Patterns of Inheritance Practice Test

Dominant and Recessive Alleles Questions

1. Several black guinea pigs of the same genotype were mated and they produced 29 black and 9 white offspring. What would you predict the genotypes of the parents to be?
2. If a black female guinea pig is testcrossed and produces at least one white offspring, determine (a) the genotype and phenotype of the sire (male parent) that produced the white offspring, (b) the genotype of this female.
3. In Drosophila, sepia-colored eyes are due to a recessive allele s and wild type (red eye color) to its dominant allele s⁺. If sepia-eyed females are crossed to pure wild-type males, what phenotypic and genotypic ratios are expected if the F₂ males are backcrossed to the sepia-eyed parental females?
4. The lack of pigmentation, called albinism, in humans is the result of a recessive allele a and normal pigmentation is the result of its dominant allele A. Two normal parents have an albino child. Determine the probability that (a) the next child is albino, (b) the next two children are albinos. (c) What is the chance of these parents producing two children, one albino and the other normal?
5. Short hair is due to a dominant gene L in rabbits, and long hair to its recessive allele l. A cross between a short-haired female and a long-haired male produces a litter of one long-haired and seven short-haired bunnies. (a) What are the genotypes of the parents? (b) What phenotypic ratio was expected in the offspring generation? (c) How many of the eight bunnies were expected to be long-haired?
6. Black wool of sheep is due to a recessive allele b and white wool to its dominant allele B. A white ram (male) is crossed to a white ewe (female), both animals carrying the allele for black. They produce a white male lamb that is then backcrossed to the female parent. What is the probability of the backcross offspring being black?
7. In foxes, silver-black coat color is governed by a recessive allele b and red color by its dominant allele B. Determine the genotypic and phenotypic ratios expected from the following matings: (a) pure red ×carrier red, (b) carrier red × silver-black, (c) pure red × silver-black.

Codominance and Incomplete Dominance Questions

8. When chickens with splashed white feathers are crossed with black-feathered birds, their offspring are all slate blue (Blue Andalusian). When Blue Andalusians are crossed among themselves, they produce splashed white, blue, and black offspring in the ratio of 1 : 2 : 1, respectively. (a)Howare these feather traits inherited? (b) Using any appropriate symbols, indicate the genotypes for each phenotype.
9. The shape of radishes may be long (S^LS^L), round (S^RS^R), or oval (S^LS^R). If long radishes are crossed to oval radishes and the F₁ then allowed to cross at random among themselves, what phenotypic ratio is expected in the F₂?
10. Apalomino horse is a hybrid exhibiting a golden color with lighter mane and tail.Apair of codominant alleles (D¹ and D²) is known to be involved in the inheritance of these coat colors. Genotypes homozygous for the D¹ allele are chestnut-colored (reddish), heterozygous genotypes are palomino-colored, and genotypes homozygous for the D² allele are almost white and called cremello. (a) From matings between palominos, determine the expected palomino : nonpalomino ratio among the offspring. (b) What percentage of the nonpalomino offspring in part (a) will breed true? (c) What kind of mating will produce only palominos?

Lethal Alleles Questions

11. Chickens with shortened wings and legs are called "creepers." When creepers are mated to normal birds they produce creepers and normals with equal frequency. When creepers are mated to creepers they produce two creepers to one normal. Crosses between normal birds produce only normal progeny. How can these results be explained?
12. In the Mexican Hairless breed of dogs, the hairless condition is produced by the heterozygous genotype (Hh). Normal dogs are homozygous recessive (hh). Puppies homozygous for the H allele are usually born dead, with abnormalities of the mouth and absence of external ears. If the average litter size at weaning is six in matings between hairless dogs, what would be the average expected number of hairless and normal offspring at weaning from matings between hairless and normal dogs?
13. A pair of codominant alleles regulate cotyledon leaf color in soybeans. The homozygous genotype C^GC^G produces dark-green leaves, the heterozygous genotype C^GC^Y produces light-green leaves, and the other homozygous genotype C^YC^Y produces yellow leaves so deficient in chloroplasts that seedlings do not grow to maturity. If dark-green plants are pollinated only by light-green plants and the F¹ crosses are made at random to produce an F², what phenotypic and genotypic ratios would be expected in the mature F² plants?
14. Thalassemia is a hereditary disease of the blood of humans that results in anemia. Severe anemia (thalassemia major) is found in homozygotes (T^MT^M) and a milder form of anemia (thalassemia minor) is found in heterozygotes (T^MT^N). Normal individuals are homozygous T^NT^N. If all individuals with thalassemia major die before sexual maturity (a) what proportion of the adult F₁ from marriages of thalassemia minors to normals would be expected to be normal, (b) what fraction of the adult F₁ from marriages of minors to minors would be expected to be anemic?

