Genetics The List 2 END OF CHAPTER

5.23 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. b. What is the probability that their next child will be Rh-positive?

0.75

7.21 e. For the most likely mode of inheritance in each pedigree, calculate the likelihood that child #2 will not be affected by the trait. Pedigree A

100%

7.21 e. For the most likely mode of inheritance in each pedigree, calculate the likelihood that child #2 will not be affected by the trait. Pedigree C

100%

5.15 Angora female × angora male = all angora Angora female × short male = all short Short female × angora male = 4 short and 5 angora 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?

100% (genotypes: 50% SS and 50% Ss)

7.14A 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. b. What will be the cross-progeny of this mating if lon-2 is dominant?

100% of the females will be long and 100% of the males will be normal length.

7.14A 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. c. In fact, lon-2 is recessive. One of the F1 hermaphrodites is mated with wild-type males. What are the expected progeny among the offspring of this mating, and in what proportions will they be found.

100% of the females will be normal length and 50% of the males will be long.

7.14 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?

100% of the progeny (male and female) will be normal length.

7.21 e. For the most likely mode of inheritance in each pedigree, calculate the likelihood that child #2 will not be affected by the trait. Pedigree 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.

5.14 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. These F1 plants are grown. c. Approximately how many seeds have to be planted in order to be confident (at a 95% level) that at least one seedling has a potato leaf?

11- Think of it this way: if one seed is planted, there is a 25% chance that it will have a potato leaf and 75% chance that it will be non-potato leaf. If two seeds are planted, there is 0.75 x 0.75 (or 0.752) that neither of them will have a potato leaf. So the general formula is the probability of not having a potato leaf raised the power of the number of seeds. (0.75)11 = 0.0422; If 11 seeds are planted then there is a less than 5% chance (4.22% to be precise) that at 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 (but generally accurate) 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.

5.19 A family has three children, none of whom are twins. 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?

2(0.5*0.5) = 0.5. The next two children could be either a girl then a boy, or a boy followed by a girl; each of those has a probability of 0.25.

5.14 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. These F1 plants are grown. b. The F1 plants are self-fertilized, and their seeds are collected. These seeds are planted to create an F2 generation. What fraction of the F2 plants are expected to have potato leaves?

25% of the F2 plants are expected to have potato leaves (cc).

7.11 c. For each of the following individuals, calculate the probability that their first son will be affected by the trait. ii. III-5

25%. Her mother II-4 is heterozygous (and her father II-3 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.

5.19 A family has three children, none of whom are twins. a. What is the probability that they have two daughters and a son?

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 one-eighth.

7.11 c. For each of the following individuals, calculate the probability that their first son will be affected by the trait. i. III-3

50%

7.19 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. c. Irene's cousin Alice had a hemophiliac son, Viscount Tremation, as well as a daughter Mary, and a son who died as a newborn. What is the probability that this son was hemophiliac? (This is not known.)

50%

7.15 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 cream-colored male. b. What are the expected results if cream is autosomal and recessive?

50% cream, 50% wild-type gray-brown

7.11 c. For each of the following individuals, calculate the probability that their first son will be affected by the trait. iv. III-10

Again, assuming that the trait is rare, there is a 0% chance that III-10's son will be affected (remember that males pass on their Y-chromosome, not their X-chromosome to male offspring).

5.1 Define the following terms, and discuss how they are used in this chapter. 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.

6.1 Define the following terms as they are used in this chapter. f. Synaptonemal complex

A ladder-like structure that forms between two homologues as they synapse and has an evolutionarily conserved width.

5.1 Define the following terms, and discuss how they are used in this chapter. a. Locus

A locus refers to the location of a gene on the chromosome, either from a physical or genetic map. The term is also often used as a general term for a gene.

6.1 Define the following terms as they are used in this chapter. g. SPO11

An evolutionarily conserved enzyme among eukaryotes that creates double-stranded breaks for crossing over.

5.15 Angora female × angora male = all angora Angora female × short male = all short Short female × angora male = 4 short and 5 angora b. What are the genotypes of each of the parents?

