Discuss the genetic basis of wrinkled phenotype of pea seed.

NCERT Exemplar · Q31

Q. Discuss the genetic basis of wrinkled phenotype of pea seed.

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Concept understanding — Mendel's Experimental Design

Let’s begin with something you already know from everyday life. Suppose you mix red and white paint — you get pink. Now imagine you cross a tall pea plant with a short one. Common sense might say you’ll get a medium-sized plant. But Mendel found that you don’t. You get all tall plants in the first generation, and then in the next generation, short plants reappear as if they had never been mixed. That is the puzzle Mendel set out to solve, and his experimental design was the tool that cracked it.

The core idea: controlled, large-scale, and trait-by-trait

Mendel’s experimental design was not just “grow some peas and see what happens.” It was a carefully planned method that had three key features:

He chose a single trait at a time — for example, only seed shape (round vs wrinkled), not seed shape plus flower colour plus plant height all at once. This kept the results clear and interpretable.
He used pure-breeding lines — plants that, when self-pollinated, always produced offspring identical to themselves for that trait. A pure round-seeded plant always gave round seeds; a pure wrinkled-seeded plant always gave wrinkled seeds. This gave him a known starting point.
He counted large numbers — not just a handful of plants, but hundreds or thousands. This allowed patterns to emerge that would be invisible in small samples.

Important

Mendel’s genius was not in discovering that traits are inherited — farmers knew that. His breakthrough was in designing an experiment that could reveal the rules of inheritance by controlling variables, using pure lines, and counting systematically. This is why he is called the father of genetics.

The step-by-step plan Mendel followed

Mendel worked with the garden pea (Pisum sativum) for practical reasons: it was easy to grow, had a short generation time, and — crucially — its flowers normally self-pollinate, so he could control exactly which plants were crossed.

His typical experiment went like this:

Take two pure-breeding parents that differ in one trait — say, a pure tall plant and a pure short plant. These are the P generation (parental generation).
Cross them artificially — he removed the male parts (anthers) from one flower and dusted pollen from the other parent onto its stigma. This ensured no accidental self-pollination.
Grow the offspring — these are the F₁ generation (first filial generation). He observed that all F₁ plants were tall. The short trait had vanished.
Let the F₁ plants self-pollinate — he did not cross them again; he simply let them produce seeds on their own. These seeds grew into the F₂ generation (second filial generation).
Count and record — in the F₂ generation, both tall and short plants appeared, in a ratio of roughly three tall to one short. The short trait had reappeared.

He repeated this exact procedure for each of the seven traits he studied — seed shape, seed colour, flower colour, pod shape, pod colour, flower position, and stem height. Every time, the pattern was the same: one trait dominated in F₁, and both traits reappeared in a 3:1 ratio in F₂.

Why this design was revolutionary

Before Mendel, people believed in blending inheritance — the idea that parental traits mix like paint. Mendel’s design proved that inheritance is particulate: traits are carried by discrete units (what we now call genes) that remain intact across generations. The short trait did not disappear; it was merely hidden in the F₁ plants and reappeared unchanged in F₂.

Note

The NCERT textbook emphasises that Mendel’s success came from his scientific approach: he chose the right organism, studied one trait at a time, kept accurate records, and used mathematics to analyse his results. This was completely new for biology in the 1860s.

A few key terms you will encounter …

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The wrinkled phenotype of pea seeds arises from a recessive allele that disrupts the normal synthesis of starch, leading to reduced water retention and subsequent shrinkage upon maturation.

Gregor Mendel, often hailed as the “Father of Genetics,” meticulously chose the garden pea plant (Pisum sativum) for his groundbreaking experiments. His success stemmed from several factors, including the availability of distinct, contrasting traits that were easy to observe. Among the seven pairs of contrasting characters he studied, seed shape—round versus wrinkled—was particularly significant, as it clearly demonstrated the principles of dominance and recessiveness. While Mendel observed these patterns, the underlying molecular and genetic basis for these phenotypes was elucidated much later.

