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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 Genotype of Parents
Answer
To determine the chance of having a child affected by Phenylketonuria (PKU), we first need to identify the genotypes of the parents II-1 and II-2. Assuming one parent is a carrier (heterozygous, represented as 'Aa') and the other is affected (homozygous recessive, 'aa'), we can proceed with a Punnett square.
Step 2
Punnett Square Analysis
Answer
Using a Punnett square for 'Aa' (carrier) and 'aa' (affected), we get:
| a | a | |
|---|---|---|
| A | Aa | Aa |
| a | aa | aa |
From this, the potential offspring genotypes are:
- Aa (carrier): 2 out of 4 (50%)
- aa (affected): 2 out of 4 (50%) Thus, there is a 50% chance of the child being affected by PKU.
Step 3