Numerical Problems: Organisms and Populations

Problem 1: Exponential Population Growth

A bacterial population starts with 100 individuals and doubles every hour. How many bacteria will be present after 5 hours, assuming unlimited resources?

Solution:

This is an example of exponential growth. The formula for exponential growth is N_t = N_0 * 2^t, where:

• N_t = population size after time t
• N_0 = initial population size
• t = number of doubling periods

Given:

• N_0 = 100
• t = 5 hours

N_5 = 100 * 2^5 N_5 = 100 * 32 N_5 = 3200

Therefore, after 5 hours, there will be 3200 bacteria.

Problem 2: Population Density Calculation

In a forest, there are 500 deer in an area of 10 square kilometers. Calculate the population density of deer in this forest.

Solution:

Population density is calculated as the number of individuals per unit area or volume.

Population Density = (Number of individuals) / (Area)

Given:

• Number of individuals = 500 deer
• Area = 10 km²

Population Density = 500 deer / 10 km² = 50 deer/km²

Therefore, the population density of deer in this forest is 50 deer per square kilometer.

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Question 1: Population Density

In a 200 square kilometer forest, there are 400 deer. What is the population density of the deer?

Solution:

• Total number of individuals (N): 400
• Area (S): 200 km²
• Population Density (D) = N / S
  • D = 400 deer / 200 km² = 2 deer/km²

Therefore, the population density of the deer is 2 deer per square kilometer.

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Question 2: Population Growth Rate

A population of 500 insects has a birth rate of 100 insects per year and a death rate of 40 insects per year. Assuming no immigration or emigration, what is the net increase in the population in one year?

Solution:

• Initial Population (N): 500
• Births (B): 100
• Deaths (D): 40
• Net Increase = Births - Deaths
  • Net Increase = 100 - 40 = 60

Therefore, the net increase in the population in one year is 60 insects.

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Question 3: Exponential Growth

A population of bacteria is growing exponentially. The initial population size is 1000, and the intrinsic rate of natural increase (r) is 0.2 per hour. What will be the population size after 5 hours?

Solution:

• Initial Population (N₀): 1000
• Intrinsic rate of increase (r): 0.2
• Time (t): 5 hours
• Exponential Growth Formula: N_t = N₀ * e^rt
  • N₅ = 1000 * e^{(0.2 * 5)}
  • N₅ = 1000 * e¹
  • N₅ = 1000 * 2.71828
  • N₅ ≈ 2718

Therefore, the population size after 5 hours will be approximately 2718 bacteria.

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Question 4: Logistic Growth

A population of fish has a carrying capacity (K) of 2000. The intrinsic rate of natural increase (r) is 0.3 per year. If the current population size (N) is 500, what is the population growth rate (dN/dt)?

Solution:

• Carrying Capacity (K): 2000
• Intrinsic rate of increase (r): 0.3
• Current Population Size (N): 500
• Logistic Growth Formula: dN/dt = rN * ((K - N) / K)
  • dN/dt = 0.3 * 500 * ((2000 - 500) / 2000)
  • dN/dt = 150 * (1500 / 2000)
  • dN/dt = 150 * 0.75
  • dN/dt = 112.5

Therefore, the population growth rate is 112.5 fish per year.

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Question 5: Allele Frequencies (Hardy-Weinberg)

In a population of butterflies, the color brown (B) is dominant over the color white (b). 40% of all butterflies are white. Given this information, calculate the percentage of butterflies in the population that are heterozygous.

Solution:

• This problem can be solved using the Hardy-Weinberg principle.
• Frequency of white butterflies (bb or q²): 40% or 0.40
• Frequency of the recessive allele (b or q):
  • q = √0.40 ≈ 0.632
• Frequency of the dominant allele (B or p):
  • p = 1 - q = 1 - 0.632 = 0.368
• Frequency of heterozygous butterflies (Bb or 2pq):
  • 2pq = 2 * 0.368 * 0.632 ≈ 0.465
• Percentage of heterozygous butterflies: 0.465 * 100 = 46.5%

Therefore, approximately 46.5% of the butterflies in the population are heterozygous.