D3.2 Inheritance

D3.2 Inheritance

Overview

Inheritance is the transmission of genetic information from parents to offspring. This topic covers the fundamental principles of heredity discovered by Gregor Mendel, as well as extensions to Mendelian genetics including codominance, sex linkage, and the use of statistical tests to analyse genetic data.

Understanding inheritance patterns is essential for predicting offspring phenotypes, genetic counselling, and understanding the genetic basis of traits and diseases.

Key Genetic Terminology

Mendel's Laws of Inheritance

Gregor Mendel (1822–1884) was an Augustinian friar who discovered the fundamental laws of inheritance through experiments with pea plants.

Why Peas Were Good Experimental Organisms

Mendel's First Law: Law of Segregation

Statement: The two alleles of a gene segregate (separate) during gamete formation, so that each gamete carries only one allele for each gene.

Explanation:

• Diploid organisms have two alleles for each gene (one from each parent)

• During meiosis I, homologous chromosomes separate

• Each gamete receives only one allele

• At fertilisation, the diploid number is restored

Evidence: Mendel's monohybrid crosses showed 3:1 ratios in F2 generation, consistent with segregation of two alleles.

Mendel's Second Law: Law of Independent Assortment

Statement: Alleles of different genes assort independently of each other during gamete formation, provided the genes are on different chromosomes (or far apart on the same chromosome).

Explanation:

• During metaphase I of meiosis, bivalents line up randomly

• The orientation of one bivalent does not affect another

• Creates new combinations of alleles in gametes

Evidence: Mendel's dihybrid crosses showed 9:3:3:1 ratios in F2 generation, consistent with independent assortment of two genes.

Limitation: Genes on the same chromosome (linked genes) do not assort independently unless separated by crossing over.

Monohybrid Inheritance

Monohybrid cross: A cross examining the inheritance of a single gene with two alleles.

Performing a Monohybrid Cross

Example: Pea plant height (T = tall, t = short)

P generation: True-breeding tall (TT) × True-breeding short (tt)

Step 1: Identify parental genotypes

• Tall parent: TT (homozygous dominant)

• Short parent: tt (homozygous recessive)

Step 2: Determine gametes

• TT produces only T gametes

• tt produces only t gametes

Step 3: Set up Punnett square

F1 genotypes: All Tt (100% heterozygous) F1 phenotypes: All tall (100%)

F1 × F1 cross: Tt × Tt

Gametes: Each parent produces T and t gametes (50% each)

F2 genotypes:

• TT: 25% (1/4)

• Tt: 50% (2/4)

• tt: 25% (1/4)

Genotypic ratio: 1 TT : 2 Tt : 1 tt

F2 phenotypes:

• Tall (TT + Tt): 75% (3/4)

• Short (tt): 25% (1/4)

Phenotypic ratio: 3 tall : 1 short (3:1)

Test Cross

A test cross determines whether an individual showing the dominant phenotype is homozygous (TT) or heterozygous (Tt).

Method: Cross the unknown individual with a homozygous recessive individual (tt)

If unknown is TT:

Result: All offspring tall (100% Tt)

If unknown is Tt:

Result: 50% tall (Tt), 50% short (tt) Ratio: 1:1

Summary of Monohybrid Ratios

Dihybrid Inheritance

Dihybrid cross: A cross examining the inheritance of two genes with two alleles each.

Performing a Dihybrid Cross

Example: Pea seed shape and colour

• Round (R) dominant to wrinkled (r)

• Yellow (Y) dominant to green (y)

P generation: RRYY × rryy

F1 generation: All RrYy (round, yellow)

F1 × F1 cross: RrYy × RrYy

Step 1: Determine gametes Each parent can produce 4 types of gametes:

• RY, Ry, rY, ry (each 25%)

Step 2: Set up 4×4 Punnett square

F2 phenotypes (16 squares total):

Phenotypic ratio: 9:3:3:1

Understanding the 9:3:3:1 Ratio

The ratio can be understood as the product of two independent 3:1 ratios:

• For shape: 3 round : 1 wrinkled

• For colour: 3 yellow : 1 green

Combined:

• 3 round × 3 yellow = 9 round, yellow

• 3 round × 1 green = 3 round, green

• 1 wrinkled × 3 yellow = 3 wrinkled, yellow

• 1 wrinkled × 1 green = 1 wrinkled, green

Dihybrid Test Cross

Cross: RrYy × rryy

Gametes:

• RrYy produces: RY, Ry, rY, ry (25% each)

• rryy produces: only ry (100%)

Phenotypic ratio: 1:1:1:1 (equal proportions of all four phenotypes)

Summary of Dihybrid Ratios

Codominance

Codominance occurs when both alleles are fully expressed in the heterozygote, resulting in a phenotype that shows features of both alleles.

Key Features

• Neither allele is dominant over the other

• Heterozygotes show BOTH phenotypes simultaneously

• Different from incomplete dominance (blending)

Notation for Codominance

Codominant alleles are written with a superscript to distinguish them:

• Gene symbol in capital letter

• Allele designations as superscripts

• Example: Iᴬ, Iᴮ for ABO blood groups

Example: ABO Blood Groups

The ABO blood group system demonstrates both codominance and multiple alleles.

