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Answer: A (AA x aa)
Explanation:
In option (A) both of the parent mussels are homozygotes, for alleles "a" and "A" respectively. Therefore, every single one (100%) of their offspring would be expected to be Aa heterozygotes. Combinations (B) and (D) yield a lower proportion of heterozygotes (50%), and combination (C) yields none at all. Punnett squares follow:
(A)
<u> | A A </u>
<u>a | aA aA </u>
<u>a | aA aA </u>
<u />
(B)
<u> | A a </u>
<u>A | AA Aa </u>
<u>a | aA aa </u>
<u />
(C)
<u> | A A </u>
<u>A | AA AA </u>
<u>A | AA AA </u>
<u />
(D)
<u> | A a </u>
<u>a | aA aa </u>
<u>a | aA aa </u>
<u />
Answer:
B. 50%
Explanation:
The yellow trait is dominant (uppercase Y), so the heterozygous pairs (Yy) as well as the homozygous dominant pairs (YY) would be yellow. There are 2 pairs of Yy there for 50% of the offspring would have a yellow pod.
Answer:
2
Explanation: Oxygen has 8 electrons.In the first level it has 2 because its the maximum number of electrons that can be held in the first shell or level