A testcross with a heterozygote b+bc+c produces a total of 700 offspring. The following are the...
A testcross with a heterozygote b+bc+c produces a total of 700 offspring. The following are the resulting phenotypes of the progeny and corresponding numbers:
b+c+ = 270; bc = 255; b+c=83; bc+=92
1) Based on the data, do the genes b and c appear to be linked or are they assorting independently? How did you determine your answer? If the genes are linked, which of the above genotypes are the non-recombinant progeny? Also, is the original configuration of the loci on the heterozygote parent chromosomes cis or trans? If the genes are indeed linked, what is the map distance between the two genes?
also 2) If you were to confirm your results with the chi square test, state the null hypothesis that you would test, including the genetic assumptions. What are the expected progeny numbers of each phenotype for the Chi Square test analysis? The Chi square value of the above data is 136. Using the p value table, what you would be the conclusion(s) of the Chi square test for this problem.
Homework Answers
Answer:
1).
Hint: In linkage problems, always parental combinantions are more numbered than recombinant combinantion. Hence, parental combination is b+c+/bc. This is cis configuration.
According to Mendel experiments, Dihybrid testcross progeny would be 1:1:1:1 ratio if the two genes are assorted independently. But the numbers of each genotype would not fit into the ratio. Hence, the two genes are not assorted independently and the two genes are linked.
Non-recombinants are b+c+ & bc and recombinants are b+c & bc+.
Recombinantion frequency (RF) = (no. of recombinants / Total progeny)100
RF= (83+92/700)100
= 25%
RF (%) = Distance between genes (mu)
Distance between the two genes = 25 m.u.
2).
As the Dihybrid testcross ratio should be 1:1:1:1. Each progeny has an equal expected number, 700/4.
|
Phenotype |
Observed(O) |
Expected (E) |
O-E |
(O-E)2 |
(O-E)2/E |
|
b+c+ |
270 |
175.00 |
95.00 |
9025.00 |
51.57 |
|
b+c |
83 |
175.00 |
-92.00 |
8464.00 |
48.37 |
|
bc+ |
92 |
175.00 |
-83.00 |
6889.00 |
39.37 |
|
bc |
255 |
175.00 |
80.00 |
6400.00 |
36.57 |
|
Total |
700 |
700.00 |
175.87 |
Chi-square value = 175.87
Degrees of freedom = no. of categories – 1
DF = 4-1
P value = 7.815
The chi-square value of 175.87 is greater than the p value of 7.815. Hence, the null hypothesis is rejected and the two genes are not assorted independently and the two genes are linked.
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