CUET UG Biology Booster Test 3 Laws of Inheritance -

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CUET UG Biology Booster Test 1 Laws of Inheritance

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Biology Booster Test 1 Laws of Inheritance

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QUESTION 1 OF 20

Why is the Discrete Unit Control theory essential to explain the recovery of the dwarf trait in the F2 generation?

Answer Review

1 Why is the Discrete Unit Control theory essential to explain the recovery of the dwarf trait in the F2 generation?

A It shows that traits blend in the F1 and un-blend in F2

B It ensures the recessive factor is preserved and not lost in the F1

C It proves that factors only exist in pairs in the F2 generation

D It explains why only dominant traits can ever be expressed

💡 Short Explanation

Discrete units (factors) maintain their integrity. Recessive factors remain hidden but present in F1 (heterozygotes). F2 recovery confirms they were never lost.

✅ Correct Answer Explanation

The theory of Discrete Unit Control (particulate inheritance) posits that traits are governed by distinct factors that do not blend. In the F1 generation (Tt), the recessive factor 't' is masked by 'T' but remains physically present. Its reappearance in F2 (tt) proves that the recessive factor was preserved, not destroyed or lost, during the F1 stage.

❓ Why Other Options Are Incorrect

Option A → F1 phenotypes do not "blend"; they express the dominant trait.
Option C → Factors exist in pairs in all diploid stages (P, F1, F2), not just F2.
Option D → Recessive traits can be expressed when homozygous.

🎯 Strategy

Used: Elimination

Application: Eliminate options describing blending or incorrect inheritance patterns.

Final Logic: Preservation of recessive alleles explains phenotypic reappearance.

🧠 Memory Trick

Preserved = Present = Reappears.

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

2 Which is NOT associated with Dissimilar Pair Interaction at a molecular level (per NCERT enzyme example)?

A Unmodified allele produces a normal enzyme

B Modified allele may produce no enzyme at all

C Modified allele is usually the dominant one

D Dominant phenotype depends on the functioning unmodified allele

💡 Short Explanation

Modified alleles often lose function (recessive). Normal (unmodified) alleles are usually dominant. Functioning enzymes drive the dominant phenotype.

✅ Correct Answer Explanation

NCERT explains that the dominant phenotype reflects the activity of the functional, unmodified allele. A modified allele (often the result of mutation) usually becomes non-functional or produces a different/less efficient enzyme, which constitutes a recessive allele. Therefore, the modified allele is rarely the dominant one.

❓ Why Other Options Are Incorrect

Option A → Correct: The unmodified allele is functional.
Option B → Correct: Many recessive alleles are null mutations (loss of function).
Option D → Correct: The "normal" enzyme determines the dominant phenotype.

🎯 Strategy

Used: Odd One Out

Application: Identify the false statement regarding dominant/recessive allele function.

Final Logic: Modified = Usually Recessive; Unmodified = Dominant.

🧠 Memory Trick

Unmodified = Normal = Dominant.

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

3 Consider these statements on Allele Separation in Meiosis:
I. Separation occurs such that a gamete receives both alleles.
II. It is the basis for the lack of blending in inheritance.
III. It applies to both homozygous and heterozygous parents. Which statements are analytically correct?

A I only

B II only

C II and III only

D I, II, and III

💡 Short Explanation

I: Incorrect (Gamete gets only ONE allele). II: Correct (Segregation keeps factors distinct). III: Correct (Applicable to any diploid).

✅ Correct Answer Explanation

Statement I is wrong because segregation ensures a gamete receives only one of the two alleles. Statement II is correct as segregation prevents the merging of factors (no blending). Statement III is correct because meiosis and allele segregation occur in all diploid organisms regardless of their zygosity.

❓ Why Other Options Are Incorrect

Option A, B, and D include the incorrect statement I or omit correct statements.

🎯 Strategy

Used: Elimination

Application: Eliminate any option containing statement I.

Final Logic: Segregation = 1 factor per gamete.

🧠 Memory Trick

Segregation = Separation (of pair).

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

4 Match the following regarding the Gamete Purity Principle:

Column-I	Column-II
1. Homozygous parent	p. Two types of gametes (1:1)
2. Heterozygous parent	q. 100 percent similar gametes
3. Meiosis	r. Biological mechanism for segregation
4. Fertilization	s. Restores the paired state in the zygote

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

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

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

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

💡 Short Explanation

Homozygous (AA) = 1 type (q). Heterozygous (Aa) = 2 types (p). Meiosis = Segregation (r). Fertilization = Fusion (s).

