Chapter 3 - Independent Assortment of Genes

• The principle of independent assortment states that the alleles of two heterozygous gene pairs show independent assortment during gamete formation and fertilization, meaning the offspring will obtain the desired combination of alleles.
• Organelle genes show their own special mode of inheritance called uniparental inheritance: progeny inherit organelle genes exclusively from one parent but not the other.
• Independent assortment of genes is also useful in providing a basic mechanism of inheritance for traits that display continuous phenotypes.
• Polygenes, which are traits influenced by multiple genes, can produce a continuous phenotypic distribution among progeny.
• In an analysis of the progeny from the self of a multiply heterozygous individual, the histogram showing the proportion of each phenotype approximates a bell-shaped curve typical of continuous variation.
• The small subsets of the genome found in mitochondria and chloroplasts are inherited independently of the nuclear genome.
• Mutants in these organelle genes often show maternal inheritance, along with the cytoplasm, which is the location of these organelles.
• The expected phenotype frequencies that would result if the two genes are assorting independently can be demonstrated in a self- and testcross of a F dihybrid.
• The sum rule states that the probability of either one or the other of two mutually exclusive events occurring is the sum of their individual probabilities.
• The product rule is used to determine the probability of observing both outcome A and outcome B.
• The sum rule is used to determine the probability of observing either outcome A or outcome B.
• In genetics, the probability of a multigene genotype or phenotype can be obtained by multiplying the probabilities of the genotype or phenotype for each of the individual genes.
• The probability of at least one success is calculated by subtracting the probability of total failure from the probability of all possible outcomes.
• To satisfy the 95 percent confidence level, the probability of at least one success must be equal to 0.95.
• The rules of probability can be used to predict the number of genotypes or phenotypes in the progeny of complex parental strains.
• In a self of the “tetrahybrid” A / a; B / b; C / c; D / d, there will be three genotypes for each gene pair.
• In a testcross of such a tetrahybrid, there will be two genotypes for each gene pair and a total of 2^4 = 16 genotypes in the progeny.
• The chi-square test, or χ test, is used to assess whether observed results are consistent with a hypothesis.
• The general situation in which the chi-square test is applicable is one in which observed results are compared with those predicted by a hypothesis.
• In a simple genetic example, the chi-square test is used to assess whether the results of a testcross of a heterozygote with a tester of genotype a / a constitute the expected 1 : 1 ratio.
• The chi-square test is a way of quantifying the various deviations expected by chance if a hypothesis is true.
• The probability value in a chi-square test is the probability of observing a deviation from the expected results at least as large (not exactly this deviation) on the basis of chance if the hypothesis is correct.
• The fact that our results have “passed” the chi-square test because p > 0.05, does not mean that the hypothesis is true; it merely means that the results are compatible with that hypothesis.
• If we had obtained a value of p < 0.05, we would have been forced to reject the hypothesis.
• Differences in survival would affect the sizes of the various classes.
• The outcome of the chi-square test depends heavily on sample sizes (numbers in the classes).
• Pure lines are among the essential tools of genetics.
• Only these fully homozygous lines will express recessive alleles, but the main need for pure lines is in the maintenance of stocks for research.
• If they are on the same chromosome, the alleles on one homolog are written adjacently with no punctuation and are separated from those on the other homolog by a slash—for example, AB/ab or Ab/aB.
• If a hypothesis is true, there is a probability of 5 percent of observing a deviation from expectations at least as large as the one actually observed.
• The χ test quantifies the probability of various deviations expected by chance if a hypothesis is true.
• The χ test is used to decide whether or not an observed experimental deviation is reasonably compatible with a working hypothesis.
• In the context of testing a hypothesis, one possibility is to predict the results of a testcross.
• Mendel’s law of equal segregation predicts that a testcross should result in 50 percent A/a and 50 percent a/a.
• If the results of a testcross deviate from the expected ratio, the χ test can be used to determine the probability of the deviation.
• The χ test is performed by calculating χ by using the formula χ2 = ∑ (O − E)2 / E for all classes in which E is the expected number in a class, O is the observed number in a class, and Σ means “sum of.”
• The rows in a table list different values of degrees of freedom (df), which is the number of independent variables in the data.
• In the context of testing a hypothesis, the number of independent variables is the number of phenotypic classes minus 1.
• In the present context, the number of independent variables is 2, so the df is 2–1 = 1.
• The χ value of 0.84 lies somewhere between the columns marked 0.5 and 0.1, indicating a probability of between 50 percent and 10 percent.