Angora female X angora male: ss and ss Angora female X short male: ss and SS Short female X angora male: Ss and ss

7.11 c. For each of the following individuals, calculate the probability that their first son will be affected by the trait. iii. III-9

Assuming the trait is rare (so that II-6 is not a carrier) then the chance that III-9's son will be affected is 0%

6.15 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 non-disjunction event. b. The bottom part of the figure shows two different single-nucleotide polymorphisms (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 gel, 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 non-disjunction event?

Crossing over did not occur between the mother's chromosomes.

7.13A Drosophila female that is missing cross veins on the wing is mated to a male with normal wings. The female F1 offspring 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?

Crossveinless is X-linked and recessive.

5.1 Define the following terms, and discuss how they are used in this chapter. b. Diploid

Diploid means having two copies of every chromosome (one inherited from each parent).

5.21 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 "?" c. Is this trait inherited as a dominant trait or a recessive trait? What is the best evidence for your answer?

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 affected individuals are expected to have an affected parent.

7.18 (Challenging) As was 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 the 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) so not migrate or mix very much during development after they form so that the cells with the same inactivated X chromosome 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 other chromosomes.

7.2 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).

5.21 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 I-2?

I-2 is the maternal grandmother of the unborn child

5.21 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 "?" d. What is the probability that the unborn child will be a boy and not affected by the trait?

II-2 is aa, while II-3 is Aa. (genotypes: Aa, Aa, aa, and aa) So have of their children are expected to be affected by the 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 one-fourth.

5.22 Figure Q5.2 shows a pedigree for a rare trait in humans. Affected individuals are indicated by filled-in symbols. The woman II-2 is pregnant, with II-3 being the father. Her unborn child is represented by the "?". b. What is the probability that both II-2 and II-3 are heterozygous for the trait?

II-3 must be heterozygous (Aa) since I-3 is affected (aa). Since I-1 and I-2 are both carriers there is a 67% chance that II-2 is heterozygous. It is not one-half since we know that II-2 is not affected and has the genotype A_. P(both are heterozygous) = 1*0.67 = 0.67.

5.21 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 "?" b. What is the relationship between the unborn child and individual II-4?

II-4 is the aunt of the unborn child.

7.3 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 an express an X-linked recessive trait then the male offspring will express the trait and the females will not, as shown in this Punnett square.

5.18 The nematode Caenorhabditis 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 allowed to self-fertilize to produce an F2 offspring are capable of movement. One of the F1 generation. d. Test your hypothesis about the results in (c) by using a χ2 test.

If the incomplete dominance hypothesis is correct, then any worms that have Mm genotype should move sluggishly. Therefore, the following inheritance would be expected for a population of 278 worms: MM (normal movement): 69.5 (25% ) Mm (sluggish): 139 (50%) Mm (paralyzed): 69.5 (25%) Chi Square = (observed-expected)^2/expected MM: (73-69.5)2/69.5 = 0.1763 Mm: (139-139) 2/139 = 0 mm: (66-69.5) 2/69.5 = 0.1763 Chi Square = 0.1763 + 0 +0.1763 = 0.3526 There are 2 degrees of freedom; at a probability (p-value) of 0.05, the value of the chi-squared statistic is 5.991. Since the chi-squared value calculated is less than 5.991, the chance of seeing this much deviation from expected is greater than 0.05 (5%). The probability of seeing the observed deviation is between 0.8-0.9. Therefore, the hypothesis that the sluggish worms are the heterozygotes cannot be rejected.

7.16 Some parakeets and other birds of the parrot family have a phenotype known as lutino. Lutino is a much-prized beautiful coloration pattern in which the feathers on the body and head are yellow, with silver patches on the cheeks, and light yellow tail feathers. (These birds are pretty enough that you might want to look at a picture of them on-line.) The birds also have red or pink irises in the eyes rather than the standard black irises. b. One of the male offspring in part A is mated to a lutino female. What are the expected outcomes of this mating?

In this cross the male was ZL/Zl and female was Zl/W.

