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Chapter 6: Problem 57

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
The results do not support the hypothesis that the plant is \( B/b \) because the Chi-Square test shows a significant deviation.

Step by step solution

01

Define the hypothesis

We begin by understanding the genetic cross. The hypothesis is that the plant is heterozygous, i.e., its genotype is \( Bb \). When such a plant self-pollinates, the expected progeny ratio is 3:1 for a dominant to recessive phenotype, according to Mendelian genetics. Here, yellow (\( B \)) is the dominant trait and bronze (\( b \)) is the recessive trait. We need to check if the observed ratio of yellow to bronze (280:120) aligns with this expected 3:1 ratio.
02

State the expected ratio and calculate expected numbers

If the plant is \( Bb \), then upon self-pollination, we expect 75% of the progeny to be yellow (because they can be \( BB \), \( Bb \), or \( Bb \)) and 25% to be bronze (\( bb \)). If there are 400 progeny in total (280 yellow + 120 bronze), the expected number of yellow plants is \( 0.75 \times 400 = 300 \), and the expected number of bronze plants is \( 0.25 \times 400 = 100 \).
03

Use the Chi-Square test

To determine if the observed numbers (280 yellow, 120 bronze) significantly deviate from the expected numbers, perform a Chi-Square test. The formula for the Chi-Square statistic \( \chi^2 \) is: \[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \] where \( O_i \) is the observed number and \( E_i \) is the expected number. Calculate for each phenotype and add them up: \[ \chi^2 = \frac{(280 - 300)^2}{300} + \frac{(120 - 100)^2}{100} \]
04

Calculate Chi-Square value

Substitute the observed and expected numbers into the \( \chi^2 \) formula: \[ \chi^2 = \frac{(280 - 300)^2}{300} + \frac{(120 - 100)^2}{100} = \frac{400}{300} + \frac{400}{100} = 1.33 + 4 = 5.33 \]
05

Compare with critical value

The degrees of freedom (df) for this test is \( n - 1 = 2 - 1 = 1 \). Look up the critical value from the Chi-Square distribution table for df=1, usually at the 0.05 significance level, which is 3.841. Since the calculated \( \chi^2 = 5.33 \) is greater than 3.841, we reject the null hypothesis.
06

Conclusion

Our hypothesis (that the plant is \( Bb \)) is not supported because the observed data (\( \chi^2 = 5.33 \)) significantly deviates from the expected 3:1 ratio for a heterozygous cross at the 0.05 level of significance.

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Key Concepts

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

Chi-Square Test
The Chi-Square test is a statistical method used to compare observed data with expected data to determine if differences are due to chance or some other factor. It's particularly useful in genetics when assessing results from Mendelian crosses.

Here's a step-by-step explanation on how the Chi-Square test is used in genetics, such as in the case of testing for a heterozygous plant attribute :
  • **Formulate Hypotheses**: We propose a null hypothesis, which in our context supposes the plant is heterozygous for the particular gene combination tested.
  • **Determine Expected Ratios**: Based on Mendel's laws, if a plant is heterozygous, the expected phenotypic ratio following self-pollination should follow a 3:1 pattern (dominant : recessive).
  • **Apply the Chi-Square Formula**: The formula used is: \[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \] where \( O_i \) stands for observed values and \( E_i \) for expected values.
  • **Interpret Results**: Compare the calculated \( \chi^2 \) value to a chi-square distribution table using the appropriate degrees of freedom to accept or reject the hypothesis.
This test helps us decide if the observed outcomes (such as the number of yellow and bronze plants) occur as predicted or if other factors might be influencing the results.
Dominant and Recessive Traits
In Mendelian genetics, understanding dominant and recessive traits is pivotal. These concepts help explain how offspring inherit specific traits from their parents.
  • **Dominant Traits**: These are traits expressed when an individual has at least one dominant allele ( like yellow in our plant example). One dominant allele in the genotype is enough for the dominant trait to be seen in the phenotype.
  • **Recessive Traits**: For a recessive trait (such as bronze color) to manifest, an organism must possess two recessive alleles. If a dominant allele is present, the recessive trait will not appear in the phenotype.
Yellow and bronze in our plant example demonstrate these principles effectively. Recognizing how these traits get passed on offers insight into genetic probabilities and can be used to predict future generations' phenotypes.
Genotypic Ratio
The genotypic ratio depicts the proportion of different genotypes resulting from a genetic cross. It's separate from the phenotypic ratio, which describes the appearance of traits.
  • **Expected Genotypic Ratio**: In a heterozygous self-cross (i.e., \( Bb \times Bb \)), each parent can pass on either of its alleles. This leads to a genotypic ratio of 1:2:1 for \( BB \), \( Bb \), and \( bb \).
  • **Calculation Example**: After crossing, the possible genotype outcomes are:
    • 1 \( BB \)
    • 2 \( Bb \)
    • 1 \( bb \)
  • **Observation & Prediction**: Although the genotypic ratio differs from the phenotypic ratio (often 3:1), both are essential for predicting inheritance patterns and probabilities accurately.
Understanding these genotypic ratios provides a clearer picture of how traits are inherited, beyond what you visually observe through phenotypes.

