Genetics practice problems new (practice problem)

🧬 BIOS 120 — GENETICS PRACTICE PROBLEMS

📚 Complete Exam Study Guide (6 Pages)

🎯 Goal: Understand and solve all 22 multiple-choice questions and all 3 open-ended problems from your professor's practice worksheet.

I'm keeping the same format as your previous notes: simple explanations, emojis, step-by-step calculations, shortcuts, exam traps, and practice questions. Everything is copy-and-paste friendly.

This worksheet tests genetics, inheritance, independent assortment, incomplete dominance, environmental effects, sex determination, and lethal alleles.

🧬 PART 1 — BASIC GENETICS (Questions 1–4)

🧬 1. Heterozygous at Two Loci (Question 1)

Question: Which genotype is heterozygous at two loci?

Answer: D — AaBb

💡 Simple terms: Heterozygous means having two different alleles at a gene locus.

• AA = Homozygous.

• Aa = Heterozygous.

• BB = Homozygous.

• Bb = Heterozygous.

AaBb is heterozygous at both loci.

⭐ EXAM SHORTCUT: Count how many gene pairs have one uppercase and one lowercase letter.

🌎 2. Gene-by-Environment Interactions (Question 2)

Answer: C — The statement that either nature or nurture dominates most traits is incorrect.

Gene-by-environment interaction: Different genotypes respond differently to environmental conditions.

💡 Simple terms: Genes and environment work together to influence traits.

Examples:

• 🌱 Plant height depends on genes and nutrients.

• 🧠 Human development depends on genes and environmental conditions.

• 🩺 Many medical conditions have genetic and environmental influences.

🚨 EXAM TRAP: Do not assume traits are controlled entirely by genetics OR entirely by environment.

🍌 3. Triploid Bananas (Question 3)

Answer: A — They cannot undergo normal meiosis and generally do not reproduce sexually.

Triploid: Having three sets of chromosomes (3n).

The Cavendish banana has 33 chromosomes.

Normal banana chromosome set:

n = 11.

Triploid banana:

3n = 33.

Why is triploidy important?

Three chromosome sets make normal chromosome pairing and separation during meiosis difficult.

Therefore, Cavendish bananas are generally sterile and propagated vegetatively.

⭐ MEMORIZE: Triploidy often causes sterility because meiosis cannot proceed normally.

🐭 4. Albino Mouse Cross (Question 4)

Answer: B — aa × AA → All Aa offspring.

The albino mouse is homozygous recessive:

aa

The brown mouse is homozygous dominant:

AA

Punnett square:

Result:

• 100% Aa.

• 100% brown phenotype.

🚨 EXAM TRAP: Alleles at the same locus must use the same letter, such as A and a, not A and B.

Important: The all-brown offspring are consistent with this cross, but observing 12 brown offspring alone does not prove the father was AA.

🎲 PART 2 — INDEPENDENT ASSORTMENT (Questions 5 and 10)

🧬 5. Mendel's Law of Independent Assortment (Question 5)

Answer: B — Exceptions to independent assortment were discovered after Mendel.

Independent assortment: Alleles at different loci are inherited independently when those loci assort independently.

💡 Simple terms: Inheriting one allele does not affect which allele you inherit at another unlinked locus.

Example:

AaBb can produce:

AB, Ab, aB, ab.

Each occurs at 25% if the loci assort independently.

🔗 Linkage

Genetic linkage: Genes close together on the same chromosome tend to be inherited together.

This can violate independent assortment.

🚨 EXAM TRAP: Not all genes on the same chromosome violate independent assortment. Genes far apart can behave as unlinked.

🔢 6. Number of Gametes (Question 10)

Question: How many gamete types can AaBbCCDD produce?

Answer: C — 4.

Formula:

\[ 2^n \]

Where n = Number of heterozygous loci.

Aa = Heterozygous (1).

Bb = Heterozygous (1).

CC = Homozygous (0).

DD = Homozygous (0).

n = 2.

2² = 4 gamete types.

They are:

ABCD, AbCD, aBCD, abCD.

⭐ EXAM SHORTCUT: Count heterozygous pairs, then calculate 2ⁿ.

🩸 PART 3 — BLOOD TYPES (Question 6)

🩸 7. ABO Blood Type Inheritance

Question: AB father × O mother. What is the probability of a type A child?

Answer: A — 1/2 (50%).

AB blood genotype:

IᴬIᴮ

O blood genotype:

ii

🧮 Punnett Square

Results:

• 50% Type A.

• 50% Type B.

• 0% Type AB.

• 0% Type O.

🧬 Codominance

Codominance: Both alleles are expressed in a heterozygote.

Example:

IᴬIᴮ → Type AB blood.

Both A and B antigens are expressed.

⭐ MEMORIZE: A and B are codominant; both are dominant over O.

🚻 PART 4 — SEX DETERMINATION (Questions 7–8)

🧬 8. Intersex Variations (Question 7)

Answer: E — B and C.

