Chapter 3 Practice Problems

Question 1

the alleles of one gene mask the effects of alleles of another gene

Answer 1

epistasis

Question 2

gene whose alleles alter phenotypes produced by the action of other genes

Answer 2

modifier genes

Question 3

a genotype that is lethal in some situations but viable in others

Answer 3

conditional Lethal

Question 4

environmental condition that allows conditional lethals to live

Answer 4

permissive conditions

Question 5

less than 100% of the individuals possessing a particular genotype express it in their phenotype

Answer 5

reduced penetrance

Question 6

a trait produced by the interaction of alleles of at least two genes or from interactions between gene and environment

Answer 6

multifactorial trait

Question 7

the heterozygote resembles neither homozygote

Answer 7

incomplete dominance

Question 8

both parental phenotypes are expressed in the F1 hybrids

Answer 8

codominance

Question 9

a heritable change in a gene

Answer 9

mutation

Question 10

one gene affecting more than one phenotype

Answer 10

pleiotropy

Question 11

individuals with the same genotype have related phenotypes that vary in intensity

Answer 11

variable expressivity

Question 12

In four o’clocks, the allele for red flowers is incompletely dominant to the allele for white flowers, so heterozygotes have pink flowers. What ratios of flower colors would you expect among the offspring of the following crosses: a) pink x pink b) white x pink c) red x red d) red x pink e) white x white f) red x white? If you specifically wanted pink flowers which of these crosses would be most efficient?

Answer 12

a. Diagram the cross: FrFw × FrFw → 1/4 FrFr (red) : 1/2 FrFw (pink) : 1/4 FwFw (white).
b. FwFw × FrFw → 1/2 FrFw (pink) : 1/2 FwFw (white).
c. FrFr × FrFr → 1 FrFr (red).
d. FrFr × FrFw → 1/2 FrFr (red) : 1/2 FrFw (pink).
e. FwFw × FwFw → 1 FwFw (white).
f. FrFr × FwFw → 1 FrFw (pink).
The cross shown in part (f) is the most efficient way to produce pink flowers, because all the progeny will be pink.

Question 13

The Aa heterozygous snapdragons are pink while AA homozygotes are red. However, Mendel’s Pp heterozygous pea flowers were every bit as purple as those of PP homozygotes. Assuming that the A allele and the P allele specify functional enzymes, and the a and p alleles specify no protein at all, explain why the alleles of gene A and the alleles of gene P interact so differently.

Answer 13

In Mendel’s Pp heterozygotes, the amount of enzyme leading to purple pigment is sufficient to produce purple color as intense as the purple color in PP homozygotes.the intensity of the phenotype is proportional to the dosage of functional alleles (1 dose in the Aa heterozygote; 2 doses in the AA homozygote)

Question 14

Mendel’s R gene specifies an enzyme called Sbe1 that forms branched starches. The dominant allele R makes protein and the recessive allele r is nonfunctional. When considering the phenotype of round or wrinkled peas, R is completely dominant to r. Imagine that the phenotype described is instead the average number of Sbe1 protein molecules in a pea. How would you describe the dominance relation between R and r in this case

Answer 14

Presumably, because r is nonfunctional, RR peas have twice the number of Sbe1 protein molecules as Rr peas, while rr peas have zero Sbe1 protein molecules. If we describe the phenotype as the number of Sbe1 molecules, the R and r alleles would exhibit incomplete dominance because the phenotype of the heterozygote is in between that of dominant and recessive homozygotes.

Question 15

In the fruit fly Drosophila melanogaster, very dark body color is determined by the e allele. The e+ allele produces the normal wild type, honey colored body. In heterozygotes for the two alleles (but not the e+e+ homozygotes), a dark making called the trident can be seen on the thorax, but otherwise the body is honey colored. the e+ and e alleles are thus considered to be incompletely dominant.

a) when female e+e+ flies are crossed to male e+e flies, what is the probability that progeny will have the dark trident marking?
b) animals with the trident marking mate among themselves. Of 300 progeny, how many would be expected to have a trident, how many ebony bodies and how many honey colored bodies?

