30. In beetles, pygmy (py) is recessive to normal size (py+), and red color (r)
ID: 259116 • Letter: 3
Question
30. In beetles, pygmy (py) is recessive to normal size (py+), and red color (r) recessive to brown (r+). A beetle heterozygous for these characteristics was test crossed to a beetle homozygous for pygmy and red. The following are progeny phenotypes from this testcross: (14 pts) normal size, brown color 180 normal size, red color pygmy size, brown color 19 pygmy size, red color191 Total 412 You want to determine if the genes that determine size and color are assorting independently. Carry out a chi-square test to test your hypothesis. (Note- do NOT do the math, but rather write out the FUL, equatioriate mumerical waies) a) Calculate the expected number of progeny for each combination of genotypes. (8 pts) b) Now write out the FULL chi-square equation you would use to evaluate whether these two loci are sorting independently. (Note- write out the equation you would use and fill in the numerical values were appropriate. You do not have to solve the equation.) (4 pts) c) How many degrees of freedom exist in this example? (2 pts)Explanation / Answer
a)
Expected ratio = Total x appropriate proportion.
Category
Normal, brown
Normal, red
Pygmy, brown
pygmy, red
Total
Observed values (O)
180
22
19
191
412
Exptected Ratio (ER)
1
1
1
1
4
Propotion of each category
1/4
1/4
1/4
1/4
1
Exprected Values (E)
412*1/4= 103
412*1/4= 103
412*1/4= 103
412*1/4= 103
412
b)
Null hypothesis: The observed values are not deviating from the expected values.
Category
Normal, brown
Normal, red
Pygmy, brown
pygmy, red
Total
Observed values (O)
180
22
19
191
412
Exptected Ratio (ER)
1
1
1
1
4
Exprected Values (E)
103
103
103
103
Deviation (O-E)
77
-81
-84
88
D^2
5929
6561
7056
7744
D^2/E
57.56311
63.69903
68.50485
75.18447
264.9515
X^2
264.9515
Degrees of freedom
4
-
1
3
Inference: As the calcuclated chisquare test i.e. 264.95 is greater than the table value i.e. 7.82 at 3 DF and 0.05 probability, hence the null hypothesis is rejected. Which means those are not assoreted independently, those are linked genes.
c) Degrees of freedom = No of categories – 1 = 4-1= 3.
Category
Normal, brown
Normal, red
Pygmy, brown
pygmy, red
Total
Observed values (O)
180
22
19
191
412
Exptected Ratio (ER)
1
1
1
1
4
Propotion of each category
1/4
1/4
1/4
1/4
1
Exprected Values (E)
412*1/4= 103
412*1/4= 103
412*1/4= 103
412*1/4= 103
412
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