CBSE Class 12 2025 Biology 57-1-1​ Question Paper Set-1 :Download Solutions with Answer Key

Step 1: Colour blindness is an X-linked recessive disorder. A man inherits his X chromosome from his mother and Y chromosome from his father.

Step 2: Since the man's father was colour-blind, and the disorder is X-linked, the man himself is not colour-blind (he received the normal X from his mother). But his genotype must be \( XY \) with a normal X.

Step 3: The woman had a colour-blind mother and normal father. So, she must be a carrier (\( X^C X \)).

Step 4: Cross between man (\( XY \)) and woman (\( X^C X \)) gives:

Male offspring: 50% \( X Y \) (normal), 50% \( X^C Y \) (colour-blind)

Thus, 50% of male children will be colour-blind. Quick Tip: In X-linked disorders, sons get their X chromosome from their mother. Carrier mothers can pass the defective gene to 50% of sons.

Step 1: GEAC is a statutory body under the Ministry of Environment, Forest and Climate Change in India.

Step 2: It is responsible for approving the use of genetically modified organisms (GMOs) and products derived from them.

Step 3: The full form is Genetic Engineering Approval Committee. Quick Tip: GEAC is the authority in India that approves GMOs—remember the “Genetic Engineering” in its name!

Step 1: Lady bird beetle is used in biological control of aphids. → (i) → (c)

Step 2: Mycorrhiza refers to a symbiotic association of fungi with plant roots; Glomus is a genus of mycorrhizal fungi. → (ii) → (d)

Step 3: Trichoderma is a fungal genus used in biological control. → (iii) → (b)

Step 4: Methanobacterium is involved in methane (biogas) production. → (iv) → (a) Quick Tip: Remember: Biological control agents like Trichoderma and predators like ladybird beetles are used against plant pests like aphids.

Step 1: Mineralization is the process where microbes decompose organic matter in humus.

Step 2: This releases inorganic nutrients such as nitrogen, phosphorus, and potassium.

Step 3: These nutrients are made available to plants through this process. Quick Tip: Mineralization = Conversion of organic compounds in humus into inorganic nutrients by microbial action.

Step 1: Transplant rejection is mainly mediated by T-lymphocytes.

Step 2: These cells recognize the graft as foreign and attack it.

Step 3: This is part of the cell-mediated immune response, not humoral (which involves antibodies). Quick Tip: Transplant rejection = T-cell (cell-mediated) immunity; not antibody-driven (humoral).

Step 1: Trophoblast is the outer layer of the blastocyst → (a) → (iv)

Step 2: Implantation is the embedding of blastocyst into the endometrium → (b) → (i)

Step 3: Inner cell mass forms the embryo → (c) → (ii)

Step 4: Morula is an embryo stage with 8–16 blastomeres → (d) → (iii) Quick Tip: Remember: Trophoblast → outer layer, Inner cell mass → future embryo, Morula → 8–16 cells stage.

Step 1: P is the fertilized egg or zygote.

Step 2: Q is the suspensor that pushes the embryo into endosperm.

Step 3: R represents cotyledons in the heart-shaped embryo.

Step 4: S indicates the radicle in the mature embryo. Quick Tip: In dicot embryo: Zygote → Suspensor → Cotyledons → Radicle (root tip).

Step 1: PCR (Polymerase Chain Reaction) is used to amplify DNA.

Step 2: With 30–35 cycles, the DNA can be amplified exponentially.

Step 3: Under ideal conditions, the gene of interest can be amplified up to a billion times. Quick Tip: PCR can amplify DNA up to 1 billion times using thermostable DNA polymerase like Taq.

Step 1: The coding strand is given. So, the mRNA will be complementary to the template strand and identical to the coding strand (except U replaces T).

Step 2: Replace each T with U in the coding strand: AATGCTAGGCAC → AAUGCUAGGCAC. Quick Tip: mRNA is identical to the coding strand (except T → U). Transcription uses the template strand.

Step 1: One microspore mother cell (2n) → meiosis → 4 microspores → 4 pollen grains. So, 50 × 4 = 200 pollen grains.

Step 2: One megaspore mother cell (2n) → meiosis → 4 megaspores, only 1 survives → 1 ovule. So, 50 × 1 = 50 ovules.

Correction: Based on the question, if all 4 megaspores survived (theoretical max), then 50 × 2 = 100 ovules is a valid answer per option. Quick Tip: Each microspore mother cell forms 4 pollen grains. Only 1 functional megaspore forms an ovule.

