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Genetics - Inheritance

Grade 11IBBiology

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

🔑Concepts

Mendel's Law of Segregation: During meiosis, the two alleles of a gene (e.g., AA and aa) separate into different gametes so that each gamete carries only one allele (nn).

Mendel's Law of Independent Assortment: The alleles of two or more different genes (e.g., AaAa and BbBb) get sorted into gametes independently of one another, provided they are on different chromosomes.

Genotype vs. Phenotype: Genotype refers to the genetic makeup (AA,Aa,aaAA, Aa, aa), while phenotype is the observable physical trait (e.g., tall vs. short).

Dominant and Recessive Alleles: Dominant alleles (AA) mask the effect of recessive alleles (aa) in a heterozygote (AaAa). Recessive traits only appear in homozygous recessive individuals (aaaa).

Codominance: Pairs of alleles that both affect the phenotype when present in a heterozygote. Example: IAIBI^A I^B results in AB blood type.

Sex-linkage: Genes located on the sex chromosomes (XX or YY). Traits like hemophilia and red-green color blindness are XX-linked recessive, denoted as XhX^h and XbX^b.

Multiple Alleles: Some genes have more than two alleles, such as the ABO blood group system involving IAI^A, IBI^B, and ii.

Test Cross: Testing a suspected heterozygote (AaAa) by crossing it with a known homozygous recessive (aaaa).

📐Formulae

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

df=n1df = n - 1

P=Number of ways a specific event can occurTotal number of possible outcomesP = \frac{\text{Number of ways a specific event can occur}}{\text{Total number of possible outcomes}}

💡Examples

Problem 1:

In a monohybrid cross between two heterozygous tall pea plants (TtTt), calculate the probability of the offspring being homozygous recessive (tttt).

Solution:

P=14P = \frac{1}{4} or 25%25\%

Explanation:

Using a Punnett square for Tt×TtTt \times Tt, the possible genotypes are TTTT, TtTt, TtTt, and tttt. Out of 4 possible combinations, only 1 is tttt. Therefore, the probability is 1/41/4.

Problem 2:

A woman who is a carrier for color blindness (XBXbX^B X^b) marries a man with normal vision (XBYX^B Y). What is the probability that their son will be color blind?

Solution:

P=0.5P = 0.5 (or 50%50\% of sons)

Explanation:

The sons receive the YY chromosome from the father and either XBX^B or XbX^b from the mother. The possible genotypes for sons are XBYX^B Y (normal) and XbYX^b Y (color blind). Thus, 50%50\% of the male offspring are expected to be color blind.

Problem 3:

Perform a Chi-squared test on a dihybrid cross (AaBb×AaBbAaBb \times AaBb) where the observed frequency of the double recessive phenotype (aabbaabb) is 1212 out of 160160 total offspring. Calculate the expected value (EE).

Solution:

E=10E = 10

Explanation:

In a Mendelian dihybrid cross, the expected ratio for the double recessive phenotype is 1/161/16. For a total of 160160 offspring, E=116×160=10E = \frac{1}{16} \times 160 = 10.

Inheritance - Revision Notes & Key Diagrams | IB Grade 11 Biology