CUET UG Biology Booster Test 2 Chromosomal Inheritance -

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CUET UG Biology Booster Test 2 Chromosomal Inheritance

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Biology Booster Test 2 Chromosomal Inheritance

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QUESTION 1 OF 20

📄 Passage

Morgan and his group knew that the genes were located on the X chromosome and saw quickly that when the two genes in a dihybrid cross were situated on the same chromosome, the proportion of parental gene combinations were much higher than the non-parental type. Morgan attributed this due to the physical association or linkage of the two genes and coined the term linkage to describe this physical association... and the term recombination to describe the generation of non-parental gene combinations.

Why did Morgan observe a higher frequency of parental types compared to the 9:3:3:1 ratio in his Drosophila crosses?

Answer Review

📄 Passage

Why did Morgan observe a higher frequency of parental types compared to the 9:3:3:1 ratio in his Drosophila crosses?

A Because the genes were located on different chromosomes

B Because the genes were physically associated on the same chromosome

C Because the genes were following the law of independent assortment

D Because of the presence of the Y chromosome in females

💡 Short Explanation

Linkage refers to genes being on the same chromosome. Linked genes do not assort independently. They tend to be inherited together as a parental block.

✅ Correct Answer Explanation

The 9:3:3:1 ratio is the hallmark of independent assortment (genes on different chromosomes). Morgan found that genes for body color and eye color did not follow this. Because these genes were physically linked on the same X chromosome, they were inherited together as parental combinations much more frequently than random chance would dictate.

❓ Why Other Options Are Incorrect

Option A → This would lead to a 9:3:3:1 ratio, not the observed deviation.
Option C → Independent assortment is exactly what these genes failed to do.
Option D → The Y chromosome presence in males (not females) is irrelevant to the linkage of X-linked genes in this cross context.

🎯 Strategy

Used: Elimination Application: Eliminate options that describe independent assortment to identify the cause of deviation (linkage). Final Logic: Linkage = physical association = parental preference.

🧠 Memory Trick

Linkage = Less non-parental = Long-term association.

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

📄 Passage

If two genes are situated very close to each other on the same chromosome, what is the expected outcome in terms of parental vs. non-parental combinations?

A Non-parental types will be 50%

B Parental types will be much higher than non-parental types

C Non-parental types will be much higher than parental types

D There will be no parental types

💡 Short Explanation

Proximity increases linkage strength. Higher linkage means less crossing over between genes. Less crossing over leads to fewer recombinants (non-parentals).

✅ Correct Answer Explanation

The strength of linkage is inversely proportional to the physical distance between genes. When genes are close, the probability of a crossover event occurring between them is very low. Consequently, the alleles remain together, ensuring that parental combinations dominate the offspring phenotype.

❓ Why Other Options Are Incorrect

Option A → 50% is for unlinked (independent) genes.
Option C → This would require frequent crossing over, implying distant genes.
Option D → Parental types are always present unless 100% recombination occurs, which is biologically impossible.

🎯 Strategy

Used: Contextual/Tonal Matching Application: Apply the rule that distance and recombination frequency are directly related. Final Logic: Close distance = Low recombination = High parental frequency.

🧠 Memory Trick

Close = Conservative (Parental).

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

3 Match the scientist to their contribution or observation.

List I	List II
1. de Vries	p. Independent rediscovery of Mendelism
2. Henking	q. Observed X body in insects
3. Sutton	r. Noted chromosome behavior parallel to genes
4. Langdon Down	s. Described Trisomy 21

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

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

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

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

💡 Short Explanation

de Vries was one of the three who rediscovered Mendel. Henking is associated with the 'X-body'. Sutton is the co-proposer of Chromosomal Theory.

✅ Correct Answer Explanation

Option A correctly matches the historical facts: de Vries (rediscovery), Henking (X-body observation), Sutton (parallel chromosome/gene behavior), and Langdon Down (clinical description of what is now known as Down Syndrome/Trisomy 21).

❓ Why Other Options Are Incorrect

Options B, C, and D are incorrect as they swap the specific historical contributions of these scientists.

🎯 Strategy

Used: Substitution Application: Confirm one known match (e.g., Sutton = r) to narrow choices, then verify others. Final Logic: All pairings in A are historically accurate.

🧠 Memory Trick

Sutton = Segregation parallelism; Henking = X-body.

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Chromosomal Theory/Genetic Disorders

4 The fact that Mendel's work remained unrecognised until 1900 was primarily because:

A His mathematical approach was too simple for biologists

B Communication was difficult and his work lacked wide publicity

C He only used three pea plants for his experiments

D Chromosomes had not yet been discovered

💡 Short Explanation

Limited circulation of scientific papers. Concept of "factors" (genes) was ahead of its time. Lack of peer recognition/publicity.

