Punnett square.

Why are male calico cats both rare and usually sterile?

In order for a cat to be calico it must be heterozygous for the B allele, which is on the X chromosome. Therefore, the cat must have two X chromosomes. In order for a male to be heterozygous for an X-linked trait, he has to have the genotype XXY. These males are not common because they are the result of nondisjunction of the X chromosome. Males of this genotype will be sterile because they have 3 sex chromosomes.

A parent pea plant with genotype Rr is equally as likely to produce an r gamete as an R gamete, due to which of Mendel's laws?

Law of Segregation

What is the difference between midpoint rooting and outgroup rooting of a phylogenetic tree?

Midpoint roots are placed between the 2 tips with the longest distance between them. Outgroup roots are a taxon known to be more distantly related to the others than they are to each other is added to the alignment to give the directionality.

What would be expected of a Y chromosome with large deletions at each end?

No crossover would occur

7.22 You identify a new species of fish, which clearly has 2 separate sexes, male and female. However, all of the chromosome pairs appear identical in the two sexes so it is not apparent if sex determination depends on an X/Y system, a Z/W system, or a single mating type locus. All of these systems of sex determination are known to occur in some species of fish, so any of them could be at work in this new species. You have found a locus that might be able to distinguish these possibilities. Using PCR, you amplify this sequence from a male and a female, and from two of their offspring, one female and one male. You separate the PCR products on an agarose gel with the results shown below. What is the most likely form of sex determination in this fish?

Notice that this is a DNA sequence rather than a gene so it may or may not be coding for a protein. The male parent is heterozygous while the female parent is either homozygous (if she has 2 sex chromosomes) or hemizygous (if she has a single sex chromosome). The male offspring is not informative since no matter what system is involved, it will get one chromosome from the father and one from the mother, which is the result seen. The key is the female offspring, which has only one band at the same time as the one from her father; she has apparently not inherited the locus from her mother at all. These results are most easily explained if females are the heterogametic sex.

What molecular sequence did Carl Woese analyze to generate his tree showing that archaea form a separate lineage from bacteria?

Ribosomal RNA

A third grader decided to breed guinea pigs for her science project. She went to a pet store and bought a male with smooth black fur and a female with rough white fur. She wanted to study the inheritance patterns of these traits. A cross between these two parents resulted in two litters with a total of 15 F1 progeny, all of which were rough and black. Soon, the F1 litters began to produce F2 progeny, and before long there were 125 F2 guinea pigs. 8 were smooth and white, 25 were smooth and black, 23 were rough and white, 69 were rough and black. Draw the Punnett square for the cross between the rough black F1s. How are the coat color and texture traits inherited? What phenotypes and proportions of offspring should the girl expect if she mates one of the smooth white F2 females with an F1 male?

Rough and black is dominant. It is autosomal. Black and rough are 9, black and smooth are 3, white and rough are 3, white and smooth are 1.

Every eukaryote that has been studied has an orthodox of SPO11, but it is very difficult to obtain and maintain a mutation that knocks out the function of SPO11. What do you predict is the phenotype of a SPO11 mutant that will make it so hard to maintain and why?

SPO11 is the enzyme that creates double stranded breaks in the homologous chromosomes during prophase I of meiosis to facilitate crossing over. The homologous chromosomes for an SPO11 mutant would therefore be unable to crossover, so nondisjunction would occur for every chromosome. SPO11 mutants would therefore be unable to produce offspring. It would be difficult to maintain a strain of SPO11 mutants since they will be unable to produce offspring.

While cleaning out the storage room in the basement of a biology department you encounter a box of microscope slides. Most of the labels are illegible, but you can read that the slides are taken from the ovaries of some animal that only has two pairs of chromosomes (n=2). The slides appear to be stained with some fluorescent dye that allows you yo see the chromosomes. Three of the slides are shown in Figure Q6-1. What stage of meiosis is shown in each slide and how can you tell?

Slide A is the beginning of Meiosis II because there is only one chromosome of each type per cell, and not two like in Meiosis I. Slide B is in the middle of Meiosis I, either at the end of Prophase I or the beginning of Metaphase I. The multiple copies of chromosomes indicates this slide must be in Meiosis I, and the synapsing of the chromosomes means that this stage is before the chromosomes are pulled apart, but after the chromatin has condensed. Slide C is close to the end of Meiosis II. This cell only has one chromatid of each type, indicating that this is after each chromosome has been split, but before the decondensation of the DNA.

What are two important differences between the processes of spermatogenesis and oogenesis in animals?

Spermatogenesis results in four viable sperm being generated from one primary spermocyte. Oogenesis produces cells of different sizes: three polar bodies and one (much larger) egg cella's the cytoplasm is not devided evenly. All oocytes are made during gestation and then pause during Prophase I in meitiotic (dictyate) arrest until puberty. When ovulation occurs during the reproductive years, the primary oocytes complete meiosis I and then arrest again at metaphase of meiosis II, which is only completed if the egg is fertilized.

See Question 20 on list.

Taa (+1) aag (-1) GCA to CGT

Compare the orientation of the Kinetochore during mitosis with the orientations during meiosis I and meiosis II. How does this change in orientation affect these divisions?

