In a population of tomato plants, mean fruit weight is \(60 \mathrm{g}\) and \(h^{2}\) is \(0.3 .\) Predict the mean weight of the progeny if tomato plants whose fruit averaged \(80 \mathrm{g}\) were selected from the original population and interbred.

Short Answer

Expert verified
Answer: The predicted mean weight of the progeny is 66g.

Step by step solution

01

Find the selection differential (S)

To calculate the selection differential, subtract the mean fruit weight of the original population from the mean fruit weight of the selected parents. $$S = 80 - 60 = 20 \mathrm{g}$$
02

Calculate the selection response (R)

Now, use the selection differential (S) and heritability equation to calculate the selection response (R). $$R = h^{2} \cdot S = 0.3 \cdot 20 = 6 \mathrm{g}$$
03

Predict the mean weight of the progeny

To predict the mean weight of the progeny, add the selection response (R) to the mean fruit weight of the original population. $$60 + 6 = 66 \mathrm{g}$$ The predicted mean weight of the progeny is \(66 \mathrm{g}\).

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 Vaia!

Key Concepts

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

Selection Differential
Understanding the selection differential is essential when discussing evolutionary change or selective breeding in a population. It quantifies the strength of selection by measuring the difference between the average trait of the entire population and the average trait of the selected parents, or the breeding individuals.

For instance, if we consider the mean weight of fruits in a population of tomato plants, selection differential is calculated by subtracting the population mean from the mean weight of the selected plants. In our exercise, the selection differential (\( S \)) was found by simply taking the mean weight of the selected tomato plants (\( 80 \text{g} \) ) and subtracting the mean fruit weight of the original population (\( 60 \text{g} \) ), which equals to a 20g difference.

This differential is not just a numerical value; it reflects the potential for selection to cause a shift in the population's traits. If all other factors are equal, a higher selection differential suggests a greater potential for change.
Heritability of Traits
The concept of heritability is foundational in genetics and breeding. It represents the proportion of variation in a trait that is attributable to genetic differences among individuals within a population. This value is often denoted as (\( h^{2} \) ) and is usually expressed as a value between 0 and 1.

Heritability plays a pivotal role in predicting whether and to what extent a trait will respond to selection. A high heritability implies that a large part of the variation seen in a trait, like fruit weight in tomato plants, is due to genetic differences. In our case, the heritability of fruit weight is (\( 0.3 \) ), meaning 30% of the variation is due to genetic factors.

It's crucial to grasp that heritability does not indicate the degree to which a trait is genetically determined; rather, it signifies how much of the variation in the trait can be passed on to the next generation. Traits with higher heritability are more easily modified by selective breeding since a larger proportion of the trait difference is due to genetic factors.
Progeny Mean Weight Prediction
By integrating our understanding of selection differential and heritability, we can foresee changes in traits across generations. This process is commonly applied in agricultural breeding programs, such as the one depicted in our tomato plant example. The principle lies in combining the selection differential with the heritability to predict the response to selection, and thus, the mean trait value of the progeny.

When we predict progeny mean weight, we are essentially estimating the expected average trait of the offspring based on the selection we have applied. To determine this, we calculated the selection response (\( R \) ) by multiplying the heritability (\( h^{2} = 0.3 \) ) by the selection differential (\( S = 20 \text{g} \) ). The result gave us a 6g increase in mean fruit weight for the progeny. Finally, by adding the original population mean (\( 60 \text{g} \) ) to the selection response (\( R = 6 \text{g} \) ), we predict that the progeny will have an average fruit weight of 66g.

This prediction is an essential tool for breeders, as it helps set realistic expectations for the outcomes of their selective breeding programs and informs decisions about which individuals to select for breeding the next generation of plants or animals.

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

Height in humans depends on the additive action of genes. Assume that this trait is controlled by the four loci \(R, S, T,\) and \(U\) and that environmental effects are negligible. Instead of additive versus nonadditive alleles, assume that additive and partially additive alleles exist. Additive alleles contribute two units, and partially additive alleles contribute one unit to height. (a) Can two individuals of moderate height produce offspring that are much taller or shorter than either parent? If so, how? (b) If an individual with the minimum height specified by these genes marries an individual of intermediate or moderate height, will any of their children be taller than the tall parent? Why or why not?

An inbred strain of plants has a mean height of \(24 \mathrm{cm}\). A second strain of the same species from a different peopraphic region also has a mean height of \(24 \mathrm{cm}\). When plants from the two strains are crossed together, the \(\mathrm{F}_{1}\) plants are the same height as the parent plants, However, the \(\mathrm{F}_{2}\) generation shows a wide range of heights;the majority are like the \(P_{1}\) and \(Y_{1}\) plants, but approximately 4 of 1000 are only \(12 \mathrm{cm}\) high and about 4 of 1000 are \(36 \mathrm{cm}\) high. (a) What mode of inheritance is occurring here? (b) How many gene pairs are involved? (c) How much does each gene contribute to plant height? (d) Indicate one possible set of genotypes for the original \(P_{1}\) parents and the \(P_{1}\) plants that could account for these results. (e) Indicate three possible genotypes that could account for \(\mathrm{F}_{2}\) plants that are \(18 \mathrm{cm}\) high and three that account for \(\mathrm{F}_{2}\) plants that are \(33 \mathrm{cm}\) high.

In a cross between a strain of large guinea pigs and a strain of small guinea pigs, the \(\mathrm{F}_{1}\) are phenotypically uniform, with an average sire about intermediate between that of the two parental strains. Among \(1014 \mathrm{F}_{2}\) individuals, 3 are about the same size as the small parental strain and 5 are about the same size as the large parental strain. How many gene pairs are involved in the inheritance of size in these strains of guinea pigs?

In a series of crosses between two true-breeding strains of peaches, the \(F_{1}\) generation was uniform, producing 30 -g peaches. The \(F_{2}\) fruit mass ranges from 38 to 22 g at intervals of 2 g. (a) Using these data, determine the number of polygenic loci involved in the inheritance of peach mass. (b) Using gene symbols of your choice, give the genotypes of the parents and the \(F_{1}\)

These all center around quantitative inheritance and the study and analysis of polygenic traits. Write a short essay that dis- cusses the difference between the more traditional Mendelian and neo-Mendelian modes of inheritance (qualitative inheritance) and quantitative inheritance.

See all solutions

Recommended explanations on Biology Textbooks

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.

Sign-up for free