Fun with Punnett Squares

Tay-Sachs Disease (Autosomal Recessive)

Tay-Sachs is an autosomal recessive disorder. The normal allele is T, and the disease allele is t.

A woman who is a carrier (Tt) has children with a man who is also a carrier (Tt).

Using a Punnett square, determine which of the following statements is

correct about their potential children:

• 25% of their children will have Tay-Sachs, 50% will be carriers, and 25% will be unaffected.

• 50% of their children will have Tay-Sachs, and 50% will be carriers.

• 100% of their children will be carriers.

• None of their children will inherit Tay-Sachs.

Answer explanation

• TT (normal): inherits a normal allele from both parents

• Tt (carrier): inherits one normal and one disease allele

• tt (affected): inherits two copies of the disease allele

Since Tay-Sachs is recessive, a child must inherit two recessive alleles (tt) to have the disease. A single T allele is enough to prevent symptoms, even if the child carries one t allele.

• 1 TT = Unaffected, not a carrier (25%)

• 2 Tt = Carriers (do not have the disease) (50%)

• 1 tt = Has Tay-Sachs (25%)

Red-green colorblindness is an X-linked recessive trait.

The normal vision allele is Xᴺ, and the colorblind allele is Xᶜ.

A woman who is a carrier (XᴺXᶜ) has children with a man who has normal vision (XᴺY).

Use the blank canvas to draw a Punnett square and illustrate the statements below:

• The possible genotypes of the daughters are: __________ and __________

• The possible genotypes of the sons are: __________ and __________

• The percentage of daughters who are carriers is: __________%

• The percentage of sons who will be colorblind is: __________%

• The percentage of children overall who will be unaffected (not colorblind and not carriers) is: __________%

Answer explanation

• The possible genotypes of the daughters are: XᴺXᴺ and XᴺXᶜ

• The possible genotypes of the sons are: XᴺY and XᶜY

• The percentage of daughters who are carriers is: 50%

• The percentage of sons who will be colorblind is: 50%

• The percentage of children overall who will be unaffected (not colorblind and not carriers) is: 25%

  • (These are the XᴺXᴺ daughters and XᴺY sons)

A woman with blood type A (IAi) has children with a man who has blood type B (IBi).

True or False:

It is possible for their children to have any of the four blood types: A, B, AB, or O.

Hint: Use a Punnett Square to determine the possible blood types.

• True
• False

Answer explanation

Offspring possibilities:

• IAIB = Type AB

• IAi = Type A

• IBi = Type B

• ii = Type O

Since all four blood type genotypes are present in the Punnett square, the children could have any of the four blood types. Hence, the statement is True!

If one parent is heterozygous and

the other is homozygous recessive...

What is the probability of a heterozygous offspring?

• 25%

• 0%

• 50%

• 100%

Answer explanation

• Genotypes:

  • 2 Tt (heterozygous)

  • 2 tt (homozygous recessive)

• Probability of heterozygous offspring (Tt):
  2 out of 4 = 50%