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If you could show the work on the punnent square too, that would be amaing. I am

ID: 212683 • Letter: I

Question

If you could show the work on the punnent square too, that would be amaing. I am just very lost.

Suppose you identified two new mutations in Spartan fruit flies (Drosophila spartanensis). One mutation, 'Spartan body', results in the entire body being green with a white 'S' marking on the thorax, whereas the body color of wildtype flies is tan and without markings. A second mutation, hopper legs, shows very muscular legs, whereas the legs of wildtype flies are thin. Let's also say that you were able to culture true-breeding populations of both mutations and then you tested the inheritance of each by performing a set of reciprocal crosses with true-breeding flies. Below are the F1 and F2 progeny of these crosses Offspring Offspring Phenotype Combinations Tan & Thin Tan & Hopper Spartan & Thin Spartan & Hopper F Generation F2 Generation Parental Crosses Total 283 104 944 286 1617 MalesFe males 137 49 488 140 814 Males Females Total 146 Tan-bodied & Hopper- legged Males x Spartan bodied & Thin-legged Females 456 146 803 100* 105 205 205 Spartan-bodied & Thin- legged Males x Tan-bodied & Hopper-legged Females Tan & Thin Tan & Hopper Spartan & Thin Spartan & Hopper 161 47 450 160 818 1491 310 100 953 305 1668 53 503 145 850 Reverse 151 322 322

Explanation / Answer

Null hypothesis: The observed values are not deviating from the Dihybrid F2 ratio i.e. 9:3:3:1.

Test static – Chisquare test:

Category

Spartan, thin

Tan, thin

Spartan, hooper

Tan, hooper

Total

Observed values (O)

953

310

305

100

1668

Exptected Ratio (ER)

9

3

3

1

16

Exprected Values (E)

938.25

312.75

312.75

104.25

Deviation (O-E)

14.75

-2.75

-7.75

-4.25

D^2

217.5625

7.5625

60.0625

18.0625

D^2/E

0.231881

0.024181

0.192046

0.173261

0.62137

X^2

0.62137

Degrees of freedom

4

-

1

3

Inference : The calculated chisquare value i.e. 0.62 is less than the table value i.e. 7.82 at 3 DF and 0.05 porbability, hence the null hypothesis is accepted. Which means those two genes assorted independently.

Find the below image for punner square...

Category

Spartan, thin

Tan, thin

Spartan, hooper

Tan, hooper

Total

Observed values (O)

953

310

305

100

1668

Exptected Ratio (ER)

9

3

3

1

16

Exprected Values (E)

938.25

312.75

312.75

104.25

Deviation (O-E)

14.75

-2.75

-7.75

-4.25

D^2

217.5625

7.5625

60.0625

18.0625

D^2/E

0.231881

0.024181

0.192046

0.173261

0.62137

X^2

0.62137

Degrees of freedom

4

-

1

3