Multiple Alleles Questions

15. A multiple allelic series is known in the Chinese primrose where A (A lexandria type = white eye) > aⁿ (normal type = yellow eye) > a (Primrose Queen type = large yellow eye). List all of the genotypes possible for each of the phenotypes in this series.
16. Plumage color in mallard ducks is dependent upon a set of three alleles: M^R for restricted mallard pattern, M for mallard, and m for dusky mallard. The dominance hierarchy is M^R > M > m. Determine the genotypic and phenotypic ratios expected in the F₁ from the following crosses: (a)M^RM^R × M^RM, (b) M^RM^R × M^Rm, (c)M^RM × M^Rm, (d) M^Rm × Mm, (e) Mm × mm.
17. A number of self-incompatibility alleles are known in clover such that the growth of a pollen tube down the style of a diploid plant is inhibited when the latter contains the same self-incompatibility allele as that in the pollen tube. Given a series of self-incompatibility alleles S¹, S², S³, S⁴, what genotypic ratios would be expected in embryos and in endosperms of seeds from the following crosses?



18. The coat colors of many animals exhibit the "agouti" pattern, which is characterized by a yellow band of pigment near the tip of the hair. In rabbits, a multiple allelic series is known where the genotypes E^DE^D and E^De produce only black (nonagouti), but the heterozygous genotype E^DE produces black with a trace of agouti. The genotypes EE or Ee produce full color, and the recessive genotype ee produces reddish-yellow. What phenotypic and genotypic ratios would be expected in the F₁ and F₂ from the cross (a) E^DE^D × Ee, (b) E^De × ee?
19. The inheritance of coat colors of cattle involves a multiple allelic series with a dominance hierarchy as follows: S > s^h > s^c > s. The S allele puts a band of white color around the middle of the animal and is referred to as a Dutch belt; the s^h allele produces Hereford-type spotting; solid color is a result of the s^h allele; and Holstein-type spotting is due to the s allele. Homozygous Dutch-belted males are crossed to Holstein-type spotted females. The F₁ females are crossed to a Hereford-type spotted male of genotype s^hs^c. Predict the genotypic and phenotypic frequencies in the progeny.
20. The genetics of the ABO human blood groups was presented in Example 2.19. A man of blood group B is being sued by a woman of blood group A for paternity. The woman's child is blood group O. (a) Is this man the father of this child? Explain. (b) If this man actually is the father of this child, specify the genotypes of both parents. (c) If it was impossible for this group B man to be the father of a typeOchild, regardless of the mother's genotype, specify his genotype. (d) If a man was blood group AB, could he be the father of a group O child?

Statistical Distributions Questions

21. Black hair in the guinea pig is dominant to white hair. In families of five offspring where both parents are heterozygous black, with what frequency would we expect to find (a) three whites and two blacks, (b) two whites and three blacks, (c) one white and four blacks, (d) all whites?
22. Assuming that boys and girls are equally frequent in a population, in families of five children, what is the probability of finding (a) three or more boys, (b) three or more boys or three or more girls?
23. A dozen strains of corn are available for a cross-pollination experiment. How many different ways can these strains be paired?
24. Five coat colors in mice are agouti, cinnamon, black, chocolate, and albino. (a) List all of the possible crosses between different phenotypes. (b) Verify the number of different crosses by applying formula (2.1).
25. In mice litters of size eight, determine (a) the most frequently expected number of males and females, (b) the term of the binomial that part (a) represents, (c) the percentage of all litters of size eight expected to have four males and four females.