The genetic basis for the wrinkled phenotype lies in a specific gene that controls the synthesis of starch within the pea seed. This gene exists in two allelic forms: a dominant allele, represented as 'B', and a recessive allele, 'b' — the same B/b pair used elsewhere in this chapter to show that dominance depends on the gene product, not the allele itself. These alleles dictate the production and efficiency of a crucial enzyme known as Starch Branching Enzyme I (SBEI), which is vital for proper starch synthesis.

In pea plants with the dominant 'B' allele (genotypes BB or Bb), the gene produces a functional and efficient Starch Branching Enzyme I. This enzyme facilitates the proper synthesis of starch, leading to the formation of large, well-branched starch grains within the developing seed. These abundant starch grains allow the seed to absorb and retain a significant amount of water. As the seed matures, it remains plump and full, resulting in the characteristic round phenotype.

Important

The presence of at least one dominant 'B' allele ensures the production of functional SBEI, leading to efficient starch synthesis and round seeds.

Conversely, the wrinkled phenotype is observed in pea plants that are homozygous for the recessive 'b' allele (genotype bb). In these plants, the recessive 'b' allele is a mutated form of the gene. This mutation results in the production of an inefficient or entirely non-functional Starch Branching Enzyme I. Without a properly functioning SBEI, the process of starch synthesis is significantly impaired.

Here's a breakdown of what happens in 'bb' seeds:

Inefficient Starch Synthesis: The lack of functional SBEI means that starch is not synthesized efficiently, and the starch grains that do form are smaller and fewer in number.
Accumulation of Sucrose: Due to the impaired conversion of sugar into starch, there is a higher concentration of unutilized sucrose (a disaccharide sugar) within the developing seed.
Osmotic Effects: This higher sucrose concentration affects the osmotic balance within the seed. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
$_2$
›Reveal solutionSolution
When the phenotype of a heterozygote is intermediate between the two homozygotes, the genotypic ratio (1:2:1) directly becomes the phenotypic ratio — this happens in incomplete dominance, making the two ratios identical.

The key idea is simple: the F $_2$ phenotypic ratio mirrors the genotypic ratio only when the heterozygote has a distinct, third phenotype. In complete dominance, the heterozygote looks like the dominant homozygote, so the 1:2:1 genotypic ratio collapses into a 3:1 phenotypic ratio. In epistasis and hypostasis, gene interactions further modify ratios. Only in incomplete dominance do the two ratios match.

Let’s walk through each option.

Complete dominance — The dominant allele completely masks the recessive. A cross of two heterozygotes (Aa × Aa) gives a genotypic ratio of 1 AA : 2 Aa : 1 aa. But because AA and Aa look the same, the phenotypic ratio is 3 (dominant) : 1 (recessive). Not the same.

Incomplete dominance — Here, the heterozygote shows an intermediate phenotype (e.g., red × white → pink). The same cross gives the same genotypic ratio 1:2:1, but now each genotype has a distinct appearance: AA (red), Aa (pink), aa (white). So the phenotypic ratio is also 1:2:1. These are identical.

Epistasis — One gene masks the expression of another. For example, in recessive epistasis (9:3:4 ratio) or dominant epistasis (12:3:1), the phenotypic ratio is a modified dihybrid ratio, not 1:2:1. The genotypic ratio remains 1:2:1:2:4:2:1:2:1 (for two genes), but the phenotype count is compressed differently. Not the same. …
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TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
$Q. If a heterozygous tall plant is crossed with a homozygous dwarf plant the % of progeny having dwarf character is (A) 0 (B) 25 (C) 50 (D) 100$
›Reveal solutionSolution
A heterozygous tall plant (Tt) crossed with a homozygous dwarf plant (tt) gives a 1:1 ratio of tall to dwarf offspring, so 50% of the progeny will be dwarf.

The key here is understanding what the words "heterozygous" and "homozygous" actually mean in a genetic cross. A heterozygous tall plant has one dominant allele (T) for tallness and one recessive allele (t) for dwarfness — its genotype is Tt. A homozygous dwarf plant has two copies of the recessive allele — its genotype is tt. Since dwarfness is recessive, it only shows up when both alleles are t.

