Question

two true - breeding stocks of pea plants are crossed. one parent has red axial flowers, and the other has white terminal flowers; all f1 individuals have red axial flowers. the genes for flower color and location assort independently. if 1,000 f2 offspring resulted from the cross, approximately how many of them would you expect to have red terminal flowers?
65
190
250
565
question 10
1 pts
a woman who has blood type a positive has a daughter who is type o positive and a son who is type b negative. rh positive is a trait that shows simple dominance over rh negative and is designated by the alleles r and r, respectively. which of the following is a possible genotype for the son?
ibib
iaib
ii
ibi

Explanation:

Step1: Analyze the first cross

Let \(R\) represent the allele for red flowers (dominant) and \(r\) for white flowers (recessive), \(A\) for axial flowers (dominant) and \(a\) for terminal flowers (recessive). The \(P\) generation is \(RRAA\times rraa\), \(F_1\) is \(RrAa\). When \(F_1\) self - crosses (\(RrAa\times RrAa\)), using the Punnett square or the product rule:
The probability of getting red flowers (\(R-\)) is \(\frac{3}{4}\) (\(RR:Rr = 1:2\)), and the probability of getting terminal flowers (\(aa\)) is \(\frac{1}{4}\).

Step2: Calculate the expected number

Using the formula \(N = n\times p\), where \(n = 1000\) (total number of \(F_2\) offspring) and \(p=\frac{3}{4}\times\frac{1}{4}=\frac{3}{16}\) (probability of red terminal flowers).
\(N=1000\times\frac{3}{16}\approx190\)

Step3: Analyze the second problem (blood - type inheritance)

For blood - type, \(I^A\) and \(I^B\) are codominant, \(i\) is recessive. The woman has blood type \(A\) positive (\(I^A-\ R-\)). The daughter is \(O\) positive (\(ii\ R-\)), so the woman must have \(I^Ai\). The son is \(B\) negative (\(I^B-\ rr\)). Since the mother has \(i\) allele (from daughter's \(ii\)), the son's genotype for blood - type must have \(I^Bi\) (because the mother can contribute \(i\) and the father contributes \(I^B\) for blood - type \(B\) in the son, and for Rh factor, since the son is \(rr\), he gets one \(r\) from each parent).

Answer:

For the first question: 190.
For the second question: \(I^Bi\)