Biology Flashcards: Apply Math To Energy Flow

What this deck covers

This deck focuses on Apply Math To Energy Flow, giving you a quick way to review the definitions, rules, and examples that matter most for Biology.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: Calculate efficiency if $E_{p} = 3600\ \text{kJ}$ and $E_{n} = 540\ \text{kJ}$.

Answer: $15\%$. Use efficiency formula: $\frac{540}{3600} \times 100\% = 15\%$.

Flashcard 2: Find the energy at the next trophic level if producers have $20000\ \text{kJ}$ and efficiency is $10\%$.

Answer: $2000\ \text{kJ}$. Apply $10\%$ rule: $20000 \times 0.1 = 2000$.

Flashcard 3: Find energy lost if $E_{p} = 2500\ \text{kJ}$ and $E_{n} = 125\ \text{kJ}$.

Answer: $2375\ \text{kJ}$. Subtract transferred from previous: $2500 - 125 = 2375$.

Flashcard 4: How much energy remains after two $10\%$ transfers starting from $10000\ \text{kJ}$?

Answer: $100\ \text{kJ}$. Two $10\%$ transfers: $10000 \times (0.1)^2 = 100$.

Flashcard 5: How much energy remains after three $10\%$ transfers starting from $50000\ \text{kJ}$?

Answer: $50\ \text{kJ}$. Three $10\%$ transfers: $50000 \times (0.1)^3 = 50$.

Flashcard 6: How much energy remains after four $10\%$ transfers starting from $200000\ \text{kJ}$?

Answer: $20\ \text{kJ}$. Four $10\%$ transfers: $200000 \times (0.1)^4 = 20$.

Flashcard 7: Calculate $\text{NPP}$ if $\text{GPP} = 1500$ and producer respiration $R = 600$ (same units).

Answer: $900$. Use NPP formula: $1500 - 600 = 900$.

Flashcard 8: Calculate $\text{GPP}$ if $\text{NPP} = 700$ and producer respiration $R = 300$ (same units).

Answer: $1000$. Rearrange NPP formula: $700 + 300 = 1000$.

Flashcard 9: Calculate producer respiration $R$ if $\text{GPP} = 2200$ and $\text{NPP} = 1600$ (same units).

Answer: $600$. Rearrange NPP formula: $2200 - 1600 = 600$.

Flashcard 10: If producers store $12000\ \text{kJ}$ as NPP, estimate energy available to primary consumers using $10\%$.

Answer: $1200\ \text{kJ}$. Apply $10\%$ rule to NPP: $12000 \times 0.1 = 1200$.

Flashcard 11: If primary consumers have $900\ \text{kJ}$, estimate energy available to secondary consumers using $10\%$.

Answer: $90\ \text{kJ}$. Apply $10\%$ rule: $900 \times 0.1 = 90$.

Flashcard 12: If tertiary consumers have $8\ \text{kJ}$ and transfer is $10\%$, what was secondary consumer energy?

Answer: $80\ \text{kJ}$. Reverse $10\%$ rule: $8 \div 0.1 = 80$.

Flashcard 13: If secondary consumers have $60\ \text{kJ}$ and transfer is $10\%$, what was primary consumer energy?

Answer: $600\ \text{kJ}$. Reverse $10\%$ rule: $60 \div 0.1 = 600$.

Flashcard 14: If a top predator has $2\ \text{kJ}$ after three $10\%$ transfers, what producer energy was required?

Answer: $2000\ \text{kJ}$. Reverse three transfers: $2 \div (0.1)^3 = 2000$.

Flashcard 15: If producers have $30000\ \text{kJ}$, what percent remains at secondary consumers after two $10\%$ transfers?

Answer: $1\%$. Two $10\%$ transfers give $1\%$ of original energy.

Flashcard 16: If $E_{p} = 4000\ \text{kJ}$ and efficiency is $12\%$, what is $E_{n}$?

Answer: $480\ \text{kJ}$. Calculate $12\%$ of previous level: $4000 \times 0.12 = 480$.

Flashcard 17: If $E_{p} = 7500\ \text{kJ}$ and efficiency is $8\%$, what is $E_{n}$?

Answer: $600\ \text{kJ}$. Calculate $8\%$ of previous level: $7500 \times 0.08 = 600$.

Flashcard 18: Calculate efficiency if $E_{p} = 2500\ \text{kJ}$ and $E_{n} = 50\ \text{kJ}$.

Answer: $2\%$. Use efficiency formula: $\frac{50}{2500} \times 100\% = 2\%$.

Flashcard 19: Find the number of trophic transfers if energy drops from $10000\ \text{kJ}$ to $10\ \text{kJ}$ at $10\%$ each step.

Answer: 3 transfers. Energy ratio is $(0.1)^3 = 0.001$, so 3 transfers.

Flashcard 20: Find the number of trophic transfers if energy drops from $100000\ \text{kJ}$ to $100\ \text{kJ}$ at $10\%$ each step.

Answer: 3 transfers. Energy ratio is $(0.1)^3 = 0.001$, so 3 transfers.

Flashcard 21: What is the general formula for energy after $n$ transfers with $10\%$ efficiency from $E_0$?

Answer: $E_n = E_0 \times (0.1)^n$. General exponential decay formula with $10\%$ efficiency.

Flashcard 22: Using $E_n = E_0 \times (0.1)^n$, find $E_2$ if $E_0 = 6000\ \text{kJ}$.

Answer: $60\ \text{kJ}$. Two transfers: $6000 \times (0.1)^2 = 60$.

Flashcard 23: Using $E_n = E_0 \times (0.1)^n$, find $E_1$ if $E_0 = 4300\ \text{kJ}$.

Answer: $430\ \text{kJ}$. One transfer: $4300 \times 0.1 = 430$.

Flashcard 24: Using $E_n = E_0 \times (0.1)^n$, find $E_3$ if $E_0 = 9000\ \text{kJ}$.

Answer: $9\ \text{kJ}$. Three transfers: $9000 \times (0.1)^3 = 9$.

Flashcard 25: What is the quantitative reason energy pyramids are always upright (never inverted)?

Answer: Energy decreases each transfer due to heat loss; $E_{n} < E_{p}$. Energy loss from metabolism makes higher levels smaller.

Flashcard 26: What is the formula for trophic transfer efficiency if $E_{n}$ is next level energy and $E_{p}$ is previous?

Answer: $\text{Efficiency} = \frac{E_{n}}{E_{p}} \times 100\%$. Standard formula for calculating energy transfer between levels.

Flashcard 27: What is the definition of a trophic level in an ecosystem energy pyramid?

Answer: A feeding position based on energy transfer (producer, consumer levels). Defines organisms' position in energy flow hierarchy.

Flashcard 28: What is the primary energy source for most ecosystems that drives primary productivity?

Answer: Sunlight (solar energy). Photosynthesis converts solar energy into chemical energy for producers.

Flashcard 29: Identify the correct inequality for energy across trophic levels from producers to top predators.

Answer: $E_{\text{producers}} > E_{\text{primary}} > E_{\text{secondary}} > E_{\text{tertiary}}$. Energy decreases at each successive trophic level.

Flashcard 30: What does the $10\%$ rule state about energy transfer between trophic levels?

Answer: About $10\%$ of energy transfers; about $90\%$ is lost as heat and metabolism. Standard rule describing energy transfer efficiency in ecosystems.