When you cross these two, you're essentially asking: what fraction of the offspring will inherit a t from both parents? The tall parent can give either T or t with equal probability; the dwarf parent can only give t. So the only way to get a dwarf offspring is to get t from the tall parent — and that happens half the time.

Write the parental genotypes

Tall parent: Tt (heterozygous)

Dwarf parent: tt (homozygous recessive)

Determine the gametes each parent produces

The Tt plant makes two types of gametes: T and t, each with 50% frequency.

The tt plant makes only one type of gamete: t.

Set up the Punnett square

t (from dwarf)
T Tt (tall)
t tt (dwarf)

Two equally likely outcomes: Tt and tt.

Count the dwarf progeny …
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TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
$Q. If a red eyed male Drosophila and white eyed female Drosophila are crossed, the eye color in their offspring is (A) Females red eyed and males white eyed (B) All are red eyed (C) All are white eyed (D) Female white eyed and males red eyed$
›Reveal solutionSolution
This is an X-linked recessive cross: the red-eyed male (X\textsuperscript{R}Y) and white-eyed female (X\textsuperscript{w}X\textsuperscript{w}) produce daughters that inherit the red allele from their father (so they are red-eyed) and sons that inherit the white allele from their mother (so they are white-eyed). The correct option is (A).

The key concept here is sex-linked inheritance, specifically X-linked recessive traits. In Drosophila (fruit flies), eye color is a classic example: the gene for eye color is located on the X chromosome, and the red allele (R) is dominant over the white allele (w). Males have only one X chromosome (XY), so they express whatever allele is on that single X. Females have two X chromosomes (XX), so they need two copies of the recessive white allele to show white eyes.

Now, let’s work through the cross step by step.

Assign genotypes to the parents.

The red-eyed male has the genotype X\textsuperscript{R}Y — his single X carries the dominant red allele.

The white-eyed female has the genotype X\textsuperscript{w}X\textsuperscript{w} — both her X chromosomes carry the recessive white allele.

Determine the gametes each parent produces.

The male produces two types of sperm: half carry X\textsuperscript{R}, half carry Y.

The female produces only one type of egg: all carry X\textsuperscript{w}.

Set up the Punnett square.

Combine each sperm with each egg:

X\textsuperscript{R} sperm + X\textsuperscript{w} egg → X\textsuperscript{R}X\textsuperscript{w} (female)

Y sperm + X\textsuperscript{w} egg → X\textsuperscript{w}Y (male) …
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TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ
$Q. Who experimentally proved that "inheritance of acquired characters" suggested by Lamarck to be false (A) Darwin (B) August Weismann (C) De vries (D) Haeckel$
›Reveal solutionSolution
The concept is the germ-plasm theory — August Weismann cut off mice tails for generations and showed the trait was never inherited, disproving Lamarck's idea that acquired characters pass to offspring.

The question tests a classic moment in the history of evolutionary biology. Lamarck proposed that organisms could pass on traits they acquired during their lifetime — like a giraffe stretching its neck and then having longer-necked young. This idea was intuitive but needed experimental testing.

The key insight is that acquired characters affect the body (soma) but not the reproductive cells (germ-plasm). If you cut off a mouse's tail every generation, the tail-loss is an acquired injury, not a genetic change. If Lamarck were right, eventually tailless mice would be born naturally. The experiment that tested this directly was done by August Weismann.

Weismann's experiment: He cut off the tails of mice for 22 consecutive generations. In every generation, the newborn mice had normal-length tails. This showed that an acquired physical change (lost tail) was never inherited. …
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TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
$Q. Mutations in Oenothera lamarckiana plants was first identified by (A) Hugo de Vries (B) Morgan (C) Sutton (D) Boveri$
›Reveal solutionSolution
Hugo de Vries first identified mutations in Oenothera lamarckiana (evening primrose), leading to his Mutation Theory. The correct option is (A).

The question asks to identify the scientist who first observed mutations in Oenothera lamarckiana plants. This is a direct recall question from the history of genetics, specifically concerning the development of the mutation theory. Understanding the contributions of key scientists in genetics helps to place this discovery in its proper context.

