Chapter 1. Equations

1. (Problem 1) Why was Mendel’s approach to the study of heredity so successful?

(Problem 2) What is the difference between genotype and phenotype?

(Problem 3) What is the principle of segregation? Why is it important?

(Problem 4) How are Mendel’s principles different from the concept of blending inheritance discussed in Chapter 1?

(Problem 5) What is the concept of dominance?

6) (Problem 6) What are the addition and multiplication rules of probability and when should they be used?

(Problem 7, part 1) Which of the following crosses would likely produce a genotypic ratio of 1:2:1?

(Problem 7, part 2) Which of the following crosses would likely produce a genotypic ratio of 1:1?

(Problem 8, part 1) The chromosome theory of heredity states that

10) (Problem 8, part 2) Why was the chromosome theory of heredity important?

(Problem 9) According to the principle of independent assortment,

12) (Problem 10, part 1) In which phase of meiosis are the principles of segregation and independent assortment at work?

(Problem 10, part 2) In which phase of mitosis are the principles of segregation and independent assortment at work?

(Problem 11, part 1) How is the chi-square goodness-of-fit test used to analyze genetic crosses?

(Problem 11, part 2) What does the probability associated with a chi-square value indicate about the results of a cross?

(Problem 12, part 1) The inheritance of red hair was discussed in the introduction to this chapter. At times in the past, red hair in humans was thought to be a recessive trait and, at other times it was thought to be a dominant trait. What features of heritance would red hair be expected to exhibit as a recessive trait?

(Problem 12 part 2) The inheritance of red hair was discussed in the introduction to this chapter. At times in the past, red hair in humans was thought to be a recessive trait and at other times, it was thought to be a dominant trait. What features of heritance would red hair be expected to exhibit as a dominant trait?

(Problem 13, part 1) Which of the following is a characteristic of an organism that would make it suitable for studies of the principles of inheritance?

19) (Problem 13, part 2) Which of the following organisms does not have the characteristics that would make it suitable for studies of the principles of inheritance?

(Problem 14a) In cucumbers, orange fruit color (R) is dominant over cream fruit color (r). A cucumber plant homozygous for orange fruit is crossed with a plant homozygous for cream fruit. The F1 are intercrossed to produce the F2. What are genotypes of the parents, the F1, and the F2?

(Problem 14b) In cucumbers, orange fruit color (R) is dominant over cream fruit color (r). A cucumber plant homozygous for orange fruit is crossed with a plant homozygous for cream fruit. Give the genotypes and phenotypes of the offspring of a backcross between the F1 and the orange parent.

(Problem 14c) In cucumbers, orange fruit color (R) is dominant over cream fruit color (r). A cucumber plant homozygous for orange fruit is crossed with a plant homozygous for cream fruit. Give the genotypes and phenotypes of a backcross between the F1 and the cream parent.

23) (Problem 15) Figure 1.1 (below) shows three girls, one of whom has albinism. Could the three girls shown in the photograph be sisters? Why or why not?

(Problem 16a) J. W. McKay crossed a stock melon plant that produced tan seeds with a plant that produced red seeds and obtained the following results (J. W. McKay, 1936. Journal of Heredity 27:110–112)./What is the best explanation for the inheritance of the tan and red seeds in this plant?

25) (Problem 16b) J. W. McKay crossed a stock melon plant that produced tan seeds with a plant that produced red seeds and obtained the following results (J. W. McKay, 1936. Journal of Heredity 27:110–112)./Give genotypes for all the individual plants in this cross. As symbols for the alleles, represent the dominant allele with R and the recessive allele with r.

(Problem 17a) White (w) coat color in guinea pigs is recessive to black (W). In 1909, W. E. Castle and J. C. Phillips transplanted an ovary from a black guinea pig into a white female whose ovaries had been removed. They then mated this white female with a white male. All the offspring from the mating were black in color (W. E. Castle and J. C. Phillips, 1909, Science 30:312–313). Which of the following statements explains the results of this cross?