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Most popular questions from this chapter

The petals of the plant Collinsia parviflora are normally blue, giving the species its common name, blueeyed Mary. Two pure-breeding lines were obtained from color variants found in nature; the first line had pink petals, and the second line had white petals. The following crosses were made between pure lines, with the results shown: a. Explain these results genetically. Define the allele symbols that you use, and show the genetic constitution of the parents, the \(\mathrm{F}_{1}\), and the \(\mathrm{F}_{2}\) in each cross. b. \(A\) cross between a certain blue \(F_{2}\) plant and a certain white \(\mathrm{F}_{2}\) plant gave progeny of which \(\frac{3}{8}\) were blue, \(\frac{1}{8}\) were pink, and \(\frac{1}{2}\) were white. What must the genotypes of these two \(\mathrm{F}_{2}\) plants have been?

A yeast geneticist irradiates haploid cells of a strain that is an adenine- requiring auxotrophic mutant, caused by mutation of the gene ade1. Millions of the irradiated cells are plated on minimal medium, and a small number of cells divide and produce prototrophic colonies. These colonies are crossed individually with a wild-type strain. Two types of results are obtained: (1) prototroph \(\times\) wild type:progeny all prototrophic (2) prototroph \(\times\) wild type : progeny \(75 \%\) prototrophic, \(25 \%\) adenine-requiring auxotrophs a. Explain the difference between these two types of results. b. Write the genotypes of the prototrophs in each case. c. What progeny phenotypes and ratios do you predict from crossing a prototroph of type 2 by the original ade1 auxotroph?

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 \(\mathrm{F}_{1}\), their gametes, and the \(\mathrm{F}_{2}\). 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 wild-type 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 \(\mathrm{F}_{1}\) 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?

In sweet peas, the synthesis of purple anthocyanin pigment in the petals is controlled by two genes, \(B\) and \(D\) The pathway is a. What color petals would you expect in a purebreeding plant unable to catalyze the first reaction? b. What color petals would you expect in a purebreeding plant unable to catalyze the second reaction? c. If the plants in parts \(a\) and \(b\) are crossed, what color petals will the \(F_{1}\) plants have? d. What ratio of purple:blue:white plants would you expect in the \(\mathrm{F}_{2}\) ?

You have been given a virgin Drosophila female. You notice that the bristles on her thorax are much shorter than normal. You mate her with a normal male (with long bristles) and obtain the following \(\mathrm{F}_{1}\) progeny: \(\frac{1}{3}\) short-bristled females, \(\frac{1}{3}\) long-bristled females, and \(\frac{1}{3}\) long-bristled males. A cross of the \(\mathrm{F}_{1}\) long- bristled females with their brothers gives only long-bristled \(\mathrm{F}_{2}\). A cross of short-bristled females with their brothers gives \(\frac{1}{3}\) short-bristled females, \(\frac{1}{3}\) long-bristled females, and \(\frac{1}{3}\) long-bristled males. Provide a genetic hypothesis to account for all these results, showing genotypes in every cross.
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