Question

mexican hairless dogs are hairless because of a dominant allele of a particular gene. homozygous dominant dogs die in utero (before being born). what is the probability of two hairless dogs having a hairy puppy? 0 1/4 1/2 2/3 3/4

Explanation:

Step1: Determine the genotypes of hairless dogs

Since homozygous dominant (\(HH\)) dogs die in utero, hairless dogs must be heterozygous (\(Hh\)).

Step2: Set up the Punnett square

The cross is \(Hh\times Hh\). The possible gametes from each parent are \(H\) and \(h\).

Gametes	\(H\)	\(h\)
\(h\)	\(Hh\) (hairless)	\(hh\) (hairy)

Step3: Calculate the probability

Out of the viable offspring (\(Hh : hh= 2:1\)), the probability of getting a \(hh\) (hairy) puppy is \(\frac{1}{4}\) (because \(HH\) is non - viable. The total viable genotypes are \(Hh + hh\), and the proportion of \(hh\) is \(\frac{1}{3}\) of viable, but if we consider all possible zygotes before accounting for \(HH\) death, from \(Hh\times Hh\) cross \(HH:Hh:hh = 1:2:1\). Since \(HH\) dies, among the non - dead, the probability of \(hh\) is \(\frac{1}{3}\) of non - dead. But if we calculate the probability at the time of conception (before \(HH\) death consideration for the purpose of Mendelian ratio calculation in terms of allele segregation), the probability of getting \(hh\) from \(Hh\times Hh\) is \(\frac{1}{4}\) as per the Punnett square of allele segregation (\(H\times H\), \(H\times h\), \(h\times H\), \(h\times h\) with equal probability of \(\frac{1}{4}\) each).

Answer:

B. \(1/4\)