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two cats are both heterozygous for hair length, ss, and heterozygous fo…

Question

two cats are both heterozygous for hair length, ss, and heterozygous for white hair, ww. short hair and white hair are dominant to long hair and colored hair, respectively. what are the odds that the pairs offspring would show the following phenotypes (proportions can be used more than once)? short, white hair short, colored hair long, colored hair long, white hair

Explanation:

Step1: Determine the gametes

Each cat (SsWw) can produce 4 types of gametes: SW, Sw, sW, sw.

Step2: Use Punnett square or probability rules

For hair length (Ss x Ss):

  • Probability of short hair (S -)=$\frac{3}{4}$ (SS: $\frac{1}{4}$, Ss: $\frac{2}{4}$)
  • Probability of long hair (ss)=$\frac{1}{4}$

For hair color (Ww x Ww):

  • Probability of white hair (W -)=$\frac{3}{4}$ (WW: $\frac{1}{4}$, Ww: $\frac{2}{4}$)
  • Probability of colored hair (ww)=$\frac{1}{4}$

Step3: Calculate probabilities for each phenotype

  • Short, white hair:

Use the multiplication rule (for independent events). Probability = $\frac{3}{4}$ (short hair) $\times\frac{3}{4}$ (white hair)=$\frac{9}{16}$

  • Short, colored hair:

Probability = $\frac{3}{4}$ (short hair) $\times\frac{1}{4}$ (colored hair)=$\frac{3}{16}$

  • Long, colored hair:

Probability = $\frac{1}{4}$ (long hair) $\times\frac{1}{4}$ (colored hair)=$\frac{1}{16}$

  • Long, white hair:

Probability = $\frac{1}{4}$ (long hair) $\times\frac{3}{4}$ (white hair)=$\frac{3}{16}$

Answer:

short, white hair: $\frac{9}{16}$; short, colored hair: $\frac{3}{16}$; long, colored hair: $\frac{1}{16}$; long, white hair: $\frac{3}{16}$