Two different crosses were set up between carrots (Daucus carota \()\) of different colors and carotenoid content (Santos, Carlos A. F. and Simon, Philipp W. 2002. Horticultura Brasileira 20). Analyses of the \(\mathrm{F}_{2}\) generations showed that four loci are associated with the \(\alpha\) carotene content of carrots, with a broad-sense heritability of \(90 \% .\) How many distinct phenotypic categories and genotypes would be seen in each \(\mathrm{F}_{2}\) generation, and what does a broad-sense heritability of \(90 \%\) mean for carrot horticulture?

Short Answer

Expert verified
Answer: There are 5 distinct phenotypic categories in this carrot population. A broad-sense heritability of 90% means that 90% of the observed variation in alpha carotene content is due to genetic factors, indicating that selective breeding would be effective in increasing the carotene content of the carrots.

Step by step solution

01

Calculate the distinct genotypic categories

To calculate the distinct genotypic categories, we have to consider that each locus can be homozygous dominant (AA), heterozygous (Aa), or homozygous recessive (aa). Since there are 4 loci associated with the trait, the total number of distinct genotypic categories can be calculated using the formula \(3^n\), where n is the number of loci. Hence, the total number of genotypic categories is \(3^4\).
02

Calculate the phenotypic categories

Here we have 4 loci associated with the carotene content, and each locus may have dominant (A) and recessive (a) alleles. The phenotypes depend on these alleles as dominant alleles (A) would lead to higher carotene content, while recessive alleles (a) would lead to lower carotene content. Since we have 4 loci, there could be a phenotypic range from 0 (all recessive alleles) to 4 (all dominant alleles) in steps of 1. Therefore, there are a total of 5 distinct phenotypic categories.
03

Calculate the number of genotypes in each F2 generation

Given that we've found a total of 81 genotypic categories (\((3^4)=81)\), then each F2 generation will have 81 distinct genotypes.
04

Explain the meaning of the broad-sense heritability of 90%

Broad-sense heritability (H2) is a measure of the proportion of phenotypic variance that can be attributed to genetic factors. In this case, the broad-sense heritability of 90% means that 90% of the observed variation in alpha carotene content in the carrot population can be explained by the genetic differences among the carrots. The remaining 10% of the variation is due to environmental factors or interactions between genes and the environment. For carrot horticulture, this high heritability implies that selective breeding will be effective in increasing the carotene content of the carrots, as the majority of the variations are determined by the carrots' genetic constitution.

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Most popular questions from this chapter

Define the following: (a) polygenic, (b) additive alleles, (c) correlation, (d) monozygotic and dizygotic twins, (e) heritability, (f) \(\mathrm{QTL},\) and \((\mathrm{g})\) continuous variation.

Corn plants from a test plot are measured, and the distribution of heights at \(10-\mathrm{cm}\) intervals is recorded in the following table: $$\begin{array}{cc}\text { Height }(\mathrm{cm}) & \text { Plants (no.) } \\\100 & 20 \\\110 & 60 \\\120 & 90 \\\130 & 130 \\\140 & 180 \\\150 & 120 \\\160 & 70 \\\170 & 50 \\\180 & 40\end{array}$$ Calculate (a) the mean height, (b) the variance, (c) the standard deviation, and (d) the standard error of the mean. Plot a rough graph of plant height against frequency. Do the values represent a normal distribution? Based on your calculations, how would you assess the variation within this population?

Erma and Harvey were a compatible barnyard pair, but a curious sight. Harvey's tail was only \(6 \mathrm{cm}\) long, while Erma's was \(30 \mathrm{cm} .\) Their \(\mathrm{F}_{1}\) piglet offspring all grew tails that were \(18 \mathrm{cm}\) When inbred, an \(\mathrm{F}_{2}\) generation resulted in many piglets (Erma and Harvey's grandpigs), whose tails ranged in \(4-\mathrm{cm}\) intervals from 6 to \(30 \mathrm{cm}(6,10,14,18,22,26, \text { and } 30) .\) Most had \(18-\mathrm{cm}\) tails, while \(1 / 64\) had \(6-\mathrm{cm}\) tails and \(1 / 64\) had \(30-\mathrm{cm}\) tails. (a) Explain how these tail lengths were inherited by describing the mode of inheritance, indicating how many gene pairs were at work, and designating the genotypes of Harvey, Erma, and their 18 -cm-tail offspring. (b) If one of the \(18-\mathrm{cm} \mathrm{F}_{1}\) pigs is mated with one of the \(6-\mathrm{cm}\) \(\mathrm{F}_{2}\) pigs, what phenotypic ratio will be predicted if many offspring resulted? Diagram the cross.

Osteochondrosis (OC) is a developmental orthopedic disorder in young, growing horses, where irregular bone formation in the joints leads to necrotic areas, resulting in chronic or recurrent lameness. Incidence of OC varies considerably among breeds, and displays a multifactorial mode of inheritance. The incidence of \(\mathrm{OC}\) is rising in the population of race horses. Discuss the reasons why the incidence of OC might be rising, and describe what can be done to detect OC susceptibility in horses with the help of QTL analysis.

A dark-red strain and a white strain of wheat are crossed and produce an intermediate, medium-red \(\mathrm{F}_{1}\). When the \(\mathrm{F}_{1}\) plants are interbred, an \(\mathrm{F}_{2}\) generation is produced in a ratio of 1 darkred: 4 medium-dark-red: 6 medium-red: 4 light-red: 1 white. Further crosses reveal that the dark-red and white \(\mathrm{F}_{2}\) plants are true breeding. (a) Based on the ratios in the \(\mathrm{F}_{2}\) population, how many genes are involved in the production of color? (b) How many additive alleles are needed to produce each possible phenotype? (c) Assign symbols to these alleles and list possible genotypes that give rise to the medium-red and light-red phenotypes. (d) Predict the outcome of the \(F_{1}\) and \(F_{2}\) generations in a cross between a true-breeding medium-red plant and a white plant.

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