4.22 1. A still on-going cholera outbreak in Haiti began after an earthquake. This is the first cholera outbreak in Hispaniola (the island comprised of Haiti and the Dominican Republic) in over a hundred years. One hypothesis is that the earthquake caused ocean perturbations that allowed cholera strains residing close to other North and South American shores to reach Haiti. An opposing hypothesis is that Vibrio cholerae was accidentally brought to Haiti by UN peace keepers flown in at short notice from cholera endemic countries in Asia. The second hypothesis caused the victims to blame and attack UN personnel. To determine which hypothesis is most likely to be correct, scientists decided to compare the sequences of Haitian isolates with those from isolates from other parts of the world. The investigators identified and sequenced 1588 genes that were present in a selection of pandemic V. cholerae strains including the Haiti 2010 outbreak strains. They used the resulting nucleotide sequence to produce the phylogenetic tree shown below. The year and country of isolation were noted for each strain. Very closely clustered branches are magnified so that you can read the text. (The highlighted boxes are of no significance, they are strains selected by the investigators for other studies). The phylogenetic tree is rooted with three pre-1920 strains that are not shown in this figure because they are very distantly related to the strains shown. a. Explain why the investigators thought it necessary to root the tree with strains isolated prior to 1920 even though the strain of interest was isolated in 2010.

Investigators were interested in the evolutionary relationships between the Haitian cholera strain and strains all over the world. To get the relationships between these things, you need a common ancestor. Therefore, looking back as far as the 1920s is a good idea, since that will be most likely to give you a common ancestor. The root of a phylogenetic tree should be old enough that it is a common ancestor to all of the organisms being compared on the tree so that it can be used to determine the order of evolution of the branches of the tree. From there, you can see how each strain evolved from the common ancestor. Trees show evolutionary time and evolutionary relationships, and can tell you which strains are most closely related.

6.2 Mitosis and meiosis are both processes of eukaryotic cell division, with the same names used to describe the different stages. c. What are the similarities and differences between the events that occur during each stage of mitosis and the corresponding stage of meiosis II?

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.

7.24 (Looking ahead). 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 (Lon), while mutations in the unc-18 gene result in worms that are Uncoordinated (Unc). 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 hermaphrodites is mated to a wild-type male.What are the expected male progeny from this cross, and in what proportions will they arise?

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. This is one of the main topics in Chapter 9.

7.25 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 red 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 others with no reaction, with no intermediate phenotypes.

Males are hemizygous for all alleles on the X chromosome. Therefore there 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.

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

Males: 50% black, 50% orange Females: 50% black, 50% calico

7.15 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 cream-colored male. a. What are the expected results if cream is X-linked and recessive?

Males: 50% cream, 50% wild-type gray-brown Females: 50% cream, 50% wild-type gray-brown

7.13A Drosophila female that is missing cross veins on the wing is mated to a male with normal wings. The female F1 offspring have normal wings, but the male F1 offspring have crossveinless wings. b. One of these F1 females is mated to a normal male. What are the expected results among the F2 offspring?

Males: 50% missing cross veins Females: 100% normal

7.11 b. Could this inheritance pattern occur if the trait were X-linked but dominant

No. I-2, II-4, II-7, and III-2 are all carriers of the trait. If the trait were dominant then it would be expressed in all of those individuals.

6.16 C. elegans has two sexes—males and hermaphrodites. Hermaphrodites are essentially females that 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.

Non-disjunction 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, making the fertilized egg male.

5.18 The nematode Caenorhabditis 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 allowed to self-fertilize to produce an F2 offspring are capable of movement. One of the F1 generation. a. Which allele is dominant, and which is recessive?

Normal movement is dominant.

6.16 C. elegans has two sexes—males and hermaphrodites. Hermaphrodites are essentially females that 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. b. Mutations that result in non-disjunction 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 one male in 500 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.

Normal segregation of the chromosomes is the outcome after all of the steps of meiosis, so mutations that affect meiosis are likely to cause non-disjunction. If the chromosomes are not being segregated properly this is likely to create gametes that are missing an X chromosome, which will produce X0 male offspring if fertilized.

7.19 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. 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?

