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5.3 Mendelian Genetics

Syllabus
2025
Topic
5.3
Level
—

Learning objectives

5.3A—Explain the inheritance of genes and traits as described by Mendel’s lawsExplain the inheritance of genes and traits as described by Mendel’s laws.• Mendel’s laws of segregation and independent assortment can be applied to genes that are on different chromosomes.• In most cases, fertilization involves the fusion of two haploid gametes, restoring the diploid number of chromosomes and increasing genetic variation in populations by creating new combinations of alleles in the zygote.- i. Rules of probability can be applied to analyze the passing of single-gene traits from parent to offspring.- ii. Monohybrid, dihybrid, and test crosses can be used to determine whether alleles are dominant or recessive.- iii. An organism’s genotype is the set of alleles inherited for one or more genes by an individual organism. An organism’s genotype can be homozygous or heterozygous for each gene.- iv. An organism’s phenotype is the observable expression of the inherited traits.

Using Mendel's Laws to Predict Inheritance

Mendelian inheritance links chromosome behavior to allele transmission. A genotype is the allele combination an organism carries; a phenotype is the observable expression of inherited traits. For a gene, two identical alleles are homozygous and two different alleles are heterozygous.

Law Chromosome basis Inheritance consequence
Segregation The two alleles for a gene separate as homologous chromosomes move into different gametes Each gamete receives one allele; a monohybrid cross predicts one-gene outcomes
Independent assortment Different homologous chromosome pairs orient and separate independently For genes on different chromosomes, one gene's allele pair does not determine which alleles of another gene enter the same gamete; a dihybrid cross combines probabilities

Mutually exclusive outcomes: $P(A\text{ or }B)=P(A)+P(B)$ Independent outcomes: $P(A\text{ and }B)=P(A)\times P(B)$

Example: in Aa×AaAa \times Aa, each parent passes allele aa with probability 1/21/2, so P(aa)=1/2×1/2=1/4P(aa)=1/2 \times 1/2=1/4. If a second gene assorting independently also has a 1/41/4 chance of a recessive genotype, the chance of both recessive genotypes is 1/4×1/4=1/161/4 \times 1/4=1/16.

Punnett squares organize possible gametes and offspring genotypes. A testcross pairs an individual showing a dominant phenotype with a homozygous recessive individual to reveal the unknown genotype. Pedigree and offspring data can indicate dominant or recessive, autosomal, sex-linked, or genetically linked inheritance patterns.

Independent assortment is not assumed for genes that are genetically linked on the same chromosome. Also, dominant means expressed in a heterozygote; it does not mean more common, stronger, or more beneficial.