Question

1. the long hair of persian cats is recessive to the short hair of siamese cats, but the black colour of persians is dominant to the black - and - tan coat of the siamese. if a pure black long - hair persian is mated to a pure black - and - tan short - haired siamese, what will be the appearance of the f1? if two of the f1 are mated, what will be the chance of obtaining a long - haired black - and - tan cat in the f2?

Explanation:

Step1: Define allele notation

Let:

• Hair length: \(S\) = short hair (dominant), \(s\) = long hair (recessive)
• Coat color: \(C\) = black-and-tan (dominant), \(c\) = black (recessive)

Step2: Identify parent genotypes

Pure black long-haired Persian: \(sscc\)
Pure black-and-tan short-haired Siamese: \(SSCC\)

Step3: Determine F₁ genotypes

Each parent can only pass one allele for each trait:
Persian passes \(s\) and \(c\); Siamese passes \(S\) and \(C\).
All F₁ offspring: \(SsCc\)
Appearance: Short-haired, black-and-tan.

Step4: Set up F₂ cross (SsCc × SsCc)

Use Punnett square for two independent traits, or product rule.
For hair length: \(Ss \times Ss\) → Probability of long hair (\(ss\)) = \(\frac{1}{4}\)
For coat color: \(Cc \times Cc\) → Probability of black-and-tan (\(CC\) or \(Cc\)) = \(\frac{3}{4}\)
Probability of long-haired black-and-tan: \(\frac{1}{4} \times \frac{3}{4} = \frac{3}{16}\)

Answer:

1. Appearance of F₁: All short-haired, black-and-tan cats.
2. Chance of long-haired black-and-tan cat in F₂: \(\frac{3}{16}\) (or 18.75%)