Question

Using the data shown below, determine the dominance relationships as well as genotypes of the parental plants by analyzing the phenotypes affecting flower colour (blue or white) or pod shape (pointed or rounded) in the progeny from each cross (i-iii). Use C and c to designate the dominant and recessive alleles for colour and S or s to designate the dominant and recessive alleles for shape. Parents: Progeny i. blue, pointed x blue, pointed 3/4 blue, pointed 1/4 white, pointed ii. blue, pointed x white, pointed 6/16 white, pointed 2/16 white, rounded 6/16 blue, pointed 2/16 blue, rounded iii. blue, pointed x white, rounded 1/4 blue, pointed 1/4 blue, rounded 1/4 white, pointed 1/4 white, rounded a. Using the appropriate symbols, state the dominant and recessive alleles for these two traits. Briefly explain your answer. b. Fill in the genotypes for the parents in each cross and briefly explain your rationale in each case. i. ________X ________ ii. ________X ________ iii. ________X ________

Answer 1

Answer:

Punnet squares for three cases are shown below :
The convention is C= dominant colour c= recessive colour and S= Dominant shape and s= recessive shape

Case (i)

Case (ii)

Case (iii)

Answer a). From the crosses above we can deduce that :

- in colours Blue is the dominant colour (C) and white is recessive (c).
- in shape Pointed is the dominant shape (S) and rounded is recessive (s)

Answer b). Genotypes of the parents in each cross following the same conventions above are:

Case (i) CcSS x CcSS OR CcSS x CcSs

In this cross all the progeny have pointed shape and as we deduced pointed is the dominant then surely allele (S) is dominant here but one of the two parents may be heterozygous for shape and carry an (s) allele. Both parents can't have (s) allele as we have all the progeny with pointed shape and there are no round shapes.

For Colour we have both parents heterozygous and that's the reason for having a 1/4 progeny with white colour (cc) as both parents contribute one recessive allele each.

Case (ii) CsSs x ccSs

In this cross we have a mixed progeny and we note that the colour distribution is same in all progeny but pointed progeny are more than the rounded. Thus we conclude that at least one parent is recessive white (cc) and other is heterozygous for colour (Cc) and both parents have to be heterozygous for shape (Ss) so as to have both rounded and pointed shapes in the progeny.

Case (iii) CcSs x ccss

In this case one parent is white rounded hence has to be homozygous recessive for both the traits (ccss). As the progeny is all produced in ratio of 1 : 1 : 1 : 1 we understand that the other parent carries heterozygous genes so as to give equal opportunity for all the traits in shape and colour to arise at same frequency.