(Problem 17b) White (w) coat color in guinea pigs is recessive to black (W). In 1909, W. E. Castle and J. C. Phillips transplanted an ovary from a black guinea pig into a white female whose ovaries had been removed. They then mated this white female with a white male. All the offspring from the mating were black in color (W. E. Castle and J. C. Phillips. 1909, Science 30:312–313). Give the genotype of the offspring from this cross.

(Problem 17c) White (w) coat color in guinea pigs is recessive to black (W). In 1909, W. E. Castle and J. C. Phillips transplanted an ovary from a black guinea pig into a white female whose ovaries had been removed. They then mated this white female with a white male. All the offspring from the mating were black in color (W. E. Castle and J. C. Phillips, 1909, Science 30:312–313). What, if anything, does this experiment indicate about the validity of the pangenesis and the germ-plasm theories discussed in Chapter 1?

29) (Problem 18) In cats, blood type A results from an allele IA that is dominant over an allele iB that produces blood type B. There is no O blood type. The blood types of male and female cats that were mated and the blood types of their kittens follow. Give the most likely genotypes for the parents of the litter.

30) (Problem 19a) Figure 3.7 shows the results of a cross between a tall pea plant and a short pea plant. What phenotypes and proportions will be produced if a tall F1 progeny is backcrossed to the short parent?

31) (Problem 19b) Figure 3.7 shows the results of a cross between a tall pea plant and a short pea plant. What phenotypes and proportions will be produced if a tall F1 progeny is backcrossed to the tall parent?

(Problem 20) Joe has a white cat named Sam. When Joe crosses Sam with a black cat, he obtains 1/2 white kittens and 1/2 black kittens. When the black kittens are interbred, they produce all black kittens. On the basis of these results, would you conclude that white or black coat color in cats is a recessive trait and why?

(Problem 21) In sheep, lustrous fleece results from an allele (L) that is dominant over an allele (l) for normal fleece. A ewe (adult female) with lustrous fleece is mated with a ram (adult male) with normal fleece. The ewe then gives birth to a single lamb with normal fleece. What are the likely genotypes of the two parents?

(Problem 22a part 1) Alkaptonuria is a metabolic disorder in which affected persons produce black urine. Alkaptonuria results from an allele (a) that is recessive to the allele for normal metabolism (A). Sally has normal metabolism, but her brother has alkaptonuria. Sally’s father has alkaptonuria, and her mother has normal metabolism. What is Sally’s genotype?

(Problem 22a part 2) In humans, alkaptonuria is a metabolic disorder in which affected persons produce black urine. Alkaptonuria results from an allele (a) that is recessive to the allele for normal metabolism (A). Sally has normal metabolism, but her brother has alkaptonuria. Sally’s father has alkaptonuria, and her mother has normal metabolism. What is the genotype of Sally’s brother and Sally’s father?

(Problem 22b) In humans, alkaptonuria is a metabolic disorder in which affected persons produce black urine. Alkaptonuria results from an allele (a) that is recessive to the allele for normal metabolism (A). Sally has normal metabolism, but her brother has alkaptonuria. Sally’s father has alkaptonuria, and her mother has normal metabolism. If Sally’s parents have another child, what is the probability that this child will have alkaptonuria?

(Problem 22c) In humans, alkaptonuria is a metabolic disorder in which affected persons produce black urine. Alkaptonuria results from an allele (a) that is recessive to the allele for normal metabolism (A). Sally has normal metabolism, but her brother has alkaptonuria. Sally’s father has alkaptonuria, and her mother has normal metabolism. If Sally marries a man with alkaptonuria, what is the probability that their first child will have alkaptonuria?

(Problem 23) Suppose that you are raising Mongolian gerbils. You notice that some of your gerbils have white spots, whereas others have solid coats. What type of cross could you carry out to determine whether white spots are due to a recessive allele?

(Problem 24) Hairlessness in American rat terriers is recessive to the presence of hair. Suppose that you have a rat terrier with hair. How can you determine whether this dog is homozygous or heterozygous for the hairy trait?

