# Investigating diversity

Genetic information, variation and relationships between organisms (AQA AS Biology) PART 6 of 6 TOPICS

 TOPICS: DNA, genes and chromosomes  Genetic diversity can arise as a result of mutation or meiosis  Genetic diversity and adaptation  Species and taxonomy  Biodiversity within a community  Investigating diversity

# Investigating diversity:

Investigating diversity enables comparisons to be made:

• between different areas – comparing the biodiversity of one section of woodland with a similar type of woodland in another area to see which requires the most protection.
• in the same area at different times – comparing the biodiversity of a group of fields before and after hedgerows have been removed or replanted, or the biodiversity of a wood in two different seasons or at two different times of the day.

Genetic diversity within or between species can be made by comparing:

• the frequency of measurable or observable characteristics
• base sequence of DNA
• base sequence of mRNA
• amino acid sequence made in protein synthesis

Species richness is a number representing the different types of species in an area. The higher this number the more ‘rich’ an area is with species. The number of each individuals is not accounted for so only one individual of a species will have the same weight as a hundred individuals in another species.

Species evenness compares the size of a population where the individuals of species are accounted for.

Species diversity is directionally proportional to species richness and species evenness where if species richness and species evenness were to increase then species diversity increases.

Comparing biodiversity can be difficult as the following example shows:

 Species Number of individuals in field 1 Number of individuals in field 2 Buttercup 300 45 Dandelion 345 35 Daisy 355 920 Total 1000 1000

A conclusion can be drawn that field 1 has more species evenness than field 2 and so it has a greater biodiversity.

Index of diversity:

How to Calculate the Standard Deviation:

1. Calculate the mean (x̅) of a set of data
2. Subtract the mean from each point of data to determine (x-x̅).  You’ll do this for each data point, so you’ll have multiple (x-x̅).
3. Square each of the resulting numbers to determine (x-x̅)^2.  As in step 2, you’ll do this for each data point, so you’ll have multiple (x-x̅)^2.
4. Add the values from the previous step together to get ∑(x-x̅)^2.  Now you should be working with a single value.
5. Calculate (n-1) by subtracting 1 from your sample size.  Your sample size is the total number of data points you collected.
6. Divide the answer from step 4 by the answer from step 5
7. Calculate the square root of your previous answer to determine the standard deviation.
8. Be sure your standard deviation has the same number of units as your raw data, so you may need to round your answer.
9. The standard deviation should have the same unit as the raw data you collected.  For example, SD = +/- 0.5 cm.

The general formula for this however is:

NB: You are not required to work out standard deviations in the exam but in experiments it is needed to be learnt.

What does standard deviation show?

It shows how much the data is spread above and below the mean. A high standard deviation shows the data is more spread about the mean meaning it is less reliable however some will have a low standard deviation meaning the data is less spread out about the mean and is more reliable.

Statistical tests: Statistical tests include chi-squared, standard error and 95% confidence limits/t-test and Spearman’s rank correlation. Statistical tests are used to see if results are due to chance and if our null hypothesis (the theory which is most likely to be incorrect) is actually right. So how would you know which statistical test to use?

• Chi-squared: In a particular experiment certain values may be expected to be recorded and obtained in particular categories being investigated. These values are known as expected values. However other values may be obtained which are known as observed values. In this situation we need to use the chi-squared equation for this which is as follows:

Where O is observed and E is expected.

NB: You are not expected to work out chi squared in the exam however a demo will is given below. An example is in an A2 resource where chi-squared is covered again named ‘Genetics, populations, evolution and ecosystems 1’.

We have to come up with our null hypothesis which is: THERE IS NO SIGNIFICANT DIFFERENCE BETWEEN THE OBSERVED AND EXPECTED RESULTS.

