1) In mice, the waltzing allele (w) that causes the mouse to run in circles due to an inner ear defect is recessive to the non-waltzing allele (W).
If a heterozygous mouse mated with a waltzing mouse,
a. 75% of the offspring would have the recessive trait
b. 100% of the offspring would have a copy of the recessive allele
c. 100% of the offspring would have a copy of the dominant allele
d. 75% of the offspring would have the dominant trait
2) Use the following information to answer the next two questions.
In mice, albinism (g) is recessive to grey coat colour (G). A grey mouse mated with an albino mouse. Over the course of several months, 38 pups were born. Out of 38 pups, 17 had a grey coat, and 21 were albino.
Which of the following rows correctly identifies the parental and F1 generation genotypes?
a. Parental Cross F1 Genotypes
G G x g g Gg
b. Parental Cross F1 Genotypes
G g x g g Gg and gg
c. Parental Cross F1 Genotypes
G g x G g GG, Gg and gg
d. Parental Cross F1 Genotypes
G g x G G GG and Gg
3) Theoretically, what is the expected result from the parental cross of a grey mouse and an albino mouse, and why does the expected result not match the result stated in the information given above?
a. Theoretically, 100% of the mice should have grey coats so all 38 offspring should have grey coats. There was an experimental error that can account for the deviation.
b. Theoretically, there should be a 1:1 ratio of grey mice to albino mice, or 19 grey coat mice and 19 albino mice. With more litters, the actual results would match the expected results more closely.
c. Theoretically, there should be a 3:1 ratio of grey mice to albino mice, or 28.5 grey coat mice and 9.5 albino mice. There was an experimental error that can account for the deviation.
d. Theoretically, there should be a 3:1 ratio of grey mice to albino mice, or 29 grey coat mice and 9 albino mice. With more litters, the actual results would match the expected results more closely.
4) Which of the following statements describes Mendel’s Law of Segregation?
a. The dominant allele is represented with an uppercase letter and the recessive allele is represented with a lowercase letter.
b. Dominant alleles are always expressed in a homozygous or heterozygous individual.
c. A Punnett square can be used to predict the outcome of a parental cross.
d. A Bb individual will produce B and b alleles, while a bb individual will only produce a b allele.
5) SEE FIRST PHOTO ATTACHED
6) Grey eyes (b) are recessive to brown eyes (B). Two brown-eyed people had a grey-eyed child.
Which of the following Punnett squares demonstrate the parental cross and the resulting F1 generation?
a. B b
B BB Bb
B BB Bb
b. B b
B Bb bb
b Bb bb
c. B b
b Bb bb
b Bb bb
d. B b
B BB Bb
b Bb bb
7) In tomatoes, red fruit (R) is dominant over yellow fruit (r).
Red tomato plants were cross pollinated with yellow tomato plants. The next generation of plants produced no yellow fruit.
Which of the following rows identifies the genotype and the phenotype of the parental red tomato plants respectively?
a. red; Rr
b. red; RR
c. Rr; red
d. RR; red
8) In Manx cats, the recessive allele for tail length is lethal. A heterozygous dominant genotype results in a shortened tail, while a homozygous dominant genotype results in a normal tail. A homozygous recessive genotype is lethal.
Which of the following rows correctly identifies the phenotypic ratio of the offspring that results from crossing two shortened tail Manx cats?
a. Normal Short Tail Lethal
2 1 1
b. Normal Short Tail Lethal
2 1 0
c. Normal Short Tail Lethal
0 2 1
d. Normal Short Tail Lethal
1 2 1
9) SEE IMAGE ATTACHED PLEASE
10) In garden pea plants, tall stem length (T) is dominant over short stem length (t). A pure breeding tall stem plant was crossed with a short stem plant.
T T
t Genotype 1 Genotype 2
t Genotype 3 Genotype 4
From the F1 generation, Genotype 3 and Genotype 4 were crossed. What is the frequency of the dominant phenotype in the F2 generation?
Record your answer as a value between 0 and 1, with two significant digits.
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