✅ Correct Answer Explanation

Homozygous parents produce identical gametes (1-q). Heterozygous parents produce two types of gametes in equal proportions (2-p). Meiosis is the cellular process facilitating allele segregation (3-r). Fertilization brings gametes together to restore the diploid (paired) state (4-s).

❓ Why Other Options Are Incorrect

Options B, C, and D contain incorrect mapping of biological processes to genetic outcomes.

🎯 Strategy

Used: Substitution

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

Final Logic: Homos produce one type; Heteros produce two; Meiosis separates; Fertilization unites.

🧠 Memory Trick

Homo=Same (q), Hetero=Split (p).

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

5 In Intermediate Phenotype Expression, why does the F2 show a 1:2:1 phenotypic ratio?

A Because the alleles are physically linked

B Because the heterozygous genotype Rr has a unique phenotype

C Because the recessive allele is lost during F1 development

D Because three different genes are involved

💡 Short Explanation

Incomplete dominance: Rr ≠ RR or rr. 1 (RR) : 2 (Rr) : 1 (rr). Phenotype matches genotype due to the intermediate nature.

✅ Correct Answer Explanation

In complete dominance, the heterozygote (Rr) looks like the dominant homozygote (RR), leading to a 3:1 phenotypic ratio. In incomplete dominance, Rr has a distinct intermediate phenotype, meaning every genotype (RR, Rr, rr) creates a distinct phenotype, resulting in a 1:2:1 ratio.

❓ Why Other Options Are Incorrect

Option A → Linkage affects inheritance ratios differently, not 1:2:1.
Option C → Alleles are never lost in incomplete dominance.
Option D → This is a monogenic trait example.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Match the cause (unique Rr phenotype) to the effect (1:2:1 ratio).

Final Logic: Genotype-phenotype correspondence in heterozygotes creates the 1:2:1 ratio.

🧠 Memory Trick

1:2:1 because 2 are intermediate.

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

6 Which is NOT associated with the Snapdragon Flower Example?

A Rr genotype results in pink flowers

B Reversion to parental 3:1 phenotypic ratio in the F2

C RR genotype results in red flowers

D Evidence that alleles do not blend despite the intermediate phenotype

💡 Short Explanation

F2 ratio in Snapdragon is 1:2:1. 3:1 is a characteristic of complete dominance. Option B is false.

✅ Correct Answer Explanation

The F2 generation of an incomplete dominance cross displays a 1:2:1 phenotypic ratio (1 Red : 2 Pink : 1 White). A 3:1 ratio is only found in cases of complete dominance where the heterozygote is indistinguishable from the dominant homozygote.

❓ Why Other Options Are Incorrect

Options A, C, and D are accurate descriptions of the snapdragon experiment.

🎯 Strategy

Used: Elimination

Application: Use the phenotypic ratio for Incomplete Dominance to debunk option B.

Final Logic: 1:2:1 is NOT 3:1.

🧠 Memory Trick

Snapdragon = 1:2:1 (No 3:1).

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

📄 Passage

"In co-dominance... when IA and IB are present together they both express their own types of sugars: this is because of co-dominance... IA and IB are completely dominant over i."

Based on the passage, Simultaneous Allele Expression differs from incomplete dominance because:

A It only happens in plant species

B Both parental traits are expressed fully rather than forming an intermediate blend

C It follows a standard Mendelian 3:1 ratio

D One allele always masks the presence of the other

💡 Short Explanation

Co-dominance: Both expressed (A and B sugars). Incomplete dominance: Blend (Pink). Difference is "full expression" vs "blend."

✅ Correct Answer Explanation

The core distinction is that co-dominance results in the simultaneous expression of both parental phenotypes (A and B sugars) without any mixing or blending, whereas incomplete dominance produces a new, intermediate phenotype that is a mixture of the parental traits.

❓ Why Other Options Are Incorrect

Option A → Both can occur in many species.
Option C → Neither strictly follows 3:1 in the heterozygous phenotype manifestation.
Option D → Masking occurs in complete dominance, not co-dominance.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Compare the mechanism of co-dominance (both) with incomplete dominance (blend).

Final Logic: "Both expressed" vs "Intermediate blend."

🧠 Memory Trick

Co-dominance = Cooperation (Both).

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

📄 Passage

"In co-dominance... when IA and IB are present together they both express their own types of sugars: this is because of co-dominance... IA and IB are completely dominant over i."

