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Biology (SSC Biology)NCTB SSC Biology (Classes 9-10)BMI & BMR (Harris-Benedict)Bioenergetics & ATP YieldPunnett Squares & Color Blindness

SSC Biology Equation & Solution Suite

All-in-one NCTB SSC Biology quantitative solver covering BMI, BMR, Daily Caloric Needs (TDEE), Respiration ATP Balance Sheet, Mendelian & Sex-Linked Punnett Squares, Lindeman's 10% Energy Pyramid, and Blood Matching.

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Physical Attributes & Parameters

Body Mass Index (BMI)Normal / Healthy Weight
20.81kg/m²

Your BMI is 20.81, categorized as "Normal / Healthy Weight". For a height of 155 cm, the healthy weight range is 44.4 kg to 59.8 kg. Your body weight is currently in the ideal healthy range.

BMI Range Spectrum (WHO & NCTB)BMI: 20.81
<18.5 (Under)18.5-24.9 (Normal)25-29.9 (Over)≥30 (Obese)
Basal Metabolic Rate
1338.8
kcal/day
Daily Caloric Needs
2075.1
kcal/day

Step-by-Step Derivation (Harris-Benedict Equations)

Active Result:
BMI: 20.81 (Normal / Healthy Weight)

Formula & Derivation

Theoretical foundation, dimensional analysis, and governing boundary conditions.

NCTB SSC Biology Quantitative Formula Reference Sheet

Achieving maximum marks in secondary biology exams requires thorough mastery over quantitative problems across Chapters 4, 5, 6, 12, and 13. Below is the standardized formula bank:

Chapter 5: Food & Nutrition (BMI & BMR)

• BMI formula: BMI = Weight (kg) / [Height (m)]²

• Male BMR: 66 + (13.7 × W) + (5 × H) - (6.8 × A)

• Female BMR: 655 + (9.6 × W) + (1.8 × H) - (4.7 × A)

• TDEE = BMR × Activity Factor (1.2 to 1.9)

• Healthy Range: 18.5 - 24.9 kg/m²

Chapter 4: Bioenergetics (ATP Respiration Sheet)

• Aerobic equation: C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O + 38 ATP

• Glycolysis: 2 direct ATP + 2 NADH (6 ATP) = 8 ATP

• Acetyl-CoA: 2 NADH (6 ATP) = 6 ATP

• Krebs cycle: 2 GTP (2 ATP) + 6 NADH (18) + 2 FADH₂ (4) = 24 ATP

• 1 mole ATP = 7.3 kcal (30.55 kJ)

Chapter 12: Heredity & Punnett Squares (Genetics)

• Monohybrid cross: Tt × Tt ⟹ Phenotypic 3:1, Genotypic 1:2:1

• Test cross: Tt × tt ⟹ 1:1 ratio

• Sex-linked recessive: X-linked (Xᴺ normal, Xⁿ mutant)

• Carrier mother (XᴺXⁿ) × Normal father (XᴺY) ⟹ 50% sons affected

• Affected mother (XⁿXⁿ) × Normal father (XᴺY) ⟹ 100% sons affected

Chapter 13: Environment (Lindeman's 10% Law)

• Trophic energy transfer: E_(n+1) = E_n × 0.10

• Level 1 Producers: 100% energy

• Level 2 Herbivores: 10% energy

• Level 3 Carnivores: 1% energy

• Level 4 Apex Predators: 0.1% energy (90% dissipated as heat)

Chapter 6: Transport in Organisms (Blood Compatibility)

• Universal Donor: O- (no A, B, or Rh surface antigens on RBCs)

• Universal Recipient: AB+ (no Anti-A, Anti-B, or Anti-Rh antibodies in serum)

• Agglutination rule: Donor RBC antigens must not match recipient plasma antibodies

Solved Textbook Examples

Three fully worked pedagogical exemplars: standard textbook, advanced edge-case, and authentic past board examination.

