A man with blood group AB (I^A I^B) marries a woman with blood group A who is heterozygous… — JAMB UTME

A man with blood group AB ( $I^A I^B$ ) marries a woman with blood group A who is heterozygous ( $I^A i$ ). What is the probability that their first child will have blood group A?

$50\%$ Answer
B
$75\%$
C
$25\%$
D
$100\%$

Answer

The probability that their child will have blood group A is

50\%

(or

1/2

The option stating 50% is correct because crossing parental genotypes

I^A I^B

and

I^A i

yields four equally likely offspring genotypes:

I^A I^A

(blood group A),

I^A i

(blood group A),

I^A I^B

(blood group AB), and

I^B i

(blood group B). Combining the two genotypes that result in blood group A (

25\% + 25\%

) gives a total probability of

50\%

Step-by-Step Solution

Identify the parental genotypes and gametes produced.

Father's genotype is

I^A I^B

(gametes:

I^A

I^B

). Mother's genotype is

I^A i

(gametes:

I^A

i

Heterozygous blood group A carries the recessive allele

i

, while blood group AB expresses both

I^A

and

I^B

codominantly.

Construct a Punnett square for the cross

I^A I^B \times I^A i

The four possible offspring genotypes are

I^A I^A

(

25\%

I^A i

(

25\%

I^A I^B

(

25\%

), and

I^B i

(

25\%

Combining male and female gametes gives all expected genetic ratios.

Determine the phenotypic expression for each genotype.

I^A I^A

and

I^A i

both express blood group A (

25\% + 25\% = 50\%

I^A I^B

expresses blood group AB (

25\%

I^B i

expresses blood group B (

25\%

Alleles

I^A

and

I^B

are codominant with each other, and both are completely dominant over the recessive allele

i

Key Concept

Codominance and Multiple Alleles in Human ABO Blood Groups

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Step-by-Step Solution

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