1 The normal eye color of Drosophila is red but strains in w

1.) The normal eye color of Drosophila is red, but strains in which all flies have brown eyes are available. Similarly, wings are normally long, but there are strains with short wings. A female from a pure line with brown eyes and short wings is crossed with a male from a normal pure line. The F1 consists of normal females and short-winged males. An F2 is then produced by intercrossing the F1. Both sexes of F2 flies show phenotypes as follows:

3/8 red eyes, long wings

3/8 red eyes, short wings

1/8 brown eyes, long wings

1/8 brown eyes, short wings

Deduce the inheritance of these phenotypes; use clearly defined genetic symbols of your own invention. State the genotypes of all three generations and the genotypic proportions of the F1 and F2.

2.) When a cell of genotype A/a ; B/b ; C/c having all the genes on separate chromosome pairs divides mitotically, what are the genotypes of the daughter cells?

3.) A presumed dihybrid in Drosophila, B/b ; F/f, is test-crossed with b/b ; f/f. (B = black body ; b = brown body; F = forked bristles; f = unforked bristles.) The results are

black, forked 230

brown, forked 240

black, unforked 210

brown, unforked 250

Use the 2 test to determine if these results fit the results expected from testcrossing the hypothesized dihybrid.

4.) In mice, dwarfism is caused by an X-linked recessive allele, and pink coat is caused by an autosomal dominant allele (coats are normally brownish). If a dwarf female from a pure line is crossed with a pink male from a pure line, what will be the phenotypic ratios in the F1 and F2 in each sex? (Invent and define your own gene symbols.)

Solution

Answer 2. If a cell divides mitotically (equational division), the genetic constitution of the daughter cells remains the same as the parents. So a parent with the genotype A/a; B/b; C/c will give rise to daughter cells with the genotype A/a; B/b; C/c.

However, if the cell was to divide meiotically (reductional division), the genetic constitution would be different as only one half of the parental chromosomes goes to the daughter cell. So the genotypes would be as follows:

ABC or AbC or ABc or Abc or aBC or abC or aBc or abc

Answer 3. For the given cross

                                                BbFf X bbff

For such a test cross the resultant phenotypic ration obtained is 1 (BF):1 (Bf):1 (bF):1 (bf).

For dihybrid cross, the degree of freedom will be 4 – 1 = 3

Chi square test is as follows

               

BF

Bf

bF

bf

Totals

Observed phenotype

230

210

240

250

930

Expected phenotype

930x ¼= 232.5

930x ¼= 232.5

930x ¼= 232.5

930x ¼= 232.5

930

Observed – expected

-2.5

-22.5

7.5

17.5

(Obs – Exp)²

Expected

6.25/232.5= 0.0268

506.25/232.5= 2.1774

56.25/232.5= 0.2419

306.25/232.5= 1.3172

3.7633

Looking at the probability value table for degree of freedom = 3,

df

.90

.70

.50

.30

.20

.10

.05

.01

3

0.58

1.42

2.37

3.67

4.64

6.25

7.82

11.35

Hence we can deduce that there is more than 20% probability for the occurrence of this event.

Answer 4. Let X^D = Normal phenotype (tall)

                           X^d = Dwarf phenotype

                           P = Pink coat

                           p = Brown coat

According to the question, the following cross is done

                                X^d X^dpp       x        X^DYPP

                                               

                               

F1

X^DP

YP

X^dp

X^DX^dPp (Tall pink female)

X^dYPp (Dwarf pink male)

On intercrossing,

                                                X^DX^dPp                  x              X^dYPp

F2

X^DP

X^Dp

X^dP

X^dp

X^dP

X^DX^dPP

Tall pink female

X^DX^dPp

Tall pink female

X^dX^dPP

Dwarf pink female

X^dX^dPp

Dwarf pink female

X^dp

X^DX^dPp

Tall pink female

X^DX^dpp

Tall brown female

X^dX^dPp

Dwarf pink female

X^dX^dpp

Dwarf brown female

YP

X^DYPP

Tall pink male

X^DYPp

Tall pink male

X^dYPP

Dwarf pink male

X^dYPp

Dwarf pink male

Yp

X^DYPp

Tall pink male

X^DYpp

Tall brown male

X^dYPp

Dwarf pink male

X^dYpp

Dwarf brown male

So the phenotypic ratios are as follows:

F1-          Females: All tall and pink; Males: All dwarf and pink

F2-          Females: 3 Tall pink: 3 dwarf pink: 1 tall brown : 1 dwarf brown

                Males:      3 Tall pink: 3 dwarf pink: 1 tall brown : 1 dwarf brown

	BF	Bf	bF	bf	Totals
Observed phenotype	230	210	240	250	930
Expected phenotype	930x ¼= 232.5	930x ¼= 232.5	930x ¼= 232.5	930x ¼= 232.5	930
Observed – expected	-2.5	-22.5	7.5	17.5
(Obs – Exp)2 Expected	6.25/232.5= 0.0268	506.25/232.5= 2.1774	56.25/232.5= 0.2419	306.25/232.5= 1.3172	3.7633