9.Tasting PTC is a dominant trait (T), while non-tasting is (t). Below is data f

Question

9.Tasting PTC is a dominant trait (T), while non-tasting is (t). Below is data from two different classes of BIO 312. Calculate the p and q allele frequencies for the total data of all classes. Carry out a Chi-square analysis to see if the total BIO 312 population is in Hardy-Weinberg equilibrium. Show your calculations and explain your answer.

Tip: if you create a table in MS Word, that can be pasted into the answer, as opposed to trying to count spaces to align text.

D.f. 1, critical value of chi2 is 3.841; D.f. 2, 5.991; D.f 3, 7.816; D.f. 4 9.488.

        TT     Tt      tt

Fall 2013     6       13     6

Fall 2014     7       15     12

Fall 2015     12     18     6

Explanation / Answer

TT

(obsr.no.)

Tt

(obsr.no.)

tt

(obsr.no.)

Expected

no.

(obser-Expec)2

    /Expec

Degree of freedom

(possible phenotypes-1)

6

p= (2*6+13)/(2*25)

=0.5

6

q=(2*6+13)/(2*25)

0.5

TT=25x1/4=6.25

Tt=25x2/4=12.5

tt=25x 1/4=6.25

TT=(6.25-6)2/6.25=0.01

Tt=(13-12.5)2/12.5=0.02

tt=(6.25-6)2/6.25=0.01

0.04

populations are same

null hypothesis accepted

7

p=(2*7+15)/(2*34)

=0.426

12

q=(2*12+15)/(2*34)

=0.574

TT=34x1/4=8.5

Tt=34x2/4=17

TT=(8.5-7)2/8.5=0.26

Tt=(17-15)2/17=0.23

tt=(12-8.5)2/8.5=1.44

1.93

more than >0.05

null hypothesis rejected

12

p=(2*12+18)/(2*36)

=0.584

6

q=(2*6+18)/(2*36)

=0.416

TT=36x1/4=9

Tt=36x2/4=18

TT=(12-9)2/9=1

Tt=(18-18)2/18=0

tt=(9-6)2/9=1

2

more than >0.05

null hypothesis rejected

TT

(obsr.no.)

Tt

(obsr.no.)

tt

(obsr.no.)

Total

Expected

no.

(obser-Expec)2

    /Expec

chi square value

Degree of freedom

(possible phenotypes-1)

Fall 2013  

6

p= (2*6+13)/(2*25)

=0.5

13

6

q=(2*6+13)/(2*25)

0.5

25

TT=25x1/4=6.25

Tt=25x2/4=12.5

tt=25x 1/4=6.25

TT=(6.25-6)2/6.25=0.01

Tt=(13-12.5)2/12.5=0.02

tt=(6.25-6)2/6.25=0.01

0.04

populations are same

null hypothesis accepted

1 Fall 2014

7

p=(2*7+15)/(2*34)

=0.426

15

12

q=(2*12+15)/(2*34)

=0.574

34

TT=34x1/4=8.5

Tt=34x2/4=17

tt=34x1/4=8.5

TT=(8.5-7)2/8.5=0.26

Tt=(17-15)2/17=0.23

tt=(12-8.5)2/8.5=1.44

1.93

more than >0.05

null hypothesis rejected

1 Fall 2015

12

p=(2*12+18)/(2*36)

=0.584

18

6

q=(2*6+18)/(2*36)

=0.416

36

TT=36x1/4=9

Tt=36x2/4=18

tt=36x1/4=9

TT=(12-9)2/9=1

Tt=(18-18)2/18=0

tt=(9-6)2/9=1

2

more than >0.05

null hypothesis rejected

1

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