Chapter 14: Mendel and the gene idea

Question

We have F = freckles, f = no freckles, curly hair = C and c= straight hair. If FFCC marries with ffcc mention the P, F1 and F2 generations with the Punnett squares and showing the fractions and percentages for phenotypes and genotypes.

Answer 1

First we know from the problem that the P or Parent generation is:

FFCC X ffcc It makes no difference which one is the male or female

These produce the gametes FC and fc

When we cross these two using a Punnett square we get:

	FC	FC	FC	FC
fc	FfCc	FfCc	FfCc	FfCc
fc	FfCc	FfCc	FfCc	FfCc
fc	FfCc	FfCc	FfCc	FfCc
fc	FfCc	FfCc	FfCc	FfCc

From this crossing we get only one Geneotype for the F1 generation: FfCc (100%)

The Phenotype of this generation is Freckled and Curly haired. (100%)

Now we can cross two of the offspring from the F1 generation:

FfCc X FfCc

These produce the gametes FC, Fc, fC, and fc

When we cross these two offspring using a Punnett square we get:

	FC	Fc	fC	fc
FC	FFCC	FFCc	FfCC	FfCc
Fc	FFCc	FFcc	FfCc	Ffcc
fC	FfCC	FfCc	ffCC	ffCc
fc	FfCc	Ffcc	ffCc	ffcc

The genotypes of the F2 generation are as follows:

FFCC = 1/16

FFCc = 2/16

FFcc = 1/16

FfCC = 2/16

FfCc = 4/16

Ffcc = 2/16

ffCC = 1/16

ffCc = 2/16

ffcc = 1/16

This creates a ratio of the Genotypes for a dihybrid crossing: 1:2:1:2:4:2:1:2:1

The phenotypic ratio created by a dihybrid crossing is: 9:3:3:1

Freckled and Curly Hair = 9/16

Freckled and Straight Hair = 3/16

No Freckles and Curly Hair = 3/16

No Freckles and Straight Hair = 1/16