To ensure that you get the right concentration of sanitizing solution, follow <span>the label instructions and use a test strip. The correct option among all the options that are given in the question is the second option or option "b". I hope that this is the answer that has come to your desired help.</span>
Answer:
3/8
Explanation:
Martha has a widow's peak (dominant trait) and attached earlobes (recessive trait).
Martha's dad had a straight hairline (ww) and unattached earlobes (Ee, because she has the recessive alleles ee and both parents give us one allele).
This tells you that martha has a mother with at least one of both alleles dominant for widow peak and at least one recesive allele of attached earlobes
So, martha's alleles are: Ww and ee.
If she marries a man that is heterozygous for both traits (Ww and Ee) the probabilitys are
Ww Ee x Ww ee: WWEe, WwEe, wWee, and wwee
Answer:
25% heterozygous tall
Explanation:
If we take the F1 generation as parents and let them self-fertilise, we have 4 crosses.
The first one for homozygous tall, then we have 100% AA.
The second and third one for heterozygous tall and we have 25% AA, 50%Aa and 25%aa for each of them.
The last one would be for dwarf, and we'll have 100%aa.
Adding all of them, we'll have
AA = 100 + 25 + 25 = 150%
Aa = 50+50 = 100%
aa = 100 + 25 + 25 = 150%
as we had 4 crosses, so dividing the total percentages by 4, we'll have,
AA = 37.5%
Aa = 25%
aa = 37.5%
:. The percentage of heterozygous tall would be 25%.
Hope it helps:)
