Question

practice with probability
in fruit flies, the gene for body color (b) is dominant over the gene for body colorless (b), and the gene for wing shape (w) is dominant over the gene for winglessness (w).
a heterozygous black - bodied, winged fruit fly (bbww) is crossed with a double recessive white - bodied, wingless fruit fly (bbww).
use the results from the punnett squares below to calculate the probability.
what is the phenotypic frequency of colorless, winged fruit fly offspring (bbww)?

Explanation:

Step1: Analyze body - color probability

From the Punnett square for body color (\(B\) - black, \(b\) - colorless), when \(Bb\times bb\), the probability of getting \(bb\) (colorless) is \(\frac{1}{2}\).

Step2: Analyze wing - shape probability

Assume for wing - shape \(Ww\times ww\) (a similar cross - principle as for body color, since the cross for wing - shape of \(Ww\) (heterozygous winged) and \(ww\) (wingless) is also a test - cross). The probability of getting \(Ww\) (winged) is \(\frac{1}{2}\).

Step3: Calculate the combined probability

Using the multiplication rule for independent events (since body - color and wing - shape genes assort independently), the probability of \(bbWw\) (colorless, winged) is \(P(bb)\times P(Ww)\). Substitute \(P(bb)=\frac{1}{2}\) and \(P(Ww)=\frac{1}{2}\), we get \(\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}\).

Answer:

\(\frac{1}{4}\)