Dihybrid Crosses with Dominant and Recessive Alleles Questions

26. The position of the flower on the stem of the garden pea is governed by a pair of alleles. Flowers growing in the axils (upper angle between petiole and stem) are produced by the action of a dominant allele T, those growing only at the tip of the stem by its recessive allele t. Colored flowers are produced by a dominant gene C, and white flowers by its recessive allele c. A dihybrid plant with colored flowers in the leaf axils is crossed to a pure strain of the same phenotype. What genotypic and phenotypic ratios are expected in the F₁ progeny?
27. In summer squash, white fruit color is governed by a dominant allele W and yellow fruit color by its recessive allele w. A dominant allele S at another locus produces disk-shaped fruit and its recessive allele s yields sphere-shaped fruit. If a homozygous white disk variety of genotype WWSS is crossed with a homozygous yellow sphere variety wwss, the F₁ are all white disk dihybrids of genotype WwSs. If the F₁ is allowed to mate at random, what would be the phenotypic ratio expected in the F₂ generation?
28. In Drosophila, ebony body color is produced by a recessive gene a and wild-type (gray) body color by its dominant allele a⁺. Vestigial wings are governed by a recessive gene vg, and normal wing size (wild type) by its dominant allele vg⁺. If wild-type dihybrid flies are crossed and produce 256 progeny, how many of these progeny flies are expected in each phenotypic class?
29. Short hair in rabbits is governed by a dominant gene L and long hair by its recessive allele l. Black hair results from the action of the dominant genotype B- and brown from the recessive genotype bb. (a) In crosses between dihybrid shorthaired, black and homozygous short-haired, brown rabbits, what genotypic and phenotypic ratios are expected among their progeny? (b) Determine the expected genotypic and phenotypic ratios in progeny from the cross LlBb × Llbb.
30. The genetic information for the following eight parts is found in Problem 2.51. (a) What phenotypic ratio is expected among progeny from crosses of LlBb × LlBb? (b) What percentage of the F₁ genotypes in part (a) breeds true (i.e., what percentage is of homozygous genotypes)? (c) What percentage of the F₁ genotypes is heterozygous for only one pair of genes? (d) What percentage of the F₁ genotypes is heterozygous at both loci? (e) What percentage of the F₁ genotypes could be used for testcross purposes (i.e., homozygous double-recessive)? (f) What percentage of the F₁ progeny could be used for testcross purposes at the B locus (i.e., homozygous recessive bb)? (g) What percentage of all short-haired F₁ individuals is expected to be brown? (h) What percentage of all black F₁ individuals will breed true for both black and short hair?
31. The presence of feathers on the legs of chickens is due to a dominant allele F and clean legs to its recessive allele f. Pea-comb shape is produced by another dominant allele P and single comb by its recessive allele p. In crosses between pure feathered-leg, single-combed individuals and pure pea-combed, clean-leg individuals, suppose that only the single-combed, feathered-leg F₂ progeny are saved and allowed to mate at random. What genotypic and phenotypic ratios would be expected among the progeny (F₃)?
32. List all the different gametes produced by the following individuals: (a) AA BB Cc (b) aa Bb Cc, (c) Aa Bb cc Dd, (d) AA Bb Cc dd Ee Ff.
33. The normal cloven-footed condition in swine is produced by the homozygous recessive genotype mm. A mule-footed condition is produced by the dominant genotype M-. White coat color is governed by the dominant allele of another locus B and black by its recessive allele b. A white, mule-footed sow (female) is mated to a black, cloven-footed boar (male) and produces several litters. Among 26 offspring produced by this mating, all were found to be white with mule feet. (a) What is the most probable genotype of the sow? (b) The next litter produced eight white, mule-footed offspring and one white, cloven-footed pig. Now, what is the most probable genotype of the sow?
34. A white, mule-footed boar (see Problem 2.55) is crossed to a sow of the same phenotype. Among the F₁ offspring there were found to be 6 white, cloven-footed : 7 black, mule-footed : 15 white, mule-footed : 3 black, cloven-footed pigs. (a) If all the black, mule-footed F₁ offspring from this type of mating were to be testcrossed, what phenotypic ratio would be expected among the testcross progeny? (b) If the sow were to be testcrossed, what phenotypic ratio of progeny would be expected?