✅ Correct Answer Explanation

Mendel's work was unrecognized primarily due to the lack of communication channels and limited publicity in the 19th century. His mathematical approach was actually considered ahead of its time, and he used thousands of plants, not three. While chromosomes were unknown, the core issue was visibility.

❓ Why Other Options Are Incorrect

Option A → His approach was novel, not "too simple."
Option C → He used a massive number of plants.
Option D → While true, this was not the primary reason for the lack of publicity/recognition.

🎯 Strategy

Used: Elimination Application: Discard inaccurate historical facts (e.g., three plants). Final Logic: Historical communication barriers were the primary bottleneck.

🧠 Memory Trick

Mendel = Missing Media (publicity).

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Mendel's Laws of Inheritance

5 Order the logical steps Boveri and Sutton used to link chromosomes to Mendel's laws.
1. Observe that genes and chromosomes both occur in pairs.
2. Note that chromosomes segregate during gamete formation.
3. Use chromosome movement to explain Mendelian principles.
4. Argue that pairing and separation leads to factor segregation.

A 1-2-4-3

B 2-1-3-4

C 4-3-2-1

D 3-4-1-2

💡 Short Explanation

Observe pairs (diploidy). Observe segregation during meiosis. Deduce that this movement explains factor segregation.

✅ Correct Answer Explanation

Sutton and Boveri observed that both genes and chromosomes exist in pairs (1) and segregate during gamete formation (2). They argued that this specific pairing and subsequent separation provided the physical mechanism (4) to explain Mendel's abstract laws of inheritance (3).

❓ Why Other Options Are Incorrect

Options B, C, and D are illogical; you must observe phenomena (1, 2) before constructing a theory (3, 4).

🎯 Strategy

Used: Substitution Application: Start with the base observation (pairs) and end with the theory formation. Final Logic: 1 and 2 are observations; 3 and 4 are the theoretical conclusion.

🧠 Memory Trick

Pairs → Segregation → Explanation (PSE).

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

6 If Column A says "Independent pairs segregate independently," and Column B says "One pair segregates independently of another pair," they both describe:

A The Law of Dominance

B Parallel behavior between chromosomes and genes

C Linked inheritance

D Cytoplasmic inheritance

💡 Short Explanation

This is the core of Sutton/Boveri theory. It highlights the similarity in movement. It justifies the Chromosomal Theory.

✅ Correct Answer Explanation

The statements in Table 4.3 (NCERT) emphasize that Mendelian factors (genes) and chromosomes behave identically in their physical distribution during gametogenesis. "Independent segregation" is a key piece of evidence for this parallelism.

❓ Why Other Options Are Incorrect

Option A → Dominance is about expression, not segregation movement.
Option C → Linkage is the opposite of independent segregation.
Option D → Cytoplasmic inheritance is non-nuclear.

🎯 Strategy

Used: Contextual/Tonal Matching Application: Recognize the definition of Chromosomal Theory as described in the NCERT table. Final Logic: "Independent segregation" = Proof of Parallelism.

🧠 Memory Trick

Independent Segregation = Parallel State (IPS).

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

7 Sequence the development of the Chromosomal Theory.
1. Working out of chromosome movement during meiosis (1902).
2. Sutton uniting chromosomal knowledge with Mendel's principles.
3. Independent rediscovery of Mendel's work (1900).
4. Identification of "colored bodies" as chromosomes.

A 3-4-1-2

B 1-2-3-4

C 4-3-2-1

D 2-4-1-3

💡 Short Explanation

1900 Rediscovery first. Microscopic identification. Meiotic study. Theoretical synthesis.

✅ Correct Answer Explanation

The timeline is: Rediscovery (1900) (3) → Visualization of Chromosomes (4) → Study of Meiotic movement (1) → Synthesis of Chromosomal Theory by Sutton and Boveri (1902) (2).

❓ Why Other Options Are Incorrect

Options B, C, and D are historically inverted or scrambled.

🎯 Strategy

Used: Elimination Application: Place the 1900 rediscovery at the very start to eliminate B, C, and D. Final Logic: The theory follows the observation of movement, which follows the discovery of chromosomes.

🧠 Memory Trick

Rediscovery → Chromosomes → Meiosis → Theory.

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Chromosomal Theory of Inheritance – -71

8 What was the critical "proof" Sutton and Boveri used to justify their theory?

A Direct observation of DNA molecules

B The parallel behavior between chromosome movement and gene segregation

C The mutation of fruit flies

D The sequence of the human genome

💡 Short Explanation

The "proof" was observational logic. They noted similarities in pairing and movement. They did not see genes directly.

