In fruit flies, straight wings are dominant and curly wings are recessive. What will the generations look like? Assume that Mendel’s method of crossing two true breeding parents with opposite traits is followed. Check all that apply. The P generation has all straight wings.

The P generation has straight wings and curly wings.

The F1 generation has all straight wings.

The F1 generation has all curly wings.

The F1 generation has straight wings and curly wings.

The F2 generation has all straight wings.

The F2 generation has straight wings and curly wings.

The P generation has straight wings and curly wings.

The F1 generation has all straight wings.

The F2 generation has straight wings and curly wings.

Explanation:

This question involves a single gene coding for wing shape in fruit flies. The allele for straight wings (S) is dominant over the allele for curly wings (s). This means that allele "S" will mask the phenotypic expression of allele "s" in a heterozygous state.

According to the question, the cross follows all Mendel's method of crossing two true breeding parents with opposite traits. Hence, the parent generation (P generation) will be between a truebreeding straight wings parent (SS) and a truebreeding curly wings parent (ss). (See attached image for full cross).

Since the straight wing allele (S) is dominant, all the F1 offsprings  will possess straight wings with the genotype: Ss, which is heterozygous.

If this F1 offsprings are self-crossed i.e. Ss × Ss, four possible offsprings in the phenotypic ratio 3 straight wings : 1 curly wing will be produced in the F2 generation. The offsprings will possess the following genotypes: SS, Ss, Ss, and ss.

Offsprings SS, Ss, and Ss will all be phenotypically straight-winged

Offsprings ss will be phenotypically curly-winged.

In overall, this means that P generation has both straight and curly wings (SS and ss), F1 generation has only straight wings (Ss), F2 generation has both straight and curly wings (SS, Ss and ss).

we solve the problem using the hardy-wineberg equation.

let us start with the homozygous recessive proportion of individuals with sickle cell anaemia,

we have the proportion as q² = 0.25.

∴ q = 0.5 (frequency of hbs in population)

further,

solving for the frequency of normal cell by using the fact that,

p + q = 1

∴ p = 1 - 0.5 = 1 - 0.5 = 0.5

now we can calculate 2pq in p² + 2pq + q², which is the frequency for heterozygous carriers of sickle cell anaemia,

∴ 2pq = 2 x 0.5 x 0.5 = 0.5

and the homozygous normal cell individuals proportion is given by the expression,

∴ p² = 0.5² = 0.25

hence, the carriers for sickle cell anaemia are in proportion, 0.5, or 50%.