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.

6.16 C. elegans has two sexes—males and hermaphrodites. Hermaphrodites are essentially females that 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. d. Although the 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

Not only do these mutants have gametes that are missing an X chromosome, many of them will have non-disjunction of the other chromosomes as well. Monosomy for the X chromosome affects sex determination, while monosomy (and possibly trisomy) for the other chromosomes is likely to be lethal.

7.22 You identify a new species of fish, which clearly has two 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 male and one female. 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? Briefly explain your reasoning.

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 two 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 size 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.

7.13 A Drosophila female that is missing cross veins on the wing is mated to a male with normal wings. The female F1 offspring have normal wings, but the male F1 offspring have crossveinless wings. 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?

One of the F2 females is homozygous and the other F2 female is a heterozygous. For the heterozygous female, 50% of the offspring (both the males and the females) will be missing cross veins.

5.19 A family has three children, none of whom are twins. b. What is the probability that they have at least one son and one daughter?

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 - (0.125+0.125) = 0.75. In other words, if they do not have three sons or three daughters, they must have at least one child of each sex.

6.1 Define the following terms as they are used in this chapter. e. synapsis

The "zipping up" of two chromosomes that have been paired together.

5.18 The nematode Caenorhabditis 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 allowed to self-fertilize to produce an F2 offspring are capable of movement. One of the F1 generation. c. When an observant student did the cross, she noticed that 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 of worms in each category, with the following results. Normal movement = 73 Paralyzed = 66 Sluggish but moving = 139 How might you explain these results? What do these results suggest about the F1 worms that are sluggish but capable of movement?

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.

6.16 C. elegans has two sexes—males and hermaphrodites. Hermaphrodites are essentially females that 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. c. What are some of the biological processes that might be encoded by a him gene?

Possible answers could be synapsis, pairing, crossing over, structures of the synaptonemal complex, checkpoints, or any of the steps in meiosis.

6.2 Mitosis and meiosis are both processes of eukaryotic cell division, with the same names used to describe the different stages. b. What are the similarities and differences between the events that occur during each stage of mitosis and the corresponding stage of meiosis I?

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.

6.2 Mitosis and meiosis are both processes of eukaryotic cell division, with the same names used to describe the different stages. a. What are the significant cellular and chromosomal events that occur during each stage of mitosis?

Prophase: Chromosomes condense and spindles begin to form Metaphase: Microtubules attached to the kinetochores 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 re-form in each cell and the chromosomes decondense.

5.22 Figure Q5.2 shows a pedigree for a rare trait in humans. Affected individuals are indicated by filled-in symbols. The woman II-2 is pregnant, with II-3 being the father. Her unborn child is represented by the "?". a. Is this trait inherited as a dominant trait or a recessive trait? What is the best evidence for your answer?

Recessive. The best evidence is individual II-1, since the trait must be recessive for two unaffected individuals to give birth to an affected individual.

5.23 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?

Rh+/Rh- and Rh+/Rh-

6.7 Every eukaryote that has been studied has an ortholog 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 cross over, so non-disjunction 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.

5.15 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 rabbit breeder has four rabbits: an angora male, an angora female, a short-haired male, and a short-haired female. (The sex of the rabbits is not relevant to the solution.) He made the following crosses with the outcomes shown. Angora female × angora male = all angora Angora female × short male = all short Short female × angora male = 4 short and 5 angora a. which phenotype is dominant?

Short hair—note that when the angora and the short are mated, the offspring have short hair.

7.19 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 two 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?

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 or 25%.

5.14 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. These F1 plants are grown. a. What is the genotype of these F1 plants? What is the phenotype of their leaves?

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.

6.10 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 ovaries of some animal that has only two pairs of chromosomes (n = 2). The slides appear to be stained with some fluorescent dye that allows you to 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 that 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 had 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

6.6 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) as the cytoplasm is not divided evenly. All oocytes are made during gestation and then pause during Prophase I in meiotic (dictyate) arrest until puberty. When ovulation occurs during 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.

