Question

One-Gene Crosses 1. In Drosophila, gray body color is dominant over black body color. By crossing a heterozygous gray-bodied fly with a black-bodied fly, one hundred offspring resulted. How many would be expected to have black bodies? 7 2. Albinism, the lack of pigment, occurs commonly in animals and is always recessive to normal pigmentation. Ten brown and eight albino mice were born to parents which were likewise brown and albino. What are the genotypes of both parents? 3. The ability to roll one's tongue into a Uis determined by a dominant allele (R). Two tongue rollers have a child who cannot roll his tongue. (a) What are the genotypes of the parents and their son? (b) What is the probability that their next child will be a non-tongue roller? When a long-tailed cat was mated to a tailless cat, all of their kittens had short tails. Wher two short-tailed cats were mated, their litter contained three kittens without tails, two with long tails, and six with short tails. Explain this pattern of inheritance and determine the genotypes for tailless, long-tailed, and short-tailed cats 4.

Answer 1

Answer:

1) Drosophila - Gray body color (G) is dominant over black body color (g)

Heterozygous gray fly (Gg) x Black bodied fly (gg)

Number of offspring = 100

The cross can be represented in the form of a punnett square:

	g	g
G	Gg -Gray	Gg - Gray
g	gg - black	gg - Black

From the cross, % of flies with black bodies = 100*(2/4) = 50%

Therefore 50 flies out of 100 progenies will have black color

2)

• Let the allele for normal pigmentation be (A) - Dominant, Brown Color
• Let the allele causing Albinism be (a) - Recessive
• Parents - Brown and Albino
• Number of mice born = 10 brown and 8 albino mice
• Total number of mice born = 18
• Percentage of brown mice = 10*100/18 = 55.6%
• Percentage of albino mice = 44.4%
• Ratio of brown:albino mice = 10:8 = 5:4
• Possible parent genotype = Aa (heterozygous) and aa (homozygous recessive)

3)

• Dominant allele (R) - allele - ability to roll one's tongue
• Recessive allele (r) - cannot roll tongue
• Child cannot roll tongue, therefore child's genotype = rr
• ​Punnett square:

• 	R	r
  R	RR	Rr
  r	Rr	rr

​a) Genotype of parents and their son

• Parent genotype = Rr and Rr
• Son genotype = rr

b) What is the probability that their next child will be a non-tongue roller?

This probability is independent of the first child birth, therefore probability that next child will be a non-tongue roller = 1/4 = 25%