PART ONE: Monohybrid Crosses/Complete Dominance
NOTE: Assume complete dominance.
1. In fruit flies, normal wings (V) is dominant over vestigial wings (v). The flies produced in one mating (i.e. the F1 generation) included 734 normal winged flies and 598 vestigial winged flies. What are the genotypes and phenotypes of the parent flies?
Vv x vv
2. In cats, short hair (S) is dominant over Angora (long) hair (s). An Angora male is mated with a short haired female. She produces a litter of eight kittens, six of which are short-haired and two of which are Angora. What was the expected phenotypic ratio for this mating?
2:2
If these same two parent cats, when mated, produced a total of fifty (50) kittens in successive litters, how would the expected ratio be different?
The expected ratio would remain the same but the actual number of kittens would more likely reflect the expected ratio.
3. Albinism is a recessive skin condition caused by a lack of skin pigments. An albino man marries a woman with normal skin pigmentation. However, this normal woman had an albino mother. Using a Punnett square, show the different phenotypes possible for their children and the relative proportions of each (i.e .a phenotypic ratio).
Aa / aaAa / aa
2:2
4. In cattle, hornless (H) is completely dominant over the horned condition (h). A hornless bull is bred with three different cows. When mated with cow A, a horned cow, a hornless calf is produced. When mated with cow B, another horned cow, a horned calf is produced. With cow C, a hornless cow, a horned calf is also produced. What are the genotypes of the bull and the three cows?
Bull Hh Cow A, hh Cow B hh Cow C Hh
5. Twenty five percent (25%) of the F1 generation produced by mating two white sheep were black. What are the genotypes of the parent sheep?
Ww x Ww
6. Rough coated guinea pigs, when crossed with smooth-coated guinea pigs, produce only rough-coated offspring in the F1 generation. What phenotypes would be expected from the following crosses? (1) Crossing two F1 offspring (2) Back crossing an F1 individual with a homozygous recessive parent.
1. rough and smooth (3:1)
2. rough and smooth (2:2)
7. A blue-eyed man, both of whose parents were brown-eyed, marries a brown-eyed woman whose father was brown-eyed and whose mother was blue-eyed. They have only one child, who is blue-eyed. List the genotypes for all of the individuals listed above.
Grandpa Bb Dad bb Child bb
Grandma Bb mom Bb
8. In Holstein cattle, black or white solid coats (S) are dominant over spotted (s) coats. Give the genotypes of the parent cattle and complete a Punnett square for each of the following crosses: (1) Spotted coat x Spotted coat
bb x bb = all bb
(2) Heterozygous solid coat x Spotted coat
Bb x bb = 2:2
(3) Heterozygous solid coat x Heterozygous solid coat
Bb x Bb = 3:1
9. In goats, black coat color (B) is dominant over white coat color (b). A black-coated goat is mated with a white-coated goat. Over the course of time, a large number of offspring are produced and all of the offspring are black. What is the most likely genotype for the black-coated goat?
BB
10. Suppose the goats from question #10 were mated but only half of the offspring were black. What is the most likely genotype for the black-coated goat?
Bb
11. A normally pigmented man whose father was an albino marries an albino woman, both of whose parents were normally pigmented. Of the resulting three children, two are normal and one is albino. List the genotypes of all eight individuals listed above.
Albino father: aa man: Aa Albino woman: aa
Parents of albino woman: Aa and Aa Three children: Aa (2) and aa (1)
Part Two: Incomplete Dominance
12. In shorthorn cattle, the heterozygous condition of one red allele (R) and one white allele (r) is roan. If two roan shorthorn cattle are mated, what chance will their offspring have of resembling their parents in coat color?
50%
13. A breeder of shorthorn cattle has cows which are white, and a bull which is roan. What proportion of the calves produced in his herd will be... White? Roan? Red?
0%, 100%, 0%
14. Of the calves produced in question #18, which coat color(s) would represent homozygous individuals?
None, all offspring are roan and a mixture of the two homozygous parents.
15. All of the offspring produced by crossing red tulips with white tulips are pink. Based upon this information, what proportions of flower colors will be produced in each of the following crosses?
