CUET UG Biology Booster Test 2 Laws of Inheritance -

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CUET UG Biology Booster Test 2 Laws of Inheritance

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Biology Booster Test 2 Laws of Inheritance

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QUESTION 1 OF 20

Which observation best supports the concept of Discrete Unit Control?

Answer Review

1 Which observation best supports the concept of Discrete Unit Control?

A Appearance of intermediate heights in F1

B Stable, unchanged passage of "factors" over generations

C Blending of parental traits in the F2 generation

D Recovery of only one parental trait in the F2 generation

💡 Short Explanation

Mendel's "factors" are discrete, particulate units. They do not blend; they remain stable. This stability is the evidence for their discrete nature.

✅ Correct Answer Explanation

Mendel proposed that traits are controlled by discrete, particulate units (factors) rather than fluids that mix. The evidence for this is that traits reappear in their original form after successive generations, proving that the factors passed through generations unchanged.

❓ Why Other Options Are Incorrect

Option A → Intermediate phenotypes suggest blending or incomplete dominance, not discrete units.
Option C → Mendel's work explicitly refuted the theory of blending inheritance.
Option D → The recovery of both parental traits (not just one) in F2 is what proves particulate inheritance.

🎯 Strategy

Used: Elimination

Application: Eliminate options that describe blending or incomplete dominance.

Final Logic: Discrete units imply particles that stay intact; stability confirms this.

🧠 Memory Trick

Discrete = Distinct = Stable.

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Law of Dominance –

2 Match the following regarding Dissimilar Pair Interaction:

Column-I	Column-II
1. Tt genotype	p. Recessive factor
2. T allele	q. Heterozygous condition
3. t allele	r. Phenotypically tall
4. TT genotype	s. Dominant factor

A 1-q, 2-s, 3-p, 4-r

B 1-r, 2-s, 3-p, 4-q

C 1-q, 2-p, 3-s, 4-r

D 1-p, 2-q, 3-r, 4-s

💡 Short Explanation

Tt is heterozygous (q). T is dominant (s). t is recessive (p). TT is phenotypically tall (r).

✅ Correct Answer Explanation

Genotype Tt represents the heterozygous condition (1-q). The 'T' allele is the dominant factor (2-s), and the 't' allele is the recessive factor (3-p). The TT genotype results in a tall phenotype (4-r).

❓ Why Other Options Are Incorrect

B, C, and D incorrectly map genotypes/alleles to their biological properties.

🎯 Strategy

Used: Substitution

Application: Map each genetic term to its definition in the table.

Final Logic: Match heterozygosity with Tt, dominance with T, recessiveness with t, and TT with tall phenotype.

🧠 Memory Trick

T = Top (Dominant); t = Tiny (Recessive).

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Law of Dominance –

3 Why is Allele Separation in Meiosis considered a random process?

A Because it only happens in homozygous parents

B Because there is a 50 percent chance of a gamete containing either allele

C Because it requires the presence of three alleles

D Because the alleles always blend before separating

💡 Short Explanation

Segregation happens in meiosis (Anaphase I). Separation is stochastic (random). Each gamete has a 1/2 probability for either allele.

✅ Correct Answer Explanation

During Anaphase I of meiosis, the homologous chromosomes (carrying the alleles) move to opposite poles. Because orientation is random, there is an equal (50%) probability for a gamete to receive either the T or the t allele from a Tt parent.

❓ Why Other Options Are Incorrect

Option A → Segregation occurs in all diploid organisms, not just homozygous ones.
Option C → Gametes receive one allele from a pair of two, not three.
Option D → Alleles do not blend; they remain distinct.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Relate the mechanism of Anaphase I to the probability of gamete content.

Final Logic: Random alignment of bivalents = 50% probability per allele.

🧠 Memory Trick

Random = Equal chance = 50%.

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Law of Segregation –

4 Arrange the following for a Tt parent based on the Gamete Purity Principle:
1. T and t separate from each other
2. Meiosis occurs in the parent cell
3. Gametes formed with 50 percent T and 50 percent t
4. Pairing of T and t in the diploid cell

A 4-2-1-3

B 2-4-1-3

C 4-1-2-3

D 1-2-3-4

💡 Short Explanation

Start with diploid pairing (4). Meiosis begins (2). Alleles segregate (1). Gametes produced (3).

