GUJCET 2026 Biology Question Paper with Solution PDF is available for download here. The GSEB conducted the GUJCET 2026 Exam for Biology in the Second Shift in PBT mode.

The GUJCET Biology Question Paper 2026 includes 40 questions to be completed in 1 hour. As per the Marking Scheme +1 mark for every correct answer and -0.25 marks will be deducted for every wrong answer

GUJCET 2026 Biology Question Paper with Solution PDF

GUJCET 2026 Biology Question Paper Download PDF Check Solution
GUJCET 2026 Biology Question Paper with Solution PDF


Question 1:

Which of the following contraceptive method can be used for male and female both?
(A) Barriers and Sterilisation
(B) Barriers and Implants
(C) IUDs and Barriers
(D) Implants and Sterilisation

Correct Answer: (A) Barriers and Sterilisation
View Solution




Step 1: Understand Barrier Methods

Barrier methods are contraceptive techniques that physically prevent sperm from reaching the ovum. These include condoms (male) and diaphragms/cervical caps (female). Hence, barrier methods are applicable to both males and females.

Step 2: Understand Sterilisation

Sterilisation is a permanent method of contraception. In males, it is called vasectomy, and in females, it is called tubectomy. Since both genders have their respective procedures, sterilisation is applicable to both.

Step 3: Evaluate Other Options

Implants and IUDs are used only in females and cannot be used by males.

Final Answer: (A) Quick Tip: Barrier and sterilisation methods are the only contraceptive methods applicable to both males and females.


Question 2:

Which term is coined for the physical association of 2 genes on the same chromosome?
(A) Gene flow
(B) Pleiotropy
(C) Linkage
(D) Recombination

Correct Answer: (C) Linkage
View Solution

Step 1: Define Linkage

Linkage refers to the phenomenon where two or more genes are located close to each other on the same chromosome and tend to be inherited together during meiosis.

Step 2: Understand Why It Happens

Since linked genes are physically close, the chances of crossing over between them are low, so they are transmitted together to the next generation.

Step 3: Eliminate Other Options

Gene flow refers to movement of genes between populations. Pleiotropy means one gene affects multiple traits. Recombination involves exchange of genetic material.

Final Answer: (C) Quick Tip: Linkage = genes close together on the same chromosome inherited together.


Question 3:

What is the full form of UTR?
(A) Uni Translated Regions
(B) Ultra Transcribed Regions
(C) Un Translated Regions
(D) Unbound Transcribed Regions

 

Correct Answer: (C) Un Translated Regions
View Solution

Step 1: Definition of UTR

UTR stands for Untranslated Regions. These are sections of mRNA that are transcribed from DNA but are not translated into proteins.

Step 2: Types of UTR

There are two types: 5' UTR and 3' UTR. These regions play important roles in regulation of translation, mRNA stability, and localization.

Step 3: Importance

Although they do not code for proteins, UTRs are essential for controlling gene expression and ensuring proper protein synthesis.

Final Answer: (C) Quick Tip: UTR = Untranslated but functionally important regions of mRNA.


Question 4:

Match Column - I and Column - II and choose correct option.

(A) (i - s), (ii - r), (iii - q), (iv - p)
(B) (i - r), (ii - p), (iii - s), (iv - q)
(C) (i - s), (ii - r), (iii - p), (iv - q)
(D) (i - q), (ii - p), (iii - s), (iv - r)

Correct Answer: (A) (i - s), (ii - r), (iii - q), (iv - p)
View Solution




Step 1: Identify Key Events

65 million years ago marks the extinction of dinosaurs.

350 million years ago corresponds to evolution of jawless fishes.

500 million years ago marks the rise of early invertebrates.

320 million years ago corresponds to the existence of sea weeds.

Step 2: Match Correctly

(i) \(\rightarrow\) (s), (ii) \(\rightarrow\) (r), (iii) \(\rightarrow\) (q), (iv) \(\rightarrow\) (p)

Step 3: Choose Option

This matching corresponds to option (A).