Assume Red, Pink, White
Pink x Pink 1:2:1 Pink x White 0:2:2
Red x White 0:4:0 Pink x Red 2:2:0
16. After several matings of tan-colored birds, the following offspring resulted: 23 white, 26 brown and 53 tan birds. Show a Punnett square that would produce all tan offspring. Label the parents.
Parents BB and WW (both homozygous)
17. In Andalusian fowl, black feathers (B) are incompletely dominant to white (b) feathers. However, the heterozygous genotype (Bb) produces a blue feather color. What offspring phenotypes may be produced in the F1 generations from the following crosses? Show your Punnett squares for each cross.
Assume: Black, Blue, White
Blue x Black Blue x Blue Blue x White
2:2:0 1:2:1 0:2:2
18. In cats, yellow coat color is due to a sex-linked gene (Y) and black color is due to its allele (y). The heterozygous genotype (Yy) produces a color known as tortoise shell. What phenotypes may be present in the F1 generation produced by crossing a black male with a tortoise shell female?
Y x Yy = ¼ female tortoise, ¼ females yellow, ¼ males yellow, ¼ males black
Part Three: Dihybrid Crosses with Complete Dominance
NOTE: Assume complete dominance for all alleles in these crosses.
19. In horses, the allele for black coats (B) is dominant to its allele for chestnut coats (b). In addition, the allele for a trotting gait (T) is dominant to its allele for a pacing gait (t). What phenotypes may be present in the F1 generation resulting from the cross of a homozygous black pacer with a homozygous chestnut trotter? Include a Punnett square with your answer.
BBtt x bbTT = 100% BbTt Black Trotter
20. Using two individuals from the F1 generation in question #25, what phenotypes would be present in the F2 generation resulting from the cross of these individuals? Include a Punnett square with your answer.
9 3 3 1
Black trotter Black Pacer Chestnut Trotter Chestnut Pacer
21. If one of the F1 offspring from question #25 were crossed with a homozygous black pacer, what phenotypes would be present in the resulting F2 generation? Include a Punnett square with your answer.
BT Bt bT bt
Bt BBTt BBtt BbTt Bbtt
8:8:0:0
22. In fruit flies, the allele for grey body color (G) is dominant over the allele for black body color (g). In addition, long wings (L) are dominant over short (l) wings. A cross is made between a homozygous gray, short-winged male and a homozygous black, long-winged individual. A male and female from the resulting F1 generation cross are mated. What different phenotypes would be expected in the resulting F2 generation? Include a phenotypic ratio with your answer. 9:3:3:1
23. In guinea pigs, pigmented (P) is dominant to albino (p) and rough coats (R) are dominant to smooth coats (r). An albino (pure white) guinea pig with a rough coat is mated to a black guinea pig with a smooth coat. In a particular cross, all of the resulting offspring were black and rough. Based on these results and the information you have, do these results match the expected phenotypic ratios? Explain using a Punnett square as part of your answer. If all the pigs were black and rough then the genotype of the albino is aaRR and the black is AArr.
24. In summer squash, white fruits (W) are dominant to yellow fruits (w) while flat fruits (F) are dominant to rounded fruits (f). The selected parents are a white, flat fruit (WWFF) and a plant with yellow and rounded fruits. What would the phenotypic and genotypic ratios be for the following:
-The F1 generation 16:0:0:0
-The F2 generation 9:3:3:1
25. In humans, brown eye color (B) is dominant over blue (b) while right-handedness (R) is dominant over left-handedness (r). A brown-eyed, right-handed man marries a blue-eyed, right-handed woman. Their first child is blue-eyed and left-handed. List the genotypes of the parents. BbRr x bbRr
26. In corn, plump grains (P) are dominant over shriveled grains (p) while black grains (B) are dominant over white grains (b). What phenotypic and genotypic ratios would results from the following crosses?