✅ Correct Answer Explanation

The cycle starts with the diploid state where T and t are paired (4). Meiosis initiates the division (2). During meiosis, the alleles segregate (1), and finally, gametes with a 50:50 distribution are formed (3).

❓ Why Other Options Are Incorrect

B, C, and D violate the chronological order of cell division and gametogenesis.

🎯 Strategy

Used: Substitution

Application: Order biological stages.

Final Logic: Pair → Divide → Separate → Result.

🧠 Memory Trick

Pairing before Parting.

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Law of Segregation –

5 Which of the following is NOT associated with Intermediate Phenotype Expression?

A Phenotypic ratio of 1:2:1

B Genotypic ratio of 1:2:1

C Complete dominance of one allele over another

D Ability to distinguish heterozygotes from homozygotes

💡 Short Explanation

Intermediate expression = Incomplete dominance. No allele is fully dominant. If dominance were complete, it would be a 3:1 ratio.

✅ Correct Answer Explanation

Incomplete dominance is defined by the absence of complete dominance. The F1 hybrid is an intermediate blend, which means neither allele fully masks the other. Statement C is therefore the false statement.

❓ Why Other Options Are Incorrect

Option A & B → Both ratios are 1:2:1 in Incomplete Dominance.

Option D → Since heterozygotes are "pink" and homozygotes are "red" or "white," they are distinguishable.

🎯 Strategy

Used: Elimination

Application: Identify the statement that describes Complete Dominance.

Final Logic: Incomplete dominance = No complete dominance.

🧠 Memory Trick

Incomplete = In-between.

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Incomplete Dominance –

6 Consider the statements regarding the Snapdragon Flower Example:
I. RR plants produce red flowers.
II. Rr plants produce pink flowers.
III. rr plants produce white flowers. Which statements correctly explain why the phenotypic ratio matches the genotypic ratio?

A I only

B II only

C I and III only

D I, II, and III because R is not completely dominant over r

💡 Short Explanation

RR = Red (Genotype 1). Rr = Pink (Genotype 2). rr = White (Genotype 3). 1:2:1 Phenotype = 1:2:1 Genotype.

✅ Correct Answer Explanation

Because the allele R is not completely dominant over r, the heterozygote (Rr) expresses a unique intermediate phenotype (pink). Thus, each genotype (RR, Rr, rr) corresponds to a unique phenotype (Red, Pink, White), causing the genotypic and phenotypic ratios to be identical (1:2:1).

❓ Why Other Options Are Incorrect

Options A, B, and C only provide partial reasons or incomplete sets of facts.

🎯 Strategy

Used: Option Grouping

Application: Group all correct statements into the final answer.

Final Logic: R being incompletely dominant creates the unique phenotypic expression for each genotype.

🧠 Memory Trick

1:2:1 matches because Rr is Pink.

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Incomplete Dominance –

7 In Simultaneous Allele Expression, why do IA IB individuals have AB blood type?

A Allele IA is dominant over IB

B Both alleles produce their own types of sugars

C The alleles blend to form a new sugar type

D Only one allele expresses in the plasma membrane

💡 Short Explanation

Co-dominance: Both alleles express. IA makes A sugar. IB makes B sugar. Result: AB.

✅ Correct Answer Explanation

In co-dominance (specifically blood group AB), neither allele is dominant or recessive. Instead, both IA and IB are fully expressed. IA produces A-type sugar and IB produces B-type sugar, resulting in the AB phenotype.

❓ Why Other Options Are Incorrect

Option A → They are co-dominant, not one dominant over the other.
Option C → No blending occurs.
Option D → Both alleles express, not just one.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Use the definition of Co-dominance (both express).

Final Logic: Co-dominance = Cooperation of expression.

🧠 Memory Trick

Co-dominance = Both display.

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Co-dominance –

Column I	Column II
1. IA	p. No sugar produced
2. IB	q. Sugar A produced
3. i	r. Sugar B produced
4. IA IB	s. Both sugars A and B produced

A 1-q, 2-r, 3-p, 4-s

B 1-p, 2-q, 3-r, 4-s

C 1-r, 2-s, 3-p, 4-q

D 1-s, 2-r, 3-q, 4-p

💡 Short Explanation

IA = A sugar (q). IB = B sugar (r). i = No sugar (p). IAIB = Both sugars (s).

