Question 1. In a population of rabbits, you find three different coat color phen

Question

Question 1. In a population of rabbits, you find three different coat color phenotypes: chinchilla (C) Himalayan (H), and albino (A). To understand the inheritance of coat colors, you cross individual rabbits with each other and note the results in the following table Cross number Parental Phenotypes of phenotypes progeny 3/4 H:1/4 A 3/4 C:1/4 H all C 3/4 C:1/4 A all H 6 9 10 all A 1/2 C:1/2 H 1/2 C:1/4 H:1/4 A What can you conclude about the inheritance of coat color in this population of rabbits? 2. Ascribe genotypes to the parents in each of the 10 crosses. 3. What kinds of progeny would you expect, and in what proportions, if you crossed the chinchillia parents in crosses #9 and #10? Note: Once you have the number of genes and alleles, make up your own nomenclature and define it before to move on to solve different parts of the problem. (Problem 13 in your book.) Question 2, Fruit flies with one allele for Curly wings (Cy) and one allele for normal wings (Cy') have curly wings. When two curly-winged flies are crossed, 203 curly-winged and 98 normal winged flies are obtained. In fact, all crosses between curly-winged flied produce the same curly : normal ratio among the progeny. (a) What is the approximate phenotypic ratio in these offspring? (b) Suggest an explanation for these data. (c) If a curly-winged fly was mated to a normal winged fly, how many flies of each type would you expect among 180 total offspring? 1 Bio 206 PS4

Explanation / Answer

Question 1

C=coat color gene

Allelic=C(chinchilla),

H(himalaya),

a(albina)

C and H are codoinant alleles and CH heterozygote has dark tips(T)

1)answer:

The inheritence of coat color in this population of rabbits is similar to inheritance of A,B and i blood types alleles

2)answer:

Genotype to the parents in each of the 10 crosses

Ha×Ha

Ha×aa

CH×CH

Ca×HH

Ca×Ca

CC×Ha

Ca×aa

aa×aa

aa×CH

CC×CH

3)answer"

H from #10 is Ha and H from #9 is Ha

Question 2

C=curly and c=normal... because it's easier.

heterogeneous genotype (Cc) produces the curly phenotype
203 ~= 200 and 98 ~= 100

a) So the ratio is 2:1

b) 2 : 1 Phenotypic Ratio
The second major modification of the classical monohybrid ratio (e.g., 3: 1) is produced by alleles whose effect is sufficiently drastic to kill the bearer of certain genotypes in the prenatal or postnatal period prior to sexual maturity. Such alleles are called lethal alleles."

It seems as though being homogeneous (CC) with the curly allele is lethal, because we can see that a cross of a heterogeneous (Cc) pair (curly) would produce two heterogeneous (Cc) (curly), one homogeneous (cc) (normal), and one homogeneous (CC) (dead) in a punnet square.

c) 90 normal (cc) and 90 curly (Cc)

note :

due to chegg policy i solved two question kindly post less question so we can answer easily

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