Unit 4 Test: Mendelian Genetics Test Shortened

10

Drag 1 black mouse and 1 white mouse into the parent squares. What is the phenotype of all of the offspring?

10

Now, lick on the box that says "Show Genotype". What is the genotype of the offspring?

10

Complete the Punnett Square.
If we were to cross a heterozygous mouse with a homozygous recessive mouse, what would the probability of offspring genotypes be?
fill in the PERCENTAGE of the probability for each genotype in the show your work section.

10

What would the genotype of the parent mice need to be to produce offspring with the genotpes: FF and Ff?
(an empty punnett square is provided to show your work)

10

What would the genotype of the parent mice need to be to produce offspring with only the genotpe: Ff?
(an empty punnett square is provided to show your work)

Hungtington’s Disease Case Study
Huntington’s disease is an autosomal dominant disorder, meaning anyone with the dominant allele will develop this disease. In Huntington’s disease there is a continuous breakdown of neurons. An individual who has Huntington’s disease has an increased difficulty in their ability to think and move, as well as behavioral issues. The disease is generally diagnosed with genetic testing.

Huntington’s disease is unique in the fact that it does not affect individuals in their early years or prevent them from living past infancy.The symptoms often generally develop later in life, often when a people are in their 30s or 40s. This means that individuals may reproduce and potentially pass on the mutation to their offspring before they are aware they have the disease. There are medications available to help manage the symptoms of Huntington’s disease, but there is not yet a cure.

Huntington’s disease is caused by a mutant huntingtin gene (HTT) where there is a type of mutation called triplet repeats. In Huntington’s disease, there are extra repetitions of the DNA sequence CAG, which encodes glutamine, on chromosome 4. In a normal individual the number of repeats of the CAG sequence is between 9 to 36. The individuals with Huntington’s disease have greater than 37 repeats of the CAG sequence. The larger the number of repeats, the more severe the condition.

Since Huntington’s is autosomal dominant, the individuals who have Huntington’s disease have at least one dominant allele. This means that individuals who are homozygous dominant (HH) and individuals who are heterozygous (Hh) will develop Huntington’s disease. Individuals who do not have Huntington’s disease are homozygous recessive (hh).

Genotype Key
Normal = hh (homozygous recessive)
Huntington’s disease = Hh (heterozygous)
Huntington’s disease = HH (homozygous dominant)

15

Two individuals who are heterozygous for Huntington’s are pregnant with a child. What is the probability that the resulting child will not inherit Huntington’s disease?

Calculate the probability. The punnett square is provided to show your work.

15

Using the punnett square that you created in question number 5, what is the probability that the resulting child will develop Huntington’s disease? Remember Huntington’s disease is an autosomal dominant disease meaning the presence of a dominant allele results in the development of the disease.

10

Using the punnett square that you created in question number 5, what is the probability that the resulting child will be heterozygous for Huntington’s disease?

10

What would the genotype of the parents need to be to produce a child that has a 50% chance of being born homozygous recessive and a 50% chance of being born heterozygous for Huntington’s disease?
An empty punnett square is provided to show your work.

Drag 1 black mouse and 1 white mouse into the parent squares. What is the phenotype of all of the offspring?

Now, lick on the box that says "Show Genotype". What is the genotype of the offspring?

Complete the Punnett Square. If we were to cross a heterozygous mouse with a homozygous recessive mouse, what would the probability of offspring genotypes be? fill in the PERCENTAGE of the probability for each genotype in the show your work section.

What would the genotype of the parent mice need to be to produce offspring with the genotpes: FF and Ff? (an empty punnett square is provided to show your work)

What would the genotype of the parent mice need to be to produce offspring with only the genotpe: Ff? (an empty punnett square is provided to show your work)

Hungtington’s Disease Case Study

Two individuals who are heterozygous for Huntington’s are pregnant with a child. What is the probability that the resulting child will not inherit Huntington’s disease?Calculate the probability. The punnett square is provided to show your work.

Using the punnett square that you created in question number 5, what is the probability that the resulting child will develop Huntington’s disease? Remember Huntington’s disease is an autosomal dominant disease meaning the presence of a dominant allele results in the development of the disease.

Using the punnett square that you created in question number 5, what is the probability that the resulting child will be heterozygous for Huntington’s disease?

What would the genotype of the parents need to be to produce a child that has a 50% chance of being born homozygous recessive and a 50% chance of being born heterozygous for Huntington’s disease?An empty punnett square is provided to show your work.

Complete the Punnett Square.
If we were to cross a heterozygous mouse with a homozygous recessive mouse, what would the probability of offspring genotypes be?
fill in the PERCENTAGE of the probability for each genotype in the show your work section.

What would the genotype of the parent mice need to be to produce offspring with the genotpes: FF and Ff?
(an empty punnett square is provided to show your work)

What would the genotype of the parent mice need to be to produce offspring with only the genotpe: Ff?
(an empty punnett square is provided to show your work)

Two individuals who are heterozygous for Huntington’s are pregnant with a child. What is the probability that the resulting child will not inherit Huntington’s disease?

Calculate the probability. The punnett square is provided to show your work.

What would the genotype of the parents need to be to produce a child that has a 50% chance of being born homozygous recessive and a 50% chance of being born heterozygous for Huntington’s disease?
An empty punnett square is provided to show your work.