Board Exam Special (Nutrition, BMI, BMR & Daily Caloric Needs)bg-emerald-500/10 text-emerald-400 border border-emerald-500/20
Shayla is a 16-year-old female student who weighs 50 kg and stands 155 cm tall. She engages in moderate exercise 3 to 5 days a week. Calculate her Body Mass Index (BMI), Basal Metabolic Rate (BMR), and Total Daily Energy Expenditure (TDEE) to maintain her current weight.
Given Parameters:
  • Biological sex: Female, Age A = 16 years
  • Body weight W = 50 kg, Height H = 155 cm = 1.55 m
  • Activity factor = Moderately active (Multiplier = 1.55)
Procedural Solution:
Part 1: Calculate Body Mass Index (BMI):
BMI = Weight (kg) / [Height (m)]² = 50 / (1.55)² = 50 / 2.4025 = 20.81 kg/m²
According to WHO and NCTB standards, a BMI of 18.5 to 24.9 is classified as "Healthy / Normal Weight". Shayla has an ideal body mass status.
Part 2: Calculate BMR (Harris-Benedict equation for females):
BMR = 655 + (9.6 × W) + (1.8 × H) - (4.7 × A)
BMR = 655 + (9.6 × 50) + (1.8 × 155) - (4.7 × 16)
BMR = 655 + 480 + 279 - 75.2 = 1414 - 75.2 = 1338.8 kcal/day.
Part 3: Compute Total Daily Energy Expenditure (TDEE):
TDEE = BMR × Activity Multiplier = 1338.8 × 1.55 = 2075.14 kcal/day.
Final Answer:Shayla has a BMI of 20.81 (Healthy Weight), a BMR of 1338.8 kcal/day, and requires 2075.1 kcal daily.
Board Exam Special (Bioenergetics & Glucose Oxidation ATP Yield)bg-amber-500/10 text-amber-400 border border-amber-500/20
How many ATP molecules and how much total usable biological energy in kilocalories (kcal) and kilojoules (kJ) are liberated during the complete aerobic respiration of 2 moles of glucose (C₆H₁₂O₆) under the classical NCTB model?
Given Parameters:
  • Quantity of glucose = 2 moles
  • Classical ATP yield per mole of glucose = 38 ATP
  • Energy released per mole of ATP = 7.3 kcal (30.55 kJ)
Procedural Solution:
Step 1: Cellular respiration stage-wise ATP distribution (per mole of glucose):
• Glycolysis: Net 2 direct ATP + 2 NADH (6 ATP) = 8 ATP
• Acetyl-CoA Formation: 2 NADH (6 ATP) = 6 ATP
• Krebs Cycle: 2 GTP (2 ATP) + 6 NADH (18 ATP) + 2 FADH₂ (4 ATP) = 24 ATP
• Total per 1 mole of glucose = 8 + 6 + 24 = 38 ATP.
Step 2: Total ATP for 2 moles of glucose:
Total ATP = 2 × 38 = 76 ATP molecules.
Step 3: Calculate usable biological energy liberated:
Energy in kilocalories = 76 × 7.3 kcal = 554.8 kcal.
Energy in kilojoules = 76 × 30.55 kJ = 2321.8 kJ.
Carbon dioxide released = 2 × 6 = 12 moles CO₂.
Final Answer:Complete aerobic respiration of 2 moles of glucose yields 76 ATP, 12 moles of CO₂, and 554.8 kcal (2321.8 kJ) of energy.
Board Exam Special (Sex-Linked Color Blindness Punnett Square)bg-cyan-500/10 text-cyan-400 border border-cyan-500/20
A woman with normal vision whose father was color-blind (making her a carrier, Xᴺ Xⁿ) marries a man with normal vision (Xᴺ Y). Using a 2 × 2 Punnett Square, predict the probabilities of their sons and daughters being color-blind.
Given Parameters:
  • Maternal genotype = Xᴺ Xⁿ (Carrier mother with normal vision)
  • Paternal genotype = Xᴺ Y (Normal vision father)
  • Xᴺ = Dominant normal allele, Xⁿ = Recessive mutant color-blind allele
Procedural Solution:
Step 1: Determine parental gametes:
Maternal ova: 50% Xᴺ and 50% Xⁿ. Paternal spermatozoa: 50% Xᴺ and 50% Y.
Step 2: Execute 2 × 2 Punnett Square cross:
• (Xᴺ × Xᴺ) ⟹ XᴺXᴺ : Normal vision daughter (25% of all offspring)
• (Xᴺ × Y) ⟹ XᴺY : Normal vision son (25% of all offspring)
• (Xⁿ × Xᴺ) ⟹ XᴺXⁿ : Carrier daughter with normal vision (25% of all offspring)
• (Xⁿ × Y) ⟹ XⁿY : Color-blind son (25% of all offspring)
Step 3: Segregate phenotypic probabilities by gender:
• Daughters: 100% normal vision (50% homozygous normal + 50% heterozygous carriers). 0% of daughters are color-blind.
• Sons: 50% normal vision and 50% color-blind.
Final Answer:0% of daughters will be color-blind (50% carriers, 50% normal), whereas 50% of sons will be color-blind.
Board Exam Special (Ecosystem Dynamics & Lindeman 10% Law)bg-green-500/10 text-green-400 border border-green-500/20
In a freshwater pond ecosystem, phytoplankton (producers) synthesize 10,000 Joules of chemical biomass from sunlight. According to Lindeman’s 10% ecological efficiency law, how much energy reaches the apex predator (chital/carnivorous fish) at Level 4, and how much is dissipated into the environment?
Given Parameters:
  • Producer level energy (Level 1) E₁ = 10,000 J
  • Trophic transfer efficiency = 10% (0.10)
  • Metabolic heat loss per step = 90% (0.90)
Procedural Solution:
Step 1: Primary consumers / Zooplankton (Level 2):
E₂ = E₁ × 0.10 = 10,000 × 0.10 = 1,000 Joules (10%).
Step 2: Secondary consumers / Small fish (Level 3):
E₃ = E₂ × 0.10 = 1,000 × 0.10 = 100 Joules (1%).
Step 3: Tertiary consumers / Apex fish (Level 4):
E₄ = E₃ × 0.10 = 100 × 0.10 = 10 Joules (0.1%).
Step 4: Calculate total dissipated thermal energy:
Total heat lost = E₁ - E₄ = 10,000 - 10 = 9,990 Joules (99.9%).
Final Answer:Only 10 Joules reaches the apex predator level, with 9,990 Joules lost as respiratory heat and metabolic waste.

