Question

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

Answer 1

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)