Two locus genetics (practice problems)

🧬 BIOS 120 — TWO-LOCUS GENETICS

📚 Complete Exam Study Guide (26 Slides)

🎯 Goal: Understand everything in your professor's PowerPoint, including dihybrid crosses, Punnett squares, independent assortment, test crosses, chi-square calculations, genetic linkage, and sex-linked inheritance.

I'm keeping the same format as your previous study guides: simple explanations, emojis, step-by-step calculations, exam shortcuts, common mistakes, and practice questions with answers.

This guide covers all 26 slides of Two locus genetics (practice problems).pdf. No guide can guarantee a 100%, but if you can solve every example without notes, you'll be well prepared for this material.

🧬 PART 1 — DIHYBRID CROSSES (Slides 1–3)

🌱 1. What Is a Dihybrid Cross?

Definition: A genetic cross that examines two different gene loci at the same time.

💡 Simple terms: Instead of looking at one trait, you're looking at TWO traits.

Example:

Pea plants have two traits:

• 🌸 Flower color: Purple or white.

• 🟡 Seed color: Yellow or green.

We'll use:

• P = Purple flowers (dominant).

• p = White flowers (recessive).

• Y = Yellow seeds (dominant).

• y = Green seeds (recessive).

🧠 2. Important Vocabulary

🌸 3. Professor's Pea Plant Example (Slide 2)

Two true-breeding parental lines:

Parent 1: Purple flowers, yellow seeds.

Genotype: PPYY

Parent 2: White flowers, green seeds.

Genotype: ppyy

Step 1 — Cross the parents

PPYY × ppyy

Parent 1 produces PY gametes.

Parent 2 produces py gametes.

All F1 offspring:

PpYy

Step 2 — Determine F1 phenotype

Pp = Purple flowers.

Yy = Yellow seeds.

✅ 100% purple flowers with yellow seeds.

Step 3 — Cross two F1 individuals

PpYy × PpYy

Each parent can produce four gametes:

• PY

• Py

• pY

• py

🧮 4. The 16-Box Punnett Square

📊 5. The Famous 9:3:3:1 Ratio

Under independent assortment and complete dominance:

⭐ MEMORIZE: 9 : 3 : 3 : 1

This is the classic dihybrid F2 phenotypic ratio.

🧠 Ultimate Shortcut

For PpYy × PpYy:

Each trait separately has:

• 3/4 dominant phenotype.

• 1/4 recessive phenotype.

Multiply:

• Purple yellow = 3/4 × 3/4 = 9/16

• Purple green = 3/4 × 1/4 = 3/16

• White yellow = 1/4 × 3/4 = 3/16

• White green = 1/4 × 1/4 = 1/16

🚨 EXAM TRAP: The 9:3:3:1 ratio is not guaranteed for every two-locus cross. It requires the appropriate parental genotypes, independent assortment, and dominance relationships.

📝 6. Professor's Polling Question (Slide 3)

Question: What is the probability that an F1 pollen grain contains the white-flower allele AND the green-seed allele?

F1 genotype = PpYy.

We want: py

Probability of p = 1/2.

Probability of y = 1/2.

Multiply:

1/2 × 1/2 = 1/4

✅ Correct answer: E — 1/4 (25%).

🎲 PART 2 — INDEPENDENT ASSORTMENT (Slides 4–6)

🧬 7. What Is Independent Assortment?

Definition: Alleles at different loci are inherited independently when the loci assort independently.

💡 Simple terms: Getting an allele from one gene does not affect which allele you receive from another independently assorting gene.

Example:

AaBb can produce:

• AB

• Ab

• aB

• ab

Each has a probability of 1/4.

🧮 8. How to Calculate Gamete Probabilities

Rule: Multiply the probabilities.

Example:

AaBb

What is the probability of an Ab gamete?

A = 1/2.

b = 1/2.

1/2 × 1/2 = 1/4.

🧬 9. Three-Locus Example

AaBbCc

What is the probability of ABc?

A = 1/2.

B = 1/2.

c = 1/2.

1/2 × 1/2 × 1/2 = 1/8.

⭐ EXAM SHORTCUT: Multiply the chance of receiving each required allele.