Three alleles:

• Iᴬ — codes for A antigen

• Iᴮ — codes for B antigen

• i — codes for no antigen (O)

Dominance relationships:

• Iᴬ and Iᴮ are codominant (both expressed in heterozygote)

• Both Iᴬ and Iᴮ are dominant over i

Possible genotypes and phenotypes:

Inheritance of ABO Blood Groups

Example cross: Type A (Iᴬi) × Type B (Iᴮi)

Offspring phenotypes:

• Type AB (IᴬIᴮ): 25%

• Type A (Iᴬi): 25%

• Type B (Iᴮi): 25%

• Type O (ii): 25%

Ratio: 1 A : 1 B : 1 AB : 1 O

Blood Transfusion Compatibility

Principle: Recipient must not have antibodies against donor's antigens

Another Example: Coat Colour in Cattle

Roan cattle: Codominance of red and white alleles

Cross: Roan × Roan (CᴿCᵂ × CᴿCᵂ)

Phenotypic ratio: 1 red : 2 roan : 1 white

Sex Linkage

Sex-linked genes are genes located on sex chromosomes (X or Y). In mammals, most sex-linked genes are on the X chromosome (X-linkage).

Sex Chromosomes

In humans:

• Females: XX (homogametic)

• Males: XY (heterogametic)

X chromosome: Large; carries many genes (>800) Y chromosome: Small; carries few genes; mainly determines maleness (SRY gene)

X-Linked Inheritance

Key features:

• Males have only one X chromosome (hemizygous)

• Males express all X-linked alleles (no second allele to mask recessive)

• Females can be homozygous or heterozygous

• Carrier females are heterozygous for recessive X-linked conditions

Notation for Sex-Linked Genes

Alleles written as superscripts on the chromosome symbol:

• Xᴴ = X chromosome with dominant allele

• Xʰ = X chromosome with recessive allele

• Y = Y chromosome (no allele for this gene)

Example: Haemophilia

Haemophilia is an X-linked recessive condition affecting blood clotting.

Alleles:

• Xᴴ = normal blood clotting (dominant)

• Xʰ = haemophilia (recessive)

Possible genotypes:

Cross: Carrier female × Normal male (XᴴXʰ × XᴴY)

Offspring:

• Normal female (XᴴXᴴ): 25%

• Carrier female (XᴴXʰ): 25%

• Normal male (XᴴY): 25%

• Haemophiliac male (XʰY): 25%

Key observations:

• 50% of males affected

• 0% of females affected (but 50% are carriers)

• Sons inherit X from mother (carrier)

• Daughters inherit X from each parent

Example: Red-Green Colour Blindness

Red-green colour blindness is an X-linked recessive condition.

Alleles:

• Xᴮ = normal colour vision (dominant)

• Xᵇ = colour blindness (recessive)

Cross: Colour-blind male × Carrier female (XᵇY × XᴮXᵇ)

Offspring:

• Carrier female (XᴮXᵇ): 25%

• Colour-blind female (XᵇXᵇ): 25%

• Normal male (XᴮY): 25%

• Colour-blind male (XᵇY): 25%

Patterns of X-Linked Recessive Inheritance

1. More males affected than females (males are hemizygous)

2. Affected males pass allele to all daughters (who become carriers)

3. Affected males cannot pass condition to sons (sons get Y from father)

4. Carrier mothers can pass allele to sons and daughters

5. For female to be affected, father must be affected AND mother must be carrier or affected

6. Condition often "skips" generations (passed through carrier females)

X-Linked Dominant Inheritance

Less common; affected fathers pass condition to ALL daughters but NO sons.

Example: Some forms of rickets

Pedigree Analysis

Pedigree diagrams show the inheritance of traits through multiple generations of a family.

Symbols Used in Pedigrees

Analysing Pedigrees

Step 1: Determine the mode of inheritance

Step 2: Assign genotypes

• Start with affected individuals (known genotype)

• Work backward to parents

• Consider all possible genotypes consistent with phenotypes

Step 3: Calculate probabilities for future offspring

Example Pedigree Analysis

Generation I:      □──●
                   │
Generation II:  ●──□    □──○    □──○
                │       │       │
Generation III: □  ○    ●  □    ○  □

Analysis:

• Affected individuals in generation I and II suggest the trait is inherited

• Affected individuals have affected parent → could be dominant

• Both males and females affected → likely autosomal

• If autosomal recessive: Generation I affected female (aa) × unaffected male (Aa) → affected children possible

Chi-Squared (χ²) Test

The chi-squared test is a statistical test used to determine whether observed results differ significantly from expected results.