✅ Correct Answer Explanation

The ABO system is biochemical: IA produces Sugar A (1-q), IB produces Sugar B (2-r), the allele i does not produce any functional sugar (3-p), and IAIB produces both (4-s).

❓ Why Other Options Are Incorrect

B, C, and D mismatch the allele to the sugar product.

🎯 Strategy

Used: Substitution

Application: Match allele to antigen produced.

Final Logic: IA=A, IB=B, i=none.

🧠 Memory Trick

i = invisible (no sugar).

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Co-dominance –

9 Consider the following regarding Population Level Variation:
I. Multiple alleles can only be detected in a group of individuals.
II. An individual human can carry IA, IB, and i simultaneously.
III. There are six possible genotypes for the ABO blood system. Which statements are correct?

A I and II only

B II and III only

C I and III only

D I, II, and III

💡 Short Explanation

I: True (Population trait). II: False (Diploid can only have two). III: True (Genotypes: IAIA, IAi, IBIB, IBi, IAIB, ii).

✅ Correct Answer Explanation

Statement I is correct because multiple alleles cannot be fully represented in one diploid individual. Statement III is correct; the genotypes are $IAIA, IAi, IBIB, IBi, IAIB, ii$ (total 6). Statement II is incorrect as an individual can only have two of the three alleles.

❓ Why Other Options Are Incorrect

Options A, B, and D include the factually incorrect statement II.

🎯 Strategy

Used: Elimination

Application: Eliminate the statement claiming diploid organisms carry three alleles.

Final Logic: Diploid = max 2 alleles; 3 alleles exist at population level.

🧠 Memory Trick

3 Alleles, 6 Genotypes.

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Multiple Alleles –

10 Which is NOT associated with the Gene I Allele Trio?

A Dominance of IA over i

B Co-dominance of IA and IB

C Presence of four alleles in a single diploid individual

D Recessive nature of i

💡 Short Explanation

There are only 3 alleles (IA, IB, i). A diploid individual has only 2. Four alleles in one person is biologically impossible.

✅ Correct Answer Explanation

Gene I has three alleles. A diploid organism has only two alleles at a locus. The claim that a single individual has "four alleles" is incorrect on two counts: the number of alleles in the system is three, and the individual can only hold two.

❓ Why Other Options Are Incorrect

Option A, B, and D are standard genetic facts for the Gene I system.

🎯 Strategy

Used: Elimination

Application: Identify the statement that contradicts the basic rules of genetics (diploidy).

Final Logic: 4 alleles is impossible in both population count (3) and diploid capacity (2).

🧠 Memory Trick

Diploid = 2, Trio = 3.

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Multiple Alleles –

11 Which is NOT associated with Two Character Inheritance?

A Crossing plants that differ in two traits

B Independent segregation of seed color and shape

C A phenotypic ratio of 3:1 in the F2 generation

D Use of symbols like RRYY and rryy

💡 Short Explanation

Two characters = Dihybrid. Ratio is 9:3:3:1. 3:1 is for monohybrid.

✅ Correct Answer Explanation

A dihybrid cross involves two characters and produces a 9:3:3:1 phenotypic ratio in the F2 generation. A 3:1 ratio is strictly associated with a monohybrid cross.

❓ Why Other Options Are Incorrect

Options A, B, and D are standard features/methods of a dihybrid cross.

🎯 Strategy

Used: Elimination

Application: Use the standard ratios for dihybrid inheritance.

Final Logic: 3:1 = Monohybrid, 9:3:3:1 = Dihybrid.

🧠 Memory Trick

Dihybrid = 4 phenotypes = 9:3:3:1.

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Dihybrid Cross –

12 Arrange the following Round-Yellow vs Wrinkled-Green genotypes in order from P to F2:
1. RrYy (F1)
2. RRYY x rryy (Parental)
3. RRYY, RrYy, rrYY, etc. (F2 combinations)
4. RY and ry gametes

A 2-4-1-3

B 4-2-1-3

C 2-1-4-3

D 1-2-3-4

💡 Short Explanation

Start: Parents (2). Gametes form (4). F1 formed (1). F2 formed (3).

