Even with their cuticles and protected embryos, bryophytes are not totally liberated from their ancestral heritage. The first factor is the dependence of the sperm for water. As the sperm must swim from the antheridium to the archegonium, the presence of a single drop of rain or dew is insufficient for this process to occur. In addition to this, most bryophytes have no vascular tissue to carry water from the soil to the aerial parts of the plant. To absorb water bryophytes must imbibe it like sponges and distribute it slowly through the rest of the plant by a slow process of diffusion.
The Capture-recapture method of the snail Helix aspersa
The method of capture, marking, release and recapture is an important procedure in animal ecology because it not only allows us to estimate the density of a population, but can also give us some idea of the 'birth-rate’ and ‘death-rate’ of that population.
To determine the population of the snail, Helix aspersa, a 40m section of wall on a common that was sectioned into four separate 10m sections. In each of these sections one of four groups counted the number of snails found. On the first day all snails found were marked with a red nail varnish and the total number marked was recorded. A few days later we returned to the site and this time we recorded the number of snails found on that section of wall. On this second day we recorded the number of snails that were found with and without nail varnish on their shells.
Results
The data collected on the two days are recorded in the table below:
A student t-Test for the mean of the population captured is as follows:
Null Hypothesis (HO) That the samples are drawn from populations with equal means
(H1) There is a statistically significant difference between the means of the two populations caught
The value of the t-Test is 0.692471; therefore we accept the Null hypothesis as there is no significant difference between the two means. This conclusion therefore means that the calculation we carry out for measuring the number of the population of snails, is taken from reliable data.
The Lincoln-Peterson method assumes that, if the sample is taken in a random way, it will contain the same proportion of marked animals as that in the whole population.
To carry out this calculation we assume the following:
No. of marked animals in sample (S1) = No. of marked animals in total population (R)
Total caught in sample (S2) Total population size (N)
S1 = R
N S2
Therefore N = S1 X S2
R
Therefore the total population size of Helix aspersa = 95 X 90
41
Therefore the total population size of Helix aspersa = 208.5
Therefore the total population size of Helix aspersa = 209 Snails
Evaluation
The first question that we must ask is whether or not the population of snails we were measuring was closed? Or do we assume that snails were free to move in and out of the area that we were measuring on the wall. Also it is worth noting that the area of the wall we were measuring was slightly shifted to the right (by about a metre) during the revisit to the site. As there was the chance for snails to move out of the 40m section, not all animals had the same chance of being recaptured. As it is was more likely for the snails in the centre to be caught as opposed to those on the extremes of the sampling site. The calculation also assumes that the overall population size does not change between taking the samples. Although we did not have any direct evidence for this assumption, many of the snails identified in the first sample had painted marks on them from a previous year. This demonstrates that the survival rate of some snails is at least over a year in length. It also shows that the procedure of marking the snails with red varnish and the method used to gain access to the snails does not alter the habitat in a way that “influences their chances of survival”. It must also be noted that the group working on the 10-20m section of wall counted on average twice as many snails as everybody else. Therefore did that group count snails that were present only on the wall or snails that were present within the flora off the wall? If this is the case then this clearly does not fit the brief and reduces the validity of the estimation of population size. However if (as it looks) the number of snails counted in the second sample included snails both on and off the wall this should remove the inaccuracy from the experiment.
From the above evaluation I conclude that the results obtained in this method are reasonably valid enough to predict the population size of the snails.
To improve the validity of the results the Jolly-Seber method can be used. It is however more complicated than the Lincoln-Peterson method, as larger samples are taken over longer periods of time. However, due to the amount of work needed for such a method, it would have been unfeasible to carry out such a method.
Woodland
Mercurialis Perennis Dogs Mercury
To determine the correlation between light intensity and the growth of dog’s Mercury, we measured the height and the percentage area covered in a quadrant, compared to the light intensity and canopy cover in a woodland. The hypothesis we were determining was whether or not Dog’s Mercury was shade tolerant.
Table 3 shows the data collected for this experiment.
To present the data I have decided to use a scatter graph which was drawn with a line of best fit.
