CBSE Class 12 2024 Biology Set 2 Question Paper (Paper Code: 57/4/2) is available for download. The exam was successfully conducted by CBSE on March 19 in the morning session from 10:30 AM to 1:30 PM. As per the student’s initial reactions, the CBSE Class 12 2024 Biology Set 2 Question Paper was reported as Moderate. The Ecology section in the CBSE Class 12 2024 Biology Set 2 Question Paper was reported as Easy to Moderate, Genetics and Evolution as Challenging, and Human Physiology as Moderate.
CBSE Class 12 2024 Biology Set 2 57/4/2 Question Paper with Answer Key PDF
Candidates can download the CBSE Class 12 Biology Question Paper with Solution and Answer Key PDFs for Set 2 Question Paper (Code: 57/4/2) using the link below.
CBSE Class 12 2024 Biology Questions with Solutions
SECTION - A
Question 1:
In a fertilized ovule of an angiosperm, the cells in which n, 2n, and 3n conditions respectively occur are:
- Antipodal, zygote, and endosperm
- Zygote, nucellus, and endosperm
- Endosperm, nucellus, and zygote
- Antipodals, synergids, and integuments
Correct Answer: 1
View Solution
- n (haploid): Found in antipodal cells.
- 2n (diploid): Found in the zygote formed by fertilization.
- 3n (triploid): Found in the endosperm formed by double fertilization.
This distribution of ploidy levels is a characteristic feature of angiosperm reproduction.
Question 2:
Turner’s syndrome in humans occurs due to:
- Aneuploidy
- Euploidy
- Polyploidy
- Autosomal abnormality
Correct Answer: 1
View Solution
- Turner’s syndrome is caused by the absence of one X chromosome, resulting in a 45, XO chromosomal pattern.
- This is a type of aneuploidy, which refers to the presence of an abnormal number of chromosomes.
Turner’s syndrome affects females and is characterized by short stature, infertility, and other physical abnormalities.
Question 3:
Which of the options has correct identification of ‘P’, ‘Q’, and ‘R’ in the illustration of ‘Central Dogma’?
![DNA replication, transcription, and translation.]()
- P – Replication, Q – rRNA, R – Transcription
- P – Translation, Q – mRNA, R – Transcription
- P – Replication, Q – mRNA, R – Translation
- P – Transcription, Q – mRNA, R – Translation
Correct Answer: 4
View Solution
- P – Transcription: The process of synthesizing mRNA from DNA.
- Q – mRNA: Serves as a template for protein synthesis.
- R – Translation: The process where the mRNA sequence is decoded to synthesize a protein.
The Central Dogma describes the flow of genetic information from DNA to RNA to Protein.
Question 4:
Study the following diagram of the transverse section of a young anther of an angiosperm. Identify the correctly labeled parts:
![Cross-section of an anther. Labelled parts: A - Epidermis, B - Pollen sacs, C - Connective tissue.]()
- A – Connective, B – Endothecium, C – Pollen grain
- A – Endothecium, B – Connective, C – Pollen grain
- A – Pollen grain, B – Connective, C – Endothecium
- A – Endothecium, B – Pollen grain, C – Connective
Correct Answer: 1
View Solution
- A – Connective: Tissue joining the two lobes of the anther.
- B – Endothecium: The outer layer of the pollen sac that provides mechanical support.
- C – Pollen grain: The male gametophyte formed within the pollen sac.
The anther is a bilobed structure that contains microsporangia, which develop into pollen sacs housing pollen grains.
Question 5:
Select the option that correctly identifies ovum, morula, and blastocyst in the diagram of human female reproduction:
![Stages of male gametophyte development in flowering plants. Labels:]()
- Ovum – B, Morula – D, Blastocyst – F
- Ovum – A, Morula – B, Blastocyst – G
- Ovum – A, Morula – E, Blastocyst – G
- Ovum – B, Morula – D, Blastocyst – G
Correct Answer: 4
View Solution
- Ovum (B): The unfertilized egg.
- Morula (D): A solid ball of cells formed after a series of mitotic divisions.
- Blastocyst (G): A hollow structure in early embryonic development that implants into the uterine wall.
The blastocyst stage is critical for implantation into the uterine lining, marking the beginning of pregnancy.
