If II-1 and II-2 have another child, the chance that they will have a child affected by PKU is
A - VCE - SSCE Biology - Question 2 - 2002 - Paper 1
Question 2
If II-1 and II-2 have another child, the chance that they will have a child affected by PKU is
A.
\( \frac{1}{4} \)
B.
\( \frac{1}{3} \)
C.
\( \frac{1}{2} \)
D.
1
Worked Solution & Example Answer:If II-1 and II-2 have another child, the chance that they will have a child affected by PKU is
A - VCE - SSCE Biology - Question 2 - 2002 - Paper 1
Step 1
Determine the Genetic Background for PKU
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Answer
Phenylketonuria (PKU) is a genetic disorder inherited in an autosomal recessive manner. For two carriers (both parents being heterozygous, represented as "Aa"), the possible combinations for their offspring can be shown in a Punnett square.
Step 2
Construct the Punnett Square
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Answer
The Punnett square for two heterozygous parents is:
A
a
A
AA
Aa
a
Aa
aa
This results in the following genotype distribution:
1 AA: unaffected and not a carrier
2 Aa: unaffected but carriers
1 aa: affected by PKU
Thus, the probability of having a child affected by PKU (aa) is ( \frac{1}{4} ).
Step 3
Final Probability Calculation
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Answer
From the Punnett square, the overall probability of II-1 and II-2 having a child affected by PKU is: ( P(aa) = \frac{1}{4} ).
Step 4
Conclusion
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Answer
Therefore, the chance that they will have a child affected by PKU is ( \frac{1}{4} ), corresponding to option A.
