Variation (CIE A Level Biology)

Cacti are plants that are adapted to live in arid conditions. They store water in thick stems which are protected from grazing by herbivores through the presence of multiple spines. The number of spines that are present on cacti may vary greatly, depending on the frequency of grazing that they experience.

Fig. 1 shows the number of spines present in a species of cacti.

Fig. 1

State the type of variation that is represented by Fig. 1 and provide two reasons for your answer.

Suggest the possible impact of environmental factors on the variation observed in Fig. 1, with regards to the number of spines on cacti.

Contrast the genetic basis of continuous and discontinuous variation with each other.

Students investigated the effect of light intensity on leaf length in the Mexican sword plant (Echinodorus palaefolius). They had two groups (A and B) consisting of ten plants each, with group A grown in a laboratory where the lights were kept on, while group B was grown under similar conditions to group A but kept under low light conditions. The plants were grown at these different light intensities for 14 days, and the average leaf length of each plant was calculated for each group.

The results of the investigation is shown in Table 1.

Table 1

Calculate the mean leaf length for group B. 

Show your working.

The students used the t-test to compare the means of groups A and B.

State two features of the data that allow for the use of the t-test.

State a null hypothesis for this investigation.

After performing the t-test, the students calculated a t-value of 29.7. Table 2 shows different t-values and the probability that the differences between the data sets are due to chance.

Table 2

Using the information in Table 2, discuss the conclusions that the students can draw from their results.

Scientists investigated the effect of light on the germination of begonia seeds. They had three groups consisting of 30 seeds each, for both light and dark conditions. The seeds were exposed to light or darkness for 72 hours before scientists calculated the mean number of seeds that germinated out of 30 seeds for each light condition. All other conditions were kept constant between the seeds.

Fig. 1 shows the results of this investigation.

Fig. 1

Calculate the mean number of seeds germinating in dark conditions as a percentage of the mean number of seeds that germinated in light conditions.

Show your working and give your answer to one decimal place.

Explain why the t-test would not be a suitable statistical test to use on the data represented in Fig.1.

The house sparrow (Passer domesticus) is a small bird commonly found in most parts of the world. They feed mainly on seeds and they show variation in beak size. Two small populations of sparrows, consisting of 20 birds each, were studied to investigate the difference in beak size displayed by each population. 

The biologists conducting the study calculated the mean beak size for each population, as well as the standard deviation for each. The results of these calculations are represented in Table 1

Table 1

Fig. 2 represents the formula for calculating the t-value.

Fig. 2

Using the information in Table 1 and Fig. 2, calculate the t-value for this data set.

Show your working and give your answer to two decimal places.

The null hypothesis for this investigation states that there is no significant difference between the beak sizes of the two populations. The t-value at a probability of 0.05 is 1.96 at 38 degrees of freedom. 

With reference to the t-value calculated in part c), state the conclusion that can be made regarding the difference in beak size between the two populations.

Fig. 1 shows a type of cell division occurring in the reproductive organs of humans.

Fig. 1

[1]

[4]

Tall tomato plants are produced by the presence of a dominant allele T. Two homozygous tall plants were crossed and seeds from this cross were planted in soils with different nutrient compositions. The mean height of each group of tomato plants was calculated and the results are shown in Table 1.

Table 1

Suggest reasons for the variation in height that was observed in the tomato plants.

Seeds from the tomato plants grown in soil type B were taken and planted in soil type D where they reached an average height of 42 cm.

Explain this observation.

Meiosis is one process that contributes to genetic variation.

(i)

State precisely the stage of meiosis where single chromosomes line up on the equator.

[1]

(ii)

Outline the events taking place during anaphase I of meiosis.

[2]

(iii)

Describe how crossing over during meiosis leads to genetic variation.

[2]

Mutation also causes genetic variation. Some populations of water hemp, Amaranthus tuberculatus, have evolved herbicide resistance as a result of a mutation. This is a problem for farmers as water hemp grows in crop fields, lowering productivity.

Two populations of water hemp were tested for resistance to the herbicide mesotrione. One was a population known to be resistant (control) and the other was a test population, whose resistance was unknown.

• Leaves were removed and immersed in a radioactively labelled solution of mesotrione.
• The leaves absorbed some mesotrione and became radioactive.
• Resistant leaves are able to degrade mesotrione by metabolism.
• The time for 50 % of absorbed mesotrione to degrade was calculated by measuring the radioactivity of the leaves.

The results are shown in Table 1.

Table 1

Explain how the results in Table 1 show that the two populations differ in their resistance to mesotrione.

[2]

(ii)

Explain why this example of genetic variation is important for natural selection in water hemp populations.

[2]

(iii)

Farmers can send in a sample of leaves of water hemp from their fields to a laboratory to be tested for resistance to mesotrione or other herbicides.

   Suggest the benefit of this to a farmer.

[1]

The null hypothesis states there is no significant difference between the mean times for 50 % of absorbed mesotrione to degrade in the two populations.

A t‐test can be carried out to compare these two means. The critical value for t at the p = 0.05 significance level is 2.23.

Use the formula in Fig. 1 to calculate the value of t.

Show your working.

Key  = mean s = standard deviation n1 = 6 (number of readings for test population) n2 = 6 (number of readings for control population)

Fig. 1

 [2]

Use your calculated value of t to explain whether the null hypothesis should be accepted or rejected.

[2]

The t-test is a statistical test used to compare two sets of biological data.

State the purpose of t-test.

A number of boy students in a school measured their maximum handspan using a ruler as shown in Fig. 1.

Fig. 1

Table 1 shows their class data:

Table 1

Calculate the mean value x̄ of the students' handspans.

[2]

Use Table 2 below to calculate and sum the values of (x-x̄)²

[3]

From the values you have found in Table 2, calculate the standard deviation S using the formula below.

[2]

In a linked experiment, groups of 10 male students from two different countries (Countries A and B) also took part in this experiment. 

All the boys taking part were of the same age group as the boys in Table 1 and also measured their maximum hand spans. 

Their data calculated out as follows:

Table 2

State a null hypothesis for this investigation. 

From the data described in part (c), calculate the t value between the data from Country A and Country B.

The formula for calculating t is as follows:

Table 3

Use the value of t that you calculated in part (d) to either reject or accept your null hypothesis from part (c).

Give a reason for your decision.