Question 6:
Study the table below and identify the correct matching of contraceptive methods with their mode of action:
| Contraceptive |
Mode of Action |
| A. The pill |
III. Inhibits ovulation |
| B. Condom |
I. Prevent sperm reaching cervix |
| C. Vasectomy |
IV. Semen contains no sperm |
| D. Copper-T |
II. Prevent implantation |
- A – III, B – II, C – I, D – IV
- A – II, B – III, C – I, D – IV
- A – III, B – I, C – IV, D – II
- A – IV, B – III, C – II, D – I
Correct Answer: 3
View Solution
- A – The pill: Inhibits ovulation (III).
- B – Condom: Prevents sperm from reaching the cervix (I).
- C – Vasectomy: Semen contains no sperm after the vas deferens is cut or tied (IV).
- D – Copper-T: Prevents implantation of the fertilized egg (II).
Question 7:
Identify the category of genetic disorder depicted in the pedigree chart below:
![Pedigree diagram representing inheritance patterns.]()
- X-Linked recessive
- X-Linked dominant
- Autosomal recessive
- Autosomal dominant
Correct Answer: 3
View Solution
- The disorder is seen in both males and females, indicating autosomal inheritance.
- It skips generations, suggesting recessive inheritance.
- Affected offspring have parents who are carriers of the recessive allele.
Autosomal recessive disorders often require both parents to contribute the recessive allele for the disorder to manifest in offspring.
Question 8:
ELISA technique is based on the principle of:
- DNA replication
- Antigen-antibody interaction
- Pathogen-antigen interaction
- Antigen-protein interaction
Correct Answer: 2
View Solution
- ELISA (Enzyme-Linked Immunosorbent Assay) detects specific antigens or antibodies using the principle of antigen-antibody interaction.
- The binding is detected through an enzyme-linked reaction, often resulting in a color change.
ELISA is extensively used in medical diagnostics, such as detecting infections like HIV and hepatitis.
Question 9:
Homologous organs indicate:
- Convergent evolution
- Divergent evolution
- Adaptive radiation
- Natural selection
Correct Answer: 2
View Solution
- Homologous organs have a similar structure but may perform different functions, e.g., human arm and bat wing.
- These similarities suggest a common ancestor and illustrate divergent evolution.
Divergent evolution demonstrates how species adapt differently to various environments while sharing a common ancestry.
Question 10:
The ‘molecular scissors’ fall in the category of:
- Cleaving enzyme
- Endonuclease
- Exonuclease
- Restriction enzymes
Correct Answer: 4
View Solution
- Restriction enzymes, also known as molecular scissors, cut DNA at specific recognition sequences.
- They are widely used in genetic engineering and molecular biology for cloning and gene editing.
Restriction enzymes revolutionized molecular biology, enabling precise DNA manipulation for research and therapeutic purposes.
Question 11:
Which of the chromosomes in a human possesses the least number of genes?
- 21st chromosome
- 16th chromosome
- X chromosome
- Y chromosome
Correct Answer: 4
View Solution
- The Y chromosome is the smallest human chromosome and contains the least number of genes.
- It carries genes essential for male sex determination and spermatogenesis.
Question 12:
Given below is a list of some commercially important products in column A, whereas in column B are names of source organisms:
| Column A (Bioactive Products) |
Column B (Source Organisms) |
| A. Cyclosporin-A |
i. Streptococcus |
| B. Statins |
ii. Trichoderma polysporum |
| C. Streptokinase |
iii. Penicillium notatum |
| D. Penicillin |
iv. Monascus purpureus |
Select the option where the product and their source organisms are correctly matched:
- A – ii, B – iii, C – ii, D – iv
- A – iii, B – iv, C – iv, D – i
- A – ii, B – iv, C – i, D – iii
- A – iv, B – ii, C – i, D – iii
Correct Answer: 4
View Solution
- Cyclosporin-A: Produced by Monascus purpureus, used as an immunosuppressant.
- Statins: Derived from Trichoderma polysporum, these lower blood cholesterol.
- Streptokinase: Obtained from Streptococcus, used to remove blood clots.
- Penicillin: Produced by Penicillium notatum, an antibiotic effective against bacterial infections.
Question 13:
Assertion (A): Plasmids are autonomously replicating circular extra-chromosomal DNA.
Reason (R): Plasmids are usually present in eukaryotic cells.
- Both (A) and (R) are true and (R) is the correct explanation of (A).