Answer 15

a.Diagram the cross:
e+e+ × e+e → 1/2 e+e+ : 1/2 e+e.
The trident marking is only found in the heterozygotes, so the probability is 1/2.
b. The offspring with the trident marking are e+e, so the cross is e+e × e+e → 1/4 ee : 1/2 e+e : 1/4 e+e+. Therefore, of 300 offspring, 75 should have ebony bodies, 150 should have the trident marking and 75 should have honey-colored bodies.

Question 16

A cross between two plants that both have yellow flowers produces 80 offspring plants, of which 38 have yellow flowers, 22 have red flowers, and 20 have white flowers. If one assumes that this variation in color is due to inheritance at a single locus, what is the genotype associated with each flower color, and how can you describe the inheritance of flower color?

Answer 16

yellow × yellow → 38 yellow : 22 red : 20 white
Three phenotypes in the progeny indicate that the yellow parents are not true-breeding. The ratio of the progeny is close to 1/2 : 1/4 : 1/4. This is the result expected for crosses between individuals heterozygous for incompletely dominant alleles. Thus:
CrCw × CrCw → 1/2 CrCw (yellow) : 1/4 CrCr (red) : 1/4 CCw (white).

Question 17

Assuming no involvement of the Bombay phenotype

a) if a girl has type O, what could be the genotypes and corresponding phenotypes of her parents?
b. if a girl has blood type B and her mother has blood type A, what genotypes and corresponding phenotypes could the other parent have?
c. if a girl has blood type AB and her mother is also AB what are the genotypes and corresponding phenotypes of any male who could not be the girl’s father?

Answer 17

a. ii (O) or IAi (A) or IBi (B)
b. either IBIB, IBi, or IBIA.
c. the O phenotype (genotype ii).

Question 18

One of your fellow students tells you that there is no way to know that the spotted and dotted patterns on the lentils are due to codominant alleles of a single gene C. He claims that the spotting could be controlled by gene S with a completely dominant allele S that directs spotting and a recessive allele s that directs no spots. Likewise, he claims that dotting could be controlled by a separate gene D, with a completely dominant allele D that directs dotting and a recessive allele d that directs no dots. is he correct?

Answer 18

If you counted a large number of F2 individuals and you failed to see lentils that were neither spotted nor dotted, you would be able to exclude the hypothesis that two genes were involved.

Question 19

In cover plants, the pattern on the leaves is determined by a single gene with multiple alleles that are related in a dominance series. The gene is not pleiotropic. Seven different alleles of this gene are known: an allele that determines the absence of a pattern is recessive to the other six alleles, each of which produces a different pattern. All heterozygous combinations of alleles show complete dominance.

a. How many different kinds of leaf patterns (including the absence of a pattern) are possible in a population of clover plants in which all seven alleles are represented?
b. what is the largest number of different genotypes that could be associated with any one phenotype? Is there any phenotype that could be represented by only a single genotype?
c. In a particular field, you find that the large majority of clover plants lack a pattern on their leaves, even though you can identify a few plants representative of all possible pattern types. Explain this finding.

Answer 19

a. 7
b. The absence of pattern is caused by just one genotype, p7p7.
c. This finding suggests that the allele determining absence of pattern (p7) is very common in these clover plants so that the p7p7 genotype is the most frequent in the population. The other alleles are present, but are much less common in this population.

Question 20

Fruit flies with one allele for curly wings (Cy) and one allele for normal wings (Cy+) have curly wings. When two curly winged flies were crossed, 203 curly winged and 98 normal winged flies were obtained. In fact all crosses between curly winged flies produce nearly the same curly: normal ratio among the progeny

a. what is the approximate phenotype ratio in these offspring?
b. suggest an explanation for these data
c. if a curly winged fly was mated to a normal winged fly how many flies of each type would you expect among 180 total offspring?