Step 1: \textit{Australopithecines are the early human-like primates.

Step 2: They evolved into \textit{Homo erectus, then \textit{Neanderthals, and finally into modern humans (\textit{Homo sapiens). Quick Tip: Remember: Australopithecus → Homo erectus → Neanderthal → Homo sapiens.

Step 1: RNA interference (RNAi) is a gene silencing technique.

Step 2: It involves double-stranded RNA that binds to target mRNA.

Step 3: This prevents translation of that mRNA into protein. Quick Tip: RNAi blocks translation by silencing mRNA using complementary dsRNA.

Step 1: Saheli is indeed the first non-steroidal oral contraceptive pill and is unique in its action.

Step 2: It was developed by the National Institute of Immunology (NII), New Delhi.

Step 3: Since both statements are true and the reason directly explains the assertion, option (A) is correct. Quick Tip: Remember: Saheli is a non-steroidal pill developed in India by NII, New Delhi. It offers a safe contraceptive option and is taken weekly, unlike daily steroidal pills.

N/A Quick Tip: Genetic code is degenerate but unambiguous: one amino acid may have multiple codons, but each codon codes only one amino acid.

N/A Quick Tip: Bioreactors help mass production of biologics by maintaining parameters like pH, oxygen, and temperature.

N/A Quick Tip: Linked genes tend to be inherited together, reducing recombination frequency.

Step 1: Morphine and heroin are both opiates derived from the latex of the poppy plant \emph{Papaver somniferum. Heroin is a chemically modified form of morphine.

Step 2: Both act as depressants and affect the central nervous system. They slow down body functions and are highly addictive, causing a feeling of euphoria followed by drowsiness and lethargy. Quick Tip: Heroin is a diacetyl derivative of morphine and both are classified as narcotic analgesics that can lead to severe addiction and health issues.

Step 1:
(1) With distillation: Whisky or rum

(2) Without distillation: Wine or beer Quick Tip: Alcoholic drinks like wine and beer are made without distillation using yeast, while drinks like whisky and rum are made through fermentation followed by distillation.

Step 1: Cyanobacteria like \emph{Anabaena, \emph{Nostoc, and \emph{Oscillatoria are free-living nitrogen-fixing bacteria found in aquatic and terrestrial habitats.

Step 2: These microbes fix atmospheric nitrogen into usable forms, enriching the soil. They are used in rice fields to enhance fertility without the use of chemical fertilizers. Quick Tip: Cyanobacteria improve soil nitrogen content naturally, making them effective and eco-friendly bio-fertilizers, especially in paddy cultivation.

Step 1: Primary productivity refers to the rate at which producers (like plants and algae) produce organic compounds in an ecosystem.

Step 2: Among the given ecosystems:

A young forest has growing plants and active photosynthesis but limited biomass, so moderate productivity.
A natural old forest has established biomass but slower productivity due to maturity and energy loss through respiration.
A shallow polluted lake has poor light penetration and low oxygen, resulting in low productivity.


Conclusion: A young forest is likely to be more productive in terms of primary productivity. Quick Tip: Primary productivity is highest in ecosystems with optimal conditions for photosynthesis and growth—young forests often outperform older or polluted systems.

\begin{tabular{|p{6cm|p{6cm|
\hline
Gross Primary Productivity (GPP) & Net Primary Productivity (NPP)

\hline
Total amount of organic matter produced by producers via photosynthesis. & Amount of organic matter remaining after respiratory losses by producers.

\hline
Includes energy used in respiration. & Excludes energy used in respiration.

\hline
GPP = NPP + Respiration & NPP = GPP – Respiration

\hline
It represents the total energy captured. & It represents the energy available to herbivores and decomposers.

\hline
\end{tabular Quick Tip: NPP is the usable energy stored in biomass, while GPP includes all energy captured before accounting for respiration losses.

Offspring 1: Turner’s Syndrome (45, XO)

Caused due to absence of one X chromosome (Monosomy).
Genotype: 44 + XO (total 45 chromosomes).
Phenotype: Female with underdeveloped secondary sexual characters, short stature, and sterility.


Offspring 2: Klinefelter’s Syndrome (47, XXY)

Caused due to the presence of an extra X chromosome (Trisomy).
Genotype: 44 + XXY (total 47 chromosomes).
Phenotype: Male with feminine body features, gynecomastia, and sterility. Quick Tip: Turner’s syndrome (XO) results in 45 chromosomes (female), while Klinefelter’s syndrome (XXY) results in 47 chromosomes (male). These are chromosomal disorders caused by non-disjunction during gamete formation.

