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1. Give the genotypes of the ABO blood groups for individuals 1 and 2 - AQA - A-Level Biology - Question 6 - 2022 - Paper 2

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1. Give the genotypes of the ABO blood groups for individuals 1 and 2. 1 Blood group A 2 Blood group AB 2. Explain one piece of evidence from Figure 3 that the... show full transcript

Worked Solution & Example Answer:1. Give the genotypes of the ABO blood groups for individuals 1 and 2 - AQA - A-Level Biology - Question 6 - 2022 - Paper 2

Step 1

Give the genotypes of the ABO blood groups for individuals 1 and 2.

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Answer

  1. Individual 1 has blood group A, which can have the genotype either IAIAI^A I^A (homozygous) or IAiI^A i (heterozygous).

  2. Individual 2 has blood group AB, which has the genotype IAIBI^A I^B.

Step 2

Explain one piece of evidence from Figure 3 that the allele for Rhesus positive is dominant.

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Answer

In Figure 3, individuals with a Rhesus positive phenotype (like individuals A and AB) can produce offspring with a Rhesus negative phenotype (like individual O). This indicates that the presence of the Rhesus positive allele can mask the expression of the Rhesus negative allele.

Step 3

Calculate the probability of individuals 1 and 2 producing a Rhesus positive son with blood group A (individual 3).

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To find the probability, we assume that individual 1 is heterozygous for the Rhesus blood group (Rr{Rr}) and individual 2 is also Rr{Rr} (since they can produce Rhesus negative offspring).

Possible Rhesus offspring:

  • RR{RR} (Rhesus positive)
  • Rr{Rr} (Rhesus positive)
  • rR{rR} (Rhesus positive)
  • rr{rr} (Rhesus negative)

This results in a probability of producing a Rhesus positive child of rac{3}{4}.

Next, consider the ABO blood groups. Individual 1 can pass on either IAI^A or ii, while individual 2 can pass on IAI^A or IBI^B. The combinations yielding blood group A are:

  1. IAIAI^A I^A (from individual 1) and ii (from individual 2)
  2. IAiI^A i (from individual 1) and IBI^B (from individual 2)

Thus, the probability of individual 3 being blood group A is rac{1}{4} (for homozygous) + rac{1}{4} (for heterozygous from individual 1's contribution).

Combining these, the total probability of producing a Rhesus positive son with blood group A is:

P(RR ext{ or } Rr) imes P(A) = rac{3}{4} imes rac{1}{2} = rac{3}{8}.

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