✅ Correct Answer Explanation

Sutton and Boveri provided no direct molecular proof (DNA was not identified as genetic material then). Their "proof" was the elegant realization of the parallelism: chromosomes behave during meiosis exactly as Mendel predicted "factors" behaved during inheritance.

❓ Why Other Options Are Incorrect

Option A → DNA structure was identified in 1953.
Option C → Morgan used fruit flies, not Sutton/Boveri.
Option D → HGP is late 20th century.

🎯 Strategy

Used: Elimination Application: Eliminate modern concepts to find the early 20th-century logic. Final Logic: Logical deduction of parallelism = their proof.

🧠 Memory Trick

Parallelism = Proof for Sutton.

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

9 Regarding meiosis and germ cell formation.
Statement I: Chromosome segregation occurs when germ cells are formed.
Statement II: In Possibility I of independent assortment, the orange chromosome always segregates with the red one.

A Both statements are true

B Statement I is true, Statement II is false

C Statement I is false, Statement II is true

D Both statements are false

💡 Short Explanation

Statement I is the basic definition of meiosis. In independent assortment, segregation is random, not "always" with a specific partner.

✅ Correct Answer Explanation

Statement I is correct; meiosis reduces chromosome number in germ cells. Statement II is false; "Possibility I" in NCERT diagrams shows specific segregation, but independent assortment dictates that the segregation of one pair is random relative to another. The pairing is not "always" the same in the overall process of assortment.

❓ Why Other Options Are Incorrect

Options A, C, and D are incorrect because they fail to identify the nuance of independent assortment as random, not fixed.

🎯 Strategy

Used: Elimination Application: If Statement I is common knowledge (True), evaluate Statement II for logical fallacy. Final Logic: "Always" is rarely correct in biological stochastic processes like independent assortment.

🧠 Memory Trick

Independent = Individual/Random (no "always" pairing).

📚 Class XII – Biology – Chapter: Principles of Inheritance and Variation – Topic: Chromosomal Theory of Inheritance – (Figure 5.9)

10 The independent assortment of chromosomes is best visualised during:

A Prophase I

B Anaphase I

C Metaphase II

D Telophase II

💡 Short Explanation

Anaphase I is where chromosomes move to opposite poles. This movement establishes the segregation. It is the visual outcome of the alignment in Metaphase I.

✅ Correct Answer Explanation

While alignment happens in Metaphase, the physical separation and assortment into daughter cells occur in Anaphase I. This is where the cell "commits" to which pole each chromosome goes, representing the visual manifestation of independent assortment.

❓ Why Other Options Are Incorrect

Option A → Chromosomes condense here, but do not separate.
Option C → Metaphase II involves sister chromatids, not homologous independent pairs.
Option D → Telophase is just the completion of the migration.

🎯 Strategy

Used: Contextual/Tonal Matching Application: Identify the stage of physical movement/segregation. Final Logic: Assortment = Segregation = Anaphase.

🧠 Memory Trick

Anaphase = Apart (moving).

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

11 About Drosophila genetic variations.
Statement I: Many hereditary variations in Drosophila can be seen with high-power electron microscopes only.
Statement II: Male and female flies are easily distinguishable by size and morphology.

A Both statements are true

B Statement I is true, Statement II is false

C Statement I is false, Statement II is true

D Both statements are false

💡 Short Explanation

Variations were visible with low-power microscopes (why they were used early on). Sexual dimorphism is a key trait of Drosophila.

✅ Correct Answer Explanation

Statement I is false because one of the benefits of using Drosophila was that variations were easily visible under low-power microscopes, making them accessible to early geneticists. Statement II is true; sexual dimorphism (males are smaller with a dark abdomen, females are larger) is very distinct.

❓ Why Other Options Are Incorrect

Option A, B, and D fail because they incorrectly label Statement I as true.

🎯 Strategy

Used: Elimination Application: Eliminate based on the historical context that early 20th-century genetics did not rely on electron microscopes. Final Logic: Simple microscopes + clear sexual traits = Ideal model.

🧠 Memory Trick

Drosophila = Distinguishable sexes.

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

12 Morgan's work on "sex-linked" genes began with the study of:

A Seed color in peas

B Eye color and body color in fruit flies

C Skin color in humans

D Starch synthesis in peas

💡 Short Explanation

Morgan is famous for fruit fly (Drosophila) crosses. His seminal work was on white-eyed mutants. He linked these to the X chromosome.

