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Genetics - Mendelian Inheritance and Punnett Squares

Grade 9IB

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

🔑Concepts

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Gregor Mendel's Law of Segregation: During gamete formation, the two alleles for a gene separate so that each gamete carries only one allele, represented as AA or aa.

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Genotype vs. Phenotype: The genotype represents the genetic constitution of an organism (e.g., TTTT, TtTt, or tttt), while the phenotype is the observable physical trait (e.g., tall or short).

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Homozygous and Heterozygous: An organism is homozygous if it has two identical alleles (AAAA or aaaa) and heterozygous if it has two different alleles (AaAa).

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Dominant and Recessive Alleles: A dominant allele (denoted by a capital letter like BB) masks the expression of a recessive allele (denoted by a lowercase letter like bb) in the phenotype.

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Monohybrid Cross: A genetic cross between parents that differ in the alleles they possess for a single gene, often visualized using a 2×22 \times 2 grid.

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Law of Independent Assortment: Genes for different traits can segregate independently during the formation of gametes, provided they are on different chromosomes.

📐Formulae

P(trait)=Number of specific genotypesTotal number of possible outcomesP(\text{trait}) = \frac{\text{Number of specific genotypes}}{\text{Total number of possible outcomes}}

Phenotypic Ratio=Dominant Traits:Recessive Traits\text{Phenotypic Ratio} = \text{Dominant Traits} : \text{Recessive Traits}

Genotypic Ratio=Homozygous Dominant:Heterozygous:Homozygous Recessive\text{Genotypic Ratio} = \text{Homozygous Dominant} : \text{Heterozygous} : \text{Homozygous Recessive}

💡Examples

Problem 1:

In pea plants, round seeds (RR) are dominant over wrinkled seeds (rr). If two heterozygous plants (Rr×RrRr \times Rr) are crossed, determine the phenotypic and genotypic ratios of the offspring.

Solution:

Genotypes: 11 RRRR, 22 RrRr, 11 rrrr. Phenotypes: 33 Round, 11 Wrinkled.

Explanation:

The Punnett square results in the following combinations: R×R=RRR \times R = RR, R×r=RrR \times r = Rr, r×R=Rrr \times R = Rr, and r×r=rrr \times r = rr. The genotypic ratio is 1:2:11:2:1 for RR:Rr:rrRR:Rr:rr. Since both RRRR and RrRr express the dominant round trait, the phenotypic ratio is 3:13:1.

Problem 2:

A homozygous dominant black guinea pig (BBBB) is crossed with a homozygous recessive white guinea pig (bbbb). What is the probability that an offspring will be white?

Solution:

0%0\% probability of white offspring.

Explanation:

All offspring from this cross will have the genotype BbBb (heterozygous). Since the black allele (BB) is dominant over the white allele (bb), 100%100\% of the offspring will exhibit the black phenotype. The probability P(white)=04=0P(\text{white}) = \frac{0}{4} = 0.