Question

Exercise 6.2 worksheet - MONOHYBRID PRACTICE PROBLEMS

1. Cystic fibrosis is an autosomal recessive genetic disorder. Ron is homozygous dominant (FF) and Nancy is a carrier (Ff ) of cystic fibrosis. Use a Punnett square to predict the genotypic and phenotypic ratios of their offspring who are normal, normal carriers, and affected. Show all your work.

Assign symbols for each allele.

Parents (P): Male ♂ =

Female ♀ =

Key

Punnett Square:

First Generation (F1)

Genotypic Ratio:

Phenotypic Ratio:

2. Patty is homozygous dominant for freckles (SS), while Charlie is homozygous for no freckles (ss). Draw a Punnett square and predict the probability that their children will have freckles.

Assign symbols for each allele.

Parents (P): Male ♂ =

Female ♀ =

Key

Punnett Square:

First Generation (F1)

Genotypic Ratio:

Phenotypic Ratio:

Probability a child will have freckles:

3. Larry and Lola Little have achondroplasia, a form of dwarfism. Both are heterozygotes. Their son, Big Bob Little, is 7'1''. (Achondroplasia is autosomal dominant, and homozygous dominant babies usually are stillborn or die shortly after birth.) Use a Punnett square to show how Big Bob got his genotype.

Assign symbols for each allele.

Big Bob Little’s Genotype =

Parents (P): Male ♂ =

Female ♀ =

Key

Punnett Square:

First Generation (F1)

Genotypic Ratio:

Phenotypic Ratio:

4. Woody Guthrie, who wrote “This Land is Your Land,” was heterozygous for Huntington’s disease (Hh). His wife was homozygous recessive and perfectly normal (hh). Huntington’s disease is caused by a latent dominant gene, meaning that it is not phenotypically (physically) expressed until later in life. Dominant disease genes are expressed in homozygous dominant and heterozygous people (HH or Hh). Draw a Punnett square for Woody and his wife and predict what percentage of their children will develop Huntington’s. Assign symbols for each allele.

Parents (P): Male ♂ =

Female ♀ =

Key

Punnett Square:

First Generation (F1)

Genotypic Ratio:

Phenotypic Ratio:

Percentage of children that will have Huntington’s disease:

Answer 1

1. Parents, Male ♂ = FF and Female ♀ = Ff (male first column, blue and female first row, pink)

	F	f
F	FF	Ff
F	FF	Ff

First Generation (F1)

Genotypic Ratio: 1:1, FF : Ff

Phenotypic Ratio:1:1 Normal : Carrier

Half normal children and half carriers

2. Parents, Male ♂ = ss and Female ♀ = SS (male first column, blue and female first row, pink)

	S	S
s	Ss	Ss
s	Ss	Ss

First Generation (F1)

Genotypic Ratio: 1, Ss

Phenotypic Ratio: 1, Freckles

All children will have freckles. Probability 100% (or probability 1)

3. Big Bob Little’s Genotype = dd

Parents, Male ♂ = Dd and Female ♀ = Dd (male first column, blue and female first row, pink)

	D	d
D	DD	Dd
d	Dd	dd

First Generation (F1)

Genotypic Ratio: 1:2:1, DD:Dd:dd

Phenotypic Ratio: 1:2:1, stillborn: Dwarf : Normal

So both parents were heterozygotes (Dd) and since achondroplasia is autosomal dominant, both of them were dwarfs. But their child is dd, a homozygous recessive and therefore, normal.

4. Parents, Male ♂ = Hh and Female ♀ = hh (male first column, blue and female first row, pink)

	h	h
H	Hh	Hh
h	hh	hh

First Generation (F1)

Genotypic Ratio: 1:1, Hh : hh

Phenotypic Ratio: 1:1, Huntington’s disease : normal

Percentage of children that will have Huntington’s disease: 50%