In tomatoes, one gene determines whether the plant has purple (G) or green (g) s
ID: 198659 • Letter: I
Question
In tomatoes, one gene determines whether the plant has purple (G) or green (g) stems, and another locus determines whether the leaves are "cut" (P) or "potato" (p). A true-breeding plant with purple stem and cut leaves is crossed with a true-breeding plant having green stem and potato leaves. The F plants are then allowed to self-pollinate. The F2 generation consisted of 310 plants with purple stem, cut leaves; 31 plants with gren stem, potato leaves; 120 plants purple stem, potato leaves; and 129 plants with green stem,cut laves Using the chi-square tes, determine if this outcome is consistent with the predicted ratios for this cross. Provide the null and alternative hypotheses, chi-square statistic, P-value, and a conclusion. (10 pts)Explanation / Answer
Null hypothesis: The observed values are not deviating from the 9:3:3:1 ratio.
Altenative hypothesis : The observed values are deviating from the 9:3:3:1 ratio.
Test static – Chisquare test:
Category
Purple Cut
Purple Potato
Green cut
green potato
Total
Observed values
310
120
129
31
590
Exptected Ratio
9
3
3
1
16
Exprected Values
331.875
110.625
110.625
36.875
Deviation
-21.875
9.375
18.375
-5.875
D^2
478.5156
87.89063
337.6406
34.51563
D^2/E
1.441855
0.794492
3.052119
0.936017
6.224482
X^2
6.224482
Degrees of freedom
4
-
1
3
Conclusion / Inference: The calculated chisquare value i.e. 6.22 is lessthan the table value i.e. 7.82 at 3 DF and 0.05 probability, hence the null hypothesis is accepted. Which means this traits following the mendilian mode of inheritance pattern.
Category
Purple Cut
Purple Potato
Green cut
green potato
Total
Observed values
310
120
129
31
590
Exptected Ratio
9
3
3
1
16
Exprected Values
331.875
110.625
110.625
36.875
Deviation
-21.875
9.375
18.375
-5.875
D^2
478.5156
87.89063
337.6406
34.51563
D^2/E
1.441855
0.794492
3.052119
0.936017
6.224482
X^2
6.224482
Degrees of freedom
4
-
1
3
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