Final Answer: (A) Quick Tip: Memorize evolutionary timeline: Invertebrates \(\rightarrow\) Fish \(\rightarrow\) Plants \(\rightarrow\) Dinosaurs extinction.


Question 5:

Which chemical causes the side-effect of enlargement of clitoris in female?
(A) Diuretics
(B) Anabolic steroids
(C) Peptide Hormone
(D) Narcotic analgesics

Correct Answer: (B) Anabolic steroids
View Solution




Step 1: Understand Anabolic Steroids

Anabolic steroids are synthetic derivatives of testosterone, the male sex hormone. These substances promote muscle growth and development of male secondary sexual characteristics.

Step 2: Effect on Females

When females use anabolic steroids, they may develop masculine traits such as deepening of voice, increased body hair, and enlargement of the clitoris (clitoromegaly).

Step 3: Eliminate Other Options

Diuretics regulate fluid balance, peptide hormones regulate body functions, and narcotic analgesics relieve pain. None of these cause such masculinizing effects.

Final Answer: (B) Quick Tip: Anabolic steroids mimic testosterone → cause masculinization effects in females.


Question 6:

Which of the following bacteria is free-living in the soil?
(A) Trichoderma
(B) Oscillatoria
(C) Rhizobium
(D) Azospirillum

Correct Answer: (D) Azospirillum
View Solution




Step 1: Identify Free-Living Bacteria

Free-living bacteria are those that do not require a host organism and can survive independently in soil.

Step 2: Analyze Options

Trichoderma is a fungus, not a bacterium. Oscillatoria is a cyanobacterium but not primarily known as a nitrogen-fixing soil bacterium. Rhizobium is symbiotic and lives in root nodules of legumes.

Step 3: Correct Choice

Azospirillum is a free-living nitrogen-fixing bacterium found in soil and is associated with plant roots but does not form true symbiotic nodules.

Final Answer: (D) Quick Tip: Azospirillum = free-living nitrogen fixer; Rhizobium = symbiotic.


Question 7:

The separated bands of DNA are cut out from the agarose gel and extracted from the gel piece. This step is known as _____.


(A) Elution
(B) Micro-injection
(C) Biolistics
(D) Extension

Correct Answer: (A)
View Solution




Step 1: Understand Gel Electrophoresis

In gel electrophoresis, DNA fragments are separated based on size by moving through an agarose gel under an electric field.

Step 2: Extraction Process

After separation, the desired DNA band is cut from the gel and purified. This process of recovering DNA from the gel is called elution.

Step 3: Eliminate Other Options

Micro-injection and biolistics are methods of gene transfer, while extension refers to DNA synthesis during PCR.

Final Answer: (A) Quick Tip: Elution = extraction of DNA from gel after electrophoresis.


Question 8:

At present, about _____ recombinant therapeutics have been approved for human-use, the world over.

(A) 10
(B) 25
(C) 30
(D) 15

Correct Answer: (C)
View Solution




Step 1: Understand Recombinant Therapeutics

Recombinant therapeutics are medicines produced using recombinant DNA technology, such as insulin, growth hormone, and vaccines.

Step 2: NCERT Reference

According to standard biology references (NCERT), approximately 30 recombinant therapeutics have been approved for human use globally.

Step 3: Importance

These therapeutics have revolutionized treatment for diseases like diabetes, cancer, and hormonal disorders.

Final Answer: (C) Quick Tip: Remember: NCERT fact → Around 30 recombinant therapeutics approved.


Question 9:

In a forest, there are initially 100 deer. Over certain period 30 are born and 10 die. Assuming food is abundant and predators are few. Calculate the intrinsic rate of natural increase for that population.

(A) 0.4
(B) 0.18
(C) 0.2
(D) 0.015

Correct Answer: (C)
View Solution




Step 1: Formula for Intrinsic Rate of Increase

The intrinsic rate of increase (r) is calculated as: \[ r = \frac{Births - Deaths}{Initial Population} \]

Step 2: Substitute Values

Births = 30, Deaths = 10, Initial population = 100 \[ r = \frac{30 - 10}{100} = \frac{20}{100} = 0.2 \]

Step 3: Interpretation

This indicates a positive growth rate under favorable environmental conditions.