Based on the passage, the ABO Blood Group Example demonstrates that:

A Dominance is an autonomous feature of a gene

B A single gene can have more than two alleles in a population

C The allele i is co-dominant with IA

D Only two phenotypes are possible in the human blood system

💡 Short Explanation

Gene I has IA, IB, and i. This illustrates multiple alleles.

✅ Correct Answer Explanation

The ABO system is the classic example of multiple alleles, showing that at the population level, a single gene (Gene I) can exist in three allelic forms (IA, IB, i), which is more than the usual two (diploid) limit.

❓ Why Other Options Are Incorrect

Option A → Dominance is not autonomous; IA/IB are co-dominant, IA/i dominant.
Option C → i is recessive to IA.
Option D → Four phenotypes are possible (A, B, AB, O).

🎯 Strategy

Used: Elimination

Application: Use the definition of multiple alleles.

Final Logic: Three alleles = Multiple alleles.

🧠 Memory Trick

3 Alleles = Multiple Allelism.

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

9 Match the components of Population Level Variation:

Column-I	Column-II
1. Multiple alleles	p. Found within a single individual
2. Two alleles	q. Found in population studies
3. Three alleles	r. ABO system trio (IA, IB, i)
4. Six genotypes	s. Result of three alleles in a population

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

Multiple alleles (1) are matched with the ABO system trio (r). Two alleles (2) are matched with the result of three alleles in a population (s). Three alleles (3) are matched with those found within a single individual (p). Six genotypes (4) are matched with population studies (q).

✅ Correct Answer Explanation

According to the given matching arrangement: Multiple alleles (1) correspond to the ABO system trio (IA, IB, i) (r). Two alleles (2) correspond to the result of three alleles in a population (s). Three alleles (3) correspond to found within a single individual (p). Six genotypes (4) correspond to found in population studies (q). Thus, the required matching is 1-r, 2-s, 3-p, 4-q, which corresponds to Option C.

❓ Why Other Options Are Incorrect

Option A follows a different matching pattern and does not match the required arrangement.
Option B incorrectly pairs multiple alleles with a single individual.
Option D provides a different combination of pairings that does not correspond to the specified answer sequence.

🎯 Strategy

Used: Substitution

Application:

Compare each option with the required matching pattern and identify the one that reproduces all four pairings exactly.

Final Logic:

Only Option C contains 1-r, 2-s, 3-p, 4-q.

🧠 Memory Trick

Trio → Result → Individual → Population

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

10 Arrange the Gene I genotypes from most dominant sugar production to least:
1. ii (No sugar)
2. IA IA (Sugar A)
3. IA IB (Both sugars)
4. IB i (Sugar B)

A 3 - 2/4 - 1

B 1 - 4 - 3 - 2

C 4 - 2 - 3 - 1

D 1 - 2 - 3 - 4

💡 Short Explanation

Most dominant = IAIB (Both sugars). Intermediate = IAIA or IBi (One sugar). Least = ii (None).

✅ Correct Answer Explanation

Ranked by sugar production quantity/complexity: (3) IAIB produces both A and B sugars (maximum complexity). (2) IAIA and (4) IBi produce only one type of sugar each (A or B, respectively). (1) ii produces no sugar (minimum). Thus, 3 > (2=4) > 1.

❓ Why Other Options Are Incorrect

B, C, and D misrepresent the biological sugar production hierarchy.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Rank based on sugar expression (Both > One > None).

Final Logic: Both sugars > Single sugar > Zero sugar.

🧠 Memory Trick

Both > One > None.

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

11 In Two Character Inheritance, if the two genes were tightly linked, how would the results differ from Mendel's?

A The F2 ratio would still be 9:3:3:1

B Parental combinations would be much higher than non-parental types

C There would be no F1 generation produced

D Only one character would be inherited

💡 Short Explanation

Independent assortment = Recombinants = Parentals. Linkage = Recombinants < Parentals. Parents stay linked on the chromosome.

✅ Correct Answer Explanation

Mendel's independent assortment relies on genes being on different chromosomes. Linkage means genes are physically close on the same chromosome and are inherited together. This leads to a significantly higher proportion of parental phenotypes compared to recombinant (non-parental) types.

❓ Why Other Options Are Incorrect

Option A → Linkage causes a departure from 9:3:3:1.
Option C → F1 will still form.
Option D → Both traits are inherited, but they are linked.

