| GENETICS PROBLEMS - Part 2 | |||||||||
| BIOL 1406 | |||||||||
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Monohybrid problems: | |||||||||
| 1. In fruit flies (Drosophila melanogaster), red-eye is dominant to white-eye. By crossing a heterozygous red-eyed male fly to a white-eyed female, 60 offspring were produced. How many would you expect to be red-eyed and how many white-eyed? | |||||||||
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50% or approximately 30 would be red-eyed 50% or approximately 30 would be white-eyed
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| 2. Albinism occurs commonly in animals and it is always recessive to the normal. Six brown and five albino mice were born to parents (one brown and one albino). What is the genotype of the brown parent? | |||||||||
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Brown parent must be heterozygous to produce approximately 50% albino offspring
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| 3. Supposing brown eyes (melanin pigment present) in
people (B) is dominant to blue (b). Could a marriage between two
blue-eyed people produce a brown-eyed child? Could a marriage
between a homozygous brown-eyed person and a blue-eyed person result in
blue-eyed children? Could two brown-eyed people have a blue-eyed
child? Explain all of your answers by supplying the proper symbols.
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2 Blue-eyed people produce brown-eyed child = no bb X bb cannot produce B_
Homozygous brown X blue produce blue-eyed child = no BB X bb cannot produce bb
2 Brown-eyed parents have a blue-eyed child = yes If they are both heterozygous Bb X Bb can produce bb
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| 4. In dogs, wire hair (S) is dominant to smooth hair (s). In a cross of a wire hair (homozygous) and smooth haired dog, what will be the hair condition of the F1 generation? By inbreeding litter mates, what phenotype ratio would you expect in the F2 generation? | |||||||||
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Parental Generation: SS X ss
F1 Generation: all Ss
F2 Generation: Ss X Ss
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3 Wirey : 1 Smooth
5. Diabetes mellitus is thought to be inherited (in some cases) as a recessive gene (d). Two normal people have a diabetic child. What are the genotypes of the parents and offspring? | |||||||||
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Parents = both Dd Child = dd
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| 6. Short hair is dominant over long hair in guinea pigs. A short-haired guinea pig, one of whose parents was long-haired, was mated with a long-haired guinea pig. What are the phenotypes expected in the F1 generation? In what ratio? | |||||||||
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H = Short Hair Short Haired Pig = Hh
h = long hair Long Haired Pig = hh
Hh X hh
1 Short Haired : 1 Long Haired
1% was long haired
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| 7. Polydactyly (presence of an extra finger) is due to a dominant factor. When one parent is polydactylous, but heterozygous, and the other parent is normal, what would be the probability of their having a normal child? | |||||||||
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Polydactylous Parent (heterozygous) = Pp
Normal Parent = pp
PP X pp
50% chance of a normal child | |||||||||
| Dyhybrid problems: | |||||||||
| 1. In cattle, polled (hornless) (H) is dominant over horned (h) and black coat (B) is dominant over yellow (b). If the parental generation is homozygous polled black bulls and horned yellow cows, what is the genotype of the resulting F1 generation? When these F1 generation cattle are permitted to inbreed (mated with one another), there are 18 polled black cattle produced. How many horned black cattle should there be (assuming a large number of cattle are produced and an ideal ratio exists)? | |||||||||
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Parental Generation: PPBB X ppbb
F1 Generation: all PpBb
F1 Cross: PpBb
Expected ratio from F1 cross:
9 Polled Black : 3 Polled Yellow : 3 Horned Black : 1 Horned Yellow
If 18 Polled Black - You should expect:
6 Polled, Yellow 6 Horned, Black 2 Horned, Yellow
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| 2. If a heterozygous polled, yellow bull is crossed with a heterozygous polled, heterozygous black cow, what are the four phenotypes possible in the F1 generation and what is the probability of each being produced? |
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Ppbb X PpBb
Phenotypes:
Polled, Black = 3/8 = 37.5% Polled, Yellow = 3/8 = 37.5% Horned, Black = 1/8 = 12.5% Horned, Yellow = 1/8 = 12.5%
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| 3. A solid-colored, short-haired female rabbit (ssLL) is mated to a spotted, long-haired male (SSll). The F1 generation offspring are allowed to interbreed and 64 offspring are produced. How many will be expected to be spotted with long hair? How many will be solid-colored with short hair? | |||||||||||||||
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ssLL X SSll
F1 Generation: SsLl
Expected Ratio:
9 Spotted, Short Hair 3 Spotted, Long Hair 3 Solid, Short Hair 1 Solid, Long Hair
If 64 Offspring Produced:
12 Spotted, Long Hair 12 Solid, Short Hair
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| 4. If a spotted, short-haired female rabbit is mated with a solid-colored, long-haired male and produces offspring that are half spotted with short hair and half solid with short hair, what is the genotype of the female rabbit? This is an example of what type of cross? | |||||||||||||||
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Female = S_L_
Male = ssll
S = Spot s = solid L = Short l = long
This is an example of TEST cross:
Offspring: 50% SsLl Spotted with Short Hair 50% ssLl Solid with Short Hair
Female Rabbit must be SsLL
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| 5. In some dogs, barking when trailing is due to a dominant gene; others do not bark. Also erect ears are dominant to drooping ears. By crossing a heterozygous erect-eared barker with a droop-eared silent trailer, what kind of pups could you expect (phenotype & percent chance)? | |||||||||||||||
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B = Barking b = no bark (silent) E = Erect Ears e = drooping ears
Heterozygous erect eared barker = EeBb Droop eared silent trailer = eebb
25% Erect-eared Barker 25% Erect-eared, Silent trailer 25% Droop-eared Barker 25% Droop-eared, Silent trailer | |||||||||||||||