The Kinetochore has an inner and outer face and the microtubules of the spindle will only attach to the outer face of the Kinetochore. The inner side of the Kinetochore is associated with the DNA of the centromere. In meiosis I, the Kinetochores of the sister chromatids are facing the same direction, and the kinetochores of homologous chromosomes are facing opposite directions. This means that the sister chromatids get pulled together, but the homologous chromosomes get separated. In meiosis II, the kinetochores of the sister chromatids are oriented in opposite directions so that mircrotubules can attach one either side and pull them apart towards the poles.

The concept of "wild-type" is not typically applied to most human traits, but there are some situations in which the idea of a wild-type is helpful in human genetics even if the term is not used. What might be such a situation when the concept of a wild-type is useful from human genetics?

The concept of "wild-type" can be useful when looking at genetic diseases, since the "unaffected" phenotype could be considered wild-type and the diseased phenotype would be considered the mutant phenotype. The wild-type concept is useful in this situation since the "unaffected" phenotype is common while the "affected" phenotype is rare.

Wrinkled peas arise from a mutation that inactivated a gene that encodes an enzyme involved in the production of starch. Why should a mutation that inactivated the gene be recessive? What does this tell you about how much of the enzyme is needed by the plant?

The fact that the gene inactivating mutation is recessive suggests that the plant only needs the amount of enzyme produced by one allele to produce enough starch for round peas. Knocking out only one allele will have no affect on the phenotype of the plant because adequate amount of starch is still being produced by the other allele. However, if both alleles are knocked out (meaning that the plant is homozygous recessive) then no starch will be produced and the peas will be wrinkled. It is very common for inactivating mutations to be recessive because often the product of only one allele is require for a wild-type phenotype. Relatively few genes are known to have a dose-dependent morphological phenotype when one vs. two wild-type copies are present.

A calico cat mates with a black male cat. What phenotypic ratios are expected among her offspring? (Both color and sex are relevant phenotypes for this question.)

The male is XB/Y and the female is XB/Xb. The males will be 50% black and 50% orange. The females will be 50% black and 50% calico.

In certain games that use dice, rolling doubles gives an advantage to the player such as an extra turn. In any given roll, what is the probability of rolling doubles of any number?

There are 36 total possible combinations when rolling two dice and 6 ways to get doubles. Therefore the probability of getting doubles on any given roll is 6/36 = 0.166.

It is easier to establish true-breeding lines for recessive traits than for dominant traits. Why?

There are two possible genotypes for the dominant phenotype: homozygous dominant (AA) and heterozygous dominant (Aa). Therefore, the genotype Aa has the same appearance as the genotype of a true-breeding line AA. For recessive traits, the only possible genotype for the recessive phenotype is aa, and all of the offspring will also be aa.

The pedigree in question 39 is for a rare autosomal allele that causes severe body odor when homozygous. Individual IIw has recently learned she's pregnant and asks you, a genetic counselor, to tell her the probability that the baby will be a daughter and have the trait ( she doesn't care if a son has the trait). What probability do you give her, assuming I4 is not a carrier?

There is a 1/2 chance the baby is a girl, a 2/3 chance the mother is a carrier and a 1/4 chance that the baby is homozygous recessive. Therefore, 2/24.

The small probability for error during replication can result in mutations that change the function of a gene. Certain mutations in the E. coli rpsL gene result in the phenotype of resistance to the antibiotic streptomycin. Similarly, specific mutations in the rpoS gene yield rifampicin-resistant varieties. The antibiotics have distinct mechanisms of action, so that a bacterial cell might be resistant to one, both, or neither. When 10^8 of a certain strain of mismatch repair deficient E. coli bacteria from a rapidly dividing culture are plated onto nutrient agar plates containing streptomycin, 119 resistant colonies ( each arising from a single cell) are obtained. When the same number of bacteria are plated out on plates containing rifampicin, 35 colonies are counted. What is the probability of obtaining a colony resistant to both drugs?

These are independent events so the probability of a double mutant is the product of their independent probabilities. (119/10^8)*(35/10^8) = 4.165*10^-13

7.23 The pedigree below shows the inheritance of a rare trait in humans. What is the probability that the individual shown by the "?" will not be affected by this trait?

This trait is not X linked, because the affected female is the daughter of an unaffected father, and it is not dominant because two unaffected parents produced an affected child. The trait must be autosomal recessive. The probability that the father of the child (?) is a carrier is 2/3, since we know he is unaffected. The probability that the mother of the child is a carrier is 1, since she must have inherited a recessive allele from her affected father. If both parents are carriers, there is a 1/4 that they will an affected offspring. Therefore, the probability of the child inheriting the trait is 2/3 x 1 x 1/4 = 1/6.

What is the most common underlying cause of trisomy 21 in humans?

Trisomy 21 is usually caused by a failure to cross over in the mother. This means that the homologous chromosome 21 pair can't be held at the metaphase plate, resulting in non-disjunction, and an egg cell with two copies of this chromosome instead of one. The disomic egg is fertilized by sperm with one copy of chromosome 21 to produce an embryo with three copies of chromosome 21.

True or false: Nondisjunction during meiosis results in nullisomic and disomic gametes.

True

True or false: During meiosis I, the Kinetochores of the sister chromatids from one homologous face in the same direction.

True, slide 25 lecture 2

You have a batch of round peas and you want to know their genotype. What do you cross them with to obtain the clearest results?

Wrinkled peas

A first year graduate student attempted to run a PCR reaction but failed to get the product he wanted. Look at the ingredients of his reaction tube and impress him by pointing out what he forgot to add - Buffer, DNA polymerase, Forward primer, dNTPs.

ddNTPs to terminate the sequence.