/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 Consider a population of individ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 39

Consider a population of individuals each of whom possesses two genes which can be either type \(A\) or type \(a\). Suppose that in outward appearance type \(A\) is dominant and type \(a\) is recessive. That is, an individual will only have the outward characteristics of the recessive gene if its pair is aa.) Suppose that the population has stabilized, and the percentages of individuals having respective gene pairs \(A A, a a\), and \(A a\) are \(p, q\), and \(r .\) Call an individual dominant or recessive depending on the outward characteristics it exhibits. Let \(S_{11}\) denote the probability that an offspring of two dominant parents will be recessive; and let \(S_{10}\) denote the probability that the offspring of one dominant and one recessive parent will be recessive. Compute \(S_{11}\) and \(S_{10}\) to show that \(S_{11}=S_{10}^{2}\). The quantities \(S_{10}\) and \(S_{11}\) are known in the genetics literature as Snyder's ratios.)

Short Answer

Expert verified
The Snyder's ratios \(S_{11}\) and \(S_{10}\) represent the probabilities of a recessive offspring from two dominant parents and between a dominant and a recessive parent, respectively. By calculating these probabilities using the given population percentages (\(p, q\), and \(r\)), we find that \(S_{11} = \frac{r^2}{(p+r)^2}\) and \(S_{10} = \frac{r}{2(p+r)}\). We have shown that \(S_{11} = S_{10}^2\), which confirms the relation between Snyder's ratios in this genetics problem.

Step by step solution

01

Compute the probability of two dominant parents having a recessive offspring #

Since we know the population percentages \(p, q\) and \(r\), let's find the probability that a dominant parent has the "Aa" gene pair (since if they have the "A A" gene pair, they cannot produce a recessive offspring). The percentage of dominant parents having an "Aa" gene pair is given by the proportion in the population i.e., \(r/(p+r)\). Now, to produce a recessive offspring, both parents must pass an "a" gene. The probability of this happening for each parent with an "Aa" gene pair is 1/2. Thus, the probability of two dominant parents with "Aa" gene pairs having a recessive offspring is: \(S_{11} = (\frac{r}{p+r})^2 \times (\frac{1}{2})^2 = \frac{r^2}{(p+r)^2}\)
02

Compute the probability of a dominant parent and a recessive parent having a recessive offspring #

Now let's find \(S_{10}\). For a dominant parent to have a recessive offspring, they must have an "Aa" gene pair and pass on the "a" gene (probability = 1/2). The recessive parent will always pass on an "a" gene since they have an "aa" gene pair. Thus, the probability of a dominant parent with an "Aa" gene pair and a recessive parent having a recessive offspring is: \(S_{10} = \frac{r}{p+r} \times \frac{1}{2} = \frac{r}{2(p+r)}\)
03

