Genetics Homework Problem 1 You are mapping genes in a speci

Genetics Homework Problem 1:

You are mapping genes in a species of fruit fly. You cross a wild type fly to a fly from a true breeding mutant strain with green eyes (g), short wings (s w), and hairy legs (hl). The F1 progeny all appear wild type (red eyes, long wings, smooth legs). You then testcross a female F1 fly to a male from the mutant strain, and obtain 200 progeny a. What are the genotypes of the two tiles in the test cross?_____ times ______ b. How many progeny of each phenotype would be expected if the genes were assorting independently? c. Can you reject independent assortment at P=0.05? Use statistical evidence to back up your claim. d. Assuming these genes are all linked on the same chromosome, draw a genetic map showing the order and map distances between the three genes.

Solution

Answer:

a). The genotypes of the two flies in the test cross: Gg Swsw Hlhl X   gg swsw hlhl

b). If the genes were assorting independently, 8 types of phenotypes are expected in equl proportions

1:1:1:1:1:1:1:1

Green, short, hair: 1

Red, long, smooth: 1

Green, long, smooth: 1

Green, short, smooth: 1

Red, short, hairy: 1

Red, long, hairy: 1

Green, long, hairy: 1

Red, short, hairy: 1

c).

Phenotype

Observed(O)

Expected (E)

O-E

(O-E)2

(O-E)2/E

Green, short, hair:

44

25

19

361.00

14.44

Red, long, smooth

42

25

17

289.00

11.56

Green, long, smooth

20

25

-5

25.00

1.00

Green, short, smooth

5

25

-20

400.00

16.00

Red, short, hairy

20

25

-5

25.00

1.00

Red, long, hairy

7

25

-18

324.00

12.96

Green, long, hairy

30

25

5

25.00

1.00

Red, short, hairy

32

25

7

49.00

1.96

200

200

59.92

Chi-square value = 59.92

The independent assortment at P=0.05 is rejected.

d).

Explanation

Shorthand

Number of progeny

g sw hl

44

G Sw Hl

42

g Sw Hl

20

g sw Hl

5

G sw hl

20

G Sw hl

7

g Sw hl

30

G sw Hl

32

Imagine the order is

G SW Hl / g sw hl

1.    If single cross over (SCO) occurs between G & Sw

Normal order= G---------Sw & g-----sw

After cross over= G-----sw & g------Sw

G-----sw recombinants are 20+ 32 = 52

g----- Sw recombinants are 20+ 30 = 50

Total recombinants = 102

RF = (102/200.)*100 =51%

2.    If single cross over (SCO) occurs between Sw & Hl

Normal order= Sw---------Hl & sw----hl

After cross over= Sw-----hl & sw------Hl

Sw-----hl recombinants are 30 + 7 = 37

sw------Hl recombinants are 32+5 = 37

Total recombinants = 74

RF = (74/200)*100 = 37%

3.    If single cross over (SCO) occurs between G & Hl

Normal order= G---------Hl & g------hl

After cross over= G-----hl & g------Hl

G-----hl recombinants are 20 + 7=27

g------Hl recombinants are 20+5=25

Total recombinants = 52

RF = (52/200)*100 = 26%

% RF = Map unit distance

The order of genes is -----

G------26 m.u.--------Hl------37 m.u.---------Sw

Gene for color of eyes and length of wings are not linked. The gene for legs is linked with gene for wings & gene for eye.

Phenotype	Observed(O)	Expected (E)	O-E	(O-E)2	(O-E)2/E
Green, short, hair:	44	25	19	361.00	14.44
Red, long, smooth	42	25	17	289.00	11.56
Green, long, smooth	20	25	-5	25.00	1.00
Green, short, smooth	5	25	-20	400.00	16.00
Red, short, hairy	20	25	-5	25.00	1.00
Red, long, hairy	7	25	-18	324.00	12.96
Green, long, hairy	30	25	5	25.00	1.00
Red, short, hairy	32	25	7	49.00	1.96
	200	200			59.92