KS

9/3 lecture

Meiosis context and alleles

  • A person in treatment/environments that are not friendly to gamete production can affect meiosis; as a result, meiosis may not happen as it should.
  • Alleles and inheritance: as the slide says, most genes have more than two alleles.
  • One well-studied gene example is used to illustrate statistics in genetics; a statistical test is used to test a hypothesis.
  • Results from the test tell us whether experimental results support or reject the hypothesis.

Probability, Mendelian ratios, and statistical testing

  • Mendelian genetics deals with the prediction of genotypic and/or phenotypic ratios of crosses; these genetic ratios are best expressed as probabilities.
  • This video demonstrates how the principles of probability are used to predict and interpret the outcomes of genetic crosses.
  • If you toss a coin, you expect outcomes to be heads or tails; there are only two possible outcomes.
  • The result of the second toss is independent of the result of the first toss (independence).
  • With a very large number of tosses, you should get heads about as often as tails.
  • For a single toss, P(H)=P(T)= frac{1}{2}.
  • When tossing a penny and a dime simultaneously, there are four possible outcomes: HH, HT, TH, TT (penny dime).
  • As the number of tosses increases, the frequency of each outcome approaches the predicted probability of one fourth: P(HH)=P(HT)=P(TH)=P(TT)= frac{1}{4}.
  • The probability of obtaining a head or tail for either coin in a given toss is P( ext{head or tail for a given coin})= frac{1}{2}.
  • The probability of getting any one of the four possible combinations (HH, HT, TH, TT) is frac{1}{4} for each individual outcome.
  • To understand the probability of obtaining one head and one tail across both coins, add the two relevant cases: P( ext{penny head, dime tail})+
    P( ext{penny tail, dime head})= frac{1}{4}+ frac{1}{4}= frac{1}{2}.
  • Binomial considerations: for independent events, use the idea that outcomes can be evaluated with the binomial framework; if a and b are the probabilities of two outcomes, you can evaluate whether a test cross matches the expected ratios.
  • As sampling size grows, the deviation from the expected result decreases (sampling error diminishes with more data).

Chi-square test: purpose, key concepts, and interpretation

  • The chi-square test evaluates how much random chance (sampling variation) explains observed data when you have categorical data (e.g., phenotypic classes).
  • It helps determine whether the observed deviations from expectations are likely due to chance or due to some other factor.
  • Formula: \chi^2 = \sumi \frac{(Oi - Ei)^2}{Ei} where Oi is observed frequency and Ei is expected frequency for class i.
  • Degrees of freedom (df): typically df = n - 1, where n is the number of classes (or groups) being compared; for two classes (parallels vs recombinants) df = 1.
  • p-value interpretation:
    • If p > 0.05, the deviation is not statistically significant and the null hypothesis (e.g., independent assortment) is not rejected.
    • If p \le 0.05, the deviation is statistically significant and the null hypothesis may be rejected.
  • Chi-square tables are used to determine the probability value corresponding to the calculated chi-square and the degrees of freedom.

Example 1: Test cross for two genes (eye color and wing shape) in Drosophila

  • Cross setup:
    • Parent 1 (true breeding): purple eyes and dumpy wings. Genotype: PRPR\; DPDP
    • Parent 2 (true breeding): red eyes and normal wings. Genotype: PR^+ PR^+\; DP^+ DP^+
  • F1 flies: because the mutant traits are recessive, F1 are red eyes and normal wings. Genotype: pr^+\; pr^+\; dp^+\; dp^+
  • Test cross: F1 × purple dumpy flies yields 360 progeny with four phenotypic classes:
    • 89 red normal
    • 97 purple dumpy
    • 94 red dumpy
    • 80 purple normal
  • Under independent assortment, the expected ratio is 1:1:1:1, so for 360 progeny, each class is expected to be \frac{360}{4} = 90. Thus, E_i = 90 for each i.
  • Chi-square calculation as described:
    • The first column lists phenotypes, grouped into parental vs recombinant phenotypes.
    • Observed numbers: 89, 97, 94, 80.
    • Expected numbers: 90, 90, 90, 90.
    • Differences: d = O − E → -1, 7, 4, -10.
    • Sum of (d^2 / E): for this example, the transcript states the chi-square value is \chi^2 = 0.4 (calculated as 0.2 + 0.2 in the transcript).
  • Degrees of freedom: as discussed, there are two classes (parental vs recombinant), so df = 1.
  • p-value interpretation: For \chi^2 = 0.4 with df = 1, the p-value falls between 0.50 and 0.70. This means that, in repeated experiments, 50–70% of the time you would observe a chi-square value of this magnitude or smaller due to chance.
  • Conclusion from the transcript: Because the p-value is greater than 0.05, there is no statistically compelling argument to reject the hypothesis of independent assortment for this cross.

Example 2: Test cross for two recessive traits (eye color and wing shape) in Drosophila

  • Cross setup:
    • Parent 1: purple eyes and curved wings. Genotype: pr\; pr\; cr\; cr
    • Parent 2: red eyes and normal length/shape. Genotype: pr^+\; pr^+\; cr^+\; cr^+
  • F1 flies: recessive traits yield red eyes and normal wings in F1. Genotype: pr^+\; pr^+\; cr^+\; cr^+
  • Test cross: F1 female × purple curved male yields 448 progeny with four phenotypic classes:
    • 181 red normal
    • 169 purple curved
    • 52 red curved
    • 46 purple normal
  • Under the hypothesis of independent assortment of the two genes, the expected ratio is 1:1:1:1, so for 448 progeny, each class is expected to be \frac{448}{4} = 112. Thus, E_i = 112 for each i.
  • Chi-square setup (as described in the transcript):
    • Observed vs. expected for each class: red normal (181), purple curved (169), red curved (52), purple normal (46).
    • The transcript shows how the chi-square table is filled, starting with parental vs recombinant phenotypes, and computing (O − E)^2 / E for each class, then summing to obtain the chi-square value. The exact chi-square value for this cross is not provided in the transcript continuation.
  • Degrees of freedom: the transcript states there are two classes (parentals vs recombinants), so df = 1.
  • p-value interpretation (as given for the previous example): the decision criterion is the same: if the resulting p-value is greater than 0.05, the deviation is not statistically significant and the hypothesis of independent assortment is not rejected.
  • Note: The transcript cuts off here, so the final chi-square value and p-value for this second example are not provided in the