The key to solving this problem is to consider the two genes separately. Focusin
ID: 34648 • Letter: T
Question
The key to solving this problem is to consider the two genes separately. Focusing on the ABO locus, we must first determine the genotype of the woman in the cross. If a woman is type A but has a type O father, she must be IAi, as her father contributed an i allele. A man that is type O will be ii. When considering the IAi x ii cross, 1/2 of offspring will be type A and 1/2 will be type O. Now considering the locus for MN antigens, individuals that are heterozygous for the MN antigen will have a genotype of LMLN. When two heterozygous MN parents mate, they are expected to produce the following offspring: 1/4 LMLM, 1/2 LMLN, and 1/4 LNLN. Because the probability of two events occurring simultaneously is equal to the product of their individual probabilities, the product law can be used to determine the proportion of various phenotypes of offspring that this couple may have.
Part A results
Type A with M antigen: 1/8
Type A with M and N antigens: 1/4
Type A with N antigen: 1/8
Type O with M antigen: 1/8
Type O with M and N antigens: 1/4
Type O with N antigen: 1/8
How would the results from Part A change if both parents are also heterozygous for the gene controlling the synthesis of the H substance (Hh)?
Type A with M antigen:
Type A with M and N antigens:
Type A with N antigen:
Type O with M antigen:
Type O with M and N antigens:
Type O with N antigen:
Options: 0, 1/32, 3/32, 5/32, 6/32, 10/32, 1
Explanation / Answer
Answer
The woman’s blood group is A so her genotype is IAIO has children with a man that has type O blood, so the man’s genotype is IOIO. The following will be the genotypes of their children:
F/M
IO
IO
IA
IA IO
IA IO
IO
IO IO
IO IO
The probability that children will be of A group is 0.5, and the probability of children of O group blood is 0.5. Both the individuals are heterozygous for MN antigen. The probability of MN antigen inheritance is as follows:
F/M
M
N
M
MM
MN
N
MN
NN
The probability of homozygous M antigen type is 0.25
The probability of heterozygous MN antigen type is 0.5
The probability of homozygous N antigen type is 0.25
The probability of ABO blood group and MN antigen inheritance is the product of the individual probabilities of these two events, since they are independent of each other.
Type A with M antigen = 0.5 × 0.25 = 0.125 = 1/8
Type A with M & N antigens = 0.5 × 0.5 = 0.25 = 1.4
Type A with N antigen = 0.5 × 0.25 = 0.125 = 1/8
Type O with M antigen = 0.5 × 0.25 = 0.125= 1/8
Type O with M and N antigens = 0.25= 1/4
Type O with N antigens = 0.125 = 1/8
If both parents are also heterozygous for the gene controlling the synthesis of the H substance (Hh) then:
Type A with M antigen: 3/32
Type A with M & N antigens: 6/32
Type A with N antigen: 3/32
Type O with M antigen: 5/32
Type O with M and N antigens: 10/32
Type O with N antigens: 5/32
F/M
IO
IO
IA
IA IO
IA IO
IO
IO IO
IO IO
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