Modified Dihybrid Ratios Questions

35. In peaches, the homozygous genotype G^OG^O produces oval glands at the base of the leaves, the heterozygous genotype G^OG^A produces round glands, and the homozygous genotype G^AG^A results in the absence of glands. At another locus, a dominant gene S produces fuzzy peach skin and its recessive allele s produces smooth (nectarine) skin. A homozygous variety with oval glands and smooth skin is crossed to a homozygous variety with fuzzy skin and lacking glands at the base of its leaves. What genotypic and phenotypic proportions are expected in the F₂?
36. In Shorthorn cattle, coat colors are governed by a codominant pair of alleles C^R and C^W. The homozygous genotype C^RC^R produces red, the other homozygote produces white, and the heterozygote produces roan (a mixture of red and white). The presence of horns is produced by the homozygous recessive genotype pp and the polled condition by its dominant allele P. If roan cows heterozygous for the horned gene are mated to a horned, roan bull, what phenotypic ratio is expected in the offspring?
37. A gene locus with codominant alleles is known to govern feather color in chickens such that the genotype F^BF^B = black, F^WF^W =splashed white, and F^BF^W = blue. Another locus with codominant alleles governs feather morphology such that M^NM^N = normal feather shape, M^NM^F = slightly abnormal feathers called "mild frizzle," and M^FM^F = grossly abnormal feathers called "extreme frizzle." If blue, mildly frizzled birds are crossed among themselves, what phenotypic proportions are expected among their offspring?
38. In the above problem, if all the blue offspring with normal feathers and all the splashed-white, extremely frizzled offspring are isolated and allowed to mate at random, what phenotypic ratio would be expected among their progeny?
39. The shape of radishes may be long (LL), round (L'L'), or oval (LL'). Color may be red (RR), white (R'R'), or purple (RR'). If a long, white strain is crossed with a round, red strain, what phenotypic proportions are expected in the F₁ and F₂?
40. Suppose that two strains of radishes are crossed (see the above problem) and produce a progeny consisting of 16 long white, 31 oval purple, 16 oval white, 15 long red, 17 oval red, and 32 long purple. What would be the phenotypes of the parental strains?
41. A dominant gene K in mice produces a kinked tail; recessive genotypes at this locus kk have normal tails. The homozygous condition of another locus AA produces a gray color called agouti; the heterozygous condition A^yA produces yellow color; the homozygous genotype A^yA^y is lethal. (a) If yellow mice, heterozygous for kinky tail, are crossed together, what phenotypic proportions are expected in their offspring? (b) What proportion of the offspring is expected to be of genotype A^yAKk? (c) If all the yellow offspring were allowed to mate at random, what would be the genotypic and phenotypic ratios among their adult progeny?
42. An incompletely dominant gene N in the Romney Marsh breed of sheep causes the fleece of homozygotes to be "hairy," i.e., containing fibers lacking the normal amount of crimp. Normal wool is produced by the homozygous genotype N'N'. Heterozygotes NN' can be distinguished at birth by the presence of large, medulated fibers called "halo-hairs" scattered over the body. A gene known as "lethal gray" causes homozygous gray fetuses (G^lG^l) to die before 15 weeks in gestation. The heterozygous genotype G¹G produces gray fleece, and the homozygous genotype GG produces black. If heterozygous halo-haired, gray individuals are mated together, (a) what would be the phenotypic proportions expected in the live progeny, (b) what proportion of the live progeny would carry the lethal gene, (c) what proportion of the live progeny with halo-hairs would carry the lethal gene, (d) what proportion of all the zygotes would be expected to be of genotype NN'G^lG^l?
43. Tay-Sachs disease is a recessive hereditary abnormality in humans causing death within the first few years of life only when homozygous (ii). The dominant condition at this locus produces a normal phenotype (I-). Abnormally shortened fingers (brachyphalangy) is thought to be due to a genotype heterozygous for a lethal gene (BB^L), the homozygote (BB>) being normal, and the other homozygote (B^LB^L) being lethal. What are the phenotypic expectations among teenage children from parents who are both brachyphalangic and heterozygous for Tay-Sachs disease?
44. In addition to the gene governing Tay-Sachs disease in the above problem, the recessive genotype of another locus (jj) results in death before age 18 due to a condition called Spielmeyer-Vogt disease. Only individuals of genotype I- J will survive to adulthood, (a) What proportion of the children from parents of genotype IiJj would probably not survive to adulthood? (b) What proportion of the adult survivors in part (a) would not be carriers of either hereditary abnormality'?
45. A genetic condition on chromosome 2 in the fruit fly Drosophila melanogaster is lethal when homozygous (Pm/Pm), but when heterozygous (Pm/Pm⁺) produces a purplish eye color called "plum." The other homozygous condition (Pm⁺/Pm⁺) produces wild-type eye color. On chromosome 3, a gene called "stubble" produces short, thick bristles when heterozygous (Sb/Sb⁺) but is lethal when homozygous (Sb/Sb). The homozygous condition of its alternative allele (Sb⁺/Sb⁺), produces bristles of normal size (wild type). (a) What phenotypic ratio is expected among progeny from crosses between plum, stubble parents? (b) If the offspring of part (a) are allowed to mate at random to produce an F₂, what phenotypic ratio is expected?
46. Feather color in chickens is governed by a pair of codominant alleles such that F^BF^B produces black, F^WF^W produces splashed white, and F^BF^W produces blue. An independently segregating locus governs the length of leg; CC genotypes possess normal leg length, CC^L genotypes produce squatty, short-legged types called "creepers," but homozygous C^LC^L genotypes are lethal. Determine the kinds of progeny phenotypes and their expected ratios that crosses between dihybrid blue creepers are likely to produce.