Casio Scientific Calculator Keystroke GuideExam Ready

Exact button sequences permitted in board and university entrance examination halls:

Casio fx-991ES PLUS / 2nd Edition
Genetic probability & Lindeman 10% energy pyramid.
1. Compute 10% Energy Transfer
1000×10SHIFT((%)=
Evaluates succeeding trophic level energy.

Practical Applications

How this scientific principle drives chemical engineering, aerospace, medicine, and research.

Board Exam Tips & Scoring Guidelines:

  • Unit Conversion for Height: When computing BMI, convert height in centimeters into meters first (divide by 100) before squaring. E.g., 155 cm = 1.55 m, (1.55)² = 2.4025.
  • BMR Sex Distinction: Pay close attention to the constants: male equations begin with 66 with a 13.7 weight multiplier, while female equations begin with 655 with a 9.6 weight multiplier.
  • Gender-Segregated Punnett Questions: When a creative question asks "What percentage of daughters are color-blind?", calculate the fraction out of total daughters (not out of total children).
  • Offline CLI Tool: For offline study or scripting, run our standalone Python CLI (scripts/ssc_biology_calculator.py) on your local machine.

Frequently Asked Questions

Answers to common conceptual misconceptions and board examination guidelines.

How many ATP molecules are produced in aerobic respiration according to NCTB?▼
Under the National Curriculum and Textbook Board (NCTB) SSC Biology curriculum, complete oxidation of 1 molecule of glucose yields 38 ATP (8 from Glycolysis, 6 from Acetyl-CoA, and 24 from Krebs Cycle). In standard international biochemistry, shuttle mechanisms may reduce this to 36 ATP or 30-32 ATP. Our calculator supports toggling between both models.
What is the fundamental difference between BMI and BMR?▼
BMI (Body Mass Index) evaluates weight relative to height (BMI = W / H²) to categorize body mass as underweight, normal, or obese. BMR (Basal Metabolic Rate) computes the minimal caloric energy expended at absolute physical and mental rest to keep vital organs functioning.
Why can human males never be carriers for color blindness?▼
Red-green color blindness is an X-linked recessive disorder. Females possess two X chromosomes (XX), so a single mutant allele (Xⁿ) is masked by a dominant normal allele (Xᴺ), producing an asymptomatic carrier. Males have only one X chromosome (XY). Any male inheriting the mutant Xⁿ allele will exhibit the disease; having no second X chromosome, males cannot be silent carriers.
Why are trophic food chains rarely longer than 4 or 5 levels?▼
Due to Lindeman’s 10% ecological efficiency rule, 90% of available energy is dissipated as respiratory heat at each successive link. By the 4th or 5th trophic level, remaining energy is too miniscule (e.g. 0.1% of original producer energy) to sustain another population of consumers.