✅ Correct Answer Explanation

The sequence is: Parental cross (RRYY x rryy) (2) → Gamete formation (RY and ry) (4) → F1 Hybrid (RrYy) (1) → F2 combinations (3).

❓ Why Other Options Are Incorrect

B, C, and D do not follow the biological progression of the cross.

🎯 Strategy

Used: Substitution

Application: Order the logical steps of the dihybrid experiment.

Final Logic: Parents → Gametes → F1 → F2.

🧠 Memory Trick

Parents, Gametes, F1, F2.

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Dihybrid Cross –

📄 Passage

"In the dihybrid cross, the phenotypes round, yellow; wrinkled, yellow; round, green and wrinkled, green appeared in the ratio 9:3:3:1... This proves that the segregation of one pair of characters is independent of the other pair."

Based on the passage, the Phenotypic Ratio 9:3:3:1 is proof of:

A Linkage between seed color and shape

B The Law of Independent Assortment

C The Law of Purity of Gametes only

D Incomplete dominance in dihybrids

💡 Short Explanation

9:3:3:1 is the classic ratio for independent assortment. If linked, the ratio would be different.

✅ Correct Answer Explanation

The 9:3:3:1 ratio is the mathematical signature of independent assortment. It shows that characters (seed color and shape) segregate independently of one another during gamete formation.

❓ Why Other Options Are Incorrect

Option A → Linkage would cause deviation from 9:3:3:1.
Option C → While related, 9:3:3:1 specifically demonstrates Independent Assortment.
Option D → This is a dihybrid cross with complete dominance.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Directly map the provided ratio to Mendel's Law.

Final Logic: Ratio 9:3:3:1 = Independent Assortment.

🧠 Memory Trick

9:3:3:1 = Independent!

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Law of Independent Assortment –

📄 Passage

Based on the passage, the series (3:1) x (3:1) implies that:

A The traits are physically attached

B Each trait behaves as if it were in a monohybrid cross

C Yellow is always linked to Round

D Green is always linked to Wrinkled

💡 Short Explanation

Mathematical expansion of (3:1)(3:1) = 9:3:3:1. This is valid only if traits are independent. Each trait follows its own 3:1 law.

✅ Correct Answer Explanation

The multiplication of the monohybrid ratios (3:1) demonstrates that the probability of inheriting one trait is independent of the other. It implies that both characters behave according to the laws established in their respective monohybrid crosses.

❓ Why Other Options Are Incorrect

Option A, C, and D suggest linkage, which is the opposite of the mathematical independence shown by the ratio multiplication.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Apply probability multiplication rules to independent events.

Final Logic: Mathematical independence = Biological independence (monohybrid behavior).

🧠 Memory Trick

3:1 * 3:1 = 9:3:3:1.

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Law of Independent Assortment –

15 Which is NOT associated with Independent Pair Segregation?

A Meiosis

B RrYy hybrids producing 4 types of gametes

C Genes located very close on the same chromosome (Tightly linked)

D 25 percent frequency for each gamete type

💡 Short Explanation

Independent assortment requires genes to be far apart or on different chromosomes. Close linkage prevents independent assortment.

✅ Correct Answer Explanation

Independent assortment occurs when genes are on different chromosomes or far apart on the same chromosome. Tightly linked genes do not assort independently; they tend to stay together during inheritance.

❓ Why Other Options Are Incorrect

Options A, B, and D are standard properties of independent assortment.

🎯 Strategy

Used: Elimination

Application: Use the definition of linkage to contradict independent assortment.

Final Logic: Linkage = Deviation from independent assortment.

🧠 Memory Trick

Linked = Together (not independent).

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Linkage and Recombination –

16 In Hybrid Permutation Diversity, if 50 percent of gametes have R and 50 percent have r, what determines if a gamete gets Y or y?

A The R allele chooses the Y allele

B The segregation of Y/y is independent of R/r

C Dominant alleles always stay together

D Recessive alleles are excluded from R-bearing gametes

💡 Short Explanation

Law of Independent Assortment. Alleles at different loci sort independently. R/r has no influence on Y/y.

✅ Correct Answer Explanation

The Law of Independent Assortment ensures that the distribution of one pair of alleles (Y/y) into gametes occurs without influence from the distribution of another pair (R/r).

