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Genetics - Inheritance

Grade 12IBBiology

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 separate so that each gamete receives only one allele. This results in a 3:13:1 phenotypic ratio in a monohybrid cross of F1F_1 heterozygotes (Aa×AaAa \times Aa).

Mendel's Law of Independent Assortment: The alleles of two different genes get sorted into gametes independently of one another. This applies to genes on different chromosomes and results in a 9:3:3:19:3:3:1 ratio in a dihybrid cross (AaBb×AaBbAaBb \times AaBb).

Genotype vs. Phenotype: The genotype is the symbolic representation of the pair of alleles (e.g., BBBB, BbBb, bbbb), while the phenotype is the physical characteristic (e.g., Brown eyes).

Codominance and Multiple Alleles: Some genes have more than two alleles (multiple alleles), such as ABO blood groups where IAI^A and IBI^B are codominant and ii is recessive. Genotypes include IAIAI^A I^A, IAiI^A i, IBIBI^B I^B, IBiI^B i, IAIBI^A I^B, and iiii.

Sex Linkage: Genes located on the sex chromosomes (XX or YY) show different inheritance patterns. Hemophilia and color blindness are XX-linked recessive traits, denoted as XHX^H, XhX^h, or XYX^Y. Males (XYXY) are more likely to express recessive XX-linked traits as they only have one XX chromosome.

Linked Genes: Genes located on the same chromosome do not assort independently unless crossing over occurs. The closer two genes are, the lower the probability of a chiasma forming between them.

Polygenic Inheritance: Characteristics like skin color or height are determined by the additive effect of two or more genes, leading to continuous variation represented by a normal distribution curve.

Chi-squared (χ2\chi^2) Test: A statistical test used to determine if there is a significant difference between the observed (OO) and expected (EE) frequencies in a genetic cross.

📐Formulae

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

df=n1df = n - 1

Recombination Frequency=Total number of recombinantsTotal number of offspring×100\text{Recombination Frequency} = \frac{\text{Total number of recombinants}}{\text{Total number of offspring}} \times 100

💡Examples

Problem 1:

In a cross between two pea plants heterozygous for purple flowers (Pp×PpPp \times Pp), what is the probability that an offspring will have white flowers (pppp)?

Solution:

The probability is 14\frac{1}{4} or 25%25\%.

Explanation:

Using a Punnett square, the possible genotypes are 1×PP1 \times PP, 2×Pp2 \times Pp, and 1×pp1 \times pp. Since white is the recessive trait, only the pppp genotype expresses the white phenotype.

Problem 2:

A woman who is a carrier for Hemophilia (XHXhX^H X^h) has a child with a non-affected male (XHYX^H Y). Calculate the probability of having an affected son.

Solution:

The probability of having an affected son is 25%25\% of all children, or 50%50\% of the male children.

Explanation:

The offspring genotypes are XHXHX^H X^H (normal female), XHXhX^H X^h (carrier female), XHYX^H Y (normal male), and XhYX^h Y (affected male). There is a 11 in 44 total chance for XhYX^h Y.

Problem 3:

Calculate the Chi-squared value for a cross where you expected 3030 tall plants and 1010 short plants, but observed 2525 tall and 1515 short.

Solution:

χ2=(2530)230+(1510)210=2530+2510=0.833+2.5=3.333\chi^2 = \frac{(25 - 30)^2}{30} + \frac{(15 - 10)^2}{10} = \frac{25}{30} + \frac{25}{10} = 0.833 + 2.5 = 3.333

Explanation:

The formula (OE)2E\sum \frac{(O-E)^2}{E} is applied to both phenotypic categories and summed. With df=1df = 1, we would compare 3.3333.333 to a critical value table to determine significance.

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