Figure 1 shows the relationship between the percentage cover of Mercurialis perennis (percentage cover) and light intensity
Although figure 1 does not show perfect negative correlation between the light intensity and growth of Dog’s Mercury, the data goes some way to identify a correlation between the two factors.
Students t-Test
Ho There is no relationship between the growth of Dog’s Mercury and light intensity.
H1 There is a correlation between the growth of Dog’s Mercury and the intensity of light.
A t-Test of the comparison of the light intensity and the percentage coverage of Dog’s Mercury gives a P value of > 0.0004. Therefore we reject the null hypothesis, and accept H1, as there is a statistically significant correlation between light intensity and the growth of Dog’s Mercury.
Evaluation
The data shows that there is a link between of Dog’s Mercury and light intensity. One explanation for the growth of Dog’s Mercury and light intensity could be explained by light compensation point; the level of illumination at which photosynthetic fixation of carbon dioxide just matches respiratory loss.
Relationship between Photosynthesis, Respiration and Light Intensity
Photosynthesis makes fixed carbon compounds and respiration burns fixed carbon compounds. At light intensities above the level of photosynthesis (light saturation range 1,200-2,000 ft-c), the rate of photosynthesis is much higher than the rate of respiration, up to 10-times higher. Thus, plants produce a great excess of fixed carbon. But, as the light intensity decreases the rate of photosynthesis goes down. Eventually, a light intensity is reached where the rates of photosynthesis and respiration are equal; this is called the light compensation point. At light intensities below the light compensation, the plant is starved because its rate of photosynthesis is less than its rate of respiration.
Figure 2 shows the relationship between photosynthesis and respiration.
Figure 3 shows the relationship between acclimatisation of plants and the rate of photosynthesis.
Why does shade or acclimatised plants grow well at low light intensities?
The relationship shown in the figure 2 above applies to sun plants, which are plants that grow best at very high light intensities. Shade plants grow best at lower light intensities, such as would be found on a forest floors. Notice in the figure 3, the shade plants have a:
- lower maximum photosynthesis rate,
- lower light saturation range, but most importantly
- lower light compensation point.
Thus, shade plants are adapted to growing best at lower light intensities which is why they make good indoor plants.
This data does not however, directly prove that light intensity is the only factor in the growth and distribution of Dog’s Mercury, and therefore other factors must be considered for example:
- Local differences in soil moisture could also play a significant factor.
- Dog’s Mercury could be intolerant to variations on humidity which are more extreme in the field as compared to the woods.
- Dog’s Mercury may be intolerant to the grazing of wild/domesticated animals
- Dog’s Mercury is intolerant to the trampling of humans/animals.
- Dog’s Mercury is excluded from the open field by competition of faster growing species that cannot grow in the woods but are able to flourish in the field.
Before the hypothesis of light intensity can be determined as the major factor in the growth of Dog’s Mercury, answers to the above possible confounding factors must be determined with separate experiments.
Animal Behavioural Study Of A Burchell’s Zebra
Before I discuss the behaviour of the Burchell’s Zebra I must consider how behaviour is to be defined. The Readers Digest pocket dictionary defines behaviour; “as the manner, conduct, way of behaving, in response to a stimulus.” Much of an animal’s behaviour does “consist of externally observable muscular activity, the act and react of this definition.” (Campbell). But what of the behavioural responses, that have no visible muscular response, such as a young bird hearing the call of its parent bird. This sound of this call is stored within the brain of the young bird. Although on hearing this song, it may change the chemical pathways of the brain, it may not produce any visible muscular response. In fact such a response may take months to occur, when the young bird begins to match this song with the memory of its parent. Therefore if we consider, behaviour as the actions of an animal and how it carries out those actions. Then this definition should include the nonmotor components of behaviour as well as an animal’s visible actions.
In this study I am looking at the activity of the Burchell's Zebra. To do this I must consider the behaviour of the zebra. The Burchell's Zebra is more active in the cooler early morning and late afternoon. Predominantly a grazer, feeding in areas with short grass, they have a strong sensitive upper lip with which it gathers herbage by collecting the grass between the lip and the lower incisors before plucking the harvest. The zebra drinks at least once a day and has a strong preference for clean water. The Burchell’s Zebra lives in small family units, which typically consist of one stallion and one mare with their foals. Non-breeding stallions occur in bachelor groups.