🔢 10. How Many Different Gametes?

Your professor gives:

Number of gamete types = 2ⁿ

Where n = number of heterozygous, independently assorting loci.

Examples

🧠 Shortcut

Count the gene pairs with two different letters.

Then calculate 2 raised to that number.

Example: AABbCc

AA = Homozygous, don't count.

Bb = Heterozygous, count 1.

Cc = Heterozygous, count 1.

n = 2.

2² = 4 gametes.

🧬 11. List the Gametes

AaBbCc

Eight possible gametes:

ABC, ABc, AbC, Abc, aBC, aBc, abC, abc.

AABBCc

Only two:

ABC and ABc.

🚨 EXAM TRAP: Do NOT count homozygous loci when calculating 2ⁿ.

🧪 PART 3 — TEST CROSSES (Slides 7–9)

🧬 12. What Is a Test Cross?

Definition: A cross with a homozygous recessive individual, used here to test independent assortment and linkage.

💡 Simple terms: Cross an organism with one that has two recessive alleles at both loci.

🪰 13. Professor's Fruit Fly Example

The lecture uses Drosophila melanogaster.

Two traits:

Eye color:

• SE = Red eyes.

• se = Sepia eyes.

Body color:

• BK = Normal body.

• bk = Black body.

Step 1 — Cross True-Breeding Parents

se se BK BK × SE SE bk bk

Step 2 — Find F1

All F1:

SE se BK bk

Phenotype:

Red eyes and normal body.

Step 3 — Perform the Test Cross

SE se BK bk × se se bk bk

The homozygous recessive parent produces only:

se bk

The heterozygous parent produces four gametes if the genes assort independently.

🧮 14. Test Cross Punnett Square

📊 15. Expected Test Cross Ratio

1 : 1 : 1 : 1

Each phenotype should represent approximately 25% of offspring.

⭐ MEMORIZE:

Dihybrid F1 × F1 → 9:3:3:1

Dihybrid test cross → 1:1:1:1

These ratios assume independent assortment and the relevant dominance relationships.

🚨 16. How to Recognize Linkage

If the genes assort independently:

Four phenotypes occur in approximately equal proportions.

If the genes are linked:

Parental combinations are more common than recombinant combinations.

💡 Simple terms: Linked genes tend to stay together because they are close to one another on the same chromosome.

📊 PART 4 — CHI-SQUARE TEST (Slides 10–18)

🧮 17. What Is a Chi-Square Test?

Definition: A statistical test comparing observed counts with expected counts.

💡 Simple terms: It helps determine whether differences between actual results and predicted results are larger than expected from random chance.

⭐ Formula to Memorize

χ² = Σ [(O − E)² / E]

Where:

• O = Observed number.

• E = Expected number.

• Σ = Add all results.

• χ² = Chi-square statistic.

🪰 18. Professor's Actual Example

400 offspring were observed.

Step 1 — Find Expected Counts

Under independent assortment:

Each phenotype should represent 1/4.

400 × 1/4 = 100 each.

Step 2 — Apply the Formula

Red, normal:

(101 − 100)² / 100

= 1/100

= 0.01

Sepia, normal:

(99 − 100)² / 100

= 1/100

= 0.01

Red, black:

(106 − 100)² / 100

= 36/100

= 0.36

Sepia, black:

(94 − 100)² / 100

= 36/100

= 0.36

Step 3 — Add Everything

χ² = 0.01 + 0.01 + 0.36 + 0.36

✅ χ² = 0.74

🔢 19. Degrees of Freedom

Your professor uses:

df = Number of categories − 1

Four categories:

df = 4 − 1

df = 3

📊 20. How to Interpret Chi-Square

For three degrees of freedom, your professor gives these reference values:

⭐ Decision Rule

If P is less than 0.05 → Reject the null hypothesis.

If P is 0.05 or greater → Fail to reject the null hypothesis.

💡 Simple terms:

• Small chi-square = Observed values are close to expected values.

• Large chi-square = Observed values differ substantially from expected values.

🧠 21. What Is the Null Hypothesis?

For this experiment:

H₀: The two loci assort independently.

