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. How would the results from Part A change if both parents are also heterozygous for the FUT1 gene controlling the synthesis of the H substance (Hh)?
Drag the correct value to the blank following each offspring type.
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:

A woman with type A blood (whose father was type O) meaning her genotype is AO mates with

Man that has type O blood (OO genotype)

Both are heterozygous for MN blood group and both also heterozygous for the FUT1 gene controlling the synthesis of the H substance (Hh)- which determines the expression of the A and B antigen.

Cross

A O M N H h

O AO OO M MM MN H HH Hh

O AO OO N MN NN h Hh hh

Type A- 1/2 O-1/2 type M- 1/4 MN-1/2 N- 1/4, type H- 3/4 h-1/4

Type A with M antigen:

1/2*1/4*3/4 = 3/32

Type A with M and N antigens:

1/2*1/2*3/4 = 3/16

Type A with N antigen:

1/2*1/4*3/4 = 3/32

Type O with M antigen:

1/2*1/4*3/4= 3/32

Type O with M and N antigens:

1/2*1/2*3/4 = 3/16

Type O with N antigen:

1/2*1/4*3/4 = 3/32.

The 3/4 value comes from the expression of Hh-3/4 (this determines if the A and B Angie will be expressed).

answer: i don't know

explanation:

you need to provide a picture, otherwise i can't tell you.

if you can provide one, i'd be more than happy to you with this!

the answer is: trichonympha lives in the gut of termites and digests cellulose eaten by the termite.

this is an example of a symbiotic association of a protozoan with its mode of association.

hope i

-lavira: )