Genetics: Calculating Progeny Genotypes Using Probability Rules

To calculate probabilities in pedigree analysis: assign genotypes based on inheritance pattern; for affected individuals in recessive disorders, assign 100% homozygous recessive probability; for unaffected individuals, calculate carrier probability based on parents' genotypes. For carrier (Aa) × affected (aa) cross: 50% carrier, 50% affected. For carrier (Aa) × normal (AA) cross: 50% carrier, 50% normal. For two carriers (Aa × Aa): 25% affected, 50% carrier, 25% normal. Multiply probabilities of independent events to find final probability. For example, if both parents have 2/3 probability of being carriers and each has 1/2 probability of passing the allele, the final probability is (2/3) × (2/3) × (1/2) × (1/2) = 1/9.

Probability calculations determine disease transmission likelihood across generations. For two heterozygous parents (Dd × Dd), the probability of affected offspring is 1/4. When healthy offspring are produced, the probability they are carriers is 2/3 (since two of three healthy genotypes are heterozygous). Individuals entering pedigrees from outside are assumed homozygous dominant (DD) unless evidence proves otherwise. These probabilities are essential for predicting disease transmission in future generations.

In pedigree analysis, carrier probability is calculated by considering the probability that an individual is a carrier (1/2) and the probability that their offspring would show the expected pattern. For females, if they have affected offspring, the carrier probability is calculated by multiplying 1/2 by the probability of affected offspring. For example, if a female carrier has two affected sons, the probability is 1/2 × 1/2 × 1/2 = 1/8. If she has two normal sons, the probability is also 1/8. This method helps determine whether an individual is likely a carrier based on their offspring's phenotypes.

In pedigree analysis for recessive disorders: (1) An affected individual must be homozygous for the disease allele, (2) Their parents must be heterozygous carriers, (3) Unaffected siblings have a 2/3 probability of being carriers (since they cannot be homozygous affected). This information is used to calculate carrier probabilities for at-risk individuals in consanguineous families.

Pedigree charts use squares for males, circles for females, and yellow coloring for affected individuals. Roman numerals indicate generations, Arabic numerals identify individuals. When two healthy parents have an affected child, both must be carriers (heterozygous). The probability of being a carrier depends on family history. For example, if parents 2-1 and 6-2 had an affected child, they are both heterozygous. The child of two heterozygous carriers has a 1/2 probability of being a carrier themselves.

Chi-Square test can test proportions or ratios. For testing if sex ratio of newborn children is 1:1 with 1000 children (600 males, 400 females): (1) Null hypothesis states ratio is 1:1, (2) Expected values = (1/2)×1000 = 500 for each category, (3) Calculate Chi-Square = 40, (4) Degrees of freedom = 2 - 1 = 1, (5) Critical value for df=1 and α=0.05 is 3.84, (6) Since 40 > 3.84, reject null hypothesis (ratio is not 1:1).

The chi-square goodness-of-fit test is a statistical tool used in genetics to determine whether observed experimental data matches expected ratios predicted by genetic crosses; the test involves calculating χ² = Σ[(O-E)²/E], where O is observed values and E is expected values, then comparing the result to a critical value from a chi-square table using degrees of freedom (number of categories minus one) and a significance level (typically p=0.05), where if the calculated χ² is less than the critical value, the null hypothesis (that observed variation is due to chance) is not rejected, but if it exceeds the critical value, the null hypothesis is rejected, indicating the observed data significantly differs from expectations.

A chi-square goodness-of-fit test determines if observed data significantly differs from expected values by calculating χ² = Σ[(O-E)²/E], where O is observed and E is expected; the test involves establishing null and alternative hypotheses, computing expected values, calculating the chi-square statistic, determining degrees of freedom (number of groups minus one), comparing the result to a critical value from the chi-square distribution table at a chosen alpha level (typically 0.05), and deciding whether to reject the null hypothesis based on whether the calculated value exceeds the critical value.

