Question

If a blonde man with blood type AB positive marries a brown haired woman (h1h2) that has a blood type O positive (++), what are the chances of getting a brown haired B positive child?

The question doesn't state the man's Rh label, so is it possible to answer the question?

Answer 1

2 cases are possible as per man's genotype.

Blonde hair = H1H1

Brown hair = H1H2

Black hair = H2H2

Where H2 is dominant to H1.

Genotype of man = H1H1, IaIb, Rh+Rh+ or H1H1, IaIb, Rh+Rh-

Gametes of man = H1IaRh+, H1IbRh+ or H1IaRh+, H1IbRh+, H1IbRh-, H1IaRh-

Genotype of woman = H1H2, IoIo, Rh+

Gametes of woman = H1IoRh+, H2IoRh+

Case 1

Man × woman (if man is H1H1, IaIb, Rh+Rh+)

Gametes	H1IoRh+	H2IoRh+
H1IaRh+	H1H1 IaIo Rh+/Rh+ (blonde, A, +)	H1H2 IaIo Rh+Rh+ (brown, A, +)
H1IbRh+	H1H1 IbIo Rh+Rh+ (blonde, B, +)	H1H2 IbIo Rh+Rh+ (brown, B, +)

Out of 4 progenies, 1 child is brown, B, +

So, probablity = 1/4

Case 2

Man × woman (man is H1H1, IaIb, Rh+Rh-)

Gametes	H1IoRh+	H2IoRh+
H1IaRh+	H1H1 IaIo Rh+ Rh+ (blonde, A, +)	H1H2 IaIo Rh+Rh+ (brown, A, +)
H1IbRh+	H1H1 IbIo Rh+Rh+ (blonde, B, +)	H1H2 IbIo Rh+Rh+ (brown, B, +)
H1IaRh-	H1H1 IaIo Rh+Rh- (blonde, A, +)	H1H2 IaIo Rh+Rh- (brown, A, +)
H1IbRh-	H1H1 IbIo Rh+Rh- (blonde, B, +)	H1H2 IbIo Rh+Rh- (brown, B, +)

Out of 8 progenies, 2 are brown, B, +.

So, probability will be 2/8 = 1/4.

Whatever the genotype of father, 25% of progenies will be brown, B, +.

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