5.18 The nematode Caenorhabditis 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 allowed to self-fertilize to produce an F2 offspring are capable of movement. One of the F1 generation. b. Among the F2 generation worms, what fraction of them do you predict will be paralyzed?

The F1 offspring must be heterozygous (Mm) since the parent genotypes are MM and mm. 25% of the F2 worms will be paralyzed. (genotypes: MM, Mm, Mm, and mm)

b. Do the data more closely support the Ocean perturbation or the UN peace-keeper hypothesis and why?

The Haitian strains are most likely from the UN peace-keepers because the tree shows that the strains are most closely related to strains from Asia--specifically Bangladesh, which supports that the Haitian strains evolved from the Bangladesh strains.

5.6 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 for 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 disease 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.

c. Do the data effectively rule out either hypothesis? Why or why not?

The data does not totally rule out either hypothesis because there is still the possibility that the Haitian strains are more closely related to a different strain from the Americas that was not put onto this phylogenetic tree. The strain that is most closely related to the Haitian strains is from 2002, leaving ample time for that strain to have been previously transmitted to South America and then left to evolve into the current strains over the next 8 years.

5.15 Angora female × angora male = all angora Angora female × short male = all short Short female × angora male = 4 short and 5 angora 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 (c) is performed, a litter of five rabbits is born. What is the probability that at least one of the five is a female?

The easiest way to determine the probability that at least one is a female is to calculate the probability that all of the rabbits are males and then subtract from 1. (0.5)^5 = 0.03125 (1-0.03125)*100% = 96.875%

5.20 The ability to taste bitterness in certain foods, such as Brussels 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. b. How does the phenotype of the daughter help you answer this question?

The fact that the 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.

5.13 Wrinkled peas arise from a mutation that inactivates a gene that encodes an enzyme involved in the production of starch. Why should a mutation that inactivates 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 an 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 required 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.

6.1 Define the following terms as they are used in this chapter. h. Non-disjunction

The failure of chromosomes to segregate properly during meiosis.

6.4 Compare the orientation of the kinetochore during mitosis with the orientations during meiosis I and meiosis II. How does this change in the 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 microtubules can attach on either side and pull them apart towards the poles.

6.15 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 non-disjunction event. 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 a 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?

The mother of this child experienced non-disjunction 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.

5.20 The ability to taste bitterness in certain foods, such as Brussels 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?

The parents' genotypes are both Tt. Therefore, the probability that their son can taste bitter foods is 0.75. (genotypes: TT, Tt, Tt, tt)

6.17 The woman II-2 shown in the pedigree in Figure Q6.5A 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 child will not be affected by this trait?

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 (⅔ x ⅔= 4/9). If they are both heterozygous, there is a 1 in 4 chance that their child will homozygous recessive, which makes the probability that both parents are heterozygous and their child is homozygous recessive 4/36 or 1/9 (4/9 x ¼ = 4/36 or 1/9). Therefore the chance that that the child does not have the trait is 8/9.

6.1 Define the following terms as they are used in this chapter. d. Cohesion

The process by which sister chromatids stay together after replication but before metaphase of mitosis or meiosis II.

6.1 Define the following terms as they are used in this chapter. c. Centromere and kinetochore

The protein structure that holds together two sister chromatids (the centromere) and the proteins around this structure that provide a place for microtubules to attach (the kinetochore).

5.20 The ability to taste bitterness in certain foods, such as Brussels 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. 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 (the grandson of the original couple) will be able to taste bitter foods?

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 100% probability that his son will be able to taste bitter foods. If the son's genotype is Tt then there is a 50% that his son will be a taster. We can combine this to find that 0.33 + (0.67*0.5) = 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.

6.1 Define the following terms as they are used in this chapter. b. Centrosome

The structures in an animal cell from which microtubules are grown and organized.

7.16 Some parakeets and other birds of the parrot family have a phenotype known as lutino. Lutino is a much-prized beautiful coloration pattern in which the feathers on the body and head are yellow, with silver patches on the cheeks, and light yellow tail feathers. (These birds are pretty enough that you might want to look at a picture of them on-line.) The birds also have red or pink irises in the eyes rather than the standard black irises. 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.