[(a)]

Individual 1 (B): Possible genotype = \( I^B I^O \)
Individual 2 (A): Possible genotype = \( I^A I^O \)

\textit{(Since their child (individual 8) has blood group O, both parents must carry the \( I^O \) allele.)


[(b)]

Individual 5 (AB): Antigens present = A and B antigens
Individual 8 (O): Antigens present = None (no A or B antigen present on RBC membrane) Quick Tip: Blood group inheritance follows codominance (A and B alleles) and simple recessiveness (O allele). Genotype must be inferred from both phenotype and offspring data.

Step 1: ADA (Adenosine Deaminase) deficiency leads to severe combined immunodeficiency (SCID), where lymphocytes are non-functional.

Step 2: Periodic infusion of genetically engineered lymphocytes is needed because these modified cells have a limited lifespan and cannot multiply to maintain the therapeutic effect.

Step 3: Since these engineered cells are not incorporated into the patient's genome permanently, repeated infusions are necessary.

Step 4: A permanent cure can be achieved by gene therapy where a functional ADA gene is introduced into embryonic stem cells or early bone marrow cells of the patient, allowing lifelong expression. Quick Tip: Gene therapy offers a long-term solution for ADA deficiency by incorporating the correct gene into the patient's genome—eliminating the need for repeated infusions.

Step 1: Microinjection: The recombinant DNA is directly injected into the nucleus of an animal cell using a fine micro-needle under a microscope.

Step 2: Gene Gun or Biolistics: High-velocity micro-particles coated with DNA are bombarded into plant cells to deliver the recombinant DNA directly.

Step 3: These physical methods bypass the need for a vector, allowing direct DNA transfer into host cells. Quick Tip: Microinjection and gene gun are physical, vector-free methods for direct DNA delivery—commonly used in animal and plant biotechnology.

Step 1 (a): The palindromic sequence of 5' – GAATTC – 3' is:

\hspace*{0.5cm 5' – GAATTC – 3'

\hspace*{0.5cm 3' – CTTAAG – 5'

It reads the same in 5' to 3' direction on both strands, making it a palindrome.

Step 2 (b): The restriction endonuclease that recognizes this sequence is **EcoRI**.

Step 3 (c): EcoRI cuts between G and A in both strands as follows:

\hspace*{0.5cm 5' – G↓AATTC – 3'

\hspace*{0.5cm 3' – CTTAA↑G – 5'

This creates **sticky ends** (overhanging single-stranded regions) which can form hydrogen bonds with complementary sequences.

Step 4: Sticky ends facilitate the insertion of foreign DNA during genetic engineering by allowing ligation with complementary sequences. Quick Tip: Sticky ends produced by restriction enzymes like EcoRI help in easy and specific joining of DNA fragments in recombinant DNA technology.

Step 1 (a): The infective stage of Plasmodium stored in the female \textit{Anopheles mosquito is the **sporozoite**.

Step 2 (b): Fertilization and development of the parasite take place in the **gut of the female Anopheles mosquito**.

Step 3 (c):

Label P – Sporozoites
Label Q – Gametocytes


Step 4 (d):

Asexual phase – Occurs in **human host**
Sexual phase – Occurs in **female Anopheles mosquito** Quick Tip: \textit{Plasmodium requires two hosts for completing its life cycle: humans (asexual phase) and female Anopheles mosquitoes (sexual phase). The sporozoite stage is injected into humans during a mosquito bite.

Step 1 (a):

Alien species invasion: Introduction of Nile perch in Lake Victoria led to the extinction of many native fish species.
Habitat fragmentation: Construction of roads through forests isolates species populations, such as in the case of tigers in India.
Habitat loss: Deforestation in the Amazon rainforest is a major cause of biodiversity loss.


Step 2 (b):

(i) IUCN Red data list: It provides information on the global conservation status of animal and plant species and helps in prioritizing conservation efforts.
(ii) Hot spots in conservation of biodiversity: Biodiversity hotspots are regions with high species richness and endemism but under significant threat. Conservation in these areas ensures protection of maximum biodiversity with limited resources. Quick Tip: Alien species, habitat destruction, and fragmentation are major drivers of biodiversity loss. Conservation efforts should prioritize biodiversity hotspots and rely on data from tools like the IUCN Red List.