- Both (A) and (R) are true and (R) is not the correct explanation of (A).
- (A) is true, but (R) is false.
- (A) is false, but (R) is true.
Correct Answer: 3
View Solution
- Plasmids are circular DNA molecules capable of autonomous replication.
- They are commonly found in prokaryotic cells like bacteria, not in eukaryotic cells.
Question 14:
Assertion (A): Patents are granted by governments to an inventor.
Reason (R): Patents prevent others from commercial use of an invention.
- Both (A) and (R) are true and (R) is the correct explanation of (A).
- Both (A) and (R) are true and (R) is not the correct explanation of (A).
- (A) is true, but (R) is false.
- (A) is false, but (R) is true.
Correct Answer: 1
View Solution
- Patents are legal rights granted to inventors for exclusive use of their inventions.
- Preventing unauthorized use is the primary reason for granting patents, ensuring inventors can commercialize their inventions without competition for a limited time.
Question 15:
Assertion (A): A given fig species can be pollinated only by its partner wasp.
Reason (R): The wasp pollinates the fig inflorescence while searching for suitable egg-laying sites.
- Both (A) and (R) are true and (R) is the correct explanation of (A).
- Both (A) and (R) are true and (R) is not the correct explanation of (A).
- (A) is true, but (R) is false.
- (A) is false, but (R) is true.
Correct Answer: 1
View Solution
- Fig and wasp share an obligate mutualistic relationship where each depends on the other for survival and reproduction.
- Wasps pollinate the fig while laying eggs, ensuring mutual benefit.
Question 16:
Assertion (A): Some aquatic ecosystems have inverted biomass pyramids.
Reason (R): More energy is required by organisms occupying higher trophic levels.
- Both (A) and (R) are true and (R) is the correct explanation of (A).
- Both (A) and (R) are true and (R) is not the correct explanation of (A).
- (A) is true, but (R) is false.
- (A) is false, but (R) is true.
Correct Answer: 2
View Solution
- Inverted biomass pyramids occur in aquatic ecosystems where the biomass of primary consumers exceeds that of producers.
- More energy is required by organisms occupying higher trophic levels, but this does not explain why inverted biomass pyramids occur in aquatic ecosystems.
SECTION - B
Question 17:
What is artificial insemination in ART? Under what conditions is the person medically advised to go for it?
View Solution
- Artificial Insemination: A technique in Assisted Reproductive Technology (ART) where semen is artificially introduced into the female reproductive tract.
- Conditions for Medical Advice:
- When the male partner has a low sperm count or poor sperm motility.
- In cases of physical or anatomical issues in natural intercourse.
Question 18:
Why does a patient of ADA-deficiency require repeated infusions of genetically engineered lymphocytes? Suggest a possible permanent remedy.
View Solution
- Repeated Infusions:
- ADA (Adenosine Deaminase) deficiency leads to severe immune system dysfunction.
- Repeated infusions of genetically engineered lymphocytes temporarily restore immune function by providing the necessary enzyme.
- Permanent Remedy:
- Gene therapy to introduce a functional ADA gene into the patient’s bone marrow cells.
- This approach ensures lifelong production of the enzyme.
Question 19:
Study the graph given below that represents the changes in the thickening of the uterine wall in women ‘X’ and women ‘Y’ over a period of one month:
![Graph depicting the change in wall thickness over time (days).]()
View Solution
- Woman X:
- Normal thickening of the uterine wall due to regular hormonal activity.
- Woman Y:
- Lack of thickening indicates hormonal imbalance or insufficient secretion of estrogen and progesterone.
Question 20(a):
Biodiversity hotspots cover less than 2% of Earth’s land area. Strict protection of these areas can reduce the rate of ongoing extinctions. Explain.
View Solution
- Biodiversity hotspots:
- Regions with high species richness and endemic species but face significant habitat destruction.
- Protecting these areas ensures species survival, maintains ecosystem stability, and preserves genetic resources for future use.
Question 20(b):
Name any two biodiversity hotspots in India.
View Solution
- Western Ghats.
- Indo-Burma region.
Question 21(a):
Differentiate between grazing food chain and detritus food chain.
View Solution
- Grazing Food Chain:
- Begins with producers (plants) consumed by herbivores.
- Example: Grass → Deer → Lion.
- Detritus Food Chain:
- Begins with decomposers breaking down organic matter.