Final Answer: (C) Quick Tip: Intrinsic rate (r) = (B − D)/N → quick formula for population growth.


Question 10:

In the given statements, select the correct option for the process of decomposition.
Statements:
(1) Moist environment favour decomposition.
(2) Low temperature and anaerobiosis inhibit decomposition.
(3) Decomposition is slower, if detritus is rich in nitrogen.
(4) Decomposition is quicker if detritus is rich in chitin.
(5) Warm environment favour decomposition.

(A) 1, 2, 3, 5 are true
(B) 1, 2, 5 are true
(C) 3, 4, 5 are true
(D) 1, 5 are true

Correct Answer: (B)
View Solution




Step 1: Analyze Each Statement

(1) Moist conditions enhance microbial activity → True.

(2) Low temperature and lack of oxygen slow decomposition → True.

(3) Nitrogen-rich detritus decomposes faster, not slower → False.

(4) Chitin is resistant to decomposition → slows process → False.

(5) Warm conditions favour microbial growth → True.

Step 2: Select Correct Combination

Correct statements are (1), (2), and (5).

Final Answer: (B) Quick Tip: Decomposition is fastest in warm, moist, oxygen-rich conditions.


Question 11:

Match Column - I and Column - II and select the correct option.

(A) (i - r), (ii - s), (iii - p), (iv - q)
(B) (i - q), (ii - r), (iii - s), (iv - p)
(C) (i - s), (ii - r), (iii - q), (iv - p)
(D) (i - q), (ii - p), (iii - s), (iv - r)

Correct Answer: (D)
View Solution




Step 1: Identify Species Origins

Dodo was native to Mauritius.

Quagga was found in Africa.

Thylacine (Tasmanian tiger) lived in Australia.

Steller’s sea cow was found near Russia (Bering Sea region).

Step 2: Match Correctly

(i) \(\rightarrow\) (q), (ii) \(\rightarrow\) (p), (iii) \(\rightarrow\) (s), (iv) \(\rightarrow\) (r)

Step 3: Choose Option

This corresponds to option (D).

Final Answer: (D) Quick Tip: Extinct species locations are commonly asked → memorize dodo (Mauritius), thylacine (Australia).


Question 12:

In an angiosperm plant endosperm cell is having 24 chromosomes, then how many chromosomes are present in MMC of the same plant?

(A) 32
(B) 16
(C) 12
(D) 8

Correct Answer: (B)
View Solution




Step 1: Understand Endosperm

Endosperm in angiosperms is triploid (3n), formed by fusion of one male gamete (n) with two polar nuclei (n + n).

Step 2: Calculate Haploid Number

Given: Endosperm = 24 chromosomes (3n) \[ 3n = 24 \Rightarrow n = 8 \]

Step 3: Determine MMC Chromosomes

Megaspore Mother Cell (MMC) is diploid (2n): \[ 2n = 2 \times 8 = 16 \]

Final Answer: (B) Quick Tip: Endosperm = 3n → find n → MMC = 2n.


Question 13:

Which of the following hormone is not produced from the placenta?

(A) Gonadotropin
(B) Human placental Lactogen
(C) Estrogen
(D) Relaxin

Correct Answer: (A)
View Solution




Step 1: Placental Hormones

The placenta produces several hormones such as human chorionic gonadotropin (hCG), human placental lactogen (hPL), estrogen, and relaxin.

Step 2: Analyze Option A

The term "gonadotropin" is a general term referring to hormones like FSH and LH, which are secreted by the pituitary gland, not the placenta.

Step 3: Clarification

Although hCG is a gonadotropin, the option given is generic "gonadotropin", which typically refers to pituitary hormones, hence considered incorrect here.

Final Answer: (A) Quick Tip: Placenta secretes hCG, hPL, estrogen, relaxin — but not FSH/LH.


