Problem 29 - Algebra and Trigonometry

Question

Genetics Huntington's disease is a hereditary ailment caused by a recessive gene. If both parents carry the gene but do not have the disease, there is a 0.25 probability that an offspring will fall victim to the condition. A newly wed couple find through genetic testing that they both carry the gene (but do not have the disease). If they intend to have four children, find the probability of each event. (a) At least one child gets the disease. (b) At least 3 of the children get the disease.

Step-by-Step Solution

Answer

(a) 0.6836; (b) 0.0508.

Step 1: Define the probability variables

We know that the probability of a child inheriting Huntington's disease (denote as 'success') is 0.25, since both parents carry the recessive gene. Hence, the probability of a child not getting the disease (denote as 'failure') is 1 - 0.25, which is 0.75.

Step 2: Understanding the probability of at least one child with the disease

The probability of at least one child having the disease is the complement of the event where no children get the disease. If the probability of no children getting the disease is found, subtracting it from 1 will provide this probability. This can be derived from binomial distribution characteristics where the number of trials n = 4 and the success probability p = 0.25.

Step 3: Calculate the probability of no child having the disease

Using the binomial probability formula, the probability that none of the 4 children have the disease can be expressed as \( P(X = 0) = ({4 \choose 0})(0.25)^0(0.75)^4 = (1)(0.75)^4 \). Calculate this to get \( (0.75)^4 = 0.3164 \).

Step 4: Calculate the probability of at least one child having the disease

The probability of at least one child having the disease is \( 1 - P(X = 0) = 1 - 0.3164 = 0.6836 \).

Step 5: Calculate the probability of at least 3 children having the disease

To find this probability, calculate the probabilities of exactly 3 and exactly 4 children having the disease and sum them. This requires using the binomial formula for those respective x values where n = 4, and p = 0.25.

Step 6: Calculate probability of exactly 3 children having the disease

Use the formula \( P(X = 3) = ({4 \choose 3})(0.25)^3(0.75)^1 \). Compute \( {4 \choose 3} = 4 \), so \( P(X = 3) = 4 \times (0.25)^3 \times (0.75)^1 = 0.0469 \).

Step 7: Calculate probability of exactly 4 children having the disease

Use the formula \( P(X = 4) = ({4 \choose 4})(0.25)^4(0.75)^0 \). Compute \( {4 \choose 4} = 1 \), so \( P(X = 4) = 1 \times (0.25)^4 = 0.0039 \).

Step 8: Sum probabilities for at least 3 children

The total probability for at least 3 children having the disease is the sum of the probabilities calculated in Step 6 and Step 7: \( P(X = 3) + P(X = 4) = 0.0469 + 0.0039 = 0.0508 \).

Key Concepts

Binomial Distribution

In probability theory, a binomial distribution is a common distribution that describes the number of successes in a fixed number of independent trials, each with the same probability of success. A helpful way to think of it is like flipping a coin several times and counting the number of heads, except with variable probabilities. In our context, the 'success' relates to a child inheriting Huntington's disease, which carries its own defined probability.To break it down:

• There are a certain number of trials (in this case, 4 children).
• Each trial has two possible outcomes: the child gets the disease (success), or the child does not (failure).
• The probability of success remains constant across trials (here, 0.25 for getting the disease).

The binomial formula used in calculating these probabilities is:\[P(X = k) = {n \choose k} p^k (1-p)^{n-k} \]where \( n \) is the number of trials, \( k \) is the number of successes, and \( p \) is the probability of success. This formula helps us determine the probability of having any specific number of affected children.

Genetics

Genetics is the study of genes, genetic variation, and heredity in living organisms. It explains how traits are passed from parents to offspring through genetic material. Each individual inherits genes from both parents, and these genes can determine a range of characteristics, from eye color to susceptibility to certain diseases. When both parents carry a recessive gene for a disease, their children may inherit this gene. This likelihood follows Mendelian genetics principles, where a combination of two recessive genes (one from each parent) leads to the manifestation of a trait or condition. In the case being considered:

• Both parents carry a recessive gene for Huntington's disease, which they can potentially pass to their children.
• The probability of a child being affected is calculated based on these genetic principles.

Genetics helps us not only understand why hereditary diseases occur but also aids in predicting the likelihood of such conditions in future generations.

Hereditary Disease

Hereditary diseases are conditions passed from parents to their children through genes. Unlike other diseases, which might be influenced by the environment or lifestyle, hereditary diseases have genetic causes. These diseases follow specific inheritance patterns, often predictable through genetic testing and family history. This predictability allows individuals and healthcare professionals to assess and understand the risk factors. Key points about hereditary diseases include:

• They can be either dominant or recessive; diseases like cystic fibrosis or sickle cell anemia are recessive, requiring both parents to carry the gene.
• Genetic testing can identify carriers of such diseases, providing insight into potential outcomes for offspring.

Understanding hereditary diseases allows families to make informed decisions about health risks and future family planning.

Huntington's Disease

Huntington's disease is a genetic disorder that affects the brain and leads to progressive degeneration of nerve cells. It is unique because it shows an autosomal dominant pattern. This means that only one copy of the defective gene, inherited from either parent, can cause the disease. Even though Huntington's disease is known for being a dominant condition, in certain exercises, like the one referenced, it may be portrayed under hypothetical recessive conditions for illustrative purposes. Here are some critical features of Huntington's disease:

• The disease often begins between ages 30 and 50, advancing over a span of 10 to 25 years.
• Symptoms include motor skill decline, cognitive deterioration, and emotional problems.
• Genetic testing can confirm a diagnosis, although there is no cure yet available.

Due to its genetic nature, Huntington's disease raises significant concerns for those with a family history of the condition. Exploring hypothetical scenarios, such as predicting its occurrence in offspring, is vital in comprehending its inheritance and impact.