Question

Using a Punnett square, work out the distribution of an autosomal phenotypic feature in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus.

Hint:

In a cross between a homozygous dominant female and a heterozygous male, 50% of the offspring will be heterozygous and 50% will be homozygous dominant, all exhibiting the dominant trait.

Answer 1

Let us assume the autosomal gene for the trait is represented by the letter "A", where "A" is the dominant allele and "a" is the recessive allele.
- The female is homozygous dominant, meaning her genotype is \(AA\).
- The male is heterozygous, meaning his genotype is \(Aa\).
We will use a Punnett square to determine the genotype and phenotype distribution in the first filial generation (F1).
Punnett Square: \[ \begin{array}{c|c c} & A & A
\hline A & AA & AA
a & Aa & Aa
\end{array} \] Step 1: Genotype Distribution:
- The genotypes of the F1 generation will be:
- \(AA\) (homozygous dominant) with a probability of 50%.
- \(Aa\) (heterozygous) with a probability of 50%.
Step 2: Phenotype Distribution:
Since \(A\) is dominant, both \(AA\) and \(Aa\) will show the dominant phenotype. Therefore, the F1 generation will show the dominant phenotype.
- 100(\%) of the F1 generation will display the dominant phenotype.
Conclusion:
In the first filial generation, 50(\% )will be homozygous dominant (\(AA\)) and 50(\% ) will be heterozygous (\(Aa\)). All offspring will exhibit the dominant phenotype due to the presence of at least one dominant allele.