Higher Combinations Questions

47. The seeds from Mendel's tall plants were round and yellow, all three characters due to a dominant gene at each of three independently assorting loci. The recessive genotypes dd, ww, and gg produce dwarf plants with wrinkled and green seeds, respectively. (a) If a pure tall, wrinkled, yellow variety is crossed with a pure dwarf, round, green variety, what phenotypic ratio is expected in the F₁ and F₂? (b) What percentage of the F₂ is expected to be of genotype DdWWgg? (c) If all the dwarf, round, green individuals in the F₂ are isolated and artificially crossed at random, what phenotypic ratio of offspring is expected?
48. The coat colors of mice are known to be regulated by several genes. The presence of a yellow band of pigment near the tip of the hair is called "agouti" pattern and is produced by the dominant allele A. The recessive condition at this locus (aa) does not have this subapical band and is termed nonagouti. The dominant allele of another locus B produces black and the recessive genotype bb produces brown. The homozygous genotype c^hc^h restricts pigment production to the extremities in a pattern called Himalayan, whereas the genotype C- allows pigment to be distributed over the entire body. (a) In crosses between pure brown, agouti, Himalayan, and pure black mice, what are the phenotypic expectations of the F₁ and F₂? (b) What proportion of the black-agouti, full-colored F₂ would be expected to be of genotype AaBBCc? (c) What percentage of all the Himalayans in the F₂ would be expected to show brown pigment? (d) What percentage of all the agoutis in the F₂ would be expected to exhibit black pigment?
49. In addition to the information given in the problem above, a fourth locus in mice is known to govern the density of pigment deposition. The genotype D- produces full color, but the recessive genotype dd produces a dilution of pigment. Another allele at this locus, dl is lethal when homozygous, produces a dilution of pigment in the genotype dd^l, and produces full color when in heterozygous condition with the dominant allele Dd^l. (a) What phenotypic ratio would be expected among the live F₂ progeny if the F₁ from the cross aabbCCDd × AABBccdd^l were allowed to mate at random? (b) What proportion of the live F₂ would be expected to be of genotype AABbccdd_l?
50. In the parental cross AABBCCDDEE × aabbccddee, (a) how many different F₁ gametes can be formed, (b) how many different genotypes are expected in the F₂, (c) how many squares of a Punnett square would be necessary to accommodate the F₂?
51. A pure strain of Mendel's peas, dominant for all seven of his independently assorting genes, was testcrossed. (a) How many different kinds of gametes could each of the parents produce? (b) How many different gametes could the F₁ produce? (c) If the F1 was testcrossed, how many phenotypes would be expected in the offspring and in what proportions? (d) How many genotypes would be expected in the F₂? (e) How many combinations of F1 gametes are theoretically possible (considering, e.g., AABBCCDDEEFFGG sperm nucleus × aabbccddeeffgg egg nucleus, a different combination than AABBCCDDEEFFGG egg nucleus × aabbccddeeffgg sperm nucleus)? ( f ) How many different kinds of matings could theoretically be made among the F₂? [Hint: See solution to Problem 2.14(c)].

Testing Genetic Ratios Questions

52. Determine the number of degrees of freedom when testing the ratios (a) 3 : 1 (b) 9 : 3 : 3 :1 (c) 1 : 2 : 1 (d ) 9 :3 : 4. Find the number of degrees of freedom in applying a chi-square test to the results from (e) testcrossing a dihybrid, ( f ) testcrossing a trihybrid, (g) trihybrid × trihybrid cross.
53. Two phenotypes appear in an experiment in the ratio 16 : 4. (a) How well does this sample fit a 3 : 1 ratio? Would a sample with the same proportional deviation fit a 3 : 1 ratio if it were (b) 10 times larger than (a), (c) 20 times larger than (a)?
54. The flowers of four o'clock plants may be red, pink, or white. Reds crossed to whites produced only pink offspring. When pink-flowered plants were crossed they produced 113 red, 129 white, and 242 pink. It is hypothesized that these colors are produced by a single-gene locus with codominant alleles. Is this hypothesis acceptable on the basis of a chisquare test?
55. A heterozygous genetic condition called "creeper" in chickens produces shortened and deformed legs and wings, giving the bird a squatty appearance. Matings between creepers produced 775 creeper : 388 normal progeny. (a) Is the hypothesis of a 3 : 1 ratio acceptable? (b) Does a 2 : 1 ratio fit the data better? (c) What phenotype is probably produced by the gene for creeper when in homozygous condition?
56. Among fraternal (nonidentical, dizygotic) twins, the expected sex ratio is 1MM: 2MF : 1 FF (M = male, F = female). A sample from a sheep population contained 50 MM, 142 MF, and 61 FF twin pairs. (a) Do the data conform within statistically acceptable limits to the expectations? (b) If identical (monozygotic) twin pairs = total pairs × (2 × MF pairs), what do the data indicate concerning the frequency of monozygotic sheep twins?
57. A total of 320 families with six children each were surveyed with the results shown below. Does this distribution indicate that boys and girls are occurring with equal frequency?



58. In guinea pigs, it is hypothesized that a dominant allele L governs short hair and its recessive allele l governs long hair. Codominant alleles at an independently assorting locus are assumed to govern hair color, such that C^yC^y = yellow, C^yC^w = cream, and C^wC^w = white. From the cross Ll C^yC^w × Ll C^yC^w, the following progeny were obtained: 50 short cream : 21 short yellow : 23 short white : 21 long cream : 7 long yellow : 6 long white. Are the data consistent with the genetic hypothesis?
59. Observations of 30 : 3 in a genetic experiment are postulated to be in conformity with a 3 : 1 ratio. Is a 3 : 1 ratio acceptable at the 5% level on the basis of (a) an uncorrected chi-square test, (b) a corrected chi-square test? [Hint: Corrected indicates use of Yates correction factor of continuity.]