-Homozygous black & plump x Homozygous white & shriveled
16:0:0:0
-Heterozygous black & plump x Heterozygous black and plump
9:3:3:1
-PPBb x Ppbb
8:8:0:0
-ppbb x ppBb
0:0:8:8
-PpBb x ppBb
6:2:6:2
27. In guinea pigs, short hair (S) is dominant over long hair (s) and black hair (B) is dominant over white hair (b). If a homozygous black, short-haired guinea pig mates with a white, long-haired pig, what phenotypes may appear in the resulting F1 generation? Show your Punnett square?
BBSS x bbss = 100% BbSs
Part Four: Sex-Linked Traits
28. Color-blindness is a sex-linked character. The gene for normal eye condition (N) is dominant over the gene for color blindness (n). In humans, the sex chromosomal condition of XX indicates a female and XY indicates a male. What phenotype-sex combinations may result in the offspring of a colorblind male and a female who lacks any recessive alleles for color blindness? Show your Punnett square with your answer.
50% women are heterozygous carriers
100% of the men are normal vision
29. Using one of the female offspring from question #38, what phenotype-sex combinations may result when this female is crossed with a normal male?
½ of the females would be normal
½ of the females would be a carrier
½ of the males would be color blind
½ of the males would be normal vision
30. Two normal visioned parents produce a color-blind son. What are the genotypes of the parents? What are the chances of their next child being a color-blind female?
Zero percent
31. In a certain family there are six girls. What are the chances of the next child being a boy? 50%
32. In humans, albinism (a) is recessive to normal pigmentation (A). Hemophelia (h) is also recessive to normal blood clotting (H). If an albino, non-hemophiliac man and a normally-pigmented, non-hemophiliac female whose father was a hemophiliac and whose mother was an albino marry and have children, what kind of phenotypes can their children have and in what proportions?
Female Phenotypes Male Phenotypes
25% normal for both traits 25% will be normal for both traits
25% normal but albino 25% normal but albino
25% carrier for hemo and normal skin 25% hemophilic normal skin
25% carrier for hemo, albino 25% hemophilic, albino
33. In fruit flies, white eyes (W) is a sex-linked gene, with red eyes (w) being the normal condition. In one cross, the following numbers of flies of four different phenotypes were produced: 273 red-eyed males, 258 white eyed males, 304 red eyed females, 197 white eyed females. Based on this information, what were the genotypes and phenotypes of the parent flies? Mom heterozygous Dad recessive
34. In another fruit fly crossing similar traits to the one in the previous item (but with different parent), the following number of flies were produced: 262 white eyed males, 0 red eyed males, 0 white eyed females and 243 red eyed females. Based on this information, what were the genotypes and phenotypes of the parent flies.?
Mom heterozygous, Dad dominant
35. Hemophilia, the "bleeder's disease", is associated with an X-linked recessive allele (h). The daughter of a hemophiliac man is not a bleeder herself. She marries a normal man. What is the probability that their child will be affected by hemophilia? Use a Punnett square in your answer.
¼ chance overall (1/2 if male)
Part Five-Multiple Alleles
39. In humans, the blood types A, B, AB and O are determined by three different alleles IA, IB and i.
What phenotypes would be anticipated in the offspring of the following marriages?
-O x A -O x B -O x AB -A x B -AB x B -AB x A
homo A B A and B AB AB B AB A
Hetero A and O B and O A and B AB and O AB, B, A AB, A, B
40. Four babies were born in a hospital on the same night, and their blood groups were later found to be O, A, B and AB. The four pairs of parent were: OxO, ABxO, AxB, BxB. Assign each baby to its parents, showing your Punnett squares with your work.
OxO parent belong to baby O
AxB parents belong to baby AB
BxB parents belong to baby B
ABxO parents belong to baby A
41. A wealthy, elderly couple died together in an accident. Soon, a man showed up to claim his inheritance, contending that he was their son who ran away from home when he was a boy. Other relatives disputed this claim. Hospital records showed that the deceased couple were types AB and O respectively. The claimant to the fortune was type O. Is the claimant an imposter? Prove your conclusion.
Yes he is an imposter, the only blood types that can be produced from this couple was A and B. He is O—no conceivable way.