Question

1.4.2 test (cst): heredity
question 25 of 25
in banded snails, shell color is controlled by a gene with a dominant allele for
brown color, b, and a recessive allele for yellow color, b. the presence of
bands is controlled by a gene with a dominant allele that produces no bands,
u, and a recessive allele that produces five bands, u. the two genes do not
occur on the same chromosome. snails with the following genotypes are
crossed:
bbuu × bbuu
what is the probability that the offspring will have the bbuu genotype?
a. 1.00
b. 0.25
c. 0.50
d. 0.75

Explanation:

Step1: Analyze the shell - color gene cross

For the shell - color gene (B and b), the cross is \(Bb\times Bb\). Using the Punnett square, the possible genotypes are \(BB:Bb : bb=1:2:1\). The probability of getting \(BB\) is \(\frac{1}{4}= 0.25\).

Step2: Analyze the band - related gene cross

For the band - related gene (U and u), the cross is \(uu\times uu\). The only possible genotype for the offspring is \(uu\) with a probability of \(1\).

Step3: Use the multiplication rule

Since the two genes assort independently (they do not occur on the same chromosome), we use the multiplication rule for independent events. The probability of \(BBuu\) is the probability of \(BB\) times the probability of \(uu\). So \(P(BBuu)=P(BB)\times P(uu)=0.25\times1 = 0.25\)

Answer:

B. 0.25