Pedigree Analysis Questions

60. The phenotypic expression of a dominant gene in Ayrshire cattle is a notch in the tips of the ears. In the pedigree below, where solid symbols represent notched individuals, determine the probability of notched progeny being produced from the matings (a) III1 × III3 (b) III2 × III3 (c) III3 × III4 (d) III1 × III5 (e) III2 × III5



61. A multiple allelic series in dogs governs the distribution of coat-color pigments. The allele A^s produces an even distribution of dark pigment over the body; the allele a^y reduces the intensity of pigmentation and produces sable or tan-colored dogs; the allele at produces spotted patterns such as tan and black, tan and brown, etc. The dominance hierarchy is A^s > a^y > a^t. Given the following family pedigree, (a) determine the genotypes of all the individuals insofar as possible, (b) calculate the probability of spotted offspring being produced by mating III1 by III2, (c) find the fraction of the dark-pigmented offspring from I1 × III3 that is expected to be heterozygous.

Answers

Matching Questions 1

1. F   2. C   3. A   4. H   5. B   6. D   7. G   8. E

Matching Questions 2

1. H   2. C   3. C   4. A   5. F   6. A   7. B   8. B

Vocabulary

1. phenotype
2. genotype
3. heterozygote (heterozygous cell)
4. homozygote (homozygous cell)
5. recessive
6. mutant type
7. codominant, incompletely dominant, partially dominant, semidominant
8. lethal gene
9. penetrance
10. expressivity

Multiple-Choice

1. c
2. e
3. d
4. b
5. b
6. c
7. a
8. e (10 genotypes)
9. d
10. c
11. e (27)
12. c
13. e (1/6)
14. e (2/9)

Dominant and Recessive Alleles

1. Bb × Bb
2. (a) bb = white   (b) Bb
3. 1/2 wild-type s⁺s : 1/2 sepia ss
4. (a) 1/4   (b) 1/16   (c) 2(3/4 × 1/4) = 3/8
5. (a) Ll female × ll male   (b) 1 short : 1 long   (c) 4
6. 1/6
7. (a) 1/2 BB : 1/2 Bb-all red   (b) 1/2 Bb = red : 1/2 bb = silver-black   (c) all Bb = red

Codominance and Incomplete Dominance

8. (a) Single pair of codominant alleles   (b) F^SF^S = splashed-white : F^SF^B = Blue Andalusian : F^BF^B = black
9. 9/16 long : 6/16 oval : 1/16 round
10. (a) 1 palomino : 1 nonpalomino   (b) 100%; D¹D¹ × D¹D¹ = all D¹D¹(chestnut); similarly D²D² × D2D2 = all D²D²(cremello)   (c) D¹D¹(chestnut) × D²D²(cremello)

Lethal Alleles

11. Creepers are heterozygous. Normal birds and lethal zygotes are homozygous for alternative alleles. One of the alleles is dominant with respect to the creeper phenotype; the other allele is dominant with respect to viability.
12. 4 normal : 4 hairless
13. 9/15 dark-green C^GC^G : 6/15 light-green C^GC^Y
14. (a) 1/2   (b) 2/3

Multiple Alleles

15. Alexandria type (white eye) = AA, Aaⁿ, Aa; normal type (yellow eye) = aⁿaⁿ, aⁿa; Primrose Queen type (large yellow eye) = aa
16. (a) 1/2 M^RM^R : 1/2 M^RM; all restricted   (b) 1/2 M^RM^R : 1/2 M^Rm; all restricted   (c) 1/4 M^RM^R : 1/4 M^Rm : 1/4 M^RM : 1/4 Mm; 3/4 restricted : 1/4 mallard   (d) 1/4 M^RM : 1/4 M^Rm : 1/4 Mm: 1/4 mm; 1/2 restricted : 1/4 mallard : 1/4 dusky   (e) = Mm = mallard : 1/2 mm = dusky
17. (a) Embryos = 1/2 S¹S³ : 1/2 S³S⁴

Endosperms = 1/2 S¹S¹S³ : 1/2 S⁴S⁴S³

(b) None

(c) Embryos = 1/4 S¹S² : 1/4 S¹S⁴ : 1/4 S³S² : 1/4 S³S⁴

Endosperms = 1/4 S¹S¹S² : 1/4 S¹S¹S⁴ : 1/4 S³S³S² : 1/4 S³S³S⁴

(d) Embryos = 1/2 S²S⁴ : 1/2 S³S⁴;