✅ Correct Answer Explanation

Morgan observed a white-eyed male Drosophila in a population of red-eyed flies. Crosses involving this eye color (and later body color) allowed him to demonstrate that these genes were located on the X chromosome, establishing the concept of sex-linkage.

❓ Why Other Options Are Incorrect

Options A and D are Mendelian pea studies.
Option C is polygenic and not related to Morgan's initial Drosophila discovery.

🎯 Strategy

Used: Substitution Application: Associate "Morgan" directly with "Fruit Fly." Final Logic: Morgan = Drosophila eye/body color.

🧠 Memory Trick

Morgan = Mutant fly eyes.

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

13 Which frequency value is NOT associated with the specific gene pairs studied by Morgan?

A 1.3 per cent

B 37.2 per cent

C 9:3:3:1 per cent

D Recombination frequency

💡 Short Explanation

1.3% and 37.2% are specific recombination frequencies found by Morgan. 9:3:3:1 is a ratio, not a frequency.

✅ Correct Answer Explanation

Morgan mapped genes by measuring recombination frequencies (like 1.3% and 37.2%). 9:3:3:1 is a phenotypic ratio expected from independent assortment; it is not a "frequency" value used in mapping nor was it the value observed for these linked pairs.

❓ Why Other Options Are Incorrect

Options A and B are the actual values Morgan reported for linkage distances.
Option D is the category of data he used.

🎯 Strategy

Used: Odd One Out Application: Recognize the outlier that is a ratio rather than a decimal frequency. Final Logic: 9:3:3:1 is a ratio; frequencies are percentages.

🧠 Memory Trick

9:3:3:1 = Ratio (not Frequency).

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

14 Tightly linked genes on the same chromosome will show:

A High recombination frequencies

B Very low recombination frequencies

C Independent assortment

D 100% recombination

💡 Short Explanation

Tight linkage = close together. Close together = low probability of crossing over. Low crossing over = low recombination frequency.

✅ Correct Answer Explanation

Genetic map distance is determined by how often crossing over occurs between two genes. If genes are "tightly linked," they are very close physically, meaning the chance of a crossover event occurring between them is near zero, resulting in a very low recombination frequency.

❓ Why Other Options Are Incorrect

Option A → High frequency means distant genes.
Option C → This only happens on separate chromosomes.
Option D → This is impossible for linked genes.

🎯 Strategy

Used: Contextual/Tonal Matching Application: Use the relationship: Distance ∝ Recombination Frequency. Final Logic: Tight linkage = small distance = low frequency.

🧠 Memory Trick

Tight = Tiny distance.

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

15 Which of the following concepts is NOT part of Sturtevant's mapping logic?

A Closer genes recombine less often

B Recombination frequency equals distance

C Genes on different chromosomes are mapped together

D Map positions are linear on the chromosome

💡 Short Explanation

Mapping is based on linkage. Linkage only occurs on the same chromosome. Genes on different chromosomes assort independently.

✅ Correct Answer Explanation

Sturtevant's genetic mapping is entirely dependent on the phenomenon of linkage. Since genes on different chromosomes assort independently and do not show the physical linkage required to measure distance via recombination frequency, they cannot be "mapped" together in this system.

❓ Why Other Options Are Incorrect

Options A, B, and D are the fundamental pillars of genetic mapping.

🎯 Strategy

Used: Elimination Application: Eliminate the option that contradicts the definition of linkage mapping. Final Logic: Linkage mapping requires being on the same chromosome.

🧠 Memory Trick

Linkage = Linear same-chromosome mapping.

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

16 If the recombination frequency between gene A and B is 1.3% and between A and C is 37.2%, which pair is physically closer?

A A and C

B A and B

C B and C

D They are the same distance

💡 Short Explanation

Recombination frequency = physical distance. 1.3% is smaller than 37.2%. Therefore, A and B are closer.

✅ Correct Answer Explanation

Sturtevant used recombination frequency as a proxy for genetic distance. A lower frequency (1.3%) indicates that the genes (A and B) are physically closer on the chromosome compared to a higher frequency (37.2%), which indicates a greater distance.

❓ Why Other Options Are Incorrect

Option A → 37.2% indicates they are far apart.
Option C → Without more info, we cannot be certain, but A and B is the definitely closer known pair.
Option D → The frequencies are different.

🎯 Strategy

Used: Dimensional/Unit Analysis Application: Directly compare the percentages (1.3 vs 37.2). Final Logic: 1.3 < 37.2; therefore, A-B is the smaller interval.

🧠 Memory Trick

Lower frequency = Less distance.

