The gametes derived from the genotype GGBB include dominant and recessive plus mixed. These could show as GG, BB, GB, and BG although there are some multiples of each.
The possible gametes produced by an individual with the genotype Ww will be W and w.
Only one; A. At least concerning this one trait.
8
It depends on the allele combinations of the other pea plant parent. If the other pea plant parent also has a genotype of RrYy, then there are 16 combinations. If the other parent has genotype RRYY, then there are only 4 allele combinations. If the other parent has a genotype of RrYY, then there are 8 allele combinations in the gametes. If the other parent has a genotype of RRYy, then there are also 8 possible combinations in the gametes. Finally, if the other parent has the genotype rryy, then there are 4 gamete combinations.
ABC ABc
The possible gametes produced by an individual with the genotype Ww will be W and w.
The possible gametes are ABDEf, aBDEf, ABDef, and aBDef.
Only one; A. At least concerning this one trait.
Ab and ab There would be about a 50/50 ratio of each.
Gk, gk
You need to make a Punnet Square Put A and a on top and B and B on the left side. These represent the possible gametes. Your results should be: AB AB Ba (aB) Ba (aB) So 50% of the genotype have the A allele.
In this case neither Pete nor Jack can be the father. A baby inherits one allele from each of its parents. This means that it will receive one G or g and one B or b from each parent. If Pete only has gb gametes - (meaning his genotype is ggbb) - then it is not possible for him to give a B to the baby. The same goes for Jack, whose gametes are Gb (meaning his genotype is GGbb). Neither of these men have a B - which means that they cannot be the father. The father would have to have BB or Bb for his child to have the genotype BB.
8
4
2 can
16
at the moment of conception when the male and female gametes fuse together. at the moment of conception when the male and female gametes fuse together.