Endosperm = 1/2 S²S²S⁴ : 1/2 S³S³S⁴

18. (a) F₁ = 1/2 E^DE (black with trace of agouti) : 1/2 E^De (nonagouti black)

F₂ = 1/4 E^DE^D : 1/4 E^De : 1/4 E^DE : 1/16 EE : 1/8 Ee : 1/16 ee

1/2 nonagouti black : 1/4 black with trace of agouti : 3/16 full color : 1/16 reddish-yellow

(b) F₁ = 1/2 E^De (nonagouti black) : 1/2 ee (reddish – yellow)

F₂ = 1/16 E^DE^D : 3/8 E^De : 9/16 ee; 7/16 nonagouti black : 9/16 reddish-yellow

19. 1/4 Ss^h : 1/4 Ss^c: 1/4 s^hs: 1/4 s^cs; 1/2 Dutch-belted : 1/4 Hereford-type spotting : 1/4 solid color
20. (a) The man could be the father, but paternity cannot be proved by blood type. In certain cases, a man may be excluded as a father of a child [see part (d)].   (b) I^Bi man × I^Ai woman   (c) I^BI^B   (d) no

Statistical Distribution

21. (a) 90/1024   (b) 270/1024   (c) 405/1024   (d) 1/1024
22. (a) 1/2 (b) 1
23. 66
24. (a) (1) agouti × cinnamon (2) agouti × black (3) agouti × chocolate (4) agouti × albino (5) cinnamon × black (6) cinnamon × chocolate (7) cinnamon × albino (8) black× chocolate (9) black × albino (10) chocolate × albino
25. (a) 4 males : 4 females   (b) 5th   (c) 27.34%

Dihybrid Crosses With Dominant and Recessive Alleles

26. 1/4 CCTT : 1/4 CCTt : 1/4 CcTT : 1/4 CcTt; all axial, colored
27. 9/16 white, disk : 3/16 white, sphere : 3/16 yellow, disk : 1/16 yellow, sphere
28. 144 wild type : 48 vestigial : 48 ebony : 16 ebony, vestigial
29. (a) 1/4 LLBb : 1/4 LlBb : 1/4 LLbb : 1/4 Llbb; 1/2 short, black : 1/2 short, brown   (b) 1/8 LLBb : 1/4 LIBb : 1/8 LLbb : 1/4 Llbb : 1/8 llBb : 1/8 llbb; 3/8 short, black : 3/8 short, brown : 1/8 long, black : 1/8 long, brown
30. (a) 9/16 short, black : 3/16 short, brown : 3/16 long, black : 1/16 long, brown   (b) 25%   c) 50%   (d) 25%   (e) 6.25%   (f) 25%   (g) 25%   (h) 8.33%
31. 4 FFpp : 4 Ffpp : 1 ffpp; 8 feathered leg, single comb : 1 clean leg, single comb
32. (a) ABC, ABc   (b) aBC, aBc, abC, abc   (c) ABcD, ABcd, AbcD, Abcd, aBcD, aBcd, abcD, abcd   (d) ABCdEF, ABCdEf, ABCdeF, ABCdef, ABcdEF, ABcdEf, ABcdeF, ABcdef, AbCdEF, AbCdEf, AbCdeF, AbCdef, AbcdEF, AbcdEf, AbcdeF, Abcdef
33. (a) BBMM   (b) BBMm
34. (a) 2 black, mule-foot : 1 black, cloven-foot   (b) 1/4 white, mule-foot : 1/4 white, cloven-foot : 1/4 black, mule-foot : 1/4 black, cloven-foot