- Example: Dead leaves → Fungi → Earthworms.
Question 21(b):
Explain Brood parasitism with the help of a suitable example.
View Solution
- Brood Parasitism: A reproductive strategy where one species lays its eggs in the nest of another species.
- Example: Cuckoo lays its eggs in crow nests, and cuckoo chicks often outcompete crow chicks for food.
SECTION - C
Question 22:
With reference to flower color, two independent crosses were made: one between true breeding garden pea plants and another between true breeding Antirrhinum plants. Write the phenotypes of their F1 progeny. Justify your answer.
View Solution
- Garden Pea: F1 progeny exhibits complete dominance, showing the dominant flower color.
- Antirrhinum: F1 progeny exhibits incomplete dominance, resulting in an intermediate flower color (e.g., pink when red and white are crossed).
Question 23:
Draw a well-labelled diagram of the sectional view of a male gametophyte/microspore of an angiosperm and write the functions of any two parts labelled.
![male gametophyte.]()
View Solution
- Diagram should include:
- Generative Cell.
- Tube Cell.
- Pollen Wall.
- Pollen Grain.
- Functions:
- Generative Cell: Divides to form two male gametes.
- Tube Cell: Grows into a pollen tube to facilitate fertilization.
Question 24(a):
(i) Name the group of drugs whose skeletal molecule is shown below.
![Chemical structure of the given compound.]()
View Solution
- Answer: Steroids.
- How they are consumed: Orally, injected, or applied topically depending on their formulation.
- Organ affected: Liver.
Question 24(b):
Draw a schematic diagram of an antibody molecule, label any 4 parts, mention their chemical nature, and name the cells which produce them.
View Solution
- Antibody Diagram: Should include:
- Light Chain.
- Heavy Chain.
- Antigen Binding Site.
- Disulfide Bond.
- Chemical Nature: Glycoproteins composed of polypeptide chains.
- Producing Cells: B-Lymphocytes or Plasma Cells.
Question 25:
Explain the role of transgenic animals in:
(a) Production of Biological Products:
View Solution
- Transgenic animals are genetically modified to produce important biological substances like proteins, enzymes, and hormones.
- Example: Transgenic goats produce antithrombin, a protein used to treat blood clotting disorders.
(b) Studying Diseases:
View Solution
- Transgenic animals are used as model organisms to study human diseases, their progression, and treatment.
- Example: Transgenic mice are used to study cancer, diabetes, and Alzheimer’s disease.
(c) Chemical Safety Testing:
View Solution
- Transgenic animals are used to test the safety and efficacy of chemicals and drugs before human use.
- Example: Toxicity testing using transgenic rodents ensures the safety of pharmaceuticals and other chemicals.
Question 26:
The schematic representation below shows the linking of two DNA fragments.
![Diagram illustrating the cleavage of DNA by EcoRI enzyme.]()
(a) Name ‘A’ and ‘B’ fragments:
View Solution
- A: Vector DNA.
- B: Foreign DNA.
(b) Write the ‘palindrome’ recognized by EcoRI:
View Solution
The palindrome sequence recognized by EcoRI is:
5'-GAATTC-3'
3'-CTTAAG-5'
(c) Where does EcoRI cut the palindrome? Write the events followed thereafter to form a recombinant DNA:
View Solution
- EcoRI cuts between G and A in the sequence, generating sticky ends.
- Events that follow:
- The sticky ends of vector DNA and foreign DNA pair through complementary base pairing.
- DNA ligase seals the nicks, forming recombinant DNA.
Question 27:
How do the following organisms act as bio-fertilizers? Explain:
(a) Mycorrhiza
View Solution
- A symbiotic association between fungi and plant roots.
- Enhances water and nutrient absorption, particularly phosphorus.
(b) Anabaena
View Solution
- A cyanobacterium capable of nitrogen fixation.
- Provides nitrogen to plants in a usable form and enhances soil fertility.
(c) Rhizobium
View Solution
- A bacterium that forms nodules on the roots of leguminous plants.
- Converts atmospheric nitrogen into ammonia, which is used by the plant.
Question 28:
(a) Whose skulls ‘A’, ‘B’, and ‘C’ are shown below? Which of the two are more similar to each other?
![Comparison of skulls labeled as ’A’, ’B’, and ’C’.]()
View Solution
- Skull A: Homo sapiens.
