Answer:
My best recommendation is C.
Explanation:
I won't explain it because you should be able to figure it out using the process of elimination, but try C.
I think it I A because it benifits
Answer:
- If they have one child, the probability that he or she will be affected is 1/4.
- If they have two children, the probability that at least one of them will be affected is 7/16.
Explanation:
A cross between two heterozygous Aa individuals will produce the followinf offspring: 1/4 AA, 2/4 Aa and 1/4 aa.
Since the disease is recessive, 1/4 of the offspring will have the <em>aa </em>genotype and 3/4 of the offspring will be unaffected.
Every time they have children new gametes were generated <u>independently</u>.
The probability of having <u>no</u> affected children both times is, according to rules of probability for independent events, 3/4 × 3/4 = 9/16 (it's the probability of having a healthy child the first time multiplied by the probability of having a healthy child the second time).
The probability of having at least one affected child is 1 - probability of no affected children = 1 - 9/16 = 7/16.
Option A Base pair substitution
