In humans, the ABO blood type is under the control of autoso mal multiple alleles. Red-green color blindness is a recessive X linked trait. If two parents who are both type A and have normal vision produce a son who is color-blind and type O, what is the probability that their next child will be a female who has normal vision and is type O?
Answers
Answer:
Answer:
1/8
Explanation:
The alleles that control the blood type are autosomal. The A and B alleles are codominant, and they are both dominant over the i allele.
The possible genotypes that determine the blood types are:
Type A: I^AI^AI
A
I
A
, I^AiI
A
i
Type B: I^BI^BI
B
I
B
, I^BiI
B
i
Type AB: I^AI^BI
A
I
B
Type O: iiii
Color blindness is a recessive X-linked trait. The possible genotypes and phenotypes are:
Normal vision woman: X^VX^VX
V
X
V
, X^VX^vX
V
X
v
,
Color-blind woman: X^vX^vX
v
X
v
Normal vision man: X^VYX
V
Y
Color-blind man: X^vYX
v
Y
The mom is type A and color-blind. She is X^vX^vX
v
X
v
and could be either I^AI^AI
A
I
A
or I^AiI
A
i .
The dad is type B and has normal vision. He is X^VYX
V
Y and could be either I^BI^BI
B
I
B
or I^BiI
B
i .
Because the daughter is type O, her genotype for blood type is ii. That means that both parents must have the recessive i allele. The mom is I^AiI
A
i and the dad is I^BiI
B
i .
To calculate the probability of having a type AB color-blind son, we need to do the Punnett Square (see attached image) and the rules of probability.
Given two independent events, the rules of probability state that the probability that both events occur is found by multiplying the probabilities of each event.
P(A and B) = P(A) x P(B)
Genes in different chromosomes assort independently, so the probability of having offspring with the genotype X^vYX
v
Y , I^AI^BI
A
I
B
is:
P= 1/4 x 1/2 = 1/8.