Question 14:

In the given assisted reproductive technologies which method is taking place inside the body for the process of fertilisation?

(A) GIFT
(B) ZIFT
(C) IUT
(D) ICSI

Correct Answer: (A)
View Solution




Step 1: Understand GIFT

GIFT (Gamete Intra-Fallopian Transfer) involves transfer of gametes (sperm and ovum) into the fallopian tube, where fertilisation occurs naturally inside the body.

Step 2: Compare Other Techniques

ZIFT involves transfer of zygote (fertilisation already done in vitro).

IUT involves embryo transfer into uterus.

ICSI involves injection of sperm into ovum in laboratory.

Step 3: Key Concept

Only GIFT allows fertilisation to occur inside the female body (in vivo).

Final Answer: (A) Quick Tip: GIFT = fertilisation inside body; IVF methods = fertilisation outside body.


Question 15:

Select the incorrect option for mutation.

(A) Results in alternation of DNA sequence
(B) According to De-Vries mutation is the reason for speciation
(C) According to Hugo-de-Vries mutation is small and directional
(D) Results in changes in genotype and phenotype

Correct Answer: (C)
View Solution




Step 1: Understand Mutation

Mutation refers to sudden heritable changes in the DNA sequence.

Step 2: De Vries Theory

Hugo de Vries proposed that mutations are large, sudden, and discontinuous changes that lead to evolution.

Step 3: Identify Incorrect Statement

Option (C) states mutation is small and directional, which contradicts De Vries’ theory.

Step 4: Verify Other Options

All other options correctly describe mutation.

Final Answer: (C) Quick Tip: De Vries → mutations are large, sudden, and random (not small/directional).


Question 16:

In Meselson and Stahl's Experiment, heavy DNA molecule could be distinguished from the normal DNA by centrifugation in a _____ density gradient.

(A) NH\(_4\)Cl
(B) CsCl
(C) Ethidium bromide
(D) Trichloro acetic acid

Correct Answer: (B)
View Solution




Step 1: Experiment Overview

Meselson and Stahl demonstrated semi-conservative replication of DNA using isotopes of nitrogen (\(^{15}\)N and \(^{14}\)N).

Step 2: Role of CsCl

They used Cesium chloride (CsCl) density gradient centrifugation to separate DNA based on density differences.

Step 3: Principle

Heavy DNA (with \(^{15}\)N) forms a band at a different position than light DNA (\(^{14}\)N), allowing clear distinction.

Final Answer: (B) Quick Tip: Meselson–Stahl experiment → CsCl density gradient centrifugation.


Question 17:

During which geological period order Gnetales of gymnosperms are evolved?

(A) Permian
(B) Triassic
(C) Jurassic
(D) Cretaceous

Correct Answer: (D)
View Solution




Step 1: Understand Gnetales

Gnetales are a group of gymnosperms that show some advanced features similar to angiosperms.

Step 2: Geological Timeline

According to evolutionary records, Gnetales appeared during the Cretaceous period.

Step 3: Eliminate Other Options

Permian, Triassic, and Jurassic periods correspond to earlier evolutionary stages.

Final Answer: (D) Quick Tip: Gnetales → advanced gymnosperms → Cretaceous period.


Question 18:

For General Students: The molecule shown is naturally obtained from which plant?

(A) Papaver somniferum
(B) Cannabis sativa
(C) Erythroxylum coca
(D) Atropa belladonna

Correct Answer: (B)
View Solution




Step 1: Identify the Compound

The given structure corresponds to cannabinoids, which are chemical compounds found in cannabis.

Step 2: Source Plant

Cannabinoids are naturally obtained from Cannabis sativa.

Step 3: Eliminate Other Options

Papaver somniferum produces opium.

Erythroxylum coca produces cocaine.

Atropa belladonna produces atropine.

Final Answer: (B) Quick Tip: Cannabis → cannabinoids; Poppy → morphine; Coca → cocaine.


