1. You mate a normal mouse (short hair, big eyes, 5 toes, solid coat color) with
ID: 61986 • Letter: 1
Question
1. You mate a normal mouse (short hair, big eyes, 5 toes, solid coat color) with one that has long hair, micropthalmia (small eyes). polydactyly (extra toes), and spotted coat color. The F1 offspring are normal (short hair, big eyes, 5 and solid coat, color). You intercross the Fi offspring. What fraction of the F2 would you expert to have micropthalmia and polydactyly but be otherwise normal? a. 3/256 b. 9/256 c. 27/256 d. 27/64 e. 3/64 2. You mate a normal mouse with one that has long ears, white fur, missing limbs (they get around OK, they really do. You just have to put the food on the cage bottom) and wavy hair. The Fi offspring are all phenotypically normal. b is the gene for ear length C is albino g is limbless e is wavy hairExplanation / Answer
1. (b) 9 / 256
Lets denote each phenotype by a allele
H : short hair h : long hair
E : big eyes e : micropthalmia
T : 5 toes t : polydactyly
C : solid coat colour c : spotted coat colour
Genotype of F1 : HhEeTtCc
The gametes produced will be : HETC, HETc, HEtC, HEtc, HeTC, HeTc, HetC, Hetc, hETC, hETc, hEtC, hEtc, heTC, heTc, hetC, hetc
You can make punnet square to get genotypes of F2 offspring which will give you 256 genotypes (16 x 16).
In other way round, just pick the gametes which show "et" alleles and make a cross among them. The genotypes which you will get are : HHeettCC, HHeettCc, HheettCC, HheettCc, HHeettCc, HHeettcc, HheettcC, Hheettcc, hHeettCc, hHeettCc, hheettCC, hheettCc, hHeettCc, hHeettcc, hheettCc, hheettcc.
So, we have 9 / 256 fraction of F2 which will have micropthalmia and polydactyly.
2. (c) BcgE
Genotype of F1 : bBcCgGeE
Gametes produced: BCGE, BCGe, BCgE, BCge, BcGE, BcGe, BcgE, Bcge, bCGE, bCGe, bCgE, bCge, bcGE, bcGe, bcgE, bcge,
3.
By definition Aunt is the sister of one's father or mother, or the wife of one's uncle. If we consider aunt as sister of ones father or mother then there is possibility that charlie will be heterozygous for the cystic fibrosis gene. Same is case for charlene i.e possibility she is also heterozygous.
In that case 1 / 4 children have possibility of having cystic fibrosis.
(remember that cystic fibrosis is a autosomal recessive disease)
4. (c) 3 olive, 6 dark green, 3 light green, 2 cobalt, 1 mauve, 1 sky blue
Genotype of dark green birds : BbDd
Gametes produced : BD, Bd, bD, bd
After mating you get following progeny
bbdd : sky blue (1)
bbDD : mauve (1)
bbDd, bbDD : cobalt (2)
BBDD, BbDD, BbDD : Olive (3)
BBDd, BbDd, BBDd, BbDd, bBDd, bBdD : dark green (6)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.