Answer:
35%
Explanation:
If two genes are 30 map units apart, 30% of the produced gametes will be recombinant.
A mating between an individual homozygous dominant for both traits (AB/AB) and one homozygous recessive for both traits (ab/ab) is conducted.
The F1 will be heterozygous for both genes: AB/ab.
The F1 progeny is then test-crossed to a homozygous recessive individual:
<h3>AB/ab X ab/ab</h3>
<u>The possible offspring will be:</u>
- Parental (70%): AB/ab and ab/ab
- Recombinant (30%): Ab/ab and aB/ab
Since 30% of all the gametes produced by the F1 individual will be recombinant, 70% will be parental. As there are two types of parental gametes, each of them will have a frequency of 35%.
<u>The offspring that will have a dominant phenotype for both traits has the genotype AB/ab with a proportion of 35%.</u>
The answer is C
Completing multiple trials during an experiment
The rest of the options are possibilities of how you can "mess up" and experiment, but if you run multiple trials doing everything right, for example, making sure all variables are controlled, using the right equipment, and making sure you cover all possible variables. If you make sure you do multiple trials and you do them right, you get successful results.
So the answer is C
I hope this was helpful :)
Goodluck!!
Answer:
my guess is C
Explanation:
All you have to do is eliminate the ones that don't make sense and c is the only one left
A person can only show a recessive trait if both of their parents carried at least one copy of each of the recessive allele. The parents do not need to show the trait, as one copy is not enough to reveal it, but they must both carry it.
Hope this helped!