Question 19:

For Blind Students: Which drug is obtained by acetylation of morphine?

(A) Charas
(B) Smack
(C) Hashish
(D) Coke

Correct Answer: (B)
View Solution




Step 1: Understand Morphine Derivatives

Morphine undergoes acetylation to form diacetylmorphine.

Step 2: Common Name

Diacetylmorphine is commonly known as heroin or smack.

Step 3: Eliminate Other Options

Charas and hashish are cannabis products, while coke refers to cocaine.

Final Answer: (B) Quick Tip: Morphine + acetylation → heroin (smack).


Question 20:

In the given column, which one is mis-matched?

Correct Answer: (C)

Correct Answer: (C)
View Solution




Step 1: Verify Each Pair

(A) Correct → Citric acid by Aspergillus niger.

(B) Correct → Streptokinase used as clot buster.

(D) Correct → Yeast produces ethanol.

Step 2: Identify Error

Cyclosporin-A is actually produced by Trichoderma polysporum, not Monascus purpureus.

Final Answer: (C) Quick Tip: Monascus → statins; Trichoderma → cyclosporin-A.


Question 21:

Enzyme used for extension process in PCR method.

(A) Taq polymerase
(B) Restriction Endonuclease
(C) DNA Helicase
(D) DNA Ligase

Correct Answer: (A)
View Solution




Step 1: PCR Steps

PCR involves denaturation, annealing, and extension.

Step 2: Role of Taq Polymerase

Taq polymerase is a heat-stable enzyme that synthesizes DNA during the extension phase.

Step 3: Other Options

Restriction enzymes cut DNA, helicase unwinds DNA, ligase joins fragments.

Final Answer: (A) Quick Tip: PCR extension step → Taq polymerase.


Question 22:

RNA interference involves silencing of specific mRNA due to complementary _____ molecule that binds to and prevents translation of the mRNA.

(A) ds-DNA
(B) ss-RNA
(C) ds-RNA
(D) ss-DNA

Correct Answer: (C)
View Solution




Step 1: Define RNA Interference

RNA interference (RNAi) is a biological process where gene expression is silenced.

Step 2: Mechanism

Double-stranded RNA (dsRNA) binds to complementary mRNA and leads to its degradation, preventing protein synthesis.

Final Answer: (C) Quick Tip: RNAi → dsRNA binds mRNA → gene silencing.


Question 23:

Whose field experiment showed that on the rocky sea coast, the larger and competitively superior barnacle excludes smaller barnacle?

(A) Connell
(B) MacArthur
(C) Gause
(D) Verhulst-Pearl

Correct Answer: (A)
View Solution




Step 1: Study Description

Connell conducted experiments on barnacles in intertidal zones.

Step 2: Observation

He observed that larger barnacles outcompete and exclude smaller ones from their habitat.

Step 3: Concept

This demonstrates competitive exclusion principle in natural ecosystems.

Final Answer: (A) Quick Tip: Connell → barnacle experiment → competitive exclusion.


Question 24:

In an ecosystem 1000 J energy is stored in producer level, then how much energy will be obtained in secondary consumer level?

(A) 1 J
(B) 100 J
(C) 0.1 J
(D) 10 J

Correct Answer: (D)
View Solution




Step 1: Apply 10% Law of Energy Transfer

According to Lindeman’s 10% law, only 10% of energy is transferred from one trophic level to the next.

Step 2: Calculate Energy Flow

Producer = 1000 J

Primary consumer = 10% of 1000 = 100 J

Secondary consumer = 10% of 100 = 10 J

Step 3: Interpretation

Energy decreases significantly at each trophic level due to heat loss and metabolic activities.

Final Answer: (D) Quick Tip: 10% law → multiply by 0.1 at each trophic level.


Question 25:

The term Biodiversity is popularised by _____.

(A) Paul Ehrlich
(B) Edward Wilson
(C) David Tilman
(D) Von Humboldt

Correct Answer: (B)
View Solution




Step 1: Identify Key Scientist

Edward O. Wilson is known for promoting the concept of biodiversity.