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.

5.16 In certain games that use dice, rolling doubles (both dice show the same number) 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.

5.22 Figure Q5.2 shows a pedigree for a rare trait in humans. Affected individuals are indicated by filled-in symbols. The woman II-2 is pregnant, with II-3 being the father. Her unborn child is represented by the "?". c. What is the overall probability that the unborn child will be unaffected by the trait and be a girl?

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 or two-thirds. Even if both parents are heterozygotes, the probability that the child will be affected is one-fourth, so the total probability that the child will be affected is 2/3 x ¼ or 1/6. So 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 that the unborn child will be unaffected by the trait and a girl is 5/6 x ½ or 5/12.

5.3 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: term-34homozygous dominant (AA) and heterozygous (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.

5.17 In Chapter 4, we saw that the small probability for error during replication can result in mutations that change the function of a gene. Certain mutations in the Escherichia 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 streptomycin and rifampicin?

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

7.24 Looking ahead). 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 (Lon), while mutations in the unc-18 gene result in worms that are Uncoordinated (Unc). c. (Looking back and ahead). 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.

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 the lon-2 and the unc-18 genes, and some will have X chromosomes carrying the both the lon-2 and the unc-18 alleles.

7.15 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 cream-colored male. c. In fact, cream is X-linked. When the F1 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.

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 fur was inactivated, resulting in the recessive allele for cream fur being expressed, resulting in cream colored fur.

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?

This pattern could occur if the trait was autosomal recessive but the trait would have to be assumed to be a common one, since II-1, II-3, and II-7, 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.

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 ⅔, since we know that 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 ¼ chance that they will have an affected offspring. Therefore, the probability of the child inheriting the trait is ⅔ x 1 x ¼ = ⅙.

6.3 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 a sperm with one copy of chromosome 21 to produce an embryo with three copies of chromosome 21.

7.25 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 red 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. b. (Challenging) 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.

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.

7.22 d. For the most likely mode of inheritance in each pedigree, calculate the likelihood that child #1 will be affected by the trait. Pedigree A

a. 50%. Since the trait is autosomal dominant, there is a 50% chance the child 1 will inherit the dominant allele and be affected.

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 the most likely? A

a. Autosomal dominant most likely. This trait is not X-linked, because a son inherits it from the father, and it is not Y-linked, because a daughter 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.

7.26 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 four daughters and a son Alexis (or Alexei) who had hemophilia. The relevant part of this pedigree for this question is also shown as Panel A below. Alexis was identified in order to confirm that the bones were from the Romanov children, the two X-linked genes for blotting clotting Factors VIII (responsible for Hemophilia A) and IX (responsible for Hemophilia B) were analyzed. 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 was 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 his sisters?

a. Genes were sequenced from the boy because the boy was known to have hemophilia, an X-linked disease, so 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 allele. Therefore, to confirm the identity of the remains, only the boy's DNA needed to be analyzed.

7.24 Looking ahead). 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 (Lon), while mutations in the unc-18 gene result in worms that are Uncoordinated (Unc). b. A Lon male is mated to a hermaphrodite that is Unc. The F1 hermaphrodites are wild-type, while the F1 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?

a. 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 the Lon-2 and Unc-18 genes on the X chromosome.

7.21 d. For the most likely mode of inheritance in each pedigree, calculate the likelihood that child #1 will be affected by the trait. Pedigree B

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 child 1 to be a carrier, then there will be a 0% chance that the child is affected.

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 the most likely? B

b. Autosomal recessive most likely. The trait is not dominant, as there are affected children of unaffected parents. The trait cannot 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.

7.21 d. For the most likely mode of inheritance in each pedigree, calculate the likelihood that child #1 will be affected by the trait. Pedigree C

c. Assuming the trait is Y-linked, there is 100% chance if child 1 being affected if he is a boy, and a 0% chance child 1 being affected if she is a girl. Since there is a 50% chance of the child being a boy, there is a 50% chance of the child being affected.

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 the most likely? C

c. Y-linked trait most likely. This trait cannot 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.