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A plant believed to be heterozygous for a pair of alleles \(B / b\) (where \(B\) encodes yellow and \(b\) encodes bronze) was selfed, and, in the progeny, there were 280 yellow and 120 bronze plants. Do these results support the hypothesis that the plant is \(B / b\) ?

Short Answer

Expert verified
No, the results do not support the hypothesis; the chi-square test suggests a significant difference.

Step by step solution

01

Understanding the Hypothesis

We hypothesize that the parental plant is heterozygous, meaning its genotype is Bb. According to Mendelian genetics, when a Bb plant is selfed, the expected genotypic ratio in the offspring is 1:2:1 (BB:Bb:bb), which phenotypically results in a 3:1 ratio for the dominant yellow color (including BB and Bb) to the recessive bronze color (bb).
02

Determine Expected Frequencies

Calculate the expected number of plants for each phenotype if the genotypic ratio is 3:1 and the total number of offspring is 400 (280 yellow + 120 bronze). The expected number of yellow plants = \(\frac{3}{4} \times 400 = 300\), and the expected number of bronze plants = \(\frac{1}{4} \times 400 = 100\).
03

Apply the Chi-Square Test

To test if the observed ratios match the expected ratios, use the chi-square formula: \[ \chi^2 = \frac{(O_{yellow} - E_{yellow})^2}{E_{yellow}} + \frac{(O_{bronze} - E_{bronze})^2}{E_{bronze}} \]Plug in the numbers: \[ \chi^2 = \frac{(280 - 300)^2}{300} + \frac{(120 - 100)^2}{100} \]\[ \chi^2 = \frac{400}{300} + \frac{400}{100} \]\[ \chi^2 = 1.33 + 4.00 = 5.33 \]
04

Interpret the Chi-Square Value

The critical value for chi-square at 1 degree of freedom (phenotypes - 1) and a significance level of 0.05 is 3.84. Compare our calculated chi-square value (5.33) to this critical value. Since 5.33 > 3.84, we reject the null hypothesis that the observed and expected ratios do not differ significantly.

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Key Concepts

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

Understanding Heterozygous Genotype
In Mendelian genetics, a genotype refers to the genetic makeup of an organism, specifically the alleles present. A heterozygous genotype consists of two different alleles for a particular gene. In this exercise, we have the alleles "B" for yellow and "b" for bronze. A heterozygous plant, therefore, has a genotype of Bb.

When dealing with heterozygous genotypes, it's essential to understand that the plant carries the trait for both colors but expresses only the dominant one. In our case, yellow is dominant over bronze. This means any plant with at least one "B" allele will appear yellow.

Studying heterozygous genotypes is crucial in predicting offspring traits. By knowing if a plant is heterozygous, we can anticipate the possible outcomes of breeding or selfing the plants. It's a foundational concept in understanding genetic variation and inheritance.
The Significance of Phenotypic Ratio
Phenotypic ratio refers to the relative number of offspring manifesting particular traits. This ratio helps us predict what features will appear in the next generation. In Mendel's pea plants, which laid the work's foundation, plants with different genotypes produced distinctive phenotypes.

In our example, when a heterozygous plant, Bb, is selfed, we expect a phenotypic ratio of 3:1 (yellow:bronze). The calculation is straightforward: The dominant trait (yellow) appears three times for every one appearance of the recessive trait (bronze).

This is calculated as follows:
  • 3 parts yellow: BB + 2Bb (dominant phenotype)
  • 1 part bronze: bb (recessive phenotype)
The expected numbers help us compare with observed data to test hypotheses about genetic makeup. This ratio is central to Mendelian genetics and allows scientists to predict how traits will pass on through generations.
Applying the Chi-Square Test
The chi-square test is a statistical method used to determine if there is a substantial difference between observed and expected data. It enables researchers to verify hypotheses about genetic distributions.

To apply the chi-square test to our plant experiment:
  • Calculate expected numbers based on the 3:1 phenotypic ratio.
  • Identify observed numbers from the data collected.
  • Use the chi-square formula to compare these values.
The formula is: \[ \chi^2 = \frac{(O_{yellow} - E_{yellow})^2}{E_{yellow}} + \frac{(O_{bronze} - E_{bronze})^2}{E_{bronze}} \]After calculations, our chi-square value is 5.33. By comparing this to the critical value of 3.84 at 1 degree of freedom and a 0.05 significance level, we determine whether to accept or reject our hypothesis.