It is best if you put your data for applying chi-squared in a table like the one below:

 Categories O E O – E (O – E)2 (O – E)2/E Probability, p Degrees of freedom 0.25 (25%) 0.20 (20%) 0.15 (15%) 0.10 (10%) 0.05 (5%) 0.02 (2%) 0.01 (1%) 1 1.32 1.64 2.07 2.71 3.84 5.41 6.63 2 2.77 3.22 3.79 4.61 5.99 7.82 9.21 3 4.11 4.64 5.32 6.25 7.81 9.84 11.34 4 5.39 5.99 6.74 7.78 9.49 11.67 13.28 5 6.63 7.29 8.12 9.24 11.07 13.39 15.09

NB: We should always use the column with the 0.05 or 5% probability highlighted in yellow as biologists always use this. The values in the table are known as critical values.

To know which degrees of freedom to use we must use the value that is 1 minus how many categories we have. You look at the corresponding critical value in the 5% probability column. If the chi-squared value is smaller than the critical value then we accept our null hypothesis saying that there is a 5% or higher probability that the results are due to chance and there is no significant difference between observed and expected values. NB: If our chi-squared value was greater than the critical value then we reject our null hypothesis and also say that there is a 5% or lower probability that the results are not due to chance and there is a significant difference between observed and expected value.

• Standard error and 95% confidence limits/t-test: If in an experiment you need to look at the differences between the means then we use the standard error and 95% confidence limits’ equation/t-test’s equation which is as follows:

Where SD is standard deviation and n is the number of results

NB: You are not expected to work out standard error and 95% confidence limits in the exam however a demo will is given below.

We have to come up with our null hypothesis which is: THERE IS NO DIFFERENCE BETWEEN THE TWO MEANS.

 Number of seeds that had germinated after 10 days Seeds touching each other Seeds placed 2 cm apart 8 10 10 5 6 12 14 16 16 12 15 16 15 9 13 12 9 10 10 12 13 16 12 10 14 9 11 7 9 14 11 15 14 8 13 14 8 10 7 10 8 7 15 11 15 14 13 13 11 10 9 12 8 9 19 11 12 17 9 17

Some working out has to take place which is as follows:

SEEDS TOUCHING EACH OTHER

 Mean (8+10+10+5+6+12+15+9+13+12+9+10+14+9+11+7+9+14+8+10+7+10+8+7+11+10+9+12+8+9)/30 = 9.73 Σx2 64+100+100+25+36+144+225+81+169+144+81+100+196+81+121+49+81+196+64+100+49+ 100+64+49+121+100+81+144+64+81 = 3010 (Σx)2 2922 = 85264 Standard deviation √(167.867/29) = 2.41 Standard error 2.41/√30 = 0.44

SEEDS PLACED 2CM APART

 Mean (14+16+16+12+15+16+10+12+13+16+12+10+11+15+14+8+13+14+15+11+15+14+13+13+19+11+12+17+9+17)/30 = 13.43 Σx2 196+256+256+144+225+256+100+144+169+256+144+100+121+225+196+64+169+196+225+121+225+196+169+169+361121+144+289+81+289 = 5607 (Σx)2 4032 = 162409 Standard deviation √(193.367/29) = 2.58 Standard error 2.58/√30 = 0.47

The 95% confidence limits is double the standard error above and below the mean which is as follows continued from the example shown above:

SEEDS TOUCHING EACH OTHER

 Upper limit (2 x 0.44) + 9.73 = 10.61 Lower limit 9.73 – (2 x 0.44) = 8.85

SEEDS PLACED 2CM APART

 Upper limit (2 x 0.47) + 13.43 = 14.37 Lower limit 13.43 – (2 x 0.47) = 12.49

These limits are represented in the graph below:

If the 95% confidence limits do overlap then we accept the null hypothesis saying there is 5% or higher probability that the difference between the two means is due to chance alone. NB: If the 95% confidence limits do not overlap then we reject the null hypothesis saying there is less than 5% probability that the difference between the means values is not due to chance. In this case we reject.

• Spearman’s rank correlation: After an experiment, if you want to see if there is any association of two sets of data then you would use this test.

Where D is the rank and n is the number of observations

NB: The Spearman’s rank correlation (Rs) value will be between either -1 and 0 or 0 and 1.

NB: You are not expected to work out Spearman’s rank correlation in the exam however a demo will is given below.

NB: The difference in rank (D) squared must add up to 6.5.