If P is less than 0.05:

Reject H₀.

If P is 0.05 or greater:

Fail to reject H₀.

🪰 Professor's Conclusion

χ² = 0.74.

df = 3.

P greater than 0.70.

Therefore:

Fail to reject independent assortment.

The results do not provide significant evidence that the genes are linked.

🚨 EXAM TRAP: Failing to reject a hypothesis does NOT prove that it is true.

Your professor uses the word "accept," but "fail to reject" is the more precise statistical wording.

📚 22. Why Sample Size Matters (Slide 17)

Your professor emphasizes that statistical significance should not be interpreted using P-values alone.

Also consider:

• Sample size.

• Size of the observed effect.

• Biological importance.

⭐ Remember: A P-value alone does not tell you how important a biological effect is.

🔗 PART 5 — GENETIC LINKAGE (Slides 19–24)

🧬 23. What Is Genetic Linkage?

Definition: Genes located close together on the same chromosome tend to be inherited together.

💡 Simple terms: Two genes are physically close, so their alleles are more likely to stay together during inheritance.

🔗 24. Linked vs. Unlinked Genes

🪰 25. Professor's Second Fruit Fly Example

This time, your professor uses:

Eye color:

• SE = Red.

• se = Sepia.

Body color:

• EB = Normal.

• eb = Ebony.

Step 1 — Cross Parents

se se eb eb × SE SE EB EB

Step 2 — Find F1

All offspring:

SE se EB eb

Phenotype:

Red eyes, normal body.

Step 3 — Perform Test Cross

Female F1 × recessive male

SE se EB eb × se se eb eb

Step 4 — Expected Results Without Linkage

Four phenotypes:

• Red, normal = 25%.

• Sepia, normal = 25%.

• Sepia, ebony = 25%.

• Red, ebony = 25%.

📊 26. Actual Observed Results (Slide 22)

🧮 27. Calculate Chi-Square

Red, normal:

(123 − 100)² / 100

= 529/100

= 5.29

Sepia, normal:

(77 − 100)² / 100

= 529/100

= 5.29

Sepia, ebony:

(119 − 100)² / 100

= 361/100

= 3.61

Red, ebony:

(81 − 100)² / 100

= 361/100

= 3.61

Step 5 — Add

χ² = 5.29 + 5.29 + 3.61 + 3.61

✅ χ² = 17.80

📊 28. Interpret the Result

Degrees of freedom:

df = 4 − 1 = 3.

Critical χ² value at P = 0.05:

7.81.

Our result:

17.80 is greater than 7.81.

Therefore:

Reject independent assortment.

The data provide strong evidence of genetic linkage.

🧠 29. Why Are Some Phenotypes More Common?

The original parental combinations were:

• Red eyes, normal body.

• Sepia eyes, ebony body.

These are more common because linked alleles tend to stay together.

The less common phenotypes are:

• Sepia eyes, normal body.

• Red eyes, ebony body.

These are recombinant combinations.

🔄 30. Crossing Over

Crossing over: Exchange of chromosome segments between homologous chromosomes during meiosis.

Crossing over can separate linked alleles and create recombinant gametes.

💡 Simple terms: Linked genes usually travel together, but crossing over can separate them.

⭐ Exam Shortcut

Parental combinations = More common when linked.

Recombinant combinations = Less common when linked.

🧮 PART 6 — RECOMBINATION FREQUENCY (Extra Practice)

Your slides explain why recombinant offspring are less common but do not explicitly teach the recombination-frequency calculation. This extra formula is useful for understanding linkage problems.

🔗 31. Recombination Frequency

Formula:

Recombination frequency = Recombinant offspring ÷ Total offspring × 100%

Using your professor's linkage example:

Recombinant offspring:

77 + 81 = 158.

Total offspring:

400.

158/400 × 100 = 39.5%

📏 32. Genetic Map Distance

For relatively short genetic distances:

1% recombination ≈ 1 centimorgan (cM).

So the observed recombination frequency is:

39.5%, corresponding to an approximate map distance of 39.5 cM under the simple mapping approximation.

🚨 Important: At larger distances, multiple crossovers can make the actual map distance greater than the observed recombination percentage.

