AP Biology : 7.12 Variations in Populations – Exam Style questions with Answer- FRQ

Question

Trichomes are hairlike outgrowths of the epidermis of plants that are thought to provide protection against being eaten by herbivores (herbivory). In a certain plant species, stem trichome density is genetically determined.
To investigate variation in stem trichome density within the plant species, a student counted the number of trichomes on the stems of six plants in each of three different populations. The student used the data to calculate the mean trichome density (numbers of hairs per square centimeter) for each population. The results are provided in the table below.
TRICHOME DENSITY IN THREE PLANT POPULATIONS (number of trichomes/cm2)

PopulationPlant 1Plant 2Plant 3Plant 4Plant 5Plant 6MeanStandard Error of the Mean (SEM)
I8119108691
II126159138111
III13179141216141

(a) On the axes provided, create an appropriately labeled graph to illustrate the sample means of the three populations to within 95% confidence (i.e., sample mean ± 2 SEM).
(b) Based on the sample means and standard errors of the means, identify the two populations that are most likely to have statistically significant differences in the mean stem trichome densities. Justify your response.
(c) Describe the independent and dependent variables and a control treatment for an experiment to test the hypothesis that higher trichome density in plants is selected for in the presence of herbivores. Identify an appropriate duration of the experiment to ensure that natural selection is measured, and predict the experimental results that would support the hypothesis.

▶️Answer/Explanation

Ans:

(b) Plant population I is most likely to have significant differences in mean stem trick homes density to population III because the weak as well as the standard error of the mean do not overload mnemonically.  There is a difference of I between the highest limit of pop I (at II) and the lowest of pop III (at 12), This means that  data without those data sets can not equal each other data in the open set and remain acceptable as data in those sets.

(c) In experiment to test for whatever  higher trachoma density is solicited for with the presence of  herbivores, the independent variable be the number trachoma/ cm2, and  a control would be an population of plants that is not exposed to herbivores. On appropriate duration would be over several generations, let us say 7 so that the affopriy of the generations begin to show the phenotype that is being selected for in a measureable percentage that can be compared to the change of the control. This change will occur if natural selection change the frequency of phenotypes.

I predict (higher Trichomes/ cm2) in the plant population that is exposed to herbivores than in the population without exposure to herbivores.

Question

Adult male guppies (Poecilia reticulata) exhibit genetically determined spots, while juvenile and adult female guppies lack spots. In a study of selection, male and female guppies from genetically diverse populations were collected from different mountain streams and placed together in an isolated environment containing no predators.
The study population was maintained for several generations in the isolated area before being separated into two groups. One group was moved to an artificial pond containing a fish predator, while a second group was moved to an artificial pond containing no predators. The two groups went through several generations in their new environments. At different times during the experiment, the mean number of spots per adult male guppy was determined as shown in the figure below. Vertical bars in the figure represent two standard errors of the mean (SEM).

(a) Describe the change in genetic variation in the population between 0 and 6 months and provide reasoning for your description based on the means and SEM.
(b) Propose ONE type of mating behavior that could have resulted in the observed change in the number of spots per adult male guppy between 6 and 20 months in the absence of the predator.
(c) Propose an evolutionary mechanism that explains the change in average number of spots between 6 and 20 months in the presence of the predator.

▶️Answer/Explanation

Ans:

In the period lasting from 0-6 months, the amount of genetic variation in the population decreased. The mean number of spots increased from 10 to 12 and the SEM decreased, which indicates that there was variation in the number of spots, and more of the individuals had a number closer to the mean.

One mating behavior that could’ve resulted in the further increase in the average number of spots in the absence of the predator is sexual selection of preference. Perhaps female guppies showed a preference for makes with a greater number of spots. This would make having spots a favorable trait in males and that increase its frequency.

An evolutionary mechanism that would explain the decrease in spots in the presence of predators is the potentially decreased fitness of makes with spots. This could be for a number of reasons, but Perhaps the spots made the guppies more visible to predators and thus covered them to be eaten more frequently. This would decrease their ability to make (they may die before they sexually mature) and hus result in a lower frequency of spots.

Question:

Leucine aminopeptidases (LAPs) are found in all living organisms and have been associated with the response of the marine mussel, Mytilus edulis, to changes in salinity. LAPs are enzymes that remove N-terminal amino acids from proteins and release the free amino acids into the cytosol. To investigate the evolution of LAPs in wild populations of M. edulis, researchers sampled adult mussels from several different locations along a part of the northeast coast of the United States, as shown in Figure 1. The researchers then determined the percent of individuals possessing a particular lap allele, lap94, in mussels from each sample site (table 1).
(a) On the axes provided, construct an appropriately labeled bar graph to illustrate the observed frequencies of the lap94 allele in the study populations.
(b) Based on the data, describe the most likely effect of salinity on the frequency of the lap94 allele in the marine mussel populations in Long Island Sound. Predict the likely lap94 allele frequency at a sampling site between site 1 and site 2 in Long Island Sound.
(c) Describe the most likely effect of LAP94 activity on the osmolarity of the cytosol. Describe the function of LAP94 in maintaining water balance in the mussels living in the Atlantic Ocean.
(d) Marine mussel larvae are evenly dispersed throughout the study area by water movement. As larvae mature, they attach to the rocks in the water. Explain the differences in lap94 allele frequency among adult mussel populations at the sample sites despite the dispersal of larvae throughout the entire study area. Predict the likely effect on distribution of mussels in Long Island Sound if the lap94 allele was found in all of the mussels in the population. Justify your prediction. 

▶️Answer/Explanation

Ans:

(b) The most likely effect on the frequency of the lap94 allele is that an increase in salinity is associated with an increase in the frequency of the lap94 allele. There is a direct relationship. The lap94 allele frequency between site 1 and site 2 is 15%. 

(c)  Lap94 activity releases amino acids in the cytosol which lower the water polenta (ψ) inside of the cell. This leads to a flow of water into the cell. The reason why lap94 activity increase as salinity increases is because the hypertense environment surrounding the cell would cause the water inside the cell to leave, eventually the cell would plasmolyze and die.  Lap94 attempts to counterbalance the effect of an increase in salinity. Attempting to create an isotones solution. 

(d) The differences in lap94 allele frequency are due to the differences of salinity at the sites where adult mussels attach themselves to rocks. A higher population of individuals with the lap94 allele will survive in areas of high salinity. That is why the frequency of the allele is different across the data presented. There would be a greater number of mussels in area of high salinity if all the mussels with the allele were in area of low salinity then water would flow into the cell causing it to burst. 

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