The harem's bonds are reinforced through social behaviour. Members of a group often perform social grooming, placing close to one another in opposite directions to gain access to the other's head, neck and shoulders. This is also a frequent resting position, since it allows them to scan in both directions for detecting a possible approach of predators.
When it does not feel threatened, zebra is very inquisitive. It is a noisy animal; its contact call is a mixture between a bray and a bark. Rival stallions can fight fiercely, kneeling down and trying to bite their opponent's legs, or standing on the hind legs and neck-wrestling; kicking with the fore legs and biting face and neck. They may kick when they are followed closely, which has mostly a dissuasive effect.
I decided to observe the zebra because of its size made it easy to observe. For this study a individual zebra was observed for five, ten-minute intervals. During this time I observed how much time the zebras spent doing each activity. The zebra was observed on a hot sunny spring day between the hour of 11 and 12. To prepare for what observations I was going to make during this study I carried out a pilot study. During this time I recorded what activities the zebra was doing, I observed the following activities:
Walking
Standing
Running
Grassing
Social interaction
I decided to make the main focus of the study to determine the amount of time the particular zebra spent in either a social context (zebra in close contact with one or more zebra’s) or alone. I also decide to determine the amount of time spent by the zebra, carrying out a physical response as shown in the below table:
Table to show the data collected over a fifty-minute period of observation.
Figure 4 shows the interaction of a Burchell’s zebra over five, ten minutes periods
Student’s tTest
H0 There is no difference in the time spent by Burchell’s zebra between social and individual interactions.
H1 there is a statistically significant difference between the social interactions of Burchell’s zebra.
The results of a student tTest show the value of P> 0.0003. Therefore we reject the null hypothesis. As there is a statistically significant difference between the means of the two sets of data, I must therefore conclude that Burchell’s zebra spend more time apart from the herd (at least from the data in this experiment).
As you can clearly see from graph 4, for the vast majority of the observed time, the zebra spent it, grazing. This is clearly a recognised characteristic of this and other Zebra families. Although this was only a short study, for much of the time the zebra that I observed was outside of the herd. This could be explained by the fact that the observed animal was a young male, and therefore excluded from the breeding herd by the lead stallion. Secondly the institutionalised environment present within a zoo could explain this lack of social interaction. The lack of stimuli from predators could have taught zebras that there is no need for the protection that is found within herds. It is safe to move around on its own such as that defined by habituation. Habituation is a simple type of learning that involves the loss of a response to unimportant stimuli.
The fact that this study was carried out for fifty minutes, on a particular day and time, could suggest that these results are only applicable to that day. Therefore to gain more insight into the behaviour of such an animal, more research needs to be done on the animal at different times of the day, but under the same weather conditions. This would determine if the time of the day has any affect on the behaviour of such an animal. It may also be advisable to determine whether the type of weather has any affect on the behaviour of such animals. As it was a particularly hot day, the effect of the temperature may have forced animals to move away from each other.
The relevance of this fieldwork trip for AS/A2 biology
There are three main exam boards for AS/A2 biology in England; these are AQA, Edexcel and OCR. To determine the suitability of this type of fieldwork trip, I must first discover the syllabus requirements of the above exam boards.
AQA requirements;
“It is expected that candidates will carry out fieldwork involving the collection of quantitative data from at least one habitat and the application of elementary statistical analysis to the results”. (Taken from www.aqa.org.uk)
Edexcel and OCR
“Candidates should be able to understand the effects of biotic and abiotic factors on the distribution of organisms in a terrestrial and an aquatic habitat; describe qualitative and quantitative field techniques, including different methods of sampling used to investigate the distribution of organisms in a specific terrestrial, freshwater or marine littoral habitat. Practical work to include the study of the distribution of plants and animals in at least one habitat and investigations of the influence of abiotic factors on them.” (Taken from www.edexcel.org and www.ocr.org).
The work that we have carried out during our field trip clearly, meets all of the requirements of the above exam boards for both AS and A level biology. The data collected also enables candidates to analyse the results via statistical means.
