Problem 58 - Calculus for Biology and Medicine

Question

A number of human traits are caused by a single pair of recessive genes and thus manifest themselves only in individuals who are homozygous for the mutant gene. An individual with one normal and one mutant gene is a carrier, but does not exhibit the trait.Calculate each of the probabilites. An attached earlobe is caused by a single pair of recessive genes \((a a) .\) For a couple consisting of a heterozygous individual \((A a)\) and an affected person \((a a)\), what is the probability that a child has an unattached earlobe?

Step-by-Step Solution

Answer

The probability that a child will have unattached earlobes is \(\frac{1}{2}\).

Step 1: Understanding Genotypes

To solve this problem, we must first understand the genotypes involved. The genotype for attached earlobes is homozygous recessive ( aa ). The heterozygous genotype ( Aa ) results in the unattached earlobe trait, as the dominant allele A masks the recessive allele a .

Step 2: Determine Parent Genotypes

The couple consists of a heterozygous individual ( Aa ) and an affected individual ( aa ). We need to use these genotypes to create a Punnett square that shows the possible genetic combinations for their offspring.

Step 3: Create Punnett Square

Draw a 2x2 square to represent the potential genotypes of the children from this couple. The heterozygous parent ( Aa ) can produce gametes with A or a alleles. The homozygous recessive parent ( aa ) will only produce gametes with a alleles. Place these gametes on the top and side of the Punnett square to find all combinations.

Step 4: Filling the Punnett Square

Fill the Punnett square with combinations: - The top row represents possible alleles from the Aa parent: A and a . - The left column represents alleles from the aa parent: a and a . - Combine these to fill in the squares: - Top left: ( Aa ) - Top right: ( aa ) - Bottom left: ( Aa ) - Bottom right: ( aa )

Step 5: Identify Unattached Earlobe Genotypes

From the Punnett square, the children can have the following genotypes: - Aa : Unattached earlobe (due to dominant allele A ) - aa : Attached earlobe (recessive homozygous) Thus, unattached earlobes occur with genotype Aa .

Step 6: Calculate Probability of Unattached Earlobes

Count the number of genotypes that lead to unattached earlobes. From the Punnett square, there are 2 Aa combinations out of 4 total combinations: - Aa (unattached): 2- aa (attached): 2The probability of a child having unattached earlobes is the number of Aa outcomes divided by the total: \( \frac{2}{4} = \frac{1}{2} \).

Key Concepts

Genotype

A genotype is the genetic makeup of an organism and determines specific traits by the combination of alleles it inherits from its parents. Alleles are variations of a gene found at the same locus on homologous chromosomes. Genotypes consist of two alleles per gene, one from each parent. In this discussion, we are focusing on human earlobe attachment, a trait influenced by genetic variation.
The key genotypes here are:

• Aa (Heterozygous Genotype): This individual carries one dominant allele ( A ) and one recessive allele ( a ). Despite having the recessive allele, the phenotype (visible trait) displays the dominant characteristic, leading to unattached earlobes.
• aa (Homozygous Recessive Genotype): This genotype contains two recessive alleles ( a ), leading to the manifestation of attached earlobes. Only when both alleles are recessive will the recessive trait appear.

The heterozygous genotype ( Aa ) masks the effect of the homozygous recessive ( aa ) due to the presence of the dominant allele. Hence, understanding genotypes help predict offspring traits using tools like the Punnett Square.

Recessive Genes

Recessive genes require two copies, one from each parent, to express a trait. If a dominant allele is present, it will mask the effect of the recessive gene. In the earlobe example, the recessive gene ( a ) must be present in both alleles ( aa ) to result in attached earlobes.

• Carrier: An individual with one dominant ( A ) and one recessive allele ( a ) is known as a carrier. Although they do not express the recessive trait, they can pass the recessive gene to their offspring.
• Expression: Only when the genotype is homozygous recessive ( aa ) will the recessive trait show. In this case, the earlobes will be attached.

Recessive genes often remain hidden in populations due to masking by dominant alleles, therefore understanding how they work is vital in genetics. In this scenario, a heterozygous parent ( Aa ) and a homozygous recessive parent ( aa ) contribute different allele combinations to their offspring.

Probability Calculation

Probability calculation in genetics uses the Punnett Square to predict offspring genotypes. It helps calculate the likelihood of specific traits appearing based on parental genotypes.

• Step 1: Identify the parents' genotypes. Here, one parent is heterozygous (Aa), and the other is recessive homozygous (aa).
• Step 2: Construct the Punnett Square. This grid outlines possible gamete combinations from both parents, considering each allele type.
  For this scenario:
  • The heterozygous parent (Aa) produces A and a gametes.
  • The homozygous (aa) parent contributes only a gametes.
• Step 3: Fill in the square with allele combinations, each pair represents a potential genotype for the offspring:
  • Aa (Unattached earlobes): 2 squares
  • aa (Attached earlobes): 2 squares
• Step 4: Calculate probability by dividing the number of desired outcomes by total outcomes. For unattached earlobes, the probability of having the Aa genotype is \( \frac{2}{4} = \frac{1}{2} \).

Thus, there's a 50% chance of producing offspring with unattached earlobes given these parent genotypes.