Question

1 .  “Chances are…” (5 pts.)

i. (Mom) AaBBccDd x (Dad) AaBbCcDD

a. What is the probability that the following cross will produce an offspring that has dominant traits for all four genes?

b. What is the chance of having 1 boy out of 4 children that have the phenotype “A””B””c””D”?

c. What is the chance of at least 2 girls out of 5 children being dominant for all four genes?

ii. (Mom) Blood Type AB x (Dad #1) O, (Dad #2) B

a. What is the probability that the cross of Mom and Dad #1 will produce a girl with blood type O?

b. What is the chance for at least 1 boy with blood type B if Mom has 2 kids with each father?

Answer 1

Answer. 1 . I) chances of having dominant trait for all four genes, as the one dominant allele can express dominant trait

So chance of having AA or Aa is 3/4

BB or Bb is 1

CC or Cc 1/2

DD or Dd is 1

so total probability is 3/4*1*1/2*1 =3/8

B) the chance of having one boy out of four is 1!/4! =or (1/1×2×3×4) =1/24 and chance of A, B, c ,D is 3/8 then total probability is 3/8*1/24 =3/192 or 1/64

C) 2 girls out of five 2!/5! = 1×2/1×2×3×4×5 = 2/120

Probability of dominant for all four gene is 3/8 so total probability = 2/120*3/8 = 1/160

Ii

A) the probability of having blood type O is zero because the mother is AB type and she doesn't have O genotype.

	A	B
B	AB	AB
B	AB	BB

	A	B
O	AO	BO
O	AO	BO

B) chance of having boy is 1/2 and chance of having B type is 1/2 in first father and for second its 1/2 then in total 1/8