Problem 57 - Calculus for Biology and Medicine

Question

A number of traits are caused by recessive genes. The traits show up only in individuals who are homozygous (i.e., have two copies of 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. The inability to roll one's tongue is caused by a single pair of recessive genes \((r r) .\) For a couple consisting of a heterozygote individual \((R r)\) and an affected person \((r r)\), what is the probability that, among their four children, at most one child is unable to roll his or her tongue?

Step-by-Step Solution

Answer

The probability is 0.3125.

Step 1: Determine Cross Outcomes

To find the probability of a specific genotype combination, we perform a Punnett square with the genotypes \(Rr\) and \(rr\). The possible combinations for offspring are: - \(Rr\): 50% chance (carrier, able to roll tongue)- \(rr\): 50% chance (affected, unable to roll tongue).

Step 2: Define Parameters for Probability

Each child's genotype is independent of the others. For each child, the probability of being affected \((rr)\) is 0.5, and the probability of not being affected \((Rr)\) is 0.5. The situation can be modeled by a binomial distribution with parameters \(n = 4\) (number of trials or children) and \(p = 0.5\) (probability of being affected for one child).

Step 3: Calculate Probability for 0 Children Affected

Using the binomial probability formula, calculate the probability of no children having the \(rr\) genotype:\[P(X = 0) = \binom{4}{0} (0.5)^0 (0.5)^4 = 1 \times 1 \times 0.0625 = 0.0625\]

Step 4: Calculate Probability for 1 Child Affected

Now, calculate the probability of exactly one child having the \(rr\) genotype:\[P(X = 1) = \binom{4}{1} (0.5)^1 (0.5)^3 = 4 \times 0.5 \times 0.125 = 0.25\]

Step 5: Combine Probabilities

The probability that at most one child will be affected (0 or 1) is the sum of the probabilities calculated in Steps 3 and 4. Thus,\[P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0625 + 0.25 = 0.3125\]

Key Concepts

Punnett square

The Punnett square is a useful tool in genetics. It helps visualize how genes from two parents might combine in their offspring.

To understand it, think of a grid. The top row and leftmost column represent the possible alleles (gene variants) from each parent. In this exercise, one parent is heterozygous, carrying one dominant gene (R) and one recessive gene (r). The other parent is homozygous recessive, having two recessive genes (rr).

By filling out the grid, we can see all possible combinations for the offspring, making it easy to predict the chance of specific traits.

• The offspring could be either carriers (Rr) or affected (rr).
• The chances are equally split with 50% for each genotype.

This visualization is particularly helpful if you are dealing with larger genotypes, as it simplifies the process by making outcomes clear.

recessive genes

Recessive genes play a crucial role in determining an individual's traits. They are represented by lowercase letters, such as "r" in this case.

For a recessive trait to show, both genes in a pair must be recessive (rr). That means the person has to inherit a recessive gene from each parent. If someone has just one recessive gene and a dominant one, they won't show the trait but can pass the recessive gene to their children.

Here's what you need to remember about recessive genes:

• They require two copies to "manifest" in an individual.
• Having one recessive and one dominant gene makes a carrier; they don't show the trait but can pass it on.
• Traits caused by recessive genes often "skip" generations if no carriers meet another carrier or affected individual.

Understanding recessive genes is key to predicting how traits can pass through a family.

genotype probability

Genotype probability involves calculating the likelihood that offspring will have a certain genotype.

In this scenario, two genotypes are possible: Rr (carrier) and rr (affected, unable to roll their tongue).

We calculate the probability using the concept of a binomial distribution because:

• Each child is an independent event like a single trial.
• The probability of each specific genotype (Rr or rr) is half (0.5).
• We repeatedly apply this over multiple trials (children) to find our desired probabilities.

To find out the probability that at most one child inherits the rr genotype: - We first calculate the chance of zero (P(X=0)), which is 6.25%. - Then, we calculate the chance of one child (P(X=1)), resulting in 25%. Adding these probabilities gives the total probability of this outcome (31.25%).

Grasping how to calculate genotype probability not only helps in predict family traits but also in understanding evolutionary patterns.