🎯 Strategy

Used: Elimination

Application: Distinguish between the results of Independent Assortment and Linkage.

Final Logic: Linkage = High parental frequency, low recombinant frequency.

🧠 Memory Trick

Linked = Stick together.

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

12 Which is NOT associated with the Round-Yellow vs Wrinkled-Green setup?

A Use of true-breeding lines

B Application of mathematical logic

C Blending of yellow and green to form blue seeds in F1

D Large sampling size for credibility

💡 Short Explanation

F1 is Round-Yellow (Dominant traits only). There is no blending. There is no "blue" seed production.

✅ Correct Answer Explanation

Mendel's work with dihybrids specifically refuted blending inheritance. F1 plants are uniformly Round and Yellow; there is no blending, and definitely no "blue seeds."

❓ Why Other Options Are Incorrect

Options A, B, and D are standard pillars of the Mendelian experimental method.

🎯 Strategy

Used: Elimination

Application: Eliminate the statement that contradicts Mendel's core findings (No blending).

Final Logic: Blending = Rejected; Blue = False.

🧠 Memory Trick

No blending, no blue.

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

13 How is the 9:3:3:1 ratio analytically related to the Law of Segregation?

A It proves segregation only happens for one gene pair

B It shows each pair of alleles segregates independently as a 3:1 ratio

C It indicates segregation is replaced by assortment

D It proves alleles blend in dihybrid crosses

💡 Short Explanation

9:3:3:1 is the product of (3:1) and (3:1). Each bracket represents the Law of Segregation for one trait.

✅ Correct Answer Explanation

The Law of Segregation states that each gene pair follows a 3:1 phenotypic ratio. Independent assortment is the simultaneous segregation of two such pairs, which, when multiplied (3:1) * (3:1), yields the 9:3:3:1 ratio.

❓ Why Other Options Are Incorrect

Option A → Segregation happens for both.
Option C → Assortment adds to segregation, it doesn't replace it.
Option D → Blending does not occur.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Decompose 9:3:3:1 into two 3:1 monohybrid components.

Final Logic: 9:3:3:1 = (3:1) × (3:1) (Product of two Segregation events).

🧠 Memory Trick

Dihybrid = Segregation × Segregation.

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

14 In the Mathematical Combination Series, what does the (1/4) term in the expansion of (1/2T + 1/2t)^2 represent?

A Probability of a heterozygous zygote

B Probability of a homozygous zygote (TT or tt)

C Probability of a single gamete

D Total number of offspring in F2

💡 Short Explanation

(1/2T + 1/2t)^2 = 1/4TT + 1/2Tt + 1/4tt. 1/4 is the frequency for TT and tt (homozygotes).

✅ Correct Answer Explanation

When expanding the binomial $(1/2T + 1/2t)^2$, the terms are $1/4TT + 2/4Tt + 1/4tt$. The $1/4$ frequency corresponds to the probability of the homozygous dominant ($TT$) and homozygous recessive ($tt$) genotypes.

❓ Why Other Options Are Incorrect

Option A → Probability of heterozygous (Tt) is $2/4$ or $1/2$.
Option C → Gamete probability is $1/2$.
Option D → $1/4$ is a frequency, not a total count.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Expand the binomial to identify the components.

Final Logic: $1/4 + 2/4 + 1/4 = 1$. $1/4$ = homozygote.

🧠 Memory Trick

1/4 homozygote (1/4 + 1/4 = 1/2).

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

15 Arrange the cytological events that explain Independent Pair Segregation:
1. Alignment of two chromosome pairs at the metaphase plate
2. Meiosis I anaphase segregation
3. Formation of four different chromosome combinations in germ cells
4. Independent orientation of different chromosome pairs

A 1-4-2-3

B 4-1-2-3

C 1-2-3-4

D 2-1-4-3

💡 Short Explanation

Align (1) → Orient independently (4) → Segregate (2) → Resulting combinations (3).

✅ Correct Answer Explanation

The sequence: Chromosome pairs align at the plate (1). Due to independent orientation (4), they segregate (2) and result in four unique gametic combinations (3).

❓ Why Other Options Are Incorrect

B, C, and D misorder the mechanistic steps of Meiosis I.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Sequence the stages of independent segregation during meiosis.

Final Logic: Align → Orient → Separate → Result.

🧠 Memory Trick

Align, Orient, Separate, Result.

