Question

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 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? (1 pt)

What are the genotypes of the two parent flies to perform a dihybrid test cross? (1 pt)

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. (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 pt).

Answer 1

It means black and smooth genes are dominant.

Let B encodes black and b encodes brown. Similarly S encodes smooth and s encodes rugose.

*Dominant allele are S encoding smooth and B encoding brown.

*The genotype of two parents flies is SsBb

As we know in dihybrid cross, the phenotypic ratio is 9:3:3:1

Black and smooth (9) : Black rugose(3) : smooth brown(3) : brown rugose(1)

*The expected number will be :

Black smooth= 9/16×1000= 562.5

Black rugose=3/16×1000=187.5

Smooth brown=3/16×1000=187.5

Brown rugose= 1/16×1000= 62.5

*Recombinant offspring represented by black rugose and smooth brown.

* the genetic distance is calculated by = (recombinant/total)

={ (187.5+187.5) /1000}

= (375/100)  

= 3.75 centi Morgan

The genetic distance between body color gene and sclupturing gene is 3.75 centi Morgan.