⭐ Remember

• 0% recombination = Complete linkage in the observed cross.

• Less than 50% = Evidence of linkage under appropriate conditions.

• 50% = Independent assortment or genes so far apart that they behave as unlinked.

🚻 PART 7 — SEX-LINKED INHERITANCE (Slides 25–26)

🧬 33. What Is Sex Linkage?

Definition: A gene is located on a sex chromosome.

Examples:

• X chromosome.

• Y chromosome.

Your professor focuses on X-linked inheritance.

💡 Simple terms: A trait is inherited differently depending on which sex chromosomes an individual has.

👨 34. Hemizygous

Hemizygous: Having only one copy of a gene rather than two.

In typical XY males, most X-linked genes have only one copy.

Therefore, a recessive X-linked allele can be expressed without a second copy.

👩 35. Female Carriers

In typical XX females, a person can have:

• Two normal alleles.

• One normal and one recessive allele.

• Two recessive alleles.

A heterozygous female may be a carrier of an X-linked recessive trait.

🩸 36. Examples of X-Linked Recessive Disorders

Your professor lists:

1. 🩸 Hemophilia.

2. 💪 Duchenne muscular dystrophy.

3. 👁 Red-green color blindness.

🧬 37. Carrier Mother × Unaffected Father

Mother:

XᴬXᵃ

Father:

XᴬY

Where:

• Xᴬ = Normal allele.

• Xᵃ = Recessive allele.

🧮 Punnett Square

📊 Results

• 👧 25% unaffected daughter (noncarrier).

• 👧 25% carrier daughter.

• 👦 25% unaffected son.

• 👦 25% affected son.

⭐ VERY IMPORTANT

Among sons:

50% affected, 50% unaffected.

Among daughters:

50% carriers, 50% noncarriers.

Among all children, assuming equal sex probabilities:

25% affected sons.

🚨 EXAM TRAP: Do not confuse the probability among sons with the probability among all children.

🧠 38. X-Linked Inheritance Rules

• Fathers pass their X chromosome to daughters, not sons.

• Fathers pass their Y chromosome to sons.

• Sons typically receive their X chromosome from their mothers.

• X-linked recessive traits are more common in XY males.

• Heterozygous XX females can be carriers.

⭐ Remember: A father cannot pass an X-linked allele directly to his son in typical XY inheritance.

🏆 PART 8 — ULTIMATE FORMULA CHEAT SHEET

🚨 PART 9 — 15 COMMON EXAM TRAPS

1. ❌ Dihybrid means one trait. ✅ Dihybrid involves two loci.

2. ❌ Every dihybrid cross produces 9:3:3:1. ✅ That ratio requires specific conditions.

3. ❌ Homozygous loci count toward 2ⁿ. ✅ Count heterozygous loci only.

4. ❌ AaBb produces two gametes. ✅ It produces four gamete types under independent assortment.

5. ❌ A test cross is always F1 × F1. ✅ It uses a homozygous recessive tester.

6. ❌ Test crosses produce 9:3:3:1. ✅ An independently assorting dihybrid test cross produces 1:1:1:1.

7. ❌ Observed and expected counts must be identical. ✅ Random variation produces differences.

8. ❌ Chi-square measures the probability that a hypothesis is true. ✅ It measures discrepancy; the P-value describes how unusual that discrepancy is under the null hypothesis.

9. ❌ A P-value greater than 0.05 proves independent assortment. ✅ It means insufficient evidence to reject it.

10. ❌ Linked genes are always inherited together. ✅ Crossing over can separate them.

11. ❌ Recombinant phenotypes are most common under linkage. ✅ Parental combinations are usually more common.

12. ❌ Genes on the same chromosome always show strong linkage. ✅ Distant genes can behave as unlinked.

13. ❌ Recombination frequency can exceed 50%. ✅ The observed recombination fraction cannot exceed 50% in the standard two-locus linkage model.

14. ❌ X-linked recessive traits affect only males. ✅ XX females can also be affected.

15. ❌ A carrier mother has a 50% chance of producing an affected child overall with an unaffected father. ✅ The chance is 25% overall, assuming equal sex probabilities, or 50% among sons.