Step 2: Contribution

He emphasized conservation of species diversity and ecosystem importance globally.

Final Answer: (B) Quick Tip: Edward Wilson = Father of Biodiversity.


Question 26:

Choose the correct option for column - I, II and III.
Column-I  Column-III
(i) Monoecious (p) Prevent both autogamy and geitonogamy (x) Caster, Maize
(ii) Dioecious  (q) Prevent autogamy but not geitonogamy (y) Papaya, Date palm

(A) I - i, II - q, III - y
(B) I - i, II - p, III - x
(C) I - ii, II - q, III - y
(D) I - i, II - q, III - x

Correct Answer: (D)
View Solution




Step 1: Monoecious Plants

Monoecious plants have both male and female flowers on the same plant → prevents autogamy but allows geitonogamy.

Step 2: Dioecious Plants

Dioecious plants have separate male and female plants → prevents both autogamy and geitonogamy.

Step 3: Examples

Caster and maize are monoecious plants.

Final Answer: (D) Quick Tip: Monoecious → same plant; Dioecious → separate plants.


Question 27:

In the given diagram, choose the correct option for labelled 'X', 'Y' and 'Z'.

(A) X - Zona pellucida, Y - Corona radiata, Z - Perivitelline space
(B) X - Corona radiata, Y - Zona pellucida, Z - Perivitelline space
(C) X - Zona pellucida, Y - Perivitelline space, Z - Corona radiata
(D) X - Corona radiata, Y - Perivitelline space, Z - Zona pellucida

Correct Answer: (A)
View Solution




Step 1: Outer Layer

Corona radiata surrounds the ovum externally.

Step 2: Middle Layer

Zona pellucida is a thick glycoprotein layer around the oocyte.

Step 3: Inner Space

Perivitelline space lies between zona pellucida and plasma membrane.

Final Answer: (A) Quick Tip: Order: Corona radiata → Zona pellucida → Perivitelline space.


Question 28:

In human during fertilisation entry of additional sperms into ovum is blocked by _____.

(A) Zona pellucida
(B) Perivitelline space
(C) Corona radiata
(D) Theca layer

Correct Answer: (A)
View Solution




Step 1: Concept of Polyspermy

Polyspermy refers to entry of more than one sperm into the ovum, which must be prevented for normal development.

Step 2: Role of Zona Pellucida

After one sperm enters, zona pellucida undergoes structural changes (zona reaction) that prevent entry of additional sperms.

Step 3: Eliminate Other Options

Corona radiata helps sperm entry, not blocking. Perivitelline space is just a gap. Theca layer is part of follicle.

Final Answer: (A) Quick Tip: Zona pellucida → blocks polyspermy (very important concept).


Question 29:

In the following STIs, which one is not completely curable?

(A) Genital warts, Chlamydiasis, HIV
(B) Trichomoniasis, HIV, Syphilis
(C) Hepatitis-B, Genital herpes, HIV
(D) Gonorrhoea, HIV, Chlamydiasis

Correct Answer: (C)
View Solution




Step 1: Curable vs Non-curable

Bacterial infections like gonorrhoea and chlamydia are curable with antibiotics.

Step 2: Viral STIs

Viral infections such as HIV, Hepatitis-B, and Genital herpes are not completely curable.

Step 3: Select Option

Option (C) contains only non-curable diseases.

Final Answer: (C) Quick Tip: Viral STIs → not curable; Bacterial STIs → curable.


Question 30:

A colour blind man marries a female with homozygous normal vision. What will be the possibility of colour blindness in male children?

(A) 100%
(B) 50%
(C) 25%
(D) 0%

Correct Answer: (D)
View Solution




Step 1: Understand Inheritance

Colour blindness is an X-linked recessive trait.

Step 2: Parental Genotype

Male (colour blind) = X\(^c\)Y

Female (normal homozygous) = X\(^C\)X\(^C\)

Step 3: Male Children

Male child receives Y from father and X from mother → X\(^C\)Y (normal).