40) (Problem 25) What is the probability of rolling one six-sided die and obtaining the following numbers?

41) (Problem 26) What is the probability of rolling two six-sided dice and obtaining the following numbers?

42) (Problem 27) In a family of seven children, what is the probability of obtaining the following numbers of boys and girls?

43) (Problem 28) Phenylketonuria (PKU) is a disease that results from a recessive gene. Two normal parents produce a child with PKU.

45) (Problem 30a) In guinea pigs, the allele for black fur (B) is dominant over the allele for brown (b) fur. A black guinea pig is crossed with a brown guinea pig, producing five F1 black guinea pigs and six F1 brown guinea pigs. How many copies of the black allele (B) will be present in each cell of an F1 black guinea pig at the following stages. Assume that no crossing over takes place.

46) (Problem 30b) In guinea pigs, the allele for black fur (B) is dominant over the allele for brown (b) fur. A black guinea pig is crossed with a brown guinea pig, producing five F1 black guinea pigs and six F1 brown guinea pigs. How many copies of the brown allele (b) will be present in each cell of an F1 brown guinea pig at the following stages. Assume that no crossing over takes place.

(Problem 31a) In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The genes for these two characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The F1 are intercrossed to produce the F2. What will be the phenotypic ratio in the F2?

(Problem 31b) In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The genes for these two characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. If an F1 plant is backcrossed with the bitter, yellow-spotted parent, what phenotypes and proportions are expected in the offspring?

(Problem 31c) In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The genes for these two characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. If an F1 plant is backcrossed with the sweet, nonspotted parent, what phenotypes and proportions are expected in the offspring?

50) (Problem 32a) Figure 3.10 shows the results of a dihybrid cross involving seed shape and seed color. What proportion of the round and yellow F2 progeny from this cross is homozygous at both loci?

51) (Problem 32b) Figure 3.10 shows the results of a dihybrid cross involving seed shape and seed color. What proportion of the round and yellow F2 progeny from this cross is homozygous at least at one locus?

(Problem 33a) In cats, curled ears result from an allele (Cu) that is dominant over an allele (cu) for normal ears. Black color results from an independently assorting allele (G) that is dominant over an allele for gray (g). A gray cat homozygous for curled ears is mated with a homozygous black cat with normal ears. All the F1 cats are black and have curled ears. If two of the F1 cats mate, what phenotypes and proportions are expected in the F2?

(Problem 33b) In cats, curled ears result from an allele (Cu) that is dominant over an allele (cu) for normal ears. Black color results from an independently assorting allele (G) that is dominant over an allele for gray (g). A gray cat homozygous for curled ears is mated with a homozygous black cat with normal ears. All the F1 cats are black and have curled ears. An F1 cat mates with a stray cat that is gray and possesses normal ears. What phenotypes and proportions of progeny are expected from this cross?

54. (Problem 34) The following two genotypes are crossed: AaBbCcddEe × AabbCcDdEe. What will be the proportion of the following genotypes be among the progeny of this cross?

(Problem 35) In mice, an allele for apricot eyes (a) is recessive to an allele for brown eyes (a+). At an independently assorting locus, an allele for tan (t) coat color is recessive to an allele for black (t+) coat color. A mouse that is homozygous for brown eyes and black coat color is crossed with a mouse having apricot eyes and a tan coat. The resulting F1 are intercrossed to produce the F2. In a litter of eight F2 mice, what is the likelihood that two will have apricot eyes and tan coats?

56) (Problem 36a) In cucumbers, dull fruit (D) is dominant over glossy fruit (d), orange fruit (R) is dominant over cream fruit (r), and bitter cotyledons (B) are dominant over nonbitter cotyledons (b). The three characters are encoded by genes located on different pairs of chromosomes. A plant homozygous for dull, orange fruit and bitter cotyledons is crossed with a plant that has glossy, cream fruit and nonbitter cotyledons. The F1 are intercrossed to produce the F2. Give the expected proportions of the following phenotypes in the F2.