Answer 20

a. This ratio is approximately 2/3 Curly : 1/3 normal.
b. The expected result for this cross is: Cy+Cy × Cy+Cy → 1/4 CyCy (?): 1/2 Cy+Cy (Curly) : 1/4 Cy+Cy+ (normal). If the Cy Cy genotype is lethal, then the expected ratio will match the observed data.
c. The cross is Cy+Cy × Cy+Cy+ → 1/2 Cy+Cy : 1/2 Cy+Cy+, so there would be approximately 90 Curly-winged and 90 normal-winged flies. (Note that because Cy Cy adults are not found, all curly-winged flies must be Cy+Cy heterozygotes.)

Question 21

A rooster with a particular comb morphology called walnut was crossed to a hen with a type of comb morphology known as single.The F1 progeny all had walnut combs. When F1 males and females were crossed, 93 walnut and 11 single combs were seen among the F2 progeny but there were also 29 birds with another new comb type called pea.

a) explain how comb morphology was inherited.
b. what progeny would result from crossing a homozygous rose combed hen with a homozygous pea combed rooster? what phenotypes and ratios would be seen in the F2 progeny?
c) a particular walnut rooster was crossed to a pea hen, and the progeny consisted of 12 walnut, 11 pea, 3 rose, and 4 single chickens. what are the likely genotypes of the parents?
d. a different walnut rooster was crossed to a rose hen, and all the progeny were walnut, what are the possible genotypes of the parents?

Answer 21

a. How many genes are involved? The four F2 phenotypes means that 2 genes are involved, A and B. Both genes affect the same structure, the comb. The F2 phenotypic dihybrid ratio among the progeny is close to 9:3:3:1, so there is no epistasis. Because walnut is the most abundant F2 phenotype, it must be the phenotype due to the A– B– genotype. Single combs are the least frequent class, and are thus aa bb. Now assign genotypes to the cross. If the walnut F2 are A– B–, then the original walnut parent must have been AA BB:
AA BB × aa bb → Aa Bb (walnut) → 9/16 A– B– (walnut) : 3/16 A– bb (rose) : 3/16 aa B– (pea) : 1/6 aa bb (single).
b. Diagram the cross, recalling that the problem states the parents are homozygous:
AA bb (rose) × aa BB (pea) → Aa Bb (walnut) → 9/16 A– B– (walnut) : 3/16 A– bb (rose) : 3/16 aa B– (pea) : 1/16 aa bb (single). Notice that this F2 is in identical proportions as the F2 generation in part (a).
c. Diagram the cross: A– B– (walnut) × aa B– (pea) → 12 A– B– (walnut) : 11 aa B– (pea) : 3 A– bb (rose) : 4 aa bb (single). Because pea and single progeny exist, you know that the walnut parent must be Aa. The 1 A– : 1 aa monohybrid ratio in the progeny also tells you the walnut parent must have been Aa. Because some of the progeny are single, you know that both parents must be Bb. In this case, the monohybrid ratio for the B gene is 3 B– : 1 bb, so both parents were Bb. The original cross must have been Aa Bb × aa Bb. You can verify that this cross would yield the observed ratio of progeny by multiplying the probabilities expected for each gene alone. For example, you anticipate that 1/2 the progeny would be Aa and 3/4 of the progeny would be Bb, so 1/2 × 3/4 = 3/8 of the progeny should be walnut; this is close to the 12 walnut chickens seen among 30 total progeny.
d. Diagram the cross: A– B– (walnut) × A– bb (rose) → all A– B– (walnut). The progeny are all walnut, so the walnut parent must be BB. No pea progeny are seen, so both parents cannot be Aa, so one of the two parents must be AA. This could be either the walnut or the rose parent or both.

Question 22

a black mare was crossed to a chestnut stallion and produced a bay son and a bay daughter. The two offspring were mated to each other several times and they produced offspring of four different coat colors: black, bay, chestnut, and liver. Crossing a liver grandson back to the black mare gave a black foal, and crossing a liver granddaughter back to the chestnut stallion gave a chestnut foal.Explain how coat color is being inherited in these horses.

