Assume that a cross is made between a heterozygous tall pea plant and a homozygo
ID: 13957 • Letter: A
Question
Assume that a cross is made between a heterozygous tall pea plant and a homozygous short pea plant. Fifty offspring are produced in the following frequency: 30= tall and 20= short.a) what frequency of tall and short plant is expected
b) If one wanted to test the goodness of fit between the observed and expected values, provide a statement of the null hypothesis.
c) compare a chi-square value associated with this appropriate test of significance.
d) How many degrees of freedom are associated with this test of significance?
Explanation / Answer
TALL allele-------T,
SHORT allele------t,
homozygous short pea plant.--------tt
heterozygous tall pea plant----Tt
a heterozygous tall pea plant and a homozygous short pea plant
tall pea plant----Tt x short pea plant.--------tt
T
t
t
Tt
Tt
t
Tt
Tt
TALL :SHORT; 3:1= 75% :25:%
ACCORDING HARDY WEINBERG LAW
P^2+2PQ+ q^2=1
q^2=recessive homozygotes
q^2=20/50= short. q =0.63
P+q=1
P=1- q =1-0.63=0.37
GENOTYPE
OBSEERVED
EXPECTED
O-E
(O-E)^2
(O-E)^2/E
P^2=TT
30
29.8
.2
0
0.0013
2PQ=2Tt
q^2=tt
20
19.84
0.06
1.81
P^2=TT=0.37 x0.37x50=6.8
2pq=2Tt=2x 0.37 x0.63x50=23
q^2=19.84
a) what frequency of tall and short plant is expected
GENOTYPE
OBSEERVED
EXPECTED
P^2=TT
30
29.8
2PQ=2Tt
q^2=tt
20
19.84
b) If one wanted to test the goodness of fit between the observed and expected values, provide a statement of the null hypothesis.
There is no difference between obserevd and expected frequencies
c) compare a chi-square value associated with this appropriate test of significance.
Calculatec value is less than table value ,null hypotheiss is acepeted
d) How many degrees of freedom are associated with this test of significance?
5% ------ test of significance
degrees of freedom= (n-1) x(c-1)=1x1=1
T
t
t
Tt
Tt
t
Tt
Tt
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