I think the probability would be 50% bc there are four options FF Ff fF and ff. Half of those are heterozygous (the child would be a carrier). So I think it's 50%. However, if we already know that the child will not have the ff combination, then we could say that there are only three other options (so maybe it's 60%?) but that's probably just me being a smartass.
Answer:
The estimated number of green seeds in the F2 generation is 2007, approximately.
Explanation:
<u>Available data:</u>
- Cross of a true-breeding plant that produced green seeds with a true-breeding plant that produced yellow seeds
- The entire F1 generation produced yellow seeds.
- Cross of the F1 offspring with each other to produce the F2 generation
- From the F2 generation, there were 6022 yellow seeds.
<u>1st Cross:</u>
Parental) YY x yy
gametes) Y Y y y
F1) 100% Yy
<u>2nd Cross</u>
Parental) Yy x Yy
Gametes) Y y Y y
Punnet square) Y y
Y YY Yy
y Yy yy
F2) 75% yellow, Y-
25% green, yy
What is a likely estimate of the number of green seeds he collected from the F2 generation?
From the whole F2 generation (100%), 75% are yellow seeds, which means that the remaining 25% are green seeds. According to this, 75% of the seeds, equal 6022 seeds. So we need to know how many seeds equal the 25% of green seeds. This is:
If 75% of the F2 generation ------- 6022 seeds
then 25% of the F2 generation -----X = 2007.333 seeds
<em>The estimated number of green seeds in the F2 generation is 2007, approximately. </em>
The parents would both have to be heterozygous for both of the traits, meaning that if the chromosome letters were "B" and "b", their genotypes would both be "Bb." If you put both parents' genotypes into a punnet square, it would be a 25% chance that their child would have blue eyes. Beatrice's genotype would end up being "bb", since blue eyes are a recessive gene in this case.