❓ Why Other Options Are Incorrect

Options A, C, and D suggest genetic "choosing" or dependency, which is false.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Apply the definition of the Law of Independent Assortment.

Final Logic: Independent means no choice/linkage involved.

🧠 Memory Trick

Independent = Autonomous.

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Law of Independent Assortment –

17 How does Graphical Probability Calculation help in genetics?

A It measures the physical size of the genes

B It calculates the frequency of all possible genotypes of offspring

C It prevents mutations from occurring

D It replaces the need for actual breeding experiments

💡 Short Explanation

Punnett Square = Graphical calculator. Predicts probability of outcomes.

✅ Correct Answer Explanation

The Punnett Square is a graphical probability tool. It allows geneticists to quickly derive all possible genotypic and phenotypic combinations and their frequencies in the offspring based on parental gametes.

❓ Why Other Options Are Incorrect

Option A → Genes are too small; squares don't measure size.
Option C → Calculation cannot change biological mutation rates.
Option D → It is a tool for breeding experiments, not a replacement.

🎯 Strategy

Used: Substitution

Application: Identify the purpose of a Punnett square from the list.

Final Logic: Graphical = Calculation of combinations/frequency.

🧠 Memory Trick

Square = Predicts.

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Punnett Square –

18 In Zygote Composition Derivation, what is the number of boxes in a dihybrid Punnett Square?

A 4

B 8

C 16

D 32

💡 Short Explanation

Dihybrid: 4 gamete types x 4 gamete types. 4 x 4 = 16.

✅ Correct Answer Explanation

In a dihybrid cross, each parent produces 4 types of gametes ($RY, Ry, rY, ry$). Arranging these in a grid creates a $4 \times 4$ Punnett square, resulting in 16 boxes.

❓ Why Other Options Are Incorrect

Options A, B, and D do not correspond to the required dimensions for a dihybrid grid.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Calculate $n^2$ where $n$ is the number of gamete types.

Final Logic: $4 \text{ gametes} \times 4 \text{ gametes} = 16 \text{ boxes}$.

🧠 Memory Trick

Dihybrid = 4x4 = 16.

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Dihybrid Cross –

19 Consider the following regarding Recessive Parent Crossing:
I. It is used to identify the genotype of a dominant-looking plant.
II. If the plant is homozygous TT, all offspring will be tall.
III. If the plant is heterozygous Tt, 50 percent of offspring will be dwarf. Which statements are correct?

A I and II only

B II and III only

C I and III only

D I, II, and III

💡 Short Explanation

I: True (Test cross purpose). II: True (TT x tt = 100% Tt). III: True (Tt x tt = 50% Tt, 50% tt).

✅ Correct Answer Explanation

Statement I describes the test cross correctly. Statement II is correct because TT x tt yields all tall offspring. Statement III is correct because Tt x tt yields 50% tall (Tt) and 50% dwarf (tt) offspring.

❓ Why Other Options Are Incorrect

Options A, B, and C are incomplete.

🎯 Strategy

Used: Option Grouping

Application: Verify all three statements as biologically correct.

Final Logic: All three are standard test cross results.

🧠 Memory Trick

Test Cross: Homo = All Tall, Hetero = 1:1 Tall/Dwarf.

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Test Cross –

20 For Unknown Genotype Prediction, if a test cross of a violet flower results in 50 percent violet and 50 percent white flowers, what is the genotype of the parent?

A VV

B Vv

C vv

D Cannot be determined

💡 Short Explanation

Result 1:1 (Violet:White). Test cross produces 1:1 if parent is Heterozygous (Vv). Test cross produces 100% Dominant if parent is Homozygous (VV).

✅ Correct Answer Explanation

In a test cross (V_ x vv), if 50% of the offspring show the recessive trait (white, vv), the dominant-looking parent must have provided a 'v' allele. Therefore, the parent must be heterozygous (Vv).

❓ Why Other Options Are Incorrect

Option A (VV) would result in 100% violet offspring.
Option C (vv) is the recessive test-parent itself.
Option D is wrong because the data is sufficient.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Use the test cross ratio to solve for the unknown parent.

Final Logic: Offspring = 50% recessive (vv) requires parent to be Vv.

🧠 Memory Trick

50/50 split = Heterozygote (Vv).

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Test Cross –

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