Question

SB13U Part 4: Genetics (Punnett Squares) 1. In pigs, erect ears (E) are dominant to drooping ears (e) and mule foot (M) is dominant to cloven foot (m). Cross a heterozygous erect eared, mule footed pig with a drooping eared, cloven footed pig. Complete the Punnett Square and list all the possible phenotypes. (10 marks) a) What are the genotypes of this cross? b) Complete the Punnett Square: List all phenotypes and how many are in above Punnett Square (NOTE: there may be phenotypes that are not included in the above cross, but still list them below and mark them as having zero cases)

Answer 1

1.
Given :-
Erect ear ‘E’ is dominant over drooping ear ‘e’.
Mule foot ‘M’ is dominant over cloven foot ‘m’.
We know that possible genotypes are EEMM, EeMm, eemm, Eemm, eeMm.
Condition for cross :- heterozygous erect eared mule footed pig should be crossed with drooping eared cloven footed pig.
Therefore the cross will happen between EeMm which is heterozygous and eemm.
Two types of alleles are involved in this cross hence the cross will be dihybrid cross which will 16 possible outcomes for this cross.
a. EeMm × eemm
b. EEMM code for erect eared mule footed pig
EeMm code for erect eared mule footed pig,
eemm code for drooping eared cloven footed pig,
Eemm code for erect eared cloven footed pig,
eeMm code for drooping eared mule footed pig.

	EM	Em	eM	em
em	EeMm	Eemm	eeMm	eemm
em	EeMm	Eemm	eeMm	eemm
em	EeMm	Eemm	eeMm	eemm
em	EeMm	Eemm	eeMm	eemm

The phenotypes in above punnet square are
1. erect eared mule footed pig
2. drooping eared cloven footed pig,
3. erect eared cloven footed pig,
4. drooping eared mule footed pig.
And 5th phenotype is erect eared mule footed pig – zero cases. EEMM cannot be observed in above punnet square because this can be observed only in case if both parent have heterozygous alleles or if both parents have homozygous dominant alleles.

2.
In this case both the alleles are dominant.
RR genotype will code for red pig.
BB genotype will code for brown pig.
RB genotype will code for red-brown pig.
Therefore the cross will occur between RR and RB
a. RR ×RB
b.

	R	R
R	RR	RR
B	RB	RB

The percentage of each phenotype will be equal that is 50% because we know that both the alleles are dominant hence both the phenotypes will be expressed in the progeny of this cross equally.