Intersex describes variations in sex characteristics, which can involve chromosomes, hormones, reproductive anatomy, or external genitalia.

Important points:

• Some intersex individuals have typical XX or XY chromosomes.

• Others have variations in sex chromosomes.

• Intersex traits do not automatically determine gender identity.

• Intersex status does not automatically require surgery.

⭐ EXAM TIP: Biological sex characteristics and gender identity are different concepts.

🐊 9. Chromosomal vs. Environmental Sex Determination (Question 8)

Answer: D — American alligators.

Chromosomal sex determination: Sex development is influenced by sex chromosomes.

Examples:

• 👨 Humans: XX/XY system.

• 🐦 Birds: ZZ/ZW system.

• 🦋 Butterflies: Often ZZ/ZW systems.

Temperature-dependent sex determination: Incubation temperature influences sex development.

Example:

🐊 American alligators.

⭐ MEMORIZE: Alligators = Temperature-dependent sex determination.

🧬 PART 5 — GENOME SIZE (Question 9)

🧬 10. Which Organism Has the Largest Genome?

Answer: C — Homo sapiens.

Among the choices in your worksheet:

• E. coli

• Drosophila melanogaster

• Homo sapiens

• Rice (Oryza sativa)

Humans have the largest genome of these four in terms of typical haploid nuclear DNA content.

📏 Approximate Genome Sizes

🚨 EXAM TRAP: Genome size does not necessarily correspond to organism complexity.

🦕 PART 6 — EVOLUTIONARY HISTORY (Questions 11–12)

🦴 11. Fossil Record (Question 11)

Answer: B — The fossil record supports evolution but also shows sudden patterns of extinction and replacement.

Fossil record: Preserved evidence of organisms from the past.

It shows:

• Evolutionary change over time.

• Extinction events.

• Appearance of new species.

• Periods of relative stability.

• Some gradual transitions.

⭐ EXAM TIP: Evolution does not always proceed at a constant rate.

👨‍🔬 12. Alfred Russel Wallace (Question 12)

Answer: A — An English naturalist who independently developed a theory of evolution by natural selection.

Alfred Russel Wallace independently developed the concept of natural selection.

His work helped prompt the joint presentation of Darwin's and Wallace's ideas in 1858.

⭐ MEMORIZE: Darwin and Wallace both developed the theory of natural selection.

🌸 PART 7 — INCOMPLETE DOMINANCE (Questions 13–16)

🌸 13. Four O'Clock Flower Genetics

Your professor gives:

• CᴿCᴿ = Red.

• CᴿCᵂ = Pink.

• CᵂCᵂ = White.

Two pink flowers are crossed:

CᴿCᵂ × CᴿCᵂ

🧮 Punnett Square

Results:

• 25% Red.

• 50% Pink.

• 25% White.

📝 Question 13

What is the probability a pollen grain contains Cᴿ?

Answer: B — 0.5.

A heterozygote produces:

50% Cᴿ gametes.

50% Cᵂ gametes.

📝 Question 14

What proportion of offspring are pink?

Answer: A — 1/2.

Two of four Punnett square boxes are pink.

📝 Question 15

Can pink flowers be true-breeding?

Answer: B — No.

Pink flowers are heterozygous.

Crossing two pink flowers produces red, pink, and white offspring.

📝 Question 16

What type of inheritance is this?

Answer: B — Incomplete dominance.

Incomplete dominance: The heterozygote has an intermediate phenotype.

💡 Simple terms: Neither allele completely masks the other, producing an intermediate appearance.

🧠 Incomplete Dominance vs. Codominance

🚨 EXAM TRAP: Incomplete dominance is not the same as codominance.

🪰 PART 8 — TWO-LOCUS FRUIT FLY GENETICS (Questions 17–20)

🧬 14. Understand the Scenario

Two fruit fly lines:

Line A: True-breeding wingless (apterous), normal eyes.

Line B: True-breeding eyeless, normal wings.

All 80 F1 offspring have normal eyes and normal wings.

🧠 Step 1 — Determine Dominance

Because all F1 have normal eyes and normal wings:

• Normal eyes = Dominant.

• Normal wings = Dominant.

• Eyeless = Recessive.

• Apterous = Recessive.

🧬 Step 2 — Assign Alleles

We'll use:

• E = Normal eyes.

• e = Eyeless.

• W = Normal wings.

• w = Apterous.

🧬 Step 3 — Parental Genotypes

Line A:

EEww

Line B:

eeWW

🧬 Step 4 — Cross Parents

EEww × eeWW

All F1:

EeWw

Phenotype:

Normal eyes and normal wings.

📝 Question 17

Which conditions are dominant?

Answer: C — Red-eyed and normal-winged, where the worksheet uses red-eyed to describe the normal-eye phenotype.

📝 Question 18

F1 × F1:

EeWw × EeWw

How many of 100 are eyeless?

Eyeless = ee.

Probability = 1/4.

100 × 1/4 = 25.

How many of 100 are eyeless AND apterous?

