A student decides to conduct a mark-recapture experiment to estimate the population size of
- 134
- 349
- 558
- 1,016
- 22,320
Questions 2–5 refer to the following table:
Age
0 100 35 1.00 0 0
1 65 ? 0.65 0 0
2 45 15 ? 3 1.35
330 20 0.301 ?
4 10 10 0.10 10.10
5 0 0 0.00 1 0.0
Introduction:
In biology, population size is the definite number of organisms in the population. The total population size is calculated by given formula:
OR
Populations are made of all individuals belonging to the same species who live together at the same time.
The density of population is the count of organisms per geographic area or volume, per unit. For example, the number of organisms per square meter, per hectare or per square ilo
Answer to Problem 1TY
The correct answer is option (b) 349.
Explanation of Solution
Explanation/justification for the correct answer:
Option (b) 349.
If,
Marked mosquito fish in 1st catch = 45 individuals
Recaptured mosquito fish in 2nd catch = 62
Marked recaptured mosquito fish in 2nd catch= 20, then according to the formula,
From the equation, the estimate total population size of mosquito fish is
So, this is the correct option.
Explanation for incorrect answer:
Option (a) 134. As explained in option (b), the estimated size of a population is not matched to given data. So, this is an incorrect option.
Option (c) 558. As explained in option (b), the estimated size of a population is not matched to given data. So, this is an incorrect option.
Option (d) 1.016. As explained in option (b), the estimated size of a population is not matched to given data. So, this is an incorrect option.
Option (e) 22.320. As explained in option (b), the estimated size of a population is not matched to given data. So, this is an incorrect option.
Want to see more full solutions like this?
Chapter 43 Solutions
Principles of Biology
- (Number marked × total number recaptured)# recaptured with mark/Number marked × total number recaptured# recaptured with mark The formula above is the equation for estimated population size using the Mark and Recapture method of population estimation. I initially catch and tag 10 rabbits. Later I recapture 50 total rabbits and 10 of those are marked. What is the estimated population size of this rabbit population? Question 22 options: 500 rabbits 20 rabbits 2 rabbits 50 Rabbitsarrow_forwardThe life table below is for a species of lizard. Use it to answer the questions below. It can be copied and pasted in excel or downloaded here. What is the intrinsic rate of increase of this population? nx bx 678 1 102 31 3 19 10 4 14 12 5 12 12 6 10 14 7 10 10 8 8 11 9 7 10 1.89 0.053 O 1.32 0.51arrow_forwardYou are contracted by the City of Ottawa to estimate the size of the population of small mouth bass in Dow's Lake. Using nets, you capture, mark, and release 48 bass. You return to Dow's Lake two weeks later and use the same techniques to sample more bass. This time, you capture 37 bass, 16 of which you had previously captured (marked ones). What is the estimated size of the population of small mouth bass in Dow's Lake?arrow_forward
- There was another research effort underway to estimate the squirrel population on the PLU campus. The first trapping effort resulted in 15 individuals being captured, marked and released. The second capture effort resulted in 15 new squirrels being captured and 5 squirrels were recaptured. What is the estimated squirrel population? What are the primary assumptions of the population estimation procedure used in this question?arrow_forwardWhich of the following equations defines population size? Question 3 options: (Pt+1) = B - I + D + I – E (Pt1) = N2 + I - D + B – E (Pt2) = Pt1 + B - D + I – E (Pt) = B - I + D – E (Pt+1) = B - I + D + I – Earrow_forwardThe mark-recapture method is used to estimate the population size of mountain yellow-legged frogs in the San Gabriel Mountains of southern California. 100 frogs are captured and marked. A week later, 50 frogs are captured, and 10 of them are marked. The total population size isarrow_forward
- QUESTION 7 You are working on an undergraduate project, studying an insect species. A recent study showed that the pigmentation pattern used for camouflage is made by an enzyme K. However, it has been shown that, when compared to the Kk individuals, the KK homozygotes would have a higher chance of dying from pigment overdose before growing up, while the kk individuals can easily fall to predation before becoming adults. In fact, when compared to the Kk individuals, the estimated selection coefficients for KK and kk are 0.25 and 0.5, respectively. You are interested in what would happen to these alleles. At present, you have conducted a population survey and have 1000 larvae in your collection. Among them, 360 are KK, 480 Kk and 160 kk. How many larvae would survive to adulthood? O 170 O 417 583 750 830 (Following up on Question 7.) What would be the KK genotype frequency at the adult stage? O 0.090 O 0.270 O 0.325 O 0.360 O 0.529 QUESTION 8arrow_forwardA population of wolves in a national park is known to be 150 individuals. As wildlife ecologist, you are tasked with assessing the accuracy of the mark and recapture technique. At the start, you mark 30 wolves, and your recapture yields 20 wolves, 5 of whom are marked. State whether the mark and recapture data results in an over or underestimation of the actual population size, and show work to support your claim.arrow_forwardYou have been tasked to estimate the number of pigeons along a lakeside in western Texas. To accomplish your assigned task, you capture 37 pigeons and mark them with PIT tags. You return to the site one week later and resample the population by capturing 52 pigeons. Of the 52, 30 are marked from your first sample. How many pigeons are in the population?arrow_forward
- How do you calculate N (population size)? Is the mark-recapture method a reliable way to estimate population size?arrow_forwardImagine we census a population of rabbits (assume females only) that has been occupying a barn for a few decades. On June 30, 2021, we census the population right after the reproductive season. We count 650 individuals and divide them into 4 age classes: 0-‐6 months, 6 months-‐1 year, 1 year to 1-‐and-‐a-‐half-‐years, and 1-‐ and-‐a-‐half to 2 years. 350 are in the first age class, 175 are in the second, 90 are in the third, and 35 are in the fourth. We repeat the census at six months and find that none of the newborns (0-‐6 months age class) from the first census have produced any offspring, and that 15% of them have died. We find that the 2nd age class had 75% survival and produced 500 offspring. We also find that only 10% of the 3rd age class has survived, and produced 180 offspring. All the rabbits in the 4th age died and produced zero offspring. (a) What is the life table for this population? (i.e., x, bx, sx, lx)arrow_forwardImagine we census a population of rabbits (assume females only) that has been occupying a barn for a few decades. On June 30, 2021, we census the population right after the reproductive season. We count 650 individuals and divide them into 4 age classes: 0-‐6 months, 6 months-‐1 year, 1 year to 1-‐and-‐a-‐half-‐years, and 1-‐ and-‐a-‐half to 2 years. 350 are in the first age class, 175 are in the second, 90 are in the third, and 35 are in the fourth. We repeat the census at six months and find that none of the newborns (0-‐6 months age class) from the first census have produced any offspring, and that 15% of them have died. We find that the 2nd age class had 75% survival and produced 500 offspring. We also find that only 10% of the 3rd age class has survived, and produced 180 offspring. All the rabbits in the 4th age died and produced zero offspring. (a) What is the life table for this population? (i.e., x, bx, sx, lx) (b) Estimate the replacement rate (R0) for the…arrow_forward
- Concepts of BiologyBiologyISBN:9781938168116Author:Samantha Fowler, Rebecca Roush, James WisePublisher:OpenStax CollegeBiology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage Learning