Identifying the plant: Oenothera lamarckiana, commonly known as the evening primrose, is a plant species that exhibits a high degree of genetic variation, including sudden, heritable changes in its characteristics. These changes were crucial for the observations made.

The discoverer of mutations: The Dutch botanist Hugo de Vries (1848-1935) conducted extensive breeding experiments on Oenothera lamarckiana in the late 19th and early 20th centuries. He observed sudden, discontinuous variations appearing in the offspring that were distinct from the parental forms and were heritable. He termed these sudden changes "mutations."

The Mutation Theory: Based on his observations in Oenothera lamarckiana, Hugo de Vries proposed his "Mutation Theory" in 1901. This theory suggested that evolution occurs through sudden, large, discontinuous changes (mutations) rather than solely through the gradual accumulation of small variations as proposed by Darwin's theory of natural selection. While later understanding refined the role of mutations, de Vries's work was pivotal in introducing the concept of mutations as a source of new variations and a driving force in evolution.

Why other options are incorrect:

(B) Morgan: Thomas Hunt Morgan is renowned for his work with the fruit fly (Drosophila melanogaster), establishing the principles of linkage, crossing over, and the chromosomal theory of inheritance. While he extensively studied mutations, his initial identification of mutations was not in Oenothera lamarckiana. …
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TG EAPCET 2022Set ap-2022-07-30-FN1 markMCQ
$Q. Which of the following character was not a part of Mendel's experiments? (A) Seed shape (B) Pod shape (C) Flower position (D) Seed size$
›Reveal solutionSolution
Mendel studied seven discrete traits in pea plants, and seed size was not among them; the correct answer is (D).

Mendel’s genius lay in choosing traits that had clear, non-overlapping forms — like round vs. wrinkled seeds or tall vs. short stems. He deliberately avoided traits that varied continuously (like size or weight) because they would have been impossible to track with simple dominant/recessive ratios. Seed size is exactly that kind of continuous trait, so it never appeared in his famous experiments.

Let’s check each option:

Seed shape (A) — Mendel studied round vs. wrinkled seeds. This was one of his seven traits.

Pod shape (B) — He studied inflated vs. constricted pods. Also one of his seven.

Flower position (C) — He studied axial vs. terminal flowers. Also one of his seven.

Seed size (D) — Mendel did not study seed size. He studied seed shape and seed color, but size varies continuously and would not produce the clear 3:1 ratios he needed. …
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TG EAPCET 2021Set ap-2021-08-09-FN1 markMCQ
$3 : 1$
›Reveal solutionSolution
When two pure-breeding (homozygous) parents are crossed—one dominant and one recessive—all F₁ offspring are heterozygous and display only the dominant phenotype, giving a ratio of 1:0 (all dominant, no recessive).

Understanding Mendelian Genetics and the F₁ Generation

This question tests your understanding of what happens in the first filial generation (F₁) when crossing two pure-breeding parents with contrasting traits. Let's think about what "pure" means and what we expect from such a cross.

In Drosophila (fruit flies), wing length follows simple Mendelian inheritance:

Long wings = dominant trait (let's call the allele L)

Vestigial wings = recessive trait (let's call the allele l)

"Pure" or "true-breeding" means homozygous—both alleles are identical.

Step-by-Step Analysis

Identify the parental genotypes

Pure long-winged parent: LL (homozygous dominant)

Pure vestigial-winged parent: ll (homozygous recessive)

Determine the gametes each parent can produce

The LL parent can only produce gametes carrying L

The ll parent can only produce gametes carrying l

Perform the cross using a Punnett square

L L
l Ll Ll
l Ll Ll

Every single offspring receives one L allele from the dominant parent and one l allele from the recessive parent.

Analyze the F₁ genotype and phenotype

Genotype: All offspring are Ll (100% heterozygous) …
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	t (from dwarf)
T	Tt (tall)
t	tt (dwarf)

	L	L
l	Ll	Ll
l	Ll	Ll

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Q.Discuss the genetic basis of wrinkled phenotype of pea seed.

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