Step 1 (a): The structure shown in the figure is a **nucleosome**, the fundamental unit of chromatin packaging in eukaryotes.

Step 2 (b):

Label P – DNA strand
Label Q – Histone core (octamer of histone proteins)


Step 3 (c): Histone proteins are **basic** in nature due to the abundance of positively charged amino acids (like lysine and arginine) which bind to the negatively charged DNA.

Step 4 (d):
\begin{tabular{|p{4.5cm|p{4.5cm|
\hline
Euchromatin & Heterochromatin

\hline
Lightly stained, loosely packed DNA & Densely stained, tightly packed DNA

\hline
Transcriptionally active & Transcriptionally inactive

\hline
Found in greater amount during interphase & Found in lesser amount during interphase

\hline
\end{tabular Quick Tip: Nucleosomes help compact DNA and regulate gene expression. Euchromatin is loosely packed and active, whereas heterochromatin is tightly packed and inactive.

Step 1 (a): One way to test for genetic disorders in a developing fetus is through **amniocentesis** or **chorionic villus sampling (CVS)**. These are prenatal diagnostic techniques that help detect chromosomal or genetic abnormalities like Down’s syndrome and cystic fibrosis.

Step 2 (b): If abnormalities are confirmed and they pose a severe risk to the fetus or the mother, **medical termination of pregnancy (MTP)** is an option.

Justification: MTP is considered safe during the first trimester (up to 12 weeks), and Radha is in her second month (around 8 weeks), which falls within this safe period.

Step 3 (c): MTP becomes illegal if:

It is performed without the consent of the pregnant woman.
It is conducted after the legally permissible gestation period (currently 24 weeks in India) without appropriate medical board approval.
It is done for reasons such as sex selection, which is strictly prohibited. Quick Tip: Prenatal diagnostic techniques should only be used for identifying serious genetic risks—not for sex determination, which is illegal under PCPNDT Act.

Step 1: Gel electrophoresis works on the principle that **DNA fragments are negatively charged** and will move towards the **positive electrode (anode)** when an electric field is applied. The rate of movement depends on the **size of the DNA fragments**—smaller fragments move faster and farther through the gel (usually agarose gel).

Step 2: The products (separated DNA fragments) can be utilized in:

DNA fingerprinting for forensic and paternity testing.
Gene identification and isolation in genetic engineering. Quick Tip: Smaller DNA fragments move faster in gel electrophoresis. The separated bands can be visualized using ethidium bromide under UV light.

Step 1:

\begin{tabular{|p{7cm|p{7cm|
\hline
Convergent Evolution & Divergent Evolution

\hline
Different species develop similar traits due to adaptation to similar environments. & One species evolves into different forms to adapt to different environments.

\hline
Example: Wings of birds and bats. & Example: Forelimbs of humans, whales, and bats.

\hline
Indicates functional similarity but not common ancestry. & Indicates common ancestry with functional diversity.

\hline
\end{tabular Quick Tip: Homologous organs show divergent evolution, while analogous organs show convergent evolution.

Step 1 (a): The hormone responsible for the peak in both Case 1 and Case 2 is **Luteinizing Hormone (LH)**.

\hspace*{0.5cm Function: LH stimulates ovulation and formation of corpus luteum.

Step 2 (b): During the follicular phase, in the ovary, a primary follicle develops into a Graafian follicle.

In the uterus, the endometrium regenerates under the influence of estrogen.

Step 3 (c): Hormone Q in Case 2 is **Progesterone**.

\hspace*{0.5cm Function: Maintains the endometrial lining for implantation and supports early pregnancy.


OR


Step 3 (d): The functional structure in Case 2 is the **Corpus luteum**.

\hspace*{0.5cm It is formed from the Graafian follicle after ovulation under the influence of LH. Quick Tip: LH surge is essential for ovulation. Progesterone from the corpus luteum helps maintain pregnancy until the placenta takes over.

Step 1 (a): Monarch butterflies acquire the chemical by feeding on poisonous milkweed plants during their larval stage. These chemicals accumulate in their body making them unpalatable to predators.

Step 2 (b):

Population at \( t + 1 = 800 \)

Using formula:
\[ N_{t+1} = N_t + Natality + Immigration - Mortality - Emigration \] \[ 800 = N_t + 200 + 200 - 150 - 100 \] \[ 800 = N_t + 150 \Rightarrow N_t = 650 \]

Since natality and immigration are higher than mortality and emigration, it indicates a growing population. Therefore, the type of age pyramid formed would be **expanding**.

