Question

In flamingos, genes A and B are known to be 21 map units apart on chromosome #5. You mate an AA:BB male with an aa:bb female to produce F1 flamingos.

a. What gametes will be made by these F1 individuals and in what proportions?

b. When the F1's are bred to each other, they produce a total of 276 offspring. How many of these offspring would you predict should have the genotype Aa:BB?

Answer 1

Ans : Since A and B are 21 map units apart, they are not linked genes. They are inherited separately. Genes are linked only when they are located 1 map unit or centimorga away from eachother.

In the homozygous dominant parent, all the gametes are heterozygous dominant AB genotype. In the homozygous recessive parent, all gametes are heterozygous recessive, ab genotype.

When these gametes mate, the F1 individuall has the genotype AaBb. The crossing is takes place as shown below.

AB   $\times$ ab

AaBb   

a. The gametes of the F1 flamingos can have the genotypes : AB, Ab, aB, ab in the ratio 1: 1: 1 : 1

b. When F1 individuals are inbred as shown below:

AABB aabb ganrites: AB ab X -> F Aa Bb -> garretes : AB, Ab, aB ab t fa Aa Bb x Aa Bb

АВ Aь ор ab Ав AA BB ААВ, Ao 20 АВЬ Ab ААВЬ AAь, Аа Вь Aa bb aB Aa BB Aa Bb - Bg aa Bь ab Aa Bb aabb Aa b, аавь

There are 16 different genotypes out of which, 1 offspring has the genotype AaBB. Therefore, out of 276 offsprings,

number of offsprings with AaBB genotype = 276/16 = 17.25