Problem

Solutions For An Introduction to Genetic Analysis Chapter 6 Problem 69P

Step-by-Step Solution

Solution 1

a) It should be noted that in cross 1 only one gene is involved and that single is dominant to double. Cross 2 which is approximately in the ration of 3:1 confirms that single is dominant to double.

It should be noted that in cross 3 and 4 a ratio of 1:1 was observed suggesting that super-double is an allele of both single and double which should be heterozygous and dominant to both single and double. When crossed with appropriate plants the super-double is capable of producing both single and double, it cannot be heterozygous for single allele. Hence, it should be heterozygous for the double allele.

Multiple allelic series detected: super-double>single>double.

Cross

Parents

Progeny

Conclusion

1

A S /A S X A D /A D

A S /A D

A S is dominant to AD

2

A S /A D X A S /A D

3 AS/-: 1 AD/AD

Supports above conclusion

3

A D /A D X A Sd /A D

1 ASd/AD: 1 AD/AD

A Sd is dominant to AD

4

A S /A S x A Sd /A D

1 ASd/AS: 1 AS/AD

A Sd is dominant to AS

5

A D /A D X A Sd /A S

1 ASd/AD: 1AD/AS

Supports conclusion of heterozygous super-double

6

A D /A D X A S /A D

1 AD/AD: 1 AD/AS

Supports conclusion of heterozygous super-double

b) The cross does not take into account about either the female sterility or the origin of super-double plant from a double-flowered variety. There are quite a few mechanisms that could be used to explain the origin of super-double from double-flowered varieties. But because of female sterility, no homozygote for super-double could be observed.

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