Since 5.33 exceeds the critical value, we reject the null hypothesis, indicating a significant difference between our expected and observed outcomes. This shows the necessity of analyzing experimental data to ensure lucky guesses or errors don't cloud our understanding of genetic patterns.

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Most popular questions from this chapter

In the nematode \(C .\) elegans, some worms have blistered cuticles due to a recessive mutation in one of the bli genes. Someone studying a suppressor mutation that suppressed bli-3 mutations wanted to know if it would also suppress mutations in \(b l i-4 .\) They had a strain that was homozygous for this recessive suppressor mutation, and its phenotype was wild type. a. How would they determine whether this recessive suppressor mutation would suppress mutations in bli- 4 ? In other words, what is the genotype of the worms required to answer the question? b. What cross(es) would they do to make these worms? c. What results would they expect in the \(\mathrm{F}_{2}\) if (1) it did act as a suppressor of bli-4? (2) it did not act as a suppressor of bli- 4 ?

Many kinds of wild animals have the agouti coloring pattern, in which each hair has a yellow band around it. a. Black mice and other black animals do not have the yellow band; each of their hairs is all black. This absence of wild agouti pattern is called nonagouti. When mice of a true-breeding agouti line are crossed with nonagoutis, the \(F_{1}\) is all agouti and the \(F_{2}\) has a 3: 1 ratio of agoutis to nonagoutis. Diagram this cross, letting \(A\) represent the allele responsible for the agouti phenotype and \(a\) nonagouti. Show the phenotypes and genotypes of the parents, their gametes, the \(F_{1}\), their gametes, and the \(F_{2}\) b. Another inherited color deviation in mice substitutes brown for the black color in the wild-type hair. Such brown-agouti mice are called cinnamons. When wildtype mice are crossed with cinnamons, all of the \(\mathrm{F}_{1}\) are wild type and the \(\mathrm{F}_{2}\) has a 3: 1 ratio of wild type to cinnamon. Diagram this cross as in part \(a\), letting \(B\) stand for the wild-type black allele and \(b\) stand for the cinnamon brown allele. c. When mice of a true-breeding cinnamon line are crossed with mice of a true- breeding nonagouti (black) line, all of the \(F_{1}\) are wild type. Use a genetic diagram to explain this result. d. In the \(F_{2}\) of the cross in part \(c,\) a fourth color called chocolate appears in addition to the parental cinnamon and nonagouti and the wild type of the \(\mathrm{F}_{1}\). Chocolate mice have a solid, rich brown color. What is the genetic constitution of the chocolates? e. Assuming that the \(A / a\) and \(B / b\) allelic pairs assort independently of each other, what do you expect to be the relative frequencies of the four color types in the \(\mathrm{F}_{2}\) described in part \(d ?\) Diagram the cross of parts \(c\) and \(d\) showing phenotypes and genotypes (including gametes). f. What phenotypes would be observed in what proportions in the progeny of a backcross of \(\mathrm{F}_{1}\) mice from part \(c\) with the cinnamon parental stock? With the nonagouti (black) parental stock? Diagram these backcrosses. g. Diagram a testcross for the \(\mathrm{F}_{1}\) of part \(c .\) What colors would result and in what proportions? h. Albino (pink-eyed white) mice are homozygous for the recessive member of an allelic pair \(C / c,\) which assorts independently of the \(A / a\) and \(B / b\) pairs. Suppose that you have four different highly inbred (and therefore presumably homozygous) albino lines. You cross each of these lines with a true-breeding wild-type line, and you raise a large \(\mathrm{F}_{2}\) progeny from each cross. What genotypes for the albino lines can you deduce from the following \(\mathrm{F}_{2}\) phenotypes? $$\begin{array}{cccccc} & {4}{c} {\text {Phenotypes of progeny}} \\ { 2 - 5 } \mathrm{F}_{2} \text { of } \text { line } & \begin{array}{c} \text { Wild } \\ \text { type } \end{array} & \text { Black } & \begin{array}{c} \text { Cinna- } \\ \text { mon } \end{array} & \begin{array}{c} \text { Choco- } \\ \text { late } \end{array} & \text { Albino } \\ \hline 1 & 87 & 0 & 32 & 0 & 39 \\ 2 & 62 & 0 & 0 & 0 & 18 \\ 3 & 96 & 30 & 0 & 0 & 41 \\ 4 & 287 & 86 & 92 & 29 & 164 \\ \hline \end{array}$$ (Adapted from A. M. Srb, R. D. Owen, and R. S. Edgar General Genetics, 2nd ed. W. H. Freeman and Company, 1965.)

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