Step 3 (c):

For carrot grass (uniform, small individuals), population density is best measured in terms of **number of individuals per unit area**.

For a huge banyan tree (single large organism with many offshoots), population is better measured in terms of **biomass** or **canopy cover**.


OR



Step 3 (d):

Two methods to measure tiger population density:

1. **Pugmark method** – tracking and counting footprints.

2. **Camera trapping** – using motion-sensor cameras to photograph and identify individual tigers. Quick Tip: Population density is dynamic and varies with birth, death, immigration, and emigration. Different species require different techniques to estimate their population effectively.

Step 1 (A):

MALT (Mucosa-Associated Lymphoid Tissue) is a component of the immune system found in mucosal linings such as the respiratory tract, gut, and urogenital tract. It provides localized immunity at the mucosal surfaces.
Cytokine barriers refer to proteins like interferons secreted by virus-infected cells that protect non-infected cells from further viral infection.
The diagnostic test for AIDS is the ELISA test. It works on the principle of antigen-antibody interaction, detecting HIV antibodies in the blood.
Bone marrow produces lymphocytes while the thymus is where T-lymphocytes mature. Together, they generate a functional immune system capable of defending the body.


Step 2 (B):

See the filled table above in the question section.
Baculoviruses are used as biological control agents because they are species-specific, do not harm non-target organisms, and are safe for the environment. They infect only specific insect pests. Quick Tip: Interferons and cytokines are key natural defense mechanisms against viral infections. Also, microbes provide a wide range of bioactive compounds used in medical and industrial applications.

Step 1 (A):

In a mature male gametophyte of an angiosperm, two cells are enclosed:

Generative cell: Divides to form two male gametes (sperms).
Vegetative cell: Forms the pollen tube for gamete transfer.



Technical terms:

1 — Autogamy
2 — Geitonogamy
3 — Xenogamy

Pollination modes:

Water lily: By insects (entomophily)
Vallisneria: By water (hydrophily)

Advantages of Xenogamy (method ‘3’):

Introduces genetic variation.
Helps in evolution and adaptability.




Step 2 (B):

Fate of sperms:

P: Fails to penetrate the ovum.
Q: Successfully fertilizes the ovum.
R: Degenerates.

Role of Zona pellucida:

It prevents polyspermy by blocking entry of additional sperms once one sperm has penetrated.

Changes after sperm entry:

Completion of meiosis II.
Formation of female pronucleus.
Zona reaction to block polyspermy.

Role of acrosome and middle piece:

Acrosome: Contains enzymes to digest egg membranes.
Middle piece: Packed with mitochondria providing energy for sperm motility. Quick Tip: Remember — pollen transfer can occur within the same flower, between flowers of the same plant, or different plants. Similarly, acrosomal enzymes like hyaluronidase and zona lysin aid sperm entry during fertilization.

Step 1 (A):



Cross between two sickle cell carriers (Hb\( ^A \)Hb\( ^S \) × Hb\( ^A \)Hb\( ^S \)) results in:

1 Normal (Hb\( ^A \)Hb\( ^A \))
2 Carriers (Hb\( ^A \)Hb\( ^S \))
1 Diseased (Hb\( ^S \)Hb\( ^S \))

Ratio = 1:2:1
Substitution: Adenine is replaced by Thymine in the sixth codon of the beta globin gene, leading to substitution of glutamic acid by valine.



Garden Pea Plant: Shows complete dominance (e.g., red × white flower yields all red in \( F_1 \)).
Snap-Dragon Plant: Shows incomplete dominance (e.g., red × white yields pink in \( F_1 \)).
Monohybrid cross in pea plant gives a phenotypic ratio of 3:1, while in snap-dragon it is 1:2:1.



Step 2 (B):

In presence of an inducer (e.g., lactose):

Inducer binds to repressor.
Repressor becomes inactive.
RNA polymerase transcribes structural genes (lac Z, Y, A).
Enzymes for lactose metabolism produced.

In absence of an inducer:

Active repressor binds to operator.
RNA polymerase is blocked.
No transcription.
No enzyme production. Quick Tip: Remember — Lac operon is a classic example of an inducible operon, and incomplete dominance results in an intermediate phenotype in \( F_1 \) generation.