Homework SourseConsider a species of dinosaurs and assume that females are
A female dinosaur with black scales and a male dinosaur with green scales have 34 baby dinosaurs. In this F1 generation, there are 9 female and 8 males that have black scales and there are 8 female and 9 males that have green scales.
What is the most likely genotype of the mother dinosaur?
a.X(B)X(B) b.X(B)X(b) c.X(b)X(b) d.X(B)Y e.X(b)Y
What genotype(s) will not be observed in the F2 generation from crosses between F1 females with black scales and F1 males with green scales?
a.X(B)X(B) and X(B)Y
b. X(b)X(b)
c. X(B)Y
d. X(B)X(B)
e. X(B)X(B) and X(B)Y
f. X(b)Y
g. X(B)X(b)
Which gender of the F2 offspring could green scales be expected?
a. both males and females
b.females only
c .males only
Solution
1. B
Because we see a phenotype of 1:1 we can think of it as a Heterozygous individual crossed with a homozygous recessive individual. In this case the homozygous recessive individual has the recessive allele paired with the y chromosome in the male parent (Xb/Y) which gives the same phenotype as a if he was doubly recessive. Therefore you can conclude that the mother was Heterozygous (XB/Xb) and the father was by. Use a punnett square crossing these genotypes (XB/Xb x Xb/y) and you will see the 1:1 of green to black in both males and females.
2. D
Using the resultant genotypes from crossing the heterozygous mother and the recessive father which is either XB/Xb or Xb/Xb for the mother crossed with Xb/y (because we know he has green scales). There is no way to produce a homozygous dominant individual crossing XB/Xb with Xb/y gives the gentoptypes XB/Xb, X/b/X/b, XB/y, and Xb/y. Corssing Xb/Xb with Xb/y gives Xb/Xb or Xb/y.
3. A
After typing out all possible F2 genotpes we can achieve homozygous recessive females and males with the recessive allel, therefore we can get males and females with black scales.