Final Answer: (D) Quick Tip: Sons get X from mother → check mother’s genotype only.


Question 31:

In human genome project among two major approaches sequencing the whole set of genome and assigning functions is called _____.

(A) Bioinformatics
(B) Sequence Annotation
(C) Expressed Sequence Tags
(D) Single nucleotide polymorphism

Correct Answer: (B)
View Solution




Step 1: Genome Sequencing

Human genome project involved sequencing DNA and identifying genes.

Step 2: Annotation

Assigning functions to identified sequences is called sequence annotation.

Final Answer: (B) Quick Tip: Sequencing = reading DNA; Annotation = assigning function.


Question 32:

In natural selection, more individuals acquire value other than the mean character value, which indicates _____.

(A) Stabilising
(B) Stabilising directional both
(C) Disruptive
(D) Directional

Correct Answer: (D)
View Solution




Step 1: Types of Natural Selection

Stabilising → favors average traits.

Disruptive → favors extremes.

Directional → shifts population toward one extreme.

Step 2: Interpretation

When individuals move away from mean toward one side → directional selection.

Final Answer: (D) Quick Tip: Directional selection → shift of mean toward one extreme.


Question 33:

Snake venom antibody is the antibody prepared against snake venom is the example of which type of immunisation?

(A) Active immunity
(B) Auto immunity
(C) Passive immunity
(D) Allergies

Correct Answer: (C)
View Solution




Step 1: Understand Passive Immunity

Passive immunity involves transfer of ready-made antibodies into the body.

Step 2: Snake Venom Case

Antivenom contains pre-formed antibodies produced in another organism and injected into patient.

Step 3: Conclusion

Since antibodies are directly provided → passive immunity.

Final Answer: (C) Quick Tip: Passive immunity = ready-made antibodies (instant protection).


Question 34:

Which group of microbes are acting as bio control agents?


(A) Baculo virus, Oscillatoria, Rhizobium
(B) Azotobacter, Trichoderma, Glomus
(C) Trichoderma, Baculo virus, Bacillus thuringiensis
(D) Azospirillum, Anabaena, Bacillus thuringiensis

Correct Answer: (C)
View Solution




Step 1: Define Biocontrol Agents

Biocontrol agents are organisms used to control pests and diseases biologically.

Step 2: Identify Correct Group

Trichoderma → fungal biocontrol agent.

Baculovirus → controls insect pests.

Bacillus thuringiensis (Bt) → insecticidal bacteria.

Final Answer: (C) Quick Tip: Bt, Baculovirus, Trichoderma = key biocontrol agents.


Question 35:

What is the correct nomenclature for restriction enzyme - I, obtained from Bacillus amyloliquefaciens H?

(A) Bam HI
(B) Bac HI
(C) Baam HI
(D) Bam IH

Correct Answer: (A)
View Solution




Step 1: Naming Rule

Restriction enzymes are named from organism:

First letter (genus) + next two letters (species) + strain.

Step 2: Apply Rule

Bacillus amyloliquefaciens → Bam

Strain H → H

Roman numeral I → first enzyme

Final Answer: (A) Quick Tip: Naming = Genus + species + strain + number → BamHI.


Question 36:

In which kind of population interaction, one species is harmed whereas the other is unaffected?

(A) Competition
(B) Amensalism
(C) Commensalism
(D) Parasitism

Correct Answer: (B)
View Solution




Step 1: Define Interactions

Competition → both harmed.

Commensalism → one benefits, other unaffected.

Parasitism → one benefits, other harmed.

Amensalism → one harmed, other unaffected.

Step 2: Match Condition

Given condition matches amensalism.

Final Answer: (B) Quick Tip: Amensalism = (-, 0) interaction.


Question 37:

Which two parts of the different plants can be fused for the formation of new plant in somatic hybridisation?


(A) Pollen grain
(B) Male - female gamete
(C) Protoplast
(D) Eggs

Correct Answer: (C)
View Solution




Step 1: Define Somatic Hybridisation

It involves fusion of somatic cells from different plants.

