Question

Suppose gene A is on the X chromosome, and genes B, C and D are on three different autosomes. Thus, A- signifies the dominant phenotype in the male or female. An equal situation holds for B-, C- and D-. The cross AA BB CC DD (female) x aY bb cc dd (male) is made.

A) probability of obtaining A- individual in F1

B) probability of obtaining an a male in the F1 progeny

C) Probability of A- B- C- D- female in F1

D) How many different F1 genotypes are there?

E) Probability of F2 individuals will be heterozygous for the four gens?

F) Determine a probability of each of the following types in the F2     individuals  

(1) A- bb CC dd (female);  

(2) aY BB Cc Dd (male);  

(3) AY bb CC dd (male);  

(4) aa bb Cc Dd (female)

Answer 1

a. 1, the female is AA, so regardless of what the male has, it will have one A and would be labelled A-.

b. 1/2, the male is aY, so there's ½ chances that the Y is given instead of a (which would result in a male).

c. If chances of getting a male is 1/2, chances of getting a female is 1/2. Similar to (a), the female is BB, CC, and DD, so regardless what the male has, the offspring will be labelled B-, C-, and D-. So, (1/2) (1) (1) (1)=1/2.

d. 2, the one female one and one male

e. F1 female (AaBbCcDd) x F1 male (AYBbCcDd)

A: offspring AA: Aa: AY: aY=1:1:1:1, so probability of Aa: 1/4

B: offspring BB: Bb: bb=1:2:1, so probability of Bb: 2/4=1/2

Similar to B, probability of Cc: 1/2 and probability of Dd: 1/2

So, (1/4) (1/2) (1/2) (1/2) = 1/32

f. Looking at the offspring possiblilty in (e),

(1) Probability of A- female: 2/4=1/2, probability of bb: 1/4, probability of CC: 1/4, probability of dd: 1/4 (1/2)(1/4)(1/4)(1/4) = 1/128

(2) Probability of aY male: 1/4, BB: 1/4, Cc: 1/2, Dd: 1/2 (1/4)(1/4)(1/2)(1/2)= 1/64

(3) Probability of AY male: 1/4, bb: 1/4, CC: 1/4, dd: 1/4 (1/4)(1/4)(1/4)(1/4)= 1/256

(4) Probability of aa is not possible so it will be 0