Question

in a group of dogs, some individuals have short fur and others have long fur. in this group, the gene for the fur length trait has two alleles. the allele for short fur (f) is dominant over the allele for long fur (f). this punnett square shows a cross between two dogs. what is the probability that a dog produced by this cross will be heterozygous for the fur length gene? \\(\frac{0}{4}\\) \\(\frac{1}{4}\\) \\(\frac{2}{4}\\) \\(\frac{3}{4}\\) \\(\frac{4}{4}\\)

Explanation:

Step1: Identify heterozygous genotype

A heterozygous genotype for the fur length gene is $Ff$ (one dominant and one recessive allele).

Step2: Count Punnett square outcomes

The Punnett square shows 4 total offspring genotypes, all of which are $ff$ (homozygous recessive). There are 0 $Ff$ genotypes.

Step3: Calculate probability

Probability = $\frac{\text{Number of heterozygous outcomes}}{\text{Total number of outcomes}} = \frac{0}{4}$

Answer:

$\frac{0}{4}$