We have to come up with our null hypothesis which is: THERE IS NO ASSOCIATION BETWEEN THE TWO SETS OF RESULTS.

 Mass of egg / g 1.37 1.49 1.56 1.7 1.72 1.79 1.93 Mass of chick on hatching / g 0.99 0.99 1.18 1.16 1.17 1.27 1.75

The first step is to start ranking the mass of egg or the mass of the chick from smallest to biggest or vice versa. NB: This is because if the two were ranked each mass of egg or chick would not be corresponding to the other value in the other set of data. Also if two values have the rank then you divide the rank they share by two, e.g. For the mass of chick there are two chicks that have a mass of 0.99. As this value is the smallest out of all the other masses of chicks it has a rank of 1. Divide 1 by two to get 0.5. The ranks are shown in the table below:

 Individual number Egg mass / g Rank Chick mass / g Rank Difference in rank (D) D2 1 1.37 1 0.99 1.5 0.5 0.25 2 1.49 2 0.99 1.5 0.5 0.25 3 1.56 3 1.18 5 2 4 4 1.70 4 1.16 3 1 1 5 1.72 5 1.17 4 1 1 6 1.79 6 1.27 6 0 0 7 1.93 7 1.75 7 0 0

In the table above the D2 values add up to 6.5. Work out the Rs value which comes to be 0.884 when n is substituted for 7 and ƩD2 is substituted for 6.5. NB: A positive Rs value shows it is more likely that there is an association. A negative Rs value shows it is more likely that there is no association. However an Rs value of zero shows that there is definitely no association. Then compare your Rs value with the critical values below:

 Number of pairs of measurements Critical value 5 1.00 6 0.89 7 0.79 8 0.74 9 0.68 10 0.65 12 0.59 14 0.54 16 0.51 18 0.48

As we have 7 pairs of data we have to look at the corresponding critical value to the number of pairs which in this case is 0.79 highlighted in yellow. Our Rs of 0.884 is bigger than the critical value and so we reject our null hypothesis by saying there is a 5% or lower probability that the results are not due to chance and there is a significant correlation between the two sets of data. NB: If our Rs value is smaller than the critical value we accept our null hypothesis by saying there is a 5% or higher probability that the results are due to chance and that there is no significant correlation between the two sets of data.

# Biodiversity within a community

 TOPICS: DNA, genes and chromosomes  Genetic diversity can arise as a result of mutation or meiosis  Genetic diversity and adaptation  Species and taxonomy  Biodiversity within a community  Investigating diversity

Genetic information, variation and relationships between organisms (AQA AS Biology) PART 5 of 6 TOPICS

# Biodiversity within a community:

Biodiversity can relate to a range of habitats, from a small local habitat to the Earth.

Species richness is a measure of the number of different species in a community known as index of diversity:

N(N-1) / ∑ n(n-1) = index of diversity where

N = total number of individuals

n = total number of organisms in a species

Framers cut shrubs, bushes and hedges to increase the farm land. However it reduces the number of habitats and the amount of food sources therefore reducing the biodiversity. Deforestation creates the same effect but also decreases the amount of rainfall meaning crops and plants will have less transpiration by photosynthesis, leads to soil erosion as the exposed soil can dry out and the nutrients from the trees are not sent back into the soil, serious flooding can occur as the rainwater will not be soaked and there will be an increase in the levels of CO2.

Conservation allows the number of reduced species to increase without destroying habitats and using pesticides. This is so biodiversity can increase. The process of using pesticides and removing hedge grows is done in a yearly cycle.

] That’s all that you need to know for biodiversity within a community at AQA [

# Genetic diversity can arise as a result of mutation or meiosis

Genetic information, variation and relationships between organisms (AQA AS Biology) PART 2 of 6 TOPICS

 TOPICS: DNA, genes and chromosomes  Genetic diversity can arise as a result of mutation or meiosis  Genetic diversity and adaptation  Species and taxonomy  Biodiversity within a community  Investigating diversity

# Genetic diversity can arise as a result of mutation or meiosis:

Gene mutations involve a change in the base sequence of chromosomes. It can happen during DNA replication or during base deletion (a base is removed) or base substitution (a base is substituted for another). Both of these cause changes to the triplets which can cause it to code for a different amino acid. Sometimes base substitution does not change the amino acid that is coded by the triplet as the genetic code is degenerate.

