An EEGGrrQq individual is mated with an EeGgRrQq individual. What is a likelihood that their one...

An EEGGrrQq individual is mated with an EeGgRrQq individual. What is a likelihood that their one offspring has genotype EeGgRrqq?

Expert Solution

1. Parents : EEGGrrQq * EeGgRrQq

2. Gametes : EGrQ EGrq EGRQ EGRq EGrQ EGrq EgRQ EgRq EgrQ Egrq

eGRQ eGRq eGrQ eGrq egRQ egRq egrQ egrq

3. Total Gametes: 2 * 16

4. Total Possible

Combinations = 2 * 16 = 32

5. We have to calculate possible combination resulting EeGgRrqq, So possibility of recombination of 16 gametes from parent 2 with EGrQ of parent 1 rules out (as it contain dominant allele of Q)

6. So, we will cross only EGrq of parent 1 with 16 gametes of parent 2

7. From below table we can get that parent 1 already have the dominant allele ' E ' , so, we can again eliminate writing recombinations with gametes having dominant allele ' E ' from Parent 2.

**Recombinations from EGrq**
Parent 1	Parent 2	Recombination
EGrq	EGRQ	--
	EGRq	--
	EGrQ	--
	EGrq	--
	EgRQ	--
	EgRq	--
	EgrQ	--
	Egrq	--
	eGRQ	EeGGRrQq
	eGRq	EeGGRrqq
	eGrQ	EeGGrrQq
	eGrq	EeGGrrqq
	egRQ	EeGgRrQq
	egRq	EeGgRrqq
	egrQ	EeGgrrQq
	egrq	EeGgrrqq

8. We have got only one recombination, which have possible genotype EeGgRrqq, out of total of 32 Genotypes.

9. So, Likelihood of one offspring having EeGgRrqq is 1/ 32.

(Gametes of Parent 2 are calculated by branching method as given below)

gladiator answered 3 years ago

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