Show that \(S_{11} = S_{10}^2\) #

We will now verify if \(S_{11} = S_{10}^2\). Computing the square of \(S_{10}\), we get: \(S_{10}^2 = (\frac{r}{2(p+r)})^2 = \frac{r^2}{(p+r)^2}\) Comparing the results from Step 1 and Step 3, we see that: \(S_{11} = S_{10}^2\) Thus, we have shown that \(S_{11} = S_{10}^2\), which confirms the relation between Snyder's ratios.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Dominant and Recessive Genes
In genetics, genes come in different types known as alleles, and these alleles can be dominant or recessive.
A dominant allele is an allele that expresses its trait even when just one copy is present, while a recessive allele only exhibits its trait when both copies in a gene pair are recessive.
For instance, if the gene for a particular trait has two alleles, 鈥淎鈥� (dominant) and 鈥渁鈥� (recessive), the outward trait, or phenotype, is determined by which alleles are present and how they interact together. - An individual with the genotype 鈥淎A鈥� has both alleles dominant, so the dominant trait shows. - An individual with 鈥淎a鈥� also shows the dominant trait because 鈥淎鈥� masks the recessive 鈥渁鈥�.
- Only individuals with 鈥渁a鈥� show the recessive trait, as there is no dominant allele 鈥淎鈥� to mask 鈥渁鈥�.
Understanding these concepts helps in predicting how traits are passed down from parents to offspring, a foundational concept in Mendelian genetics.
Probability Models
Probability models are mathematical tools used to predict how alleles combine during reproduction.
These models help in calculating the likelihood of offspring inheriting certain gene combinations.
Considering gene combinations as outcomes, probability models allow geneticists to forecast the traits of offspring based on parental genotypes. To compute these probabilities, it is crucial to understand how alleles from different parents can pair up: - If a dominant parent has the allele pair 鈥淎a鈥�, each allele has a 50% chance of being passed on. - Therefore, crossing two dominant 鈥淎a鈥� parents gives us a probability model where there's a chance for both parents to pass on 'a'. - This is exemplified in Snyder's ratios, calculating the probability of a recessive offspring from dominant parents. These probability models simplify complex genetic analyses and provide a clear methodology to approach genetics problems like the one under discussion.
Genetics Ratios
Genetics ratios, like Snyder's ratios, quantify the relationship between parental genotypes and offspring phenotypes.
They provide ratios or probabilities of possible gene pair outputs during genetic crosses.In the exercise, two essential ratios, denoted as \(S_{10}\) and \(S_{11}\), describe specific genetic probabilities:- \(S_{10}\) represents the probability that a child from a dominant 鈥淎a鈥� and recessive 鈥渁a鈥� combination will be recessive.
- \(S_{11}\) shows the chance of recessive offspring from two dominant parents, both likely 鈥淎a鈥�.Comparing these ratios reveals how different parental combinations influence the traits seen in offspring. Understanding these ratios clarifies the genetic outcomes of specific allele pair combinations.
Thus, they are central to predicting genetic variation in a population, assisting in fields like evolutionary biology and genetic counseling.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A group of \(n\) processors are arranged in an ordered list. When a job arrives, the first processor in line attempts it; if it is unsuccessful, then the next in line tries it; if it too is unsuccessful, then the next in line tries it, and so on. When the job is successfully processed or after all processors have been unsuccessful, the job leaves the system. At this point we are allowed to reorder the processors, and a new job appears. Suppose that we use the one- closer reordering rule, which moves the processor that was successful one closer to the front of the line by interchanging its position with the one in front of it. If all processors were unsuccessful (or if the processor in the first position was successful), then the ordering remains the same. Suppose that each time processor \(i\) attempts a job then, independently of anything else, it is successful with probability \(p_{i}\). (a) Define an appropriate Markov chain to analyze this model. (b) Show that this Markov chain is time reversible. (c) Find the long run probabilities.

Prove that if the number of states in a Markov chain is \(M\), and if state \(J\) can be reached from state \(i\), then it can be reached in \(M\) steps or less.

For the Markov chain with states \(1,2,3,4\) whose transition probability matrix \(\mathbf{P}\) is as specified below find \(f_{i 3}\) and \(s_{B}\) for \(i=1,2,3\). $$ \mathbf{P}=\left[\begin{array}{llll} 0.4 & 0.2 & 0.1 & 0.3 \\ 0.1 & 0.5 & 0.2 & 0.2 \\ 0.3 & 0.4 & 0.2 & 0.1 \\ 0 & 0 & 0 & 1 \end{array}\right] $$

Three white and three black balls are distributed in two urns in such a way that each contains three balls. We say that the system is in state \(l_{3}\) \(i=0,1,2,3\), if the first urn contains \(i\) white balls. At each step, we draw one ball from each urn and place the ball drawn from the first urn into the second, and conversely with the ball from the second urn. Let \(X_{n}\) denote the state of the system after the \(n\) th step. Explain why \(\left[X_{n}, n=0,1,2, \ldots\right]\) is a Markov chain and calculate its transition probability matrix.

Let \(Y_{n}\) be the sum of \(n\) independent rolls of a fair die. Find \(\lim _{n \rightarrow \infty} P\left[Y_{n}\right.\) is a multiple of 13\(\\}\) Hint: Define an appropriate Markov chain and apply the results of Excrcise \(14 .\)
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.