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Chapter 5: Problem 3

Genetics Suppose a married man and woman both carry a gene for cystic fibrosis but don't have the disease themselves. According to the laws of genetics, the probability that their first child will develop cystic fibrosis is 0.25 (a) Explain what this probability means. (b) If the couple has 4 children, is one of them guaranteed to get cystic fibrosis? Explain.

Short Answer

Expert verified
(a) Each child has a 25% chance to have cystic fibrosis. (b) No, one child is not guaranteed to have cystic fibrosis; each child has a separate 25% chance.

Step by step solution

01

Understanding the Probability

The probability of 0.25 indicates that each child independently has a 25% chance of inheriting two copies of the cystic fibrosis (CF) gene (one from each parent), which would result in the child having cystic fibrosis.
02

Situation Analysis for Multiple Children

If the couple has 4 children, the probability of any one child having cystic fibrosis is still 0.25. However, each child's genetic makeup is independently determined. This means that having multiple children does not alter the individual probability for each child.
03

Conclusion About Guaranteed Outcome

There is no guarantee that any of the children will develop cystic fibrosis because the probability of 0.25 does not accumulate or change with the birth of each child. It is still possible that none, one, or more children could have the disease, but no outcome is certain.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cystic Fibrosis
Cystic Fibrosis is a genetic disorder that affects the lungs and digestive system. It is caused by mutations in the CFTR gene, which leads to the production of thick and sticky mucus. This can result in respiratory and digestive problems.
For a child to develop cystic fibrosis, they must inherit two copies of the faulty gene, one from each parent. If both parents carry one copy of the mutated gene, there is a 25% chance their child will inherit both mutant genes and have cystic fibrosis.
It's important to note that being a gene carrier does not mean an individual will have cystic fibrosis, but they can pass the faulty gene to their offspring.
Independent Events
In the context of genetics, each child's genetic outcome is considered an independent event. This means the inheritance of genes by one child does not affect the probability of inheritance by another child.
Let's break it down:
  • Independent events are events that do not influence each other.
  • For each child born to the same parents, the probability remains consistent, regardless of the outcomes of previous children.
This principle explains why even if a couple has multiple children, the genetic probability remains constant for each child.
Probability in Genetics
Probability in genetics is used to predict the likelihood of inheriting particular traits or genetic disorders. It uses basic probability principles to estimate these chances.
For instance, when both parents carry one recessive gene for a trait like cystic fibrosis, each child has a 25% probability of inheriting the disorder.
This probability is calculated using Mendelian genetics, which provides a framework for understanding how traits are passed from generation to generation. The probability reflects the statistical nature of genetic inheritance.
Punnett Square Analysis
A Punnett Square is a helpful tool for visualizing genetic crosses. It allows us to predict the genotypes of offspring from parental genotypes.
  • It consists of a grid that helps calculate the probability of inheriting certain traits.
  • In the case of cystic fibrosis, each parent contributes one allele for the gene.
  • The square shows all possible combinations, illustrating the 25% chance for a child to inherit two faulty alleles, 50% to inherit one faulty allele (carrier), and 25% to inherit no faulty alleles.
By using a Punnett Square, we can better understand and predict genetic outcomes in a family setting.

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Most popular questions from this chapter

In an effort to find the source of an outbreak of food poisoning at a conference, a team of medical detectives carried out a study. They examined all 50 people who had food poisoning and a random sample of 200 people attending the conference who didn't get food poisoning. The detectives found that \(40 \%\) of the people with food poisoning went to a cocktail party on the second night of the conference, while only \(10 \%\) of the people in the random sample attended the same party. Which of the following statements is appropriate for describing the \(40 \%\) of people who went to the party? (Let \(F=\) got food poisoning and \(A=\) attended party. \()\) (a) \(P(F \mid A)=0.40\) (b) \(P\left(A \mid F^{C}\right)=0.40\) (c) \(P\left(F \mid A^{C}\right)=0.40\) (d) \(P\left(A^{C} \mid F\right)=0.40\) (e) \(P(A \mid F)=0.40\)

If I toss a fair coin five times and the outcomes are TITTT, then the probability that tails appears on the next toss is (a) 0.5 (b) less than 0.5 (c) greater than 0.5 . (d) \(0 .\) (e) \(1 .\)

Blood types All human blood can be typed as one of \(\mathrm{O}, \mathrm{A}, \mathrm{B},\) or \(\mathrm{AB},\) but the distribution of the types varies a bit with race. Here is the distribution of the blood type of a randomly chosen black American: $$\begin{array}{lcccc}\hline \text { Blood type: } & 0 & \mathrm{~A} & \mathrm{~B} & \mathrm{AB} \\\\\text { Probability: } & 0.49 & 0.27 & 0.20 & ? \\\\\hline\end{array}$$ (a) What is the probability of type \(A B\) blood? Why? (b) What is the probability that the person chosen does not have type \(A B\) blood? (c) Maria has type \(\mathrm{B}\) blood. She can safely receive blood transfusions from people with blood types \(\mathrm{O}\) and \(\mathrm{B}\). What is the probability that a randomly chosen black American can donate blood to Maria?

The most common bet in craps is the "pass line." \(\mathrm{A}\) pass line bettor wins immediately if either a 7 or an 11 comes up on the first roll. This is called a natural. What is the probability of a natural? (a) \(2 / 36\) (c) \(8 / 36\) (e) \(20 / 36\) (b) \(6 / 36\) (d) \(12 / 36\)

Monty Hall problem In Parade magazine, a reader posed the following question to Marilyn vos Savant and the "Ask Marilyn" column: Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say \(\\# 1,\) and the host, who knows what's behind the doors, opens another door, say # \(3,\) which has a goat. He says to you, "Do you want to pick door # \(2 ?\) Is it to your advantage to switch your choice of doors? \(^{4}\) The game show in question was Let's Make a Deal and the host was Monty Hall. Here's the first part of Marilyn's response: "Yes; you should switch. The first door has a \(1 / 3\) chance of winning, but the second door has a \(2 / 3\) chance." Thousands of readers wrote to Marilyn to disagree with her answer. But she held her ground. (a) Use an online Let's Make a Deal applet to perform at least 50 repetitions of the simulation. Record whether you stay or switch (try to do each about half the time) and the outcome of each repetition. (b) Do you agree with Marilyn or her readers? Explain.
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