Question

1. in cats, pointed ears are dominant to round ears. a homozygous recessive cat (ee) is bred with a heterozygous cat (ee)
2. in flies, long wings (w) are dominant to short wings (w). two homozygous recessive flies are crossed

Explanation:

Step1: Create Punnett square for cats

The homozygous recessive cat has genotype \(ee\) (produces only \(e\) gametes), and the heterozygous cat has genotype \(Ee\) (produces \(E\) and \(e\) gametes).

$$\begin{array}{|c|c|} \hline E & e \\ \hline e & Ee \\ \hline e & ee \\ \hline \end{array}$$

Step2: Calculate probabilities for cats

For pointed ears (\(Ee\)): There are 2 out of 4 offspring with \(Ee\) genotype. Probability \(=\frac{2}{4}\times100\% = 50\%\)
For round ears (\(ee\)): There are 2 out of 4 offspring with \(ee\) genotype. Probability \(=\frac{2}{4}\times100\%=50\%\)

Step3: Create Punnett square for flies

Both homozygous recessive flies have genotype \(ww\) (produce only \(w\) gametes)

$$\begin{array}{|c|c|} \hline w & w \\ \hline w & ww \\ \hline w & ww \\ \hline \end{array}$$

Step4: Calculate probabilities for flies

For long wings (\(WW\) or \(Ww\)): Since there is no \(W\) allele in parents (\(ww\times ww\)), probability \( = 0\%\)
For short wings (\(ww\)): All 4 offspring have \(ww\) genotype. Probability \(=\frac{4}{4}\times100\% = 100\%\)

Answer:

For cats: Pointed Ears \(50\%\), Round Ears \(50\%\)
For flies: Long Wings \(0\%\), Short Wings \(100\%\)