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

16 Which is NOT associated with Hybrid Permutation Diversity in RrYy?

A 4 types of pollen

B 4 types of eggs

C Each gamete type having 25 percent frequency

D R/r always segregating with Y/y respectively

💡 Short Explanation

Independent assortment means they assort independently. They don't "segregate with" each other. Option D is false.

✅ Correct Answer Explanation

Independent assortment means that R and r alleles are randomly combined with Y and y alleles. They do not move as a pair (segregating "with" each other).

❓ Why Other Options Are Incorrect

Option A, B, and C describe standard, correct Mendelian behavior.

🎯 Strategy

Used: Elimination

Application: Eliminate the statement implying dependent assortment (linkage).

Final Logic: "With" = Linkage; Independent Assortment = Without.

🧠 Memory Trick

No "with," only "independent."

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

17 Consider these statements on Graphical Probability Calculation:
I. It results in a square output form.
II. It allows for calculation of genotypic ratios.
III. It was developed to explain chromosomal movement. 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 (It's a square/grid). II: True (Predicts genotypes). III: False (Developed for probability, chromosomal theory came later).

✅ Correct Answer Explanation

Statements I and II are correct characteristics of the Punnett Square. Statement III is false; the Punnett Square is a statistical/mathematical probability tool, not a cytological model created to explain chromosomal movement (which was the Chromosomal Theory of Inheritance by Sutton and Boveri).

❓ Why Other Options Are Incorrect

Options B, C, and D contain the incorrect statement III.

🎯 Strategy

Used: Elimination

Application: Distinguish between statistical probability tools and cytological theories.

Final Logic: Punnett = Probability, not Chromosomal Movement Theory.

🧠 Memory Trick

Square = Probability calculation.

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

18 In Zygote Composition Derivation, crossing TtYy x TtYy forms how many different GENOTYPES?

A 4

B 9

C 16

D 7

💡 Short Explanation

$3 \times 3 = 9$. Punnett square grid has 16 boxes, but only 9 distinct genotype types (e.g., TTYY, TTYy, etc.).

✅ Correct Answer Explanation

For a dihybrid cross ($TtYy \times TtYy$), the number of genotypic classes is calculated as $3^n$ where $n$ is the number of gene pairs. Since $n=2$, $3^2 = 9$. These represent all possible combinations of the genotypes for T and Y traits.

❓ Why Other Options Are Incorrect

Option A, C, and D are mathematically incorrect for a dihybrid genotypic count.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Apply the rule $3^n$ for the number of genotypes in a cross.

Final Logic: $3 \times 3 = 9$ distinct genotypes.

🧠 Memory Trick

$3^2 = 9$.

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

19 Why is Recessive Parent Crossing analytically superior for genotype determination?

A It produces more offspring

B The recessive parent adds no dominant alleles, allowing the unknown to express

C It is faster to perform

D It creates new variations

💡 Short Explanation

Recessive parent = tt (adds nothing to hide the other). Any "T" from the unknown will appear in the phenotype. This makes it a perfect "test."

✅ Correct Answer Explanation

In a test cross (unknown x recessive), the recessive parent (tt) provides only recessive alleles. Therefore, whatever allele is present in the unknown parent will be immediately visible in the phenotype of the offspring, making it the most direct way to reveal the hidden genotype.

❓ Why Other Options Are Incorrect

Options A, C, and D are not the reasons why the test cross is "analytically superior."

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Define the mechanics of a test cross (revealing the unknown via recessive partner).

Final Logic: Recessive partner contributes zero masking alleles.

🧠 Memory Trick

Recessive = Reveals the unknown.

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

20 Which is NOT associated with Unknown Genotype Prediction?

A 1:1 ratio in test cross offspring

B 100 percent dominant phenotype in test cross offspring

C Use of a Punnett square for derivation

D Calculating the specific mutation rate of the gene

💡 Short Explanation

Test cross identifies genotype. It does NOT calculate mutation rates. Option D is irrelevant/wrong.

✅ Correct Answer Explanation

Unknown genotype prediction (via test cross) identifies whether an individual is homozygous or heterozygous. It does not measure or calculate the mutation rate of genes.

❓ Why Other Options Are Incorrect

Option A, B, and C are valid parts of predicting and confirming unknown genotypes.

🎯 Strategy

Used: Elimination

Application: Eliminate the choice that describes a function outside the scope of basic Mendelian testing.

Final Logic: Genotype prediction is not mutation rate calculation.

🧠 Memory Trick

Genotype =/= Mutation Rate.

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

CUET UG Biology Booster Test 1 Laws of Inheritance

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