- Skull B: Neanderthals.
- Skull C: Australopithecus.
- More Similar: Skulls A and B (Homo sapiens and Neanderthals).
(b) Name the (i) ape-like (ii) man-like primates that existed 1.5 million years ago:
View Solution
- (i) Ape-like: Australopithecus.
- (ii) Man-like: Homo erectus.
SECTION - D
Question 29:
Passage: Generally, in eukaryotic cells, the average length of a transcription unit along a DNA molecule is about 8,000 nucleotides, so the RNA product of the transcription is also that long. But it only takes about 1200 nucleotides from the above RNA product to translate an average-sized polypeptide of 400 amino acids.
(a) Name this RNA product transcribed from the DNA that subsequently translates into a polypeptide of 400 amino acids. Mention the enzyme responsible for transcribing this type of RNA from the DNA:
View Solution
- RNA Product: Messenger RNA (mRNA).
- Enzyme: RNA Polymerase II.
(b) Name and explain the process the RNA molecule transcribed from 8,000 nucleotide-long DNA undergoes to be able to translate a polypeptide of 400 amino acids:
View Solution
- Process: Splicing.
- Explanation:
- The pre-mRNA transcribed from DNA contains both coding (exons) and non-coding (introns) regions.
- During splicing, introns are removed, and exons are joined to form a mature mRNA, which is translated into a protein.
(c) Write the number of RNA polymerases involved in the transcription of DNA in a prokaryote and eukaryotes:
View Solution
- Prokaryotes: Have a single RNA polymerase.
- Eukaryotes: Have three RNA polymerases:
- RNA Polymerase I: Synthesizes rRNA.
- RNA Polymerase II: Synthesizes mRNA.
- RNA Polymerase III: Synthesizes tRNA and some small RNAs.
OR
(c) Mention the difference in the site of transcription in a prokaryote and eukaryote cell:
View Solution
- Prokaryotes: Transcription occurs in the cytoplasm.
- Eukaryotes: Transcription occurs in the nucleus.
- Explanation:
- Prokaryotes lack a defined nucleus, so transcription occurs directly in the cytoplasm where translation can occur simultaneously.
- Eukaryotes have a defined nucleus where transcription occurs, separating it from translation, which happens in the cytoplasm.
Question 30:
Read the passage: Populations evolve to maximise their reproductive fitness in the habitat in which they live. Ecologists suggest the life history of organisms evolves in relation to the constraints imposed by the biotic and abiotic components of the habitat. This is reflected in the population growth patterns of all organisms, including humans.
![Comparison of exponential growth (’A’) and logistic growth (’B’) models.]()
(a) Identify the growth curves ‘A’ and ‘B’ from the given graph:
View Solution
- Curve ‘A’: Exponential growth curve.
- Curve ‘B’: Logistic growth curve.
(b) What does the dotted line in the graph indicate? Explain its importance:
View Solution
- The dotted line indicates the carrying capacity (K).
- Importance: Carrying capacity represents the maximum population size that an environment can sustain indefinitely. It depends on resource availability and environmental constraints.
(c) How does growth curve ‘B’ differ from curve ‘A’? Justify:
View Solution
- Curve ‘B’ (Logistic Growth): Accounts for environmental resistance, stabilizing population size as it approaches the carrying capacity.
- Curve ‘A’ (Exponential Growth): Assumes unlimited resources, which is unrealistic in natural environments.
(d) Which curve is more realistic and why?
View Solution
- Curve ‘B’ (Logistic Growth): It reflects real-world conditions where resources are limited, and competition regulates population size.
(e) Which curve is relevant to human population trends in our country and why?
View Solution
- Exponential growth curve (Curve ‘A’): Relevant because human populations continue to grow rapidly due to advancements in healthcare and technology.
- However, environmental challenges may enforce a shift to a logistic growth model over time.
SECTION - E
Question 31(a):
Explain the following phases in the menstrual cycle of a human female:
(i) Menstruation:
View Solution
- The uterine lining sheds due to a drop in progesterone levels, resulting in menstrual bleeding.
(ii) Follicular Phase:
View Solution
- FSH stimulates the development of follicles, and estrogen levels increase, thickening the uterine lining.
(iii) Luteal Phase:
View Solution
- After ovulation, the ruptured follicle forms the corpus luteum, secreting progesterone to maintain the uterine lining for possible implantation.
