A survey was conducted to estimate the population of barnacles (figure 1) in two coastal areas (i.e. A and B). The dimension of coastal area ‘A’ is around 20 m and 26 m. The dimension of coastal area ‘B’ is around 14 m and 15 m

Sampling in both areas was conducted using 1 x 1 m quadrat. Six quadrats were sampled randomly in each area. The numbers of barnacles in each quadrat as the following.
Area Q1 Q2 Q3 Q4 Q5 Q6

A 2 1 3 1 4 2

B 12 11 5 3 6 2


1. Calculate the estimated population of barnacles in both areas A and B. Please show the details of calculations. [5]

2. Discuss your findings. [5]

Answers

Answer 1

Based on the data provided;

Estimated population of Barnacles in A is 1128 BarnaclesEstimated population of Barnacles in A is 1365 Barnacles the population density of Barnacles in B is greater than that in A.

What is a population density?

Population density is the number of organisms per unit area of a given geographical area or location.

Population density gives an idea of how densely or sparsely populated a given area is.

Population density = number of organisms/area of land.

The area of quadrant = 1 m × 1 m = 1 m^2

Total number of Barnacles in 6 quadrant sample A = 13

Average = 2.17

Population density = 2.17/1m^2 = 2.17 /m^2

Total number of Barnacles in 6 quadrant sample B = 39

Average = 6.5

Population density = 6.5/1m^2 = 6.5 /m^2

Area of A = 20 m × 26 m = 520 m^2

Estimated population of A = 520 m^2 × 2.17 /m^2

Estimated population of Barnacles in A = 1128 Barnacles

Area of B = 14 m × 15 m = 210 m^2

Estimated population of Barnacles in B = 210 m^2 × 6.5 /m^2

Estimated population of Barnacles in B = 1365 Barnacles

From the above results, the population density of Barnacles in B is greater than that in A.

Learn more about population density at: https://brainly.com/question/16894337


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Explanation: Also, what kind of question is this?

1) You have an electropherogram with a clean, three-locus profile:
STR locus #1 is 10, 14
STR locus #2 is 30 (only)
STR locus #3 is 3, 6
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Answers

Following the product probability rule and assuming independence among loci, the random match probability of this three-locus profile is 0.000853. See below the procedure for question number 2.

What does the product probability rule state?

The product probability rule allows us to calcuate the probability of occurrence of two or more events at the same time. It is about a joint probability of two or more events that might happen simultaneously, not excluding each other.

This rule is based on the dependence or independence of the events. Two events, A and B, are independent of each other if one of them does not affect the occurrence of the other one.

The rule establishes that the occurrence probability of two independent events -A and B- together is the multiplication product of the probability of occurring A by the probability of occurring B.

P(A∩B) = P(A) x P(B)

In the exposed example,

STR locus #1 has three alleles → 10, 11, 14STR locus #2 has four alleles → 27, 28, 28.2, 30STR locus #3 has three alleles → 3, 6, 7

The frequency of each allele is detailed in the table.

According to the electropherogram, the profiles are

STR locus #1 → 10, 14 → Heter0zyg0usSTR locus #2 → 30 → H0m0zyg0usSTR locus #3 → 3, 6 → Heter0zyg0us

Knowing the allele frequencies, we can get the genotypic frequencies,

   Locus               Allele frequency            Genotypic frequency

STR locus #1

10                                 0.22                    2xpxq = 2 x 0.22 x 0.18 = 0.079

14                                 0.18            

----------------------------------------------------------------------------------------------------  

STR locus #2

30                                 0.22                    p² = 0.22² = 0.048

----------------------------------------------------------------------------------------------------  

STR locus #3

3                                   0.45                    2xpxq = 2 x 0.45 x 0.25 = 0.225

3                                   0.25            

----------------------------------------------------------------------------------------------------

So, the probability, P, of getting these genotypes are,

P ₍₁₀₋₁₄₎ = 0.079P₍₃₀₋₃₀₎ = 0.048P₍₃₋₆₎ = 0.225

The random match probability of this profile is the probability of getting the three genotypes together.

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PP = P₍₁₎ x P₍₂₎ x P₍₃₎ x .... P₍ₙ₎

PP = (P₍₁₀₋₁₄₎ ) (P₍₃₀₋₃₀₎) (P₍₃₋₆₎)

PP = 0.079 x 0.048 x 0.225

PP = 0.000853

The random match probability of this three-locus profile is 0.000853.

To fill in the blank, you need to consider the population size -N- and the probability of getting this three-locus profile.  

Since I do not have the total number of individuals in the population, I will say that my population size, N = 500 individuals and I will make the calcs according to this number. You should follow the same procedure but use your N.

We know that the probability of getting this three-locus profile is 0.000853. This is the frequency of individuals carrying this genotype in the population.

To get the number of individuals with this genotype in the population, we need to multiply this frequency by the total number of individuals, N.

Nº of individuals with the three-locus profile = frequency x N

Nº of individuals with the three-locus profile = 0.000853 x 500

Nº of individuals with the three-locus profile = 0.4265 individuals.

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0.4265 individuals with the three-locus profile-------- 500 individuals

1 individual -------------------------- X = (1 x 500) / 0.4265 = 1,172.33 individuals.

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Note: Remember to use the population size provided to you in this problem. The procedure is the same, but the population size -N- will be different.

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Explanation:

hope this helps

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