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

17 In the polygenic model of human skin color (A, B, C), which genotype is NOT an intermediate color?

A AaBbCc

B AABBCC

C AABbcc

D aaBBcc

💡 Short Explanation

Polygenic model: A, B, C contribute pigment; a, b, c do not. Intermediate has a mix of dominant and recessive. Extremes are all-dominant or all-recessive.

✅ Correct Answer Explanation

In the classic polygenic model for skin color (A, B, C), the phenotype is determined by the number of dominant alleles. AABBCC (all dominant) represents the darkest extreme, not an intermediate color. Intermediate colors occur when there is a mix of alleles (e.g., AaBbCC).

❓ Why Other Options Are Incorrect

Options A, C, and D all contain a mix of dominant and recessive alleles, leading to intermediate pigment levels.

🎯 Strategy

Used: Odd One Out Application: Identify the genotype that represents a phenotypic extreme (all dominant). Final Logic: All-dominant = Extreme; Mix = Intermediate.

🧠 Memory Trick

Extreme = Every allele dominant.

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

18 Why does human skin color show a gradient rather than two distinct alternatives?

A It is controlled by a single gene with two alleles

B It is controlled by three or more genes with additive effects

C It is a sex-linked trait

D It is caused by a single base pair mutation

💡 Short Explanation

Many genes = many combinations = gradient. Single genes produce binary (distinct) traits. "Additive" means each gene adds to the color depth.

✅ Correct Answer Explanation

Polygenic inheritance involves multiple genes whose contributions are cumulative (additive). Because many allele combinations exist, the resulting phenotype is a continuous spectrum (gradient) of variations, unlike simple Mendelian traits which are binary.

❓ Why Other Options Are Incorrect

Option A → This would cause distinct alternatives (Mendelian).
Option C → Skin color is autosomal polygenic.
Option D → Mutations cause discrete disease/variation, not a continuous population gradient.

🎯 Strategy

Used: Contextual/Tonal Matching Application: Recognize the difference between monogenic (binary) and polygenic (gradient) traits. Final Logic: Many genes = Gradient.

🧠 Memory Trick

Polygenic = People's skin range.

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

19 Which symptom is NOT associated with the pleiotropic effects of Phenylketonuria?

A Mental retardation

B Reduction in hair pigmentation

C Sickle-shaped red blood cells

D Reduction in skin pigmentation

💡 Short Explanation

PKU affects phenylalanine metabolism. Sickle-cell anemia is a distinct genetic disorder caused by hemoglobin mutation. PKU does not involve sickling.

✅ Correct Answer Explanation

PKU (Phenylketonuri A) is a classic example of pleiotropy where one defective enzyme results in multiple symptoms (retardation, light pigmentation). Sickle-shaped red blood cells are the specific result of sickle-cell anemia, not PKU.

❓ Why Other Options Are Incorrect

Options A, B, and D are all well-documented clinical symptoms of PKU due to the accumulation of phenylalanine.

🎯 Strategy

Used: Elimination Application: Eliminate the symptoms known to belong to other diseases (Sickle Cell Anemi

A). Final Logic: Sickle cells ≠ PKU.

🧠 Memory Trick

PKU = Pigment & Kognitive (mental).

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

20 Match the metabolic pathway concepts.

List-I	List-II
1. Pleiotropic gene	p. Affects multiple phenotypes
2. Phenylketonuria	q. Mutation in phenylalanine hydroxylase
3. Tyrosine	r. Product of phenylalanine conversion
4. Phenylpyruvic acid	s. Accumulated derivative in PKU

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

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

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

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

💡 Short Explanation

Pleiotropic gene (1) is matched with the product of phenylalanine conversion (r). Phenylketonuria (2) is matched with the accumulated derivative in PKU (s). Tyrosine (3) is matched with affecting multiple phenotypes (p). Phenylpyruvic acid (4) is matched with mutation in phenylalanine hydroxylase (q).

✅ Correct Answer Explanation

According to the given matching arrangement: Pleiotropic gene (1) corresponds to the product of phenylalanine conversion (r). Phenylketonuria (2) corresponds to the accumulated derivative in PKU (s). Tyrosine (3) corresponds to affecting multiple phenotypes (p). Phenylpyruvic acid (4) corresponds to mutation in phenylalanine hydroxylase (q). Thus, the required matching is 1-r, 2-s, 3-p, 4-q, which matches 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 the pleiotropic gene with the enzyme mutation and rearranges the remaining concepts.
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

Gene → Product, PKU → Derivative, Tyrosine → Effects, Phenylpyruvic acid → Mutation

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

CUET UG Biology Booster Test 2 Chromosomal Inheritance

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