Math in Biology and Biochemistry for the MCAT

Key Concepts

• Autosomal Genes: Chromosomes and genes not related to determining sex (e.g., not X or Y chromosomes).
• Sex-Linked Genes: Genes on sex-determining chromosomes (X and Y) with different inheritance patterns.
• Dominant Traits: Only one copy of the gene is needed to display the phenotype.
• Recessive Traits: Both copies of the gene are needed to express the phenotype.
• Homozygous: Two copies of the same allele (either dominant or recessive).
• Heterozygous: One dominant and one recessive allele.

Inheritance Rules

• Mammals, including humans, have two copies of each gene.
• Zygote formation: One allele from each parent.
• Use Punnett squares to determine probabilities of genetic combinations.

Practice Problem 1: Hemophilia and Sex-Linked Inheritance

• Problem Setup:
  • Hemophilia is determined by an X-linked gene.
  • Man with normal clotting (X and Y; no hemophilia) marries a woman with normal clotting whose father had hemophilia.
• Solution Steps:
  • Woman is a carrier (heterozygote) because her father had hemophilia.
  • Punnett square used to determine offspring probabilities.
  • Focus on sons since they asked about sons only.
  • Probability of all three sons having hemophilia: Multiply individual probabilities (1/2 x 1/2 x 1/2 = 1/8).
• Key takeaway: Eliminate daughters from probability and focus on specified offspring.

Practice Problem 2: Flies and Multi-Generation Probability

• Problem Setup:
  • Green body (dominant) vs. brown body color in flies.
  • Two green flies crossed yielding 29 green and 8 brown offspring.
• Solution Steps:
  • Brown offspring indicate heterozygote parents.
  • Heterozygote cross yields a 1:2:1 ratio (homozygous dominant: heterozygote: homozygous recessive).
  • Focus on green flies for next cross.
  • Probability of heterozygote cross: 2/3 for male and 2/3 for female; multiply for combined probability (4/9).
• Key takeaway: Identify and eliminate non-relevant genotypes from probability calculations.

Practice Problem 3: Dihybrid Cross in Cats

• Problem Setup:
  • Short hair (dominant) vs. long hair and amber eyes (dominant) vs. green eyes.
  • Both cats are heterozygous for both traits.
• Solution Steps:
  • The goal is to find long hair (recessive, hh) and amber eyes (dominant, A?).
  • Use dihybrid cross (16-cell Punnett Square) but only fill necessary parts.
  • Eliminate cells with dominant H; only consider those with hh.
  • Identify cells with at least one dominant A.
  • Probability of desired genotype: 3 out of 16.
• Key takeaway: Simplify by eliminating unlikely genotypes and focus on desired genotype combinations.

Conclusion

• Practice problems focus on application of Mendelian genetics and Punnett squares to solve complex probability questions in genetics.
• Use provided strategies to eliminate irrelevant factors and focus on specific criteria for probability calculations.
• Further study and practice recommended for mastery.