You are investigating two traits in fruit flies controlled by one gene each. The

Question

You are investigating two traits in fruit flies controlled by one gene each. The b gene controls for body color (black or brown), while the r gene controls for body sculpture (rugose or smooth). After crossing rugose black flies with smooth brown flies, you obtain an F1 generation of 100% smooth black flies.

What are the dominant alleles?

What are the genotypes of the two parent flies to perform a dihybrid test cross?

After running a dihybrid test cross, you find 1000 offspring with the following phenotypes:

Phenotype

Observed

Expected

Black, rugose

406

Black, smooth

86

Brown, rugose

94

Brown, smooth

414

1000

Fill in the expected column with the expected number of offspring in each class of phenotypes.

What phenotype classes represent the recombinant offspring?

Calculate the relative genetic distance between the body color gene and the body sculpting gene (show your work).

Phenotype

Observed

Expected

Black, rugose

406

Black, smooth

86

Brown, rugose

94

Brown, smooth

414

1000

Explanation / Answer

Since the F1 shows black and smooth phenotype,
Black = B = Dominant
Brown = b = Recessive
Smooth = S = Dominant
Rugose = s = Recessive

Parental cross: BBss X bbSS
F1 progeny: BbSs
Dihybrid test cross: BbSs X bbss
Gametes: (BS) (Bs) (bS) (bs) X (bs)
F2 progeny: BbSs Bbss bbSs bbss
All types of progeny should appear in 1:1:1:1 ratio if the genes are unlinked.
Expected number of progeny for each genotype/phenotype = 250
Parental phenotypic classes = Black rugose and brown smooth
Recombinant phenotypic classes = Black smooth and brown rugose

Genetic distance between body colour and sculpture = (Recombinants X 100)/Total
= [(86 + 94) X 100]/1000
= 18 cM
= 18 map units

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