Name: Recitation 5 short-answer assignment You are investigating two traits in f

ID: 276119 • Letter: N

Question

Name: Recitation 5 short-answer assignment 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 rgene 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? (1Rt What are the genotypes of the two parent flies to perform a dihybrid test cross? (1pt) After running a dihybrid test cross, you find 1000 offspring with the following phenotypes Observed 406 86 94 414 Expected Black, rugose Black, smooth Brown, rugose Brown, smooth Fill in the expected column with the expected number of offspring in each class of phenotypes. (1 pt What phenotype classes represent the recombinant offspring? (1 pt each) Calculate the relative genetic distance between the body color gene and the body sculpting gene (show your work)1 t

Explanation / Answer

1. According to the question, b gene controls body color (black or brown) and r gene controls body sculpture (rugose or smooth). 100% smooth black flies were obtained in the F1 generation when rugose black flies were crossed with smooth brown flies.

This can be represented as

rugose black X smooth brown

which gave all smooth black flies.

Which means rugose is ss, smooth SS or Ss and black is BB or Bb and brown is bb.

In order to obtain all smooth black flies, we can perform the cross as follows:

ssBB (rugose black) X SSbb (smooth brown)

which gave the offspring SsBb (smooth black) in the F1 generation.

The dominant alleles here are S and B.

Hence, the dominant characters are smooth and black.

2. Test cross is done between a homozygous recessive individual and the corresponding heterozygote.

According to the question, the parents of the test cross would be ssbb (homozygous recessive, rugose brown) and SsBb (corresponding heterozygote, Smooth black).

The test cross can be represented as follows:

ssbb X SsBb

Gametes

SsBb

Smooth black

Ssbb

Smooth brown

ssBb

Rugose black

ssbb

Rugose brown

3. Dihybrid test cross ratio is 1:1:1:1. This means the expected number of offsprings would be 1000/4 = 250.

Thus, the number of expected offsprings for each class of phenotype is 250.

4. Recombinant offsprings are the ones which contain a different combination of allele from either of the parents.

Here, black rugose and brown smooth are the parental types. Their observed number is more (406 and 414) respectively.

Black smooth and brown rugose represent the recombinant offsprings. Hence, their numbers are less (86 and 94, respectively).

Thus, the phenotypes that represent recombinant offsprings are black smooth and brown rugose.

5. Since the four classes of offsprings are not produced in equal numbers, we can say that the body color gene and body sculpture gene are linked and thus they are present on the same chromosome.

In order to map the genes, we need to calculate the recombinant frequency (RF). RF is a measure of genetic linkage and is used to generate a genetic linkage map.

RF% = (Number of recombinants / Total number of offsprings) X 100

Hence, for the above results

RF% = {(86 + 94)/ 1000} x 100 = 18%

RF is not a direct measure of the distance between two genes on a chromosome, but is an estimation of the physical distance. Distance between two genes are expressed in centiMorgan (cM) or map units.

1% RF = 1cM or map unit

Hence, 18% RF = 18 cM or map units.

Therefore, the relative genetic distance between the body color gene and body sculpture gene would be 18 cM or map units.