Mutagenic agents include x-rays, gamma rays and alpha particles and includes an isotope of cobalt and caesium. UV radiation with a wavelength (λ) above 260nm causes complications in DNA replication.

Gametes (sperm and egg cells) are haploid which is the cells having one copy of each chromosome. In the end a zygote is made which is diploid meaning that half the chromosomes are from the mother and half the chromosomes are from the father. The zygote then undergoes meiosis where it starts at interphase and ends at telophase similar to mitosis. The two homologous pairs of chromosomes line themselves up on the equator of the cell. The cell align themselves up on each pole of the cell and create a spindle where each fibre attaches to each centromere. The spindle fibres contact pulling each homologous chromosome to each end of the cell. The cell then splits by cytokinesis. This process is called meiosis 1. Meiosis 2 follows through which is the same as mitosis. Four daughter cells are created at the end of the whole of meiosis which are haploid. NB: Bear in mind that meiosis 1 produces two cells which undergo meiosis 2 separately to produce another two daughter cells each totalling up to four.

Independent segregation is where the four daughter cells have different combinations of chromosomes from maternal and paternal sides. This leads to genetic variation. Crossing over of chromatids only occurs in meiosis 1 where the allele on each chromatid is swapped over on to the chromosome.

Mitosis creates two daughter cells which are genetically identical to each other and to the parent cell. Meiosis creates four daughter cells which are genetically different from one another and to the parent cell.

Random fertilization increases variation further as a random sperm can fertilise a random egg.

NB: Notice that in the picture for meiosis where the cell has homologues aligned independently, the big red chromosome has a blue allele and the big red chromosome has a red allele showing that crossing has taken place to make the alleles swap over.

# Species and taxonomy

Genetic information, variation and relationships between organisms (AQA AS Biology) PART 4 of 6 TOPICS

 TOPICS: DNA, genes and chromosomes  Genetic diversity can arise as a result of mutation or meiosis  Genetic diversity and adaptation  Species and taxonomy  Biodiversity within a community  Investigating diversity

# Species and taxonomy:

Two organisms belong to the same species if they can produce fertile offspring.

Physical and chemical makeup help distinguish members of species. Behaviours also does this too where most behaviours are genetically determined. Courtship and mating are essential when it comes to survival. Recognition of members helps to produce fertile offspring with a mate who is capable of breeding, strong and healthy as well. Whilst doing this aggression should not be triggered. Courtship can be:

• Visual: Using colours
• Behaviour: Dances or building nests
• Pheromones: Chemicals released by an organism enabling it to communicate with other members of its own species.

Phylogeny is a classification system which attempts to arrange species according to their evolutionary origins and relationships. It uses a hierarchy in which smaller groups are placed into larger groups. These groups are known as taxa and they do not overlap.

The taxons are: Domain, Kingdom, Phylum, Class, Order, Family, Genus and Species.

Each species is universally identified by a binomial consisting of the name of its genus first then the species.

NB: The 3 types of domain and the 5 types of kingdom do not need to be recalled for the AQA exam.

# Genetic diversity and adaptation

Genetic information, variation and relationships between organisms (AQA AS Biology) PART 3 of 6 TOPICS

 TOPICS: DNA, genes and chromosomes  Genetic diversity can arise as a result of mutation or meiosis  Genetic diversity and adaptation  Species and taxonomy  Biodiversity within a community  Investigating diversity

# Genetic diversity and adaptation:

As the number of alleles increases so does genetic diversity in a population.

Genetic diversity allows natural selection to occur.