A complete worked example demonstrates the Chi-Square Goodness of Fit test using Mendelian inheritance ratios. When crossing red-flowered and white-flowered plants, the expected ratio is 1:1:2 (red:white:pink). With 140 total plants observed (50 red, 30 white, 60 pink), expected frequencies are calculated as: red = 140 × (1/4) = 35, white = 140 × (1/4) = 35, pink = 140 × (2/4) = 70. The test statistic is computed as χ² = (50-35)²/35 + (30-35)²/35 + (60-70)²/70 = 6.43 + 0.71 + 1.43 = 8.57. With df = 2 and α = 0.05, the critical value is 5.991. Since 8.57 > 5.991, we reject H0, concluding the observed ratios differ significantly from the expected 1:1:2 ratio.

This section demonstrates the Chi-Square Test for goodness-of-fit in two applications. First, for uniform distribution: given 800 customers across 4 windows with observed frequencies 150, 250, 170, 230, expected frequency E_i = 800 × (1/4) = 200. Calculate G² = Σ[(O_i - E_i)² / E_i] = 34. Degrees of freedom = k - p - 1 = 2. Critical value is 7.815. Since 34 > 7.815, reject the hypothesis of uniform distribution. Second, for genetic ratios: given 64 offspring with observed 34 red, 10 black, 20 white, expected ratio 9:3:4 gives E_red=36, E_black=12, E_white=16. Calculate G² = 1.444. Degrees of freedom = 2. Critical value is 5.99. Since 1.444 < 5.99, accept the hypothesis that data is consistent with the genetic model.

Independent genes are located on different chromosomes and assort independently during gamete formation, with gene count equal to chromosome pairs. Linked genes are located on the same chromosome and tend to be inherited together. Crossing over can separate linked genes, with the probability of separation increasing with distance between genes. Gene order on chromosomes can be determined from crossing over percentages, with genes having the highest crossing over percentages placed at the ends. Mendel established that random events are independent, so previous outcomes don't affect future probabilities. The probability of two independent events occurring together equals the product of their individual probabilities. For example, flipping two coins: the probability of heads on both is 1/2 × 1/2 = 1/4.

Genetic linkage occurs when genes on the same chromosome are inherited together. The closer genes are, the stronger the linkage. Crossing over during meiosis I (prophase I) exchanges genetic material between homologous chromosomes at chiasmata, creating new allele combinations. This violates Mendel's Law of Independent Assortment. Recombination frequency is proportional to gene distance: 1% recombination = 1 map unit (centimorgan). This allows genetic mapping to determine gene order and relative positions on chromosomes.

Three indicators suggest linkage and crossing over: (1) The question explicitly states genes are 'linked' or 'on the same chromosome.' (2) A 3:1 ratio from a dihybrid cross indicates linkage without crossing over. (3) Two large ratios and two small ratios indicate both linkage and crossing over, where small ratios represent recombination events. A 1:1 ratio (50:50) indicates complete linkage with no crossing over, where genes are inherited as a unit. This differs from Mendel's 9:3:3:1 ratio for independent assortment. When two genes are linked and no crossing over occurs, dihybrid crosses produce a 3:1 ratio instead of Mendel's ratios. The deviation from Mendelian ratios indicates linkage between genes.

Linked genes are located on the same chromosome and tend to be inherited together as a single unit. Mendel deliberately chose traits on different chromosomes because linked genes produce complex inheritance patterns. When genes are unlinked (on different chromosomes), they assort independently during meiosis, producing four gamete types with equal 25% probability. When genes are on the same chromosome, they tend to be inherited together. Complete linkage occurs when genes are so close that crossing over never happens, producing only two gamete types at 50% each. Partial linkage occurs when genes are far enough apart that crossing over can happen between them, producing recombinant gametes. Crossing over is a cellular mechanism during meiosis where homologous chromosomes exchange genetic material, creating new allele combinations. The probability of crossing over depends on gene distance: genes far apart have higher recombination frequency, while genes close together have lower recombination frequency.

Mendel's Second Law of Independent Assortment applies only to genes on different chromosomes, producing 9:3:3:1 ratios. When genes are on the same chromosome (linked genes), they are inherited together as a unit. Thomas Morgan discovered this through fruit fly experiments. During meiosis, crossing over creates recombinant gametes. Recombination frequency (percentage of recombinant gametes) is directly proportional to physical distance between genes. The centimorgan (cM) unit represents 1% recombination frequency. To construct genetic maps, recombination frequencies between all gene pairs are measured, with larger distances indicating genes at chromosome ends.