A woman with type A blood (whose father was type O) has children with a man that
ID: 34372 • Letter: A
Question
A woman with type A blood (whose father was type O) has children with a man that has type O blood. Both individuals are heterozygous for the MN antigen.
Recall that MN blood group antigens are independent of the ABO locus, and that the alleles are codominant.
Determine the proportion of various phenotypes of offspring that this couple may have.
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 antigen:
Options: 0, 1/8, 1/4, 1/2, 3/4, 1
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 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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