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Mendelian Genetics
Inheritance Patterns

Master the foundational laws of heredity, from monohybrid crosses to dihybrid ratios, and learn how alleles segregate and assort independently.

MCAT Biology Foundational Concept 1C

What is Mendelian Genetics?

Mendelian genetics describes how traits are passed from parents to offspring through discrete units called genes. Gregor Mendel's experiments with pea plants revealed predictable patterns of inheritance that form the foundation of modern genetics.

Essential Vocabulary

Genotype
The genetic makeup of an organism: the specific allele combination (e.g., \(Bb\))
Phenotype
The observable physical trait expressed by the genotype (e.g., brown eyes)
Homozygous
Two identical alleles for a trait: \(BB\) (dominant) or \(bb\) (recessive)
Heterozygous
Two different alleles for a trait: \(Bb\) (carrier of recessive allele)

Why This Matters for the MCAT

The MCAT tests your ability to predict offspring ratios, identify inheritance patterns from pedigrees, and distinguish Mendelian from non-Mendelian inheritance. Foundational Concept 1C appears in both discrete questions and passage-based problems.

1
Predict genotype and phenotype ratios using Punnett squares and probability rules
2
Apply Mendel's laws (segregation and independent assortment) to genetics problems
3
Analyze pedigrees to determine dominance patterns and carrier status

Mendel's First Law: Segregation

Parent Cell
\(Bb\)
Meiosis Gamete
\(B\)
Gamete
\(b\)
Each gamete receives only one allele 50% chance of B, 50% chance of b

Mendel's Second Law: Independent Assortment

Dihybrid Parent
\(BbEe\)
\(BE\)
\(Be\)
\(bE\)
\(be\)
25% 25% 25% 25% Alleles of Gene B sort independently of Gene E Applies when genes are on different chromosomes

Dominance vs Recessiveness

Complete Dominance Incomplete Dominance Codominance
Heterozygote Phenotype Same as homozygous dominant Intermediate blend Both alleles expressed
Example \(Bb\) = Brown eyes Red + White = Pink flower \(AB\) blood type
Phenotype Ratio (Monohybrid F2) 3:1 1:2:1 1:2:1
MCAT Frequency Very High Medium Medium

The Monohybrid Cross

Cross two heterozygous parents: \(Bb \times Bb\)

\(B\)
\(b\)
\(B\)
\(b\)
\(BB\)
\(Bb\)
\(Bb\)
\(bb\)
Genotype ratio: 1 \(BB\) : 2 \(Bb\) : 1 \(bb\)
Phenotype ratio: 3 dominant : 1 recessive

How to Solve Genetics Problems

1
Identify the alleles: Assign letters to dominant (uppercase) and recessive (lowercase) alleles
2
Determine parent genotypes: Use phenotype clues and family history to deduce allele combinations
3
List possible gametes: Apply the Law of Segregation to find all gamete types for each parent
4
Set up the Punnett square: Cross gametes systematically and fill in all boxes
5
Calculate ratios: Count genotypes and phenotypes, then express as ratios or probabilities

Worked Example: The Testcross

A plant shows the dominant phenotype (purple flowers). Is it \(PP\) or \(Pp\)? Cross it with a homozygous recessive (\(pp\)) plant.

If \(PP\) \(\times\) \(pp\)
All offspring are \(Pp\) (100% purple). No recessive phenotype appears.
If \(Pp\) \(\times\) \(pp\)
Offspring: 50% \(Pp\) (purple) and 50% \(pp\) (white). Recessive phenotype appears.

Worked Example: Dihybrid Cross

Cross: \(BbEe \times BbEe\) (two traits, both heterozygous)

9
\(B\_E\_\)
Both dominant 3
\(B\_ee\)
B dom, E rec 3
\(bbE\_\)
B rec, E dom 1
\(bbee\)
Both rec : : : Classic Dihybrid Phenotype Ratio Total = 16 possible offspring combinations Shortcut: Multiply individual probabilities
\( P(B\_E\_) = P(B\_) \times P(E\_) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16} \)

The Probability Method

For multi-gene problems, the multiplication rule and addition rule are faster than drawing large Punnett squares.

Multiplication Rule (AND)
Probability of independent events both occurring: \( P(A \text{ and } B) = P(A) \times P(B) \)
Addition Rule (OR)
Probability of either event occurring: \( P(A \text{ or } B) = P(A) + P(B) \) for mutually exclusive events

Reading a Pedigree

Legend Male (unaffected) Female (unaffected) Affected I II
\(aa\)
\(Aa\)
\(Aa\)
Autosomal Recessive Pattern Affected child from unaffected parents
= both parents are carriers (\(Aa\))

Sex-Linked Inheritance

Genes on the X chromosome show distinct inheritance patterns because males have only one X chromosome (\( X^a Y \)) while females have two (\( X^A X^a \)).

X-Linked Recessive
Males are affected more often: they need only one copy of the recessive allele. Carrier mothers pass to sons.
Key Pattern
No male-to-male transmission. Affected fathers produce all carrier daughters and no affected sons.

Beyond Mendel: Extensions

Pleiotropy
One gene affects multiple phenotypes (e.g., sickle cell gene affects blood, spleen, joints)
Epistasis
One gene masks or modifies the expression of another gene (modifies 9:3:3:1 ratio)
Polygenic Traits
Multiple genes contribute to one phenotype, producing continuous variation (e.g., height, skin color)

Common Mistakes

✕ Confusing Genotype with Phenotype
\(Bb\) and \(BB\) have different genotypes but the same dominant phenotype. Don't assume they're identical.
✕ Ignoring Carrier Status
In autosomal recessive inheritance, unaffected parents can be carriers (\(Aa\)). Always consider heterozygous genotypes.
✕ Applying Independent Assortment to Linked Genes
Genes on the same chromosome do not assort independently. Linked genes require recombination frequency analysis.

Question 1 of 5

In a monohybrid cross between two heterozygous parents (\(Aa \times Aa\)), what is the expected genotype ratio among offspring?

Question 2 of 5

A woman who is a carrier for hemophilia (\( X^H X^h \)) has children with an unaffected man (\( X^H Y \)). What proportion of their sons will be affected?

Question 3 of 5

In a dihybrid cross of \(BbEe \times BbEe\), what is the probability of offspring with genotype \(bbee\)?

Question 4 of 5

Two unaffected parents have a child with an autosomal recessive disorder. What is the probability that their next child will also be affected?

Question 5 of 5

In Labrador retrievers, coat color is determined by two genes: \(B\) (black/brown) and \(E\) (pigment deposition). The \(ee\) genotype prevents all pigment deposition, resulting in a yellow coat regardless of the \(B\) genotype. This is an example of: