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91Ó°ÊÓ

To estimate the heritability of maze-learning ability in rats, a selection experiment was carried aut. From a population in which the average number of trials riecessary to learn the maze was \(10.8\), with a variance of \(4.0\), animals were selected that managed to learn the maze in an average of \(5.8\) trials. Their offspring required an average of \(8.8\) triab io learn the maze. What is the estimated narrow-sense heritability of maze-learning ability in this population?

Short Answer

Expert verified
The narrow-sense heritability is estimated to be 0.4.

Step by step solution

01

Calculate Selection Differential (S)

The selection differential \( S \) is the difference between the mean trait value of selected parents and the mean trait value of the entire population. Here, \( S = 5.8 - 10.8 = -5.0 \).
02

Calculate Response to Selection (R)

The response to selection \( R \) is the difference between the mean trait value of the offspring of the selected parents and the mean trait value of the original population. Here, \( R = 8.8 - 10.8 = -2.0 \).
03

Calculate Narrow-Sense Heritability (h²)

Narrow-sense heritability \( h^2 \) can be calculated using the formula \( h^2 = \frac{R}{S} \). By substituting the calculated values, we get \( h^2 = \frac{-2.0}{-5.0} = 0.4 \).

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Key Concepts

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

Maze-learning Ability
Maze-learning ability refers to the capability of an organism, such as a rat, to navigate and learn the fastest route through a maze. This behavior is often utilized in scientific experiments to study various aspects of learning, memory, and the impact of genetic and environmental factors on these processes.

When estimating the maze-learning ability of rats, researchers often look at the average number of trials it takes for the rats to successfully learn the maze. The fewer the trials, the better the learning ability is assumed to be. In the experiment described, the initial average number of trials needed by the population was 10.8, indicating this as a standard benchmark for the experiment.

Maze-learning ability is inherently influenced by genetic and environmental factors. The experiment aims to separate these influences and estimate the heritability, or how much of this ability can be attributed to genetics rather than other variables.
Selection Differential
The selection differential, denoted as S, is a measure used in genetics to quantify the difference between the selected breeding group's average trait value and the general population's average trait value. This concept is vital in understanding how selection pressure is applied in genetic experiments.

In the context of the maze-learning experiment, the selection differential was calculated by taking the difference between the average trials needed by the selected group of fast-learning rats (5.8 trials) and the average trials needed by the overall population (10.8 trials). Therefore, the selection differential here is \( S = 5.8 - 10.8 = -5.0 \).

The negative value indicates that the selected group of rats was significantly faster at learning the maze than the average population, showcasing a clear divergence brought about by selective breeding practices focusing on this trait.
Response to Selection
The response to selection, represented as R, is another critical concept in understanding the dynamics of genetic experiments. It measures how much the offspring of selected individuals deviate from the original population in terms of a specific trait.

In the described experiment, the response to selection is calculated by comparing the average number of trials needed by the offspring (8.8 trials) with that needed by the entire original population (10.8 trials). This leads to the calculation \( R = 8.8 - 10.8 = -2.0 \).

A negative response indicates that the offspring showed improvement in maze-learning ability compared to the general population, but not to the same extent as the selected parents. Understanding the response helps researchers determine whether selection efforts are effective in producing intended genetic changes.
Narrow-sense Heritability
Narrow-sense heritability (h²) quantifies the proportion of variance in a trait attributable to additive genetic effects. This measurement is crucial when predicting how a trait will respond to selection in breeding programs.

In this experiment, the narrow-sense heritability is calculated using the formula \( h^2 = \frac{R}{S} \). Given R = -2.0 and S = -5.0, the narrow-sense heritability is \( h^2 = \frac{-2.0}{-5.0} = 0.4 \).

A heritability value of 0.4 indicates that 40% of the variation in maze-learning ability among the rats can be attributed to genetic differences. This suggests that while some of the learning ability is influenced by genetics, a considerable portion is also affected by environmental factors or other non-additive genetic factors. Understanding the heritability helps in making informed decisions in breeding scenarios aiming to enhance certain traits.

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Most popular questions from this chapter

A mouse population has an average weight gain between ages 3 and 6 wecks of \(12 \mathrm{~g}\) and the narrow-sense heritatuility of the weight gain between 3 and 6 weeks is 20 percent. (a) What average weight gain would be expected among the offspring of parents whose average weight gain was \(16 \mathrm{~g}\) ? (b) What averake weight gain would be expected among the oftspring of parents whose average weight gain was \(8 \mathrm{~g}\) ?

The asexual unicellular protozoan Diffugida has a number of well-formed and easily counted teeth encircling the region of the cell that lanctions as the mouth. The cells also have a variable number of spiny projections. The correlation cocfticients between a parent and its offspring in a genetically heterogeneous population are \(0.956\) and \(0.287\) for tooth number and length of longest spine, respectively. (a) Estimate the broad-sence heritabilities of these traits (b) In view of the fact that parents and offspring in asextual organisms are related as identical twins in sexual organisms, how can the large difference in broadsense heritability be explained?

A quantitative trait is aflected by 20 gernetically anlinked, additive loci at which, for each loeus, the presence of 1 or 2 favorable alleles adds 1 or 2 units to the phenotype, respectively, A pepulation is segregating for all 20 loci with allele frequencies equal to I 2 for beth the favorable allele and the unfavorable allele at each locus. (a) What is the mean phenotype in this population? (b) If artificial selection were carried out until the favorable allele at each locus was fixed. what would the phenotype of the setected population be? (c) In the original population, what is the expected Irequency of the genotype that is homerysous for the favorable allele at every locus? (d) How many individuals would have to be present in the original population so that one individtal wothd toe expected that is homozygotis for all favorable alleles? (This calculation shows why artificial selecting can result in a selected population whose mean phenotypic value is much greater than that of any genotype found in the original population.)

In the \(\mathrm{F}_2\) genetation of a croxs of two cultivated varieties of tobacco, the number of leaves geer plant was distributed accoeding to a normal distritution with mean 18 and standard deviation 3. What proportion of the pogulation is expected to have the following phenotypes? (a) between 15 and 21 leaves (b) between 12 and 24 leaves (c) fewer than 15 leaves (d) more than 24 leaves (e) between 21 and 24 leaves

Distinguish between the broad-sense heritability of a quantitative trait and the narrow-sense heritability. If a population is fixed for all genes that allect a particular quantitative trait, what are the values of the narrowsense and broad-sense heritabilities?

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