Answer 22

black × chestnut → F1 bay → F2 black : bay : chestnut : liver
Four phenotypes in the F2 generation means two genes determine coat color. The F1 bay animals produce four phenotypic classes, so they must be doubly heterozygous, Aa Bb. Crossing a liver-colored horse to either of the original parents resulted in the parent’s phenotype. The liver horse’s alleles do not affect the phenotype, suggesting the recessive genotype aa bb. Though it is probable that the original black mare was AA bb and the chestnut stallion was aa BB, each of these animals produced only 3 progeny, so it cannot be concluded definitively that these animals were homozygous for the dominant allele they carry. Thus, the black mare was A– bb, the chestnut stallion was aa B–, and the F1 bay animals are Aa Bb. The F2 horses were: bay (A– B–), liver (aa bb), chestnut (aa B–), and black (A– bb).

Question 23

filled in symbols in the pedigree designate individuals who are deaf.

a. study the pedigree and explain how deafness is being inherited
b. what is the genotype of the individuals in generation V? why are they not deaf?

Answer 23

a. the trait is recessive. the trait is heterogenous,meaning that two family lines shown contain mutations in two separate genes. Furthermore, the mutant alleles of both genes determining deafness are recessive.
b. Aa Bb the product of the dominant allele of each gene is sufficient for normal function

Question 24

Two true breeding white strains of the plant Illegitimati noncarbonundrum were mated, and the F1 progeny were all white. when the F1 plants were allowed to self fertilize, 126 white flowered and 33 purple flowered F2 plants grew

a. how would you describe inheritance of flower color? Describe how specific alleles influence each other and therefore affect phenotype
b. a white F2 plant is allowed to self fertilize. of the progeny, 3/4 are white flowered and 1/4 are purple flowered. what is the genotype of the F2 plant?
c. a purple F2 plant is allowed to self fertilize. of the progeny 1/2 are purple flowered and 1/2 are white flowered. what are the two white F2 plants?
d. two white F2 plants are crossed with each other. of the progeny 1/2 are purple flowered and 1/2 are white flowered. what are the genotypes of the two white F2 plants?

Answer 24

a. The F2 phenotypic ratio is thus approximately 13 white : 3 purple. The data fit the hypothesis that two genes control color, and that the F1 are dihybrids. Assume that A– bb plants are white, and aa B– plants are purple. Our model above states that to be purple, a plant must have a B allele and no A allele. Thus, we can say that A is epistatic to B.
b. AaBB
c. aa Bb
d. aa bb (white) x Aa BB (white)

Question 25

Normally, wild violets have yellow petals with dark brown marking and erect stems. Imagine you discover a plant with white petals, no markings, and prostate stems. What experiment could you perform to determine whether the non wild type phenotypes are due to several different mutant genes or to pleiotropic effects of alleles at a single locus? Explain how your experiment would settle the question.

Answer 25

You would first self the mutant plant. If the mutant traits are dominant and the plant is heterozygous for the genes involved, it is possible that you might see some progeny displaying different combinations of the recessive wild-type traits. Such a result would suggest that different genes are responsible for the different traits.
If the mutant plant is pure-breeding, you should cross it (or its self-fertilized descendants) with a pure-breeding wild-type strain, and then self-fertilize the F1 progeny. If several genes were involved, the F2 would have several different combinations of the petal color, markings, and stem position traits. If all 3 traits were determined by an allele of one gene, the three non-wild-type or three wild-type traits would always be inherited together.

Question 26

Suppose that blue flower color in a plant species is controlled by two genes, A and B. The dominant alleles A and B specify proteins that function in the pathways shown below. The A and B proteins are both required to make blue pigment from a colorless precursor. A and B proteins also independently inhibit the production of blue pigment from a different colorless precursor; that is, the presence of either A or B is sufficient to prevent blue pigment production from precursor 2. The recessive mutant alleles a and b specify no protein. Two different true breeding mutant strains with white flowers were crossed and complementation was observed so that all the F1 were blue.

a. what are the genotypes of each white mutant strain and the F1
b. if the F1 are selfed what would be the phenotypic ratio of the F2

Answer 26

a. AA bb (white) × aa BB (white) → F1 Aa Bb (all blue).

b. You would expect in the F2 generation a ratio of 10 blue (9 A– B– + 1 aa bb) : 6 white (3 A– bb + 3 aa B– ).