Question

what is the chance that two heterozygous parents will produce a child with a heterozygous genotype?
100%
25%
0%
50%

Explanation:

Step1: Define the genotype

Let the heterozygous genotype be \( Aa \) (where \( A \) is the dominant allele and \( a \) is the recessive allele). Each parent can produce two types of gametes: \( A \) and \( a \), with equal probability (\( \frac{1}{2} \) for each gamete).

Step2: Use Punnett square

The Punnett square for the cross \( Aa\times Aa \) is:

	\( A \)	\( a \)
\( a \)	\( Aa \)	\( aa \)

From the Punnett square, the possible genotypes of the offspring are \( AA \) (homozygous dominant), \( Aa \) (heterozygous), and \( aa \) (homozygous recessive). The number of \( Aa \) (heterozygous) genotypes is 2 out of a total of 4 possible genotypes.

Step3: Calculate the probability

The probability is the number of favorable outcomes (heterozygous genotypes) divided by the total number of outcomes. So the probability \( P=\frac{2}{4} = 0.5\) or \( 50\% \).

Answer:

50% (corresponding to the option with 50%)