Modified Dihybrid Ratios

35. 1/16 G^AG^ASS : 2/16 G^AG^ASs : 1/16 G^AG^Ass : 2/16 G^A G^OSS : 4/16 G^AG^OSs : 2/16 G^AG^Oss : 1/16 G^OG^OSS : 2/16 G^OG^OSs : 1/16 G^OG^Oss; 3/16 fuzzy, glandless : 1/16 smooth, glandless : 6/16 round gland, fuzzy : 2/16 round gland, smooth : 3/16 oval gland, fuzzy : 1/16 oval gland, smooth
36. 1 red, polled : 1 red, horned : 2 roan, polled : 2 roan, horned : 1 white, polled : 1 white, horned
37. 1/16 black : 1/8 black, mildly frizzled : 1/16 black, extremely frizzled : 1/8 blue : 1/4 blue, mildly frizzled : 1/8 blue, extremely frizzled : 1/16 splashed-white : 1/8 splashed-white, mildly frizzled : 1/16 splashed-white, extremely frizzled
38. 1 black : 2 blue : 1 splashed-white : 2 blue, mildly frizzled : 2 splashed-white, mildly frizzled : 1 splashed white, extremely frizzled
39. F₁ is all oval, purple; F₂ is 1/16 long, red : 1/8 long, purple : 1/16 long, white : 1/8 oval, red : 1/4 oval, purple : 1/8 oval, white : 1/16 round, red : 1/8 round, purple : 1/16 round, white
40. Long; purple × oval; purple
41. (a) 1/2 yellow, kinky : 1/6 yellow : 1/4 agouti, kinky : 1/12 agouti   (b) 1/3   (c) 1/6 A^yAKK : 1/3 A^yAKk : 1/6 A^yAkk : 1/6 AAKK : 1/6 AAKk : 1/12 AAkk; 1/2 yellow, kinky : 1/6 yellow : 1/4 agouti, kinky : 1/12 agouti
42. (a) 1/12 black, hairy : 1/6 black, halo-haired : 1/12 black : 1/6 gray, hairy :1/3 gray, halo-haired : 1/6 gray   (b) 2/3   (c) 2/3   (d) 1/8
43. 1/3 normal : 2/3 brachyphalangic
44. (a) 7/16   (b) 1/9
45. (a) 4/9 plum, stubble : 2/9 plum : 2/9 stubble : 1/9 wild type   (b) 1 plum, stubble : 1 plum : 1 stubble : 1 wild type
46. 1/12 black : 1/6 blue : 1/12 splashed-white : 1/6 black, creeper : 1/3 blue, creeper : 1/6 splashed-white, creeper

Higher Combinations

47. (a) F₁ is all tall, round, yellow; F₂ is 27 tall, round, yellow : 9 tall, round, green : 9 tall, wrinkled, yellow : 9 dwarf, round, yellow : 3 tall, wrinkled, green : 3 dwarf, round, green : 3 dwarf, wrinkled, yellow : 1 dwarf, wrinkled, green   (b) 3.12%   (c) 8 round : 1 wrinkled
48. (a) F₁ is all agouti, black; F₂ is 27 agouti, black : 9 agouti, black, Himalayan : 9 agouti, brown : 9 black : 3 agouti, brown, Himalayan : 3 black, Himalayan : 3 brown : 1 brown, Himalayan   (b) 4/27   (c) 25%   (d) 75%
49. (a) 189 agouti, black : 216 agouti, black, dilute : 63 agouti, black, Himalayan : 72 agouti, black, Himalayan, dilute : 63 agouti, brown : 72 agouti, brown, dilute : 63 black : 72 black, dilute : 21 agouti, brown, Himalayan : 24 agouti, brown, Himalayan, dilute : 21 black, Himalayan : 24 black, Himalayan, dilute : 21 brown : 24 brown, dilute : 7 brown, Himalayan : 8 brown, Himalayan, dilute   (b) 1/120
50. (a) 2⁵ = 32   (b) 35 = 243   (^c) 45 = 1024
51. (a) One each   (b) 128   (c) 128, each with equal frequency   (d) 2187   (e) 16,384   (f) 2,392,578

Testing Genetic Ratios

52. (a) 1   (b) 3   (c) 2   (d) 2   (e) 3  (f) 7   (g) 7
53. (a) x² = 0:27; p = 0:5 – 0:7; acceptable   (b) x² = 2:67; p = 0:1 – 0:2; acceptable   (c) x² = 5:33; p = 0:01 – 0:05; not acceptable.
54. Yes; x² = 1:06; p = 0:5 – 0:7; acceptable
55. (a) x² = 43:37; p < 0:001; not acceptable   (b) x² = 0:000421; p > 0:95; a 2 : 1 ratio fits the data almost perfectly (c) lethal
56. (a) x² = 4:76; 0:10 > p > 0:05; hypothesis acceptable   (b) Monozygotic twins are estimated to be –31; the negative estimate indicates that identical sheep twins are rare provided that unlike-sex twins do not have a survival advantage over like-sex twins.
57. Yes; x² = 2:83; p = 0:8 – 0:9; the distribution is consistent with the assumption that boys and girls occur with equal frequency.
58. Yes; x² = 2:69; p = 0:7 – 0:8
59. (a) No; x² = 4:45; p < 0:05 ´ (b) Yes; x² = 3:64; p > 0:05

Pedigree Analysis

60. (a) 0   (b) 1/2  (c) 0  (d) 1/2   (e) 3/4
61. (a) I1 = A^sa^y; I2 = a^ta^t; II1 = a^ya^t; III2 = a^ya^t; III3 = A^sa^t; ^II4 = a^ya^t; III1 = a^ta^t; III2 = A^s – (A^sa^y or A^sa^t); III3 = a^yat; III4 = a^ta^t   (b) 1/4   (c) 2/3