Evolution is a change in a populations’ alleles and genotypes from generations to generations. Therefore it should be considered at a population level. Five factors affect the proportion of homozygote and heterozygote:

• Genetic drift: This is a change in the gene pool that occurs in a small population due to chance. Two situations can lead to genetic drift:
• Population Bottlenecks: This is when a large number of a population is wiped out due to disease, natural disasters or overhunting
• Founder Effect: This is when a new colony is found by a small number of individuals
• Gene flow: This is the movement of alleles from one population to another when a member moves into another population. The variety of alleles that this member has can significantly affect the gene pool of a population especially if it has good survival and mating skills.
• Mutations: These are changes to an organisms DNA. The change is transferred to gametes which immediately changes the gene pool. This is a rare event but the cumulative effect is massive. Mutations themselves play an insignificant role in changing the frequency of alleles in a population.
• Non-random mating: Homozygous individuals increases when preferred organisms mate with each other, causing frequency of genotypes to differ significantly from equilibrium values. Populations consist of individuals with different genetic make-ups. This means that Hardy-Weinberg equilibrium is not maintained. NB: Hardy-Weinberg equilibrium equation and definition does not need to be known for AS only for A2.
• Natural selection: Populations consist of individuals with different genetic make-ups. Colourful and vibrant organisms are more susceptible to predation.

Directional selection happens in bacteria for example, where the bacteria are resistant to antibiotics leading to an increase in the frequency of the allele that is resistant to antibiotics.

Stabilising selection happens in an changing environment. Stabilising selection occurs in the natural selection of birth mass in humans. Extremes of the phenotype range are selected leading to a reduction in variation.

 The thin blue line shows the distribution of mass at birth of children born in University College Hospital over a 12-year period. The thick blue line shows the percentage prenatal mortality (those failing to survive four weeks) on a logarithmic scale. ‘O’ shows the optimal mass that had the lowest prenatal mortality. ‘M’ is the mean birth mass.

# DNA, genes and chromosome

Genetic information, variation and relationships between organisms (AQA AS Biology) PART 1 of 6 TOPICS

 TOPICS: DNA, genes and chromosomes  Genetic diversity can arise as a result of mutation or meiosis  Genetic diversity and adaptation  Species and taxonomy  Biodiversity within a community  Investigating diversity

# DNA, genes and chromosome:

Prokaryotic cells have short, circular DNA molecules that are not associated with proteins known as histone.

Eukaryotic cells have long, linear DNA molecules which are associated with histone proteins. Chromosomes are formed from two DNA strands.

DNA molecules that are short and circular are found in mitochondria and DNA which are found in eukaryotic cells.

A gene is a base sequence of DNA that codes for the amino acid sequence of a polypeptide and also codes for functional RNA (transfer RNA known as tRNA).

Protein synthesis happens in two stages:

• Transcription: DNA molecule splits by DNA helicase which breaks the hydrogen bonds. RNA polymerase moves in the 5′ (5 end) to 3′ (3 end) of one DNA strand called the cistron where complementary bases of RNA are matched with bases of the DNA strand. Three bases of the DNA are called triplet and three bases of the RNA are called codon. NB: Remember that RNA does not have thymine as a base so when a complementary base of RNA is needed to pair up with adenine of DNA, uracil is used. The bases of RNA make a molecule called pre-messenger RNA or pre-mRNA. DNA reforms once pre-mRNA has finished being made. The introns are cut out of the pre-mRNA molecule leaving the exons which are bases that can code for amino acids. These are spliced together to make mRNA and this leaves the nucleus via the pores of the nuclear membrane into the cytoplasm.
• Translation: The mRNA reaches the ribosomes to make polypeptides. tRNA molecules which have three bases known as anti-codons attaches to a complementary codons on the mRNA. The tRNA molecules have an amino acid attached to them which all join by condensation reactions to form peptide bonds to create a polypeptide chain.

tRNA molecules are small in length compared to mRNA and are clover-leaf shaped because of hydrogen bonding between complementary bases on different strands to one another. mRNA molecules are single stranded. tRNA molecules have an amino acid attached to them.

Prokaryotes do not make pre-mRNA as they do not have introns.

Amino acids need activation in the form of ATP to attach to tRNA molecules.

A gene occupies a specific location on a DNA called a locus.

The genetic code has characteristics such as :

• It is universal: Every organism has this
• It is non-overlapping: Only one base is used for a triplet
• It is degenerate: Many triplets can code for an amino acids