Question

practice with probability
in rabbits, the gene for coat color (c) is dominant over the gene for albino (c), and the
gene for fur length (f) is dominant over the gene for short fur (f).
a heterozygous black - coated, long - furred rabbit (ccff) is crossed with a homozygous
recessive albino, short - furred rabbit (ccff).
use the results from the punnett squares below to calculate the probability.
what is the phenotypic frequency of colored, long furred
offspring (ccff)?
1/2
9/16
5/16

Explanation:

Step1: Analyze coat - color probability

From the Punnett square for coat color (C - dominant, c - recessive), the cross is \(Cc\times cc\). The genotypes are \(Cc\) (colored) and \(cc\) (albino). The probability of getting a colored rabbit (\(Cc\)) is \(\frac{2}{4}=\frac{1}{2}\).

Step2: Analyze fur - length probability

The cross for fur length is \(FF\times ff\). Using the Punnett square (not shown here, but for \(FF\times ff\), all offspring will have the genotype \(Ff\)). The probability of getting long - furred offspring (\(Ff\)) is \(\frac{4}{4} = 1\) (since \(F\) is dominant over \(f\) and \(FF\times ff\) gives all \(Ff\) which is long - furred).

Step3: Use the multiplication rule for independent events

Since coat color and fur length are independent traits (assuming independent assortment), the probability of a rabbit being colored (\(Cc\)) and long - furred (\(Ff\)) is the product of their individual probabilities. \(P=\frac{1}{2}\times1=\frac{1}{2}\)

Answer:

\(\frac{1}{2}\)