Purpose

• Tests the "goodness of fit" between observed and expected data

• Determines if deviations from expected ratios are due to chance or other factors

• Used to validate genetic hypotheses

The Chi-Squared Formula

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

Where:

• χ² = chi-squared value

• O = observed value

• E = expected value

• Σ = sum for all categories

Steps for Performing Chi-Squared Test

Step 1: State the null hypothesis (H₀)

• "There is no significant difference between observed and expected results"

• "The deviations are due to chance alone"

Step 2: Calculate expected values

• Based on the predicted ratio

• E = (total count × expected proportion)

Step 3: Calculate chi-squared value

• For each category: (O − E)² / E

• Sum all values

Step 4: Determine degrees of freedom $df = n - 1$ Where n = number of categories

Step 5: Find critical value

• Use chi-squared table

• Usually at p = 0.05 significance level

Step 6: Compare and conclude

• If χ² calculated < χ² critical → Accept H₀ (no significant difference)

• If χ² calculated > χ² critical → Reject H₀ (significant difference)

Example Calculation

Scenario: A monohybrid cross (Tt × Tt) should give 3:1 ratio. You observe 280 tall and 120 short plants (total 400).

Step 1: Null hypothesis H₀: There is no significant difference between observed and expected 3:1 ratio

Step 2: Calculate expected values

• Expected tall: 400 × 3/4 = 300

• Expected short: 400 × 1/4 = 100

Step 3: Calculate χ²

Step 4: Degrees of freedom df = 2 − 1 = 1

Step 5: Find critical value At p = 0.05 and df = 1, critical value = 3.84

Step 6: Conclusion χ² calculated (5.33) > χ² critical (3.84)

Reject H₀: There IS a significant difference between observed and expected results. The data does not fit the expected 3:1 ratio.

Chi-Squared Critical Values Table

Interpreting Chi-Squared Results

Applications in Genetics

• Testing Mendelian ratios (3:1, 9:3:3:1, etc.)

• Detecting linkage (results differ from independent assortment)

• Validating genetic hypotheses

• Determining if data supports a particular inheritance pattern

Genetic Diseases

Autosomal Recessive Conditions

Autosomal Dominant Conditions

X-Linked Recessive Conditions

Sickle Cell Anaemia and Heterozygote Advantage

Sickle cell trait (HbᴬHbˢ):

• Heterozygotes are carriers

• Have some sickle cells but are generally healthy

• Resistant to malaria (sickled cells destroy Plasmodium parasite)

Heterozygote advantage:

• In malaria-endemic regions, carriers have survival advantage

• Explains high frequency of sickle cell allele in these populations

• Example of balancing selection

Genetic Counselling

Genetic counselling provides information and support to individuals or families affected by or at risk of genetic disorders.

Uses of Genetic Counselling

• Assess risk of having affected children

• Explain inheritance patterns

• Discuss testing options

• Provide emotional support

• Explain treatment and management options

Calculating Risk

Example: Both parents are carriers for cystic fibrosis (Cc × Cc)

Risk for each pregnancy:

• 25% unaffected (CC)

• 50% carrier (Cc)

• 25% affected (cc)

Conditional probability: If a child is unaffected, what is the probability they are a carrier?

• 2/3 (because 2 out of 3 unaffected genotypes are Cc)

Practical Investigations

Modelling Genetic Crosses

Using coins or beads:

1. Assign alleles to coin faces (heads = A, tails = a)

2. Flip two coins to represent gamete fusion

3. Record genotype (HH = AA, HT = Aa, TT = aa)

4. Repeat many times

5. Calculate ratios and compare to expected

Analysing Genetic Data

Corn cob genetics:

1. Count kernels of different colours/textures

2. Calculate ratios

3. Hypothesise inheritance pattern

4. Perform chi-squared test

Pedigree Construction

1. Collect family history data

2. Construct pedigree using standard symbols

3. Analyse inheritance pattern

4. Assign possible genotypes

Common Exam Questions

Typical Question Types

1. Perform a monohybrid cross and state the expected ratio (4 marks)

   • Identify parental genotypes

   • Determine gametes

   • Set up Punnett square

   • State genotypic and phenotypic ratios

2. Explain codominance using ABO blood groups (4 marks)

   • Both Iᴬ and Iᴮ alleles expressed in heterozygote

   • IᴬIᴮ genotype produces both A and B antigens

   • Results in type AB blood

   • Different from complete dominance where one allele masks other

3. Explain why X-linked recessive conditions are more common in males (3 marks)

   • Males have only one X chromosome (XY)

   • Males are hemizygous for X-linked genes

   • Single recessive allele is expressed (no second allele to mask it)

   • Females need two recessive alleles to be affected

4. Perform a dihybrid cross and state the expected ratio (5 marks)

   • Identify parental genotypes

   • List all possible gametes (4 types for RrYy)

   • Set up 4×4 Punnett square

   • Count phenotypes

   • State 9:3:3:1 ratio

5. Perform a chi-squared test on given data (5 marks)

   • State null hypothesis

   • Calculate expected values

   • Calculate χ² using formula

   • Determine degrees of freedom

   • Compare to critical value

   • State conclusion (accept/reject H₀)

6. Analyse a pedigree diagram (4 marks)

   • Identify inheritance pattern (autosomal/sex-linked, dominant/recessive)

   • Explain reasoning

   • Assign genotypes to individuals

   • Calculate probability for offspring

Key Terminology Glossary

Summary Comparison Tables

Expected Ratios

Inheritance Patterns

Blood Group Genetics