57. (Problem 36b) In cucumbers, dull fruit (D) is dominant over glossy fruit (d), orange fruit (R) is dominant over cream fruit (r), and bitter cotyledons (B) are dominant over nonbitter cotyledons (b). The three characters are encoded by genes located on different pairs of chromosomes. A plant homozygous for dull, orange fruit and bitter cotyledons is crossed with a plant that has glossy, cream fruit and nonbitter cotyledons. An F1 plant is crossed with a plant that has glossy, cream fruit and nonbitter cotyledons. Give the expected proportions of the following phenotypes among the progeny of this cross.

58) (Problem 37a) A and a are alleles located on a pair of metacentric chromosomes. B and b are alleles located on a pair of acrocentric chromosomes. A cross is made between individuals having the following genotypes: Aa Bb × aa bb. Below are drawings of chromosomes as they would appear in each type of gamete produced by the individuals of this cross. Indicate which alleles belong in each gamete.

(Problem 38a) J. A. Moore investigated the inheritance of spotting patterns in leopard frogs (J. A. Moore, 1943, Journal of Heredity 34:3–7). The pipiens phenotype had the normal spots that give leopard frogs their name. In contrast, the burnsi phenotype lacked spots on its back. Moore carried out the following crosses, producing the progeny indicated./On the basis of these results, what is the most likely mode of inheritance of the burnsi phenotype?

(Problem 38b) J. A. Moore investigated the inheritance of spotting patterns in leopard frogs (J. A. Moore, 1943, Journal of Heredity 34:3–7). The pipiens phenotype had the normal spots that give leopard frogs their name. In contrast, the burnsi phenotype lacked spots on its back. Moore carried out the following crosses, producing the progeny indicated./Give the most likely genotypes of the parent in each cross using

62) (Problem 38c part 1) J. A. Moore investigated the inheritance of spotting patterns in leopard frogs (J. A. Moore, 1943, Journal of Heredity 34:3–7). The pipiens phenotype had the normal spots that give leopard frogs their name. In contrast, the burnsi phenotype lacked spots on its back. Moore carried out the following crosses, producing the progeny indicated. Use a chi-square test to evaluate the fit of the observed numbers of progeny to the number expected from the cross burnsi × burnsi that produced 39 burnsi, 6 pipiens. Assume that the burnsi allele is dominant to the pipiens allele and that the burnsi parents are heterozygous.

63) (Problem 38c part 2) J. A. Moore investigated the inheritance of spotting patterns in leopard frogs (J. A. Moore, 1943, Journal of Heredity 34:3–7). The pipiens phenotype had the normal spots that give leopard frogs their name. In contrast, the burnsi phenotype lacked spots on its back. Moore carried out the following crosses, producing the progeny indicated. Use a chi-square test to evaluate the fit of the observed numbers of progeny to the number expected from the cross burnsi × pipiens that produced 23 burnsi, 33 pipiens. Assume that the burnsi allele is dominant to the pipiens allele and that the burnsi parents are heterozygous.

64) (Problem 38c part 3) J. A. Moore investigated the inheritance of spotting patterns in leopard frogs (J. A. Moore, 1943, Journal of Heredity 34:3–7). The pipiens phenotype had the normal spots that give leopard frogs their name. In contrast, the burnsi phenotype lacked spots on its back. Moore carried out the following crosses, producing the progeny indicated. Use a chi-square test to evaluate the fit of the observed numbers of progeny to the number expected from the cross burnsi × pipiens that produced 196 burnsi, 210 pipiens. Assume that the burnsi allele is dominant to the pipiens allele and that the burnsi parents are heterozygous.

(Problem 39a) In the 1800s, a man with dwarfism who lived in Utah produced a large number of descendants: 22 children, 49 grandchildren, and 250 great-grandchildren (see the illustration of a family pedigree below), many of whom were also dwarfs (F. F. Stephens, 1943, Journal of Heredity 34:229–235). The type of dwarfism found in this family is called Schmid-type metaphyseal chondrodysplasia, although it was originally thought to be achondroplastic dwarfism. Among the families of this kindred, dwarfism appeared only in members who had one parent with dwarfism. When one parent was a dwarf, the following numbers of children were produced. With the assumption that Schmid-type metaphyseal chondrodysplasia is rare, is this type of dwarfism inherited as a dominant or recessive trait and why?

66) (Problem 39b) In the 1800s, a man with dwarfism who lived in Utah produced a large number of descendants: 22 children, 49 grandchildren, and 250 great-grandchildren (see the illustration of a family pedigree), many of whom were also dwarfs (F. F. Stephens, 1943, Journal of Heredity 34:229–235). The type of dwarfism found in this family is called Schmid-type metaphyseal chondrodysplasia, although it was originally thought to be achondroplastic dwarfism. Among the families of this kindred, dwarfism appeared only in members who had one parent with dwarfism. When one parent was a dwarf, the following numbers of children were produced. Use a chi-square test to determine if the total number of children for these families (52 normal, 40 dwarfs) is significantly different from the number expected. Assume that the trait is dominant and that one parent is heterozygous for the dwarf allele and the other parent is homozygous for the normal allele.

67) (Problem 39c part 1) In the 1800s, a man with dwarfism who lived in Utah produced a large number of descendants: 22 children, 49 grandchildren, and 250 great-grandchildren (see the illustration of a family pedigree), many of whom were also dwarfs (F. F. Stephens, 1943, Journal of Heredity 34:229–235). The type of dwarfism found in this family is called Schmid-type metaphyseal chondrodysplasia, although it was originally thought to be achondroplastic dwarfism. Among the families of this kindred, dwarfism appeared only in members who had one parent with dwarfism. When one parent was a dwarf, the following numbers of children were produced. Use a chi-square test to determine if the number of children in family C (1 normal, 6 dwarf) are significantly different from the numbers expected. Assume that the trait is dominant and that one parent is heterozygous for the dwarf allele and the other parent is homozygous for the normal allele.

68) (Problem 39c part 2) In the 1800s, a man with dwarfism who lived in Utah produced a large number of descendants: 22 children, 49 grandchildren, and 250 great-grandchildren (see the illustration of a family pedigree), many of whom were also dwarfs (F. F. Stephens, 1943, Journal of Heredity 34:229–235). The type of dwarfism found in this family is called Schmid-type metaphyseal chondrodysplasia, although it was originally thought to be achondroplastic dwarfism. Among the families of this kindred, dwarfism appeared only in members who had one parent with dwarfism. When one parent was a dwarf, the following numbers of children were produced. Use a chi-square tests to determine if the number of children in family D (6 normal, 2 dwarf) are significantly different from the numbers expected. Assume that the trait is dominant and that one parent is heterozygous for the dwarf allele and the other parent is homozygous for the normal allele.

(Problem 40a) Pink-eye and albinism are two recessive traits found in the deer mouse Peromyscus maniculatus. In mice with pink-eye, the eye is devoid of color and appears pink from the blood vessels within it. Albino mice are completely lacking color both in their fur and in their eyes. F. H. Clark crossed pink-eyed mice with albino mice; the resulting F1 had normal coloration in their fur and eyes. He then crossed these F1 mice with mice that were pink-eyed and albino and obtained the following mice. It is very hard to distinguish between mice that are albino and mice that are both pink-eyed and albino, so he combined these two phenotypes (F. H. Clark, 1936, Journal of Heredity 27:259–260).Give the expected numbers of progeny with each phenotype if the genes for pink-eye and albinism assort independently.

(Problem 40b) Pink-eye and albinism are two recessive traits found in the deer mouse Peromyscus maniculatus. In mice with pink-eye, the eye is devoid of color and appears pink from the blood vessels within it. Albino mice are completely lacking color both in their fur and in their eyes. F. H. Clark crossed pink-eyed mice with albino mice; the resulting F1 had normal coloration in their fur and eyes. He then crossed these F1 mice with mice that were pink-eyed and albino and obtained the following mice. It is very hard to distinguish between mice that are albino and mice that are both pink-eyed and albino, so he combined these two phenotypes (F. H. Clark, 1936, Journal of Heredity 27:259–260). Use a chi-square test to determine if the observed numbers of progeny fit the number expected with independent assortment.

