The question is incomplete soI am writing the complete question below:
These letters represent the genotypes and phenotypes of the fruit flies.
GG = Normal Wings EE = Red Eyes
Gg = Normal Wings Ee = Red Eyes
gg = vestigial wings ee = sepia eyes
A fly that is heterozygous for both traits (GgEe) is crossed with one that has vestigial wings and sepia eyes (ggee).
How many out of 16 are normal wings, red eyes?
Answer:
Out of 16 there are 4 flies which have normal wings and red eyes.
Step-by-step explanation:
A fly (GgEe) is crossed with (ggee). So we will use a Punnett square to calculate the alleles of their offspring.
For wings i.e. Gg x gg:
G g
g Gg gg
g Gg gg
There are 2 flies with Gg (Normal Wings) and two flies with gg (Vestigial wings. The probability of both is 2/4.
For eyes i.e. Ee x ee
E e
e Ee ee
e Ee ee
There are 2 flies with red eyes (Ee) and 2 flies with sepia eyes (ee). the probability of both is 2/4.
No. of flies with normal wings (Gg) and red eyes (Ee) = 2/4 x 2/4
No. of flies with normal wings (Gg) and red eyes (Ee) = 4/16
So, out of 16 there are 4 flies which have normal wings and red eyes.
