S5 - S1.In the garden pea, orange pods (orp) are recessive...

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S1.In the garden pea, orange pods ( orp ) are recessive to green pods ( Orp ), and sensitivity to pea mosaic virus ( mo ) is recessive to resistance to the virus ( Mo ). A plant with orange pods and sensitivity to the virus was crossed to a true- breeding plant with green pods and resistance to the virus. The F 1 plants were then testcrossed to plants with orange pods and sensitivity to the virus. The following results were obtained: 160 orange pods, virus sensitive 165 green pods, virus resistant 36 orange pods, virus resistant 39 green pods, virus sensitive 400 total A. Conduct a chi square analysis to see if these genes are linked. B. If they are linked, calculate the map distance between the two genes. Answer: A. Chi square analysis. 1. Our hypothesis is that the genes are not linked. 2. Calculate the predicted number of offspring based on the hypothesis. The testcross is The predicted outcome of this cross under our hypothesis is a 1:1:1:1 ratio of plants with the four possible phenotypes. In other words, 1/4 should have the phenotype orange pods, virus sensitive; 1/4 should have green pods, virus resistant; 1/4 should have orange pods, virus resistant; and 1/4 should have green pods, virus sensitive. Since there were a total of 400 offspring, our hypothesis predicts 100 offspring in each category. 3. Calculate the chi square. 2222 2 11 2 2 33 4 4 1234 2 222 2 2 () (160 100) (165 100) (36 100) (39 100) 100 100 100 100 36 42.3 41 37.2 156.5 OE EE EE χ −−−− =+ ++ − −−− =+++ =+ ++ = 4. Interpret the chi square value. The calculated chi square value is quite large. This indicates that the deviation between observed and expected values is very high. For three degrees of freedom in table 2.1, such a large deviation is expected to occur by chance alone less than 1% of the time. Therefore, we reject the hypothesis that the genes assort independently. As an alternative, we may infer that the two genes are linked.
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B. Calculate the map distance. Number of nonparental offspring Map distance = 100 Total number of offspring 36 39 100 36 39 160 165 18.8 mu × + ++ + = The genes are approximately 18.8 mu apart. S2. Two recessive disorders in mice, droopy ears and flaky tail, are caused by genes that are located 6 mu apart on chromosome 3. A true-breeding mouse with normal ears ( De ) and a flaky tail ( ft ) was crossed to a true-breeding mouse with droopy ears ( de ) and a normal tail ( Ft ). The F 1 offspring were then crossed to mice with droopy ears and flaky tails. If this testcross produced 100 offspring, what is the expected outcome? Answer: The testcross is The parental offspring are Dede ftft normal ears, flaky tail dede Ftft droopy ears, normal tail The recombinant offspring are dede ftft droopy ears, flaky tail Dede Ftft normal ears, normal tail Since the two genes are located 6 mu apart on the same chromosome, 6% of the offspring will be recombinants.
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This note was uploaded on 04/07/2009 for the course BIO 325 taught by Professor Saxena during the Spring '08 term at University of Texas at Austin.

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S5 - S1.In the garden pea, orange pods (orp) are recessive...

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