(Problem 41a and 41b) In the California poppy, an allele for yellow flowers (C) is dominant over an allele for white flowers (c). At an independently assorting locus, an allele for entire petals (F) is dominant over an allele for fringed petals (f). A plant that is homozygous for yellow and entire petals is crossed with a plant that is white and fringed. A resulting F1 plant is then crossed with a plant that is white and fringed, and the following progeny are produced: 54 yellow and entire, 58 yellow and fringed, 53 white and entire, and 10 white and fringed. Using a chi-square test to compare the observed numbers with those expected for the cross, what can you conclude in regard to whether the ratios observed occurred due to random variations?

(Problem 41c) In the California poppy, an allele for yellow flowers (C) is dominant over an allele for white flowers (c). At an independently assorting locus, an allele for entire petals (F) is dominant over an allele for fringed petals (f). A plant that is homozygous for yellow and entire petals is crossed with a plant that is white and fringed. A resulting F1 plant is then crossed with a plant that is white and fringed, and the following progeny are produced: 54 yellow and entire, 58 yellow and fringed, 53 white and entire, and 10 white and fringed. Suggest an explanation for the results.

73) (Problem 42) Dwarfism is a recessive trait in Hereford cattle. A rancher in western Texas discovers that several of the calves in his herd are dwarfs, and he wants to eliminate this undesirable trait from the herd as rapidly as possible. Suppose that the rancher hires you as a genetic consultant to advise him on how to breed the dwarfism trait out of the herd. What crosses would you advise the rancher to conduct to ensure that the allele causing dwarfism is eliminated from the herd?

(Problem 43) A geneticist discovers an obese mouse in his laboratory colony. He breeds this obese mouse with a normal mouse. All the F1 mice from this cross are normal in size. When he interbreeds two F1 mice, eight of the F2 mice are normal in size and two are obese. The geneticist then intercrosses two of his obese mice, and he finds that all of the progeny from this cross are obese. These results lead the geneticist to conclude that obesity in mice results from a recessive allele. A second geneticist at a different university also discovers an obese mouse in her laboratory colony. She carries out the same crosses as the first geneticist and obtains the same results. She also concludes that obesity in mice results from a recessive allele. One day, the two geneticists meet at a genetics conference, learn of each other’s experiments, and decide to exchange mice. They both find that, when they cross two obese mice from the different laboratories, all the offspring are normal. However, when they cross two obese mice from the same laboratory, all the offspring are obese. Which of the following best explains their results?

(Problem 44) Albinism is a recessive trait in humans. A geneticist studies a series of families in which both parents are normal and at least one child has albinism. The geneticist reasons that both parents in these families must be heterozygotes and that albinism should appear in 1/4 of the children of these families. To his surprise, the geneticist finds that the frequency of albinism among the children of these families is considerably greater than 1/4. Can you think of an explanation for the higher-than-expected frequency of albinism among these families?

76) (Problem 45) Two distinct phenotypes are found in the salamander Plethodon cinereus: a red form and a black form. Some biologists have speculated that the red phenotype is due to an autosomal allele that is dominant over an allele for black. Unfortunately, these salamanders will not mate in captivity; so the hypothesis that red is dominant over black has never been tested. One day a genetics student is hiking through the forest and finds 30 female salamanders, some red and some black, laying eggs. The student places each female and her eggs (about 20–30 eggs per female) in separate plastic bags and takes them back to the lab. There, the student successfully raises the eggs until they hatch. After the eggs have hatched, the student records the phenotypes of the juvenile salamanders, along with the phenotypes of their mothers. Thus, the student has the phenotypes for 30 females and their progeny, but no information is available about the phenotypes of the fathers. Explain how the student can determine whether red is dominant over black with this information on the phenotypes of the females and their offspring.