Eyeless = ee.

Apterous = ww.

Probability:

1/4 × 1/4 = 1/16.

100 × 1/16 = 6.25 expected offspring.

⭐ Important: 6.25 is an expected average, not a literal fraction of an individual fly.

📝 Question 19

F1 male × Line B female:

EeWw × eeWW

How many offspring are apterous?

Line B is WW.

Every offspring receives W.

Therefore, none can be ww.

Answer: E — 0.

📝 Question 20

The wording refers to the F2 offspring from the original F1 × F1 cross.

How many of 100 are eyeless AND apterous?

Probability:

1/4 × 1/4 = 1/16.

Answer: D — 6.25.

🚨 EXAM TRAP: If a question asks for two recessive traits, multiply their probabilities when the genes assort independently.

🌸 PART 9 — GENOTYPE × ENVIRONMENT (Questions 21–22)

🌱 15. Hydrangea Experiment 1

Your professor describes 66 genetically identical hydrangea clones.

Three nutrient environments:

📝 Question 21

What caused the differences in flower color?

Answer: D — Environmental causes.

All plants have identical genotypes.

Therefore, the flower-color differences within this clone group must reflect environmental influences rather than differences in genotype.

🌱 16. Hydrangea Experiment 2

A second hydrangea genotype is tested.

All 66 plants produce white flowers regardless of nutrient conditions.

📝 Question 22

What can we conclude from BOTH experiments?

Answer: C — Different genotypes exhibit different norms of reaction.

📈 17. Norm of Reaction

Definition: The range or pattern of phenotypes produced by a genotype across different environments.

💡 Simple terms: How one genotype responds when the environment changes.

Example:

Genotype A changes flower color depending on nutrients.

Genotype B remains white under all nutrient conditions.

🧬 18. Genotype-by-Environment Interaction

Definition: Different genotypes respond differently to environmental changes.

⭐ MEMORIZE: Same genotype + different environments can produce different phenotypes.

Different genotypes can also respond differently to the same environmental change.

🐟 PART 10 — LETHAL ALLELES (Open-Ended Question 3)

🐟 19. Blue Tilapia Saddleback Condition

Your professor gives:

• SS = Lethal.

• Ss = Saddleback.

• ss = Normal.

Two saddleback fish mate.

Ss × Ss

🧮 Punnett Square

🧬 Step 1 — Expected Offspring at Fertilization

• 25% SS → Die.

• 50% Ss → Saddleback.

• 25% ss → Normal.

🧠 Step 2 — Remove the Lethal Genotype

SS offspring do not survive to adulthood.

Surviving offspring:

• 2 Ss.

• 1 ss.

📊 Step 3 — Calculate Adult Ratio

2 : 1

Two-thirds of surviving adults have saddleback.

One-third are normal.

📝 Professor's Question

Of 100 surviving adults, how many have saddleback?

\[ 100\times\frac23=66.67 \]

✅ Answer: Approximately 67 saddleback adults.

🚨 EXAM TRAP: Do not answer 50. The question asks about surviving adults, not all fertilized eggs.

🏆 PART 11 — COMPLETE ANSWER KEY

🐟 Open-Ended Answers

1. Four O'Clock Flowers

Probability of Cᴿ pollen = 1/2.

Expected pink offspring = 1/2.

2. Fruit Flies

Dominant traits = Normal eyes and normal wings.

Out of 100 F2 offspring:

Eyeless = 25.

Eyeless and apterous = 6.25 expected.

F1 male × Line B female:

Apterous = 0.

3. Blue Tilapia

Approximately 67 of 100 surviving adults have saddleback.

🚨 PART 12 — 12 COMMON EXAM TRAPS

1. ❌ AaBB is heterozygous at two loci. ✅ Only one locus is heterozygous.

2. ❌ Environment has no influence on phenotype. ✅ Genes and environment can both affect phenotype.

3. ❌ Triploid organisms always reproduce normally. ✅ Triploidy often disrupts meiosis.

4. ❌ All genes assort independently. ✅ Linked genes can violate independent assortment.

5. ❌ AB × O produces type AB children. ✅ It produces type A or type B children.

6. ❌ Intersex always means abnormal chromosomes. ✅ Chromosomes may be typical or atypical.

7. ❌ All animals use sex chromosomes. ✅ Some use environmental sex determination.

8. ❌ Larger genome means more complex organism. ✅ Genome size and complexity are not directly proportional.

9. ❌ Pink flowers are homozygous. ✅ They are heterozygous in this example.

10. ❌ Eyeless and apterous offspring occur 25% of the time in EeWw × EeWw. ✅ They occur 1/16 = 6.25% of the time.

11. ❌ Genetically identical organisms always have identical phenotypes. ✅ Environmental differences can produce different phenotypes.

12. ❌ Lethal genotypes should be included when calculating proportions among surviving adults. ✅ Exclude offspring that died before adulthood.

🧮 PART 13 — ULTIMATE FORMULA CHEAT SHEET