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Genetics and Evolution (AHL) - Inheritance

Grade 12IBBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Mendel’s Law of Independent Assortment states that the alleles of two (or more) different genes get sorted into gametes independently of one another. This applies only to genes located on different chromosomes or very far apart on the same chromosome.

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Gene linkage occurs when gene loci are on the same chromosome and do not assort independently. Linked genes are inherited together unless crossing over occurs. Linkage notation is represented as ABab\frac{AB}{ab}.

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Recombination is the result of crossing over between non-sister chromatids during Prophase I of meiosis. It produces new combinations of alleles in gametes that differ from the parental combinations.

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The Chi-squared (χ2\chi^2) test is a statistical tool used to compare observed phenotypic ratios with expected Mendelian ratios (e.g., 9:3:3:19:3:3:1 for a dihybrid cross). It helps determine if the genes are assorting independently or are linked.

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Polygenic inheritance occurs when a single trait is controlled by two or more genes (e.g., human skin color or height). This results in continuous variation, often producing a normal distribution (bell curve) in the population.

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Thomas Hunt Morgan’s experiments with DrosophilaDrosophila melanogastermelanogaster provided evidence for gene linkage and sex-linked traits, identifying that some genes do not follow the expected Mendelian ratios because they are located on the XX chromosome.

📐Formulae

χ2=∑(O−E)2E\chi^2 = \sum \frac{(O - E)^2}{E}

df=(n−1)df = (n - 1) or (rows−1)(columns−1)(rows - 1)(columns - 1) for contingency tables

Recombination Frequency=Number of RecombinantsTotal Number of Offspring×100%\text{Recombination Frequency} = \frac{\text{Number of Recombinants}}{\text{Total Number of Offspring}} \times 100\%

💡Examples

Problem 1:

A researcher crosses a dihybrid purple-flowered, long-pollen plant (PpLlPpLl) with a red-flowered, round-pollen plant (ppllppll). The observed offspring are: Purple/Long: 435435, Red/Round: 445445, Purple/Round: 6262, Red/Long: 5858. Calculate the recombination frequency.

Solution:

Recombinants=62+58=120\text{Recombinants} = 62 + 58 = 120 Total Offspring=435+445+62+58=1000\text{Total Offspring} = 435 + 445 + 62 + 58 = 1000 Recombination Frequency=1201000×100%=12%\text{Recombination Frequency} = \frac{120}{1000} \times 100\% = 12\%

Explanation:

In a test cross involving unlinked genes, we expect a 1:1:1:11:1:1:1 ratio (25%25\% recombinants). Since the recombination frequency is 12%12\%, the genes are linked. The distance between the gene for flower color and pollen shape is 1212 centimorgans (cMcM).

Problem 2:

Perform a χ2\chi^2 test to see if the following dihybrid cross fits a 9:3:3:19:3:3:1 ratio. Observed: 315,108,101,32315, 108, 101, 32. Total: 556556. (Critical value for p=0.05,df=3p=0.05, df=3 is 7.817.81).

Solution:

  1. Expected values: E1=916×556=312.75E_1 = \frac{9}{16} \times 556 = 312.75 E2=316×556=104.25E_2 = \frac{3}{16} \times 556 = 104.25 E3=316×556=104.25E_3 = \frac{3}{16} \times 556 = 104.25 E4=116×556=34.75E_4 = \frac{1}{16} \times 556 = 34.75

  2. χ2=(315−312.75)2312.75+(108−104.25)2104.25+(101−104.25)2104.25+(32−34.75)234.75\chi^2 = \frac{(315-312.75)^2}{312.75} + \frac{(108-104.25)^2}{104.25} + \frac{(101-104.25)^2}{104.25} + \frac{(32-34.75)^2}{34.75} χ2=0.016+0.135+0.101+0.218=0.470\chi^2 = 0.016 + 0.135 + 0.101 + 0.218 = 0.470

Explanation:

Since the calculated χ2\chi^2 (0.4700.470) is less than the critical value (7.817.81), we fail to reject the null hypothesis (H0H_0). The differences between observed and expected values are statistically insignificant, and the genes assort independently.

Inheritance - Revision Notes & Key Diagrams | IB Grade 12 Biology