Step 2: Role of Protoplast

Protoplasts are cells without cell wall and can fuse easily.

Step 3: Conclusion

Fusion of protoplasts forms hybrid cells.

Final Answer: (C) Quick Tip: Somatic hybridisation = fusion of protoplasts.


Question 38:

Sparrow eats seeds. After wards it eats insects. In which trophic level, sparrow belongs in both situations respectively?

(A) Primary - Secondary
(B) Tertiary - Quaternary
(C) Secondary - Tertiary
(D) Primary - Tertiary

Correct Answer: (A)
View Solution




Step 1: When Sparrow Eats Seeds

Seeds are producers → sparrow acts as primary consumer.

Step 2: When Sparrow Eats Insects

Insects are primary consumers → sparrow becomes secondary consumer.

Final Answer: (A) Quick Tip: Plant eater → primary; animal eater → secondary.


Question 39:

Oxygen available from the nature belongs to which biodiversity conservation category?

(A) Narrowly utilitarian
(B) Intangible benefits
(C) Broadly utilitarian
(D) Ethical

Correct Answer: (C)
View Solution




Step 1: Broadly Utilitarian Values

These include ecosystem services like oxygen production, pollination, climate regulation.

Step 2: Oxygen Role

Oxygen is essential for life and provided by plants → ecosystem service.

Final Answer: (C) Quick Tip: Ecosystem services → broadly utilitarian values.


Question 40:

Assertion (A): Geitonogamy is genetically similar to autogamy.
Reason (R): In geitonogamy pollen grains come from the same plant.

(A) A and R both are correct, R is correct explanation of A
(B) A and R both are correct, R is not correct explanation
(C) A is correct but R is wrong
(D) A is wrong but R is correct

Correct Answer: (A)
View Solution




Step 1: Analyze Assertion

Geitonogamy involves transfer of pollen between different flowers of same plant → genetically similar to autogamy.

Step 2: Analyze Reason

Pollen comes from same plant → no genetic variation.

Step 3: Relation

Reason correctly explains why it is genetically similar.

Final Answer: (A) Quick Tip: Same plant → no variation → like self-pollination.


Question 41:

Identify 'X', 'Y' and 'Z' from the given diagram.

(A) X - Seed coat, Y - Cotyledon, Z - Endosperm
(B) X - Seed coat, Y - Endosperm, Z - Cotyledon
(C) X - Pericarp, Y - Root tip, Z - Endosperm
(D) X - Pericarp, Y - Endosperm, Z - Scutellum

Correct Answer: (B)
View Solution




Step 1: Outer Layer

Seed coat protects the seed.

Step 2: Endosperm

Large food storage tissue in monocots.

Step 3: Cotyledon

Small structure (scutellum) absorbs nutrients.

Final Answer: (B) Quick Tip: Monocot seed → Endosperm large, cotyledon small.


Question 42:

Only for Blind Students: In grass family the cotyledon is called _____.


(A) Hypocotyl
(B) Scutellum
(C) Epicotyl
(D) Coleoptile

Correct Answer: (B)
View Solution




Step 1: Grass Seed Structure

In monocots, cotyledon is modified into scutellum.

Final Answer: (B) Quick Tip: Monocot cotyledon = Scutellum.


Question 43:

The seminiferous tubules of the testis open into “X” through “Y”.

(A) X - Vas deferens, Y - rete testis
(B) X - rete testis, Y - epididymis
(C) X - Vas deferens, Y - epididymis
(D) X - Vasa efferentia, Y - rete testis

Correct Answer: (D)
View Solution




Step 1: Pathway

Seminiferous tubules → rete testis → vasa efferentia → epididymis.

Step 2: Match

They open into vasa efferentia through rete testis.

Final Answer: (D) Quick Tip: Flow: Seminiferous → rete testis → vasa efferentia.

GUJCET 2026 - Biology | MARATHON all PYQs + CBN IMP MCQs