(iv) How does understanding the menstrual cycle help in family planning?
View Solution
- Identifies the fertile window for conception or avoiding pregnancy.
- Helps monitor reproductive health and detect irregularities.
Question 31(b):
(i) Why does endosperm development precede embryo development in angiosperm seeds? State its role in mature albuminous seeds:
View Solution
- Reason: Endosperm development ensures a nutrient supply for the developing embryo.
- Role in albuminous seeds: In seeds like wheat and maize, the endosperm persists in mature seeds and provides nutrients during germination.
(ii) Draw a labelled diagram showing embryonic stages of a dicot plant:
View Diagram
- Stages include globular, heart-shaped, torpedo, and mature embryo stages.
- Key structures: Cotyledons, Plumule (shoot tip), Radicle (root tip).
Question 32(a):
Describe the life cycle of HIV from its entry into the human body to full-blown AIDS:
View Solution
- Entry: HIV attaches to CD4 receptors on T-cells and fuses with the cell membrane.
- Reverse Transcription: Viral RNA is converted into DNA by reverse transcriptase.
- Integration: Viral DNA integrates into the host genome via integrase.
- Replication: Host cell machinery synthesizes viral proteins and RNA.
- Release: New viruses assemble and exit, destroying T-cells over time.
- Progression: Immune system weakens, leading to opportunistic infections and AIDS.
Question 32(b):
(i) Write the symptoms of malaria in humans and explain what causes them:
View Solution
- Symptoms:
- Recurring high fever with chills.
- Sweating and shivering.
- Headache and body aches.
- Nausea and vomiting.
- Fatigue and weakness.
- Cause: Symptoms are due to the release of merozoites into the bloodstream when red blood cells rupture. Haemozoin, a toxic substance, is released, triggering the immune response.
(ii) Describe the sexual mode of reproduction in the life cycle of a malarial parasite:
View Solution
- Steps:
- Gamete Formation: Gametocytes form in humans and are ingested by a mosquito.
- Fertilization: Male and female gametocytes fuse in the mosquito's gut to form a zygote.
- Ookinete Formation: The zygote develops into a motile ookinete, penetrating the gut wall.
- Oocyst Development: The ookinete forms an oocyst on the mosquito’s gut wall.
- Sporozoite Formation: The oocyst divides to produce sporozoites, which migrate to the salivary glands.
- Completion: Sporozoites are injected into a new human host when the mosquito feeds.
Question 33(a):
Study the schematic diagram below and answer the questions:
![Diagram illustrating the translation process in protein synthesis, showing the ribosome, mRNA, tRNA, and the growing polypeptide chain.]()
(i) Identify the polarity from ‘X’ to ‘Y’ in the mRNA segment shown. Mention how many more amino acids can be added to the polypeptide and why:
View Solution
- Polarity: 5’ to 3’.
- Additional Amino Acids: Determined by the number of untranslated codons remaining in the mRNA. Each codon translates to one amino acid until a stop codon is reached.
(ii) Write the initiating codon for translation, its anticodon, and the amino acid it codes for:
View Solution
- Initiating Codon: AUG.
- Anticodon: UAC.
- Amino Acid: Methionine.
(iii) Explain the charging of an adaptor molecule. Why does this molecule need to be charged?
View Solution
- Charging Process: tRNA is charged by attaching an amino acid using aminoacyl-tRNA synthetase.
- Purpose: A charged tRNA ensures the correct amino acid is added during translation, enabling accurate protein synthesis.
Question 33(b):
Answer the following questions on sickle-cell anaemia:
(i) Why is sickle-cell anaemia named so?
View Solution
- Reason: Sickle-cell anaemia is named after the sickle-shaped red blood cells that appear under low oxygen levels.
(ii) Explain the genetic basis that results in the expression of this disorder:
View Solution
- Cause: A point mutation in the β-globin gene replaces glutamic acid with valine in haemoglobin.
- Effect: Abnormal haemoglobin (HbS) polymerizes under low oxygen, deforming red blood cells.
(iii) Work out a cross to explain how normal parents may have a sickle-cell anaemic child:
View Solution
- Parental Genotypes: HbA HbS (carriers).
- Child Genotypes: HbA HbA (normal), HbA HbS (carrier), HbS HbS (affected).
- Punnett Square: Shows a 25% probability of an affected child.