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Stop Losing Marks: Master the Maths in AQA Biology Paper 1/AS — Transcript

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0:00Hey everyone and welcome to Miss Estric
0:02Biology and this entire math skills
0:04video for AS or Alevel paper 1 AQA
0:08biology. Now this isn't literally every
0:10single math skill that is on the math
0:12skills section of the spec. Instead what
0:14I've done is gone through the
0:16specification the theory part so topics
0:181 2 3 and four and wherever they say in
0:20the spec at this point this math skill
0:22could be assessed that is what I've
0:24picked out and gone through for this
0:25entire video. So, it's all of the topic
0:28related math skills that could come up.
0:30The topics 1 2 3 and four. And if you do
0:33want even more help on the math skills,
0:34then don't forget to check out my entire
0:38math skill workbook where it goes
0:39through every skill explained model
0:41example practice questions and the
0:44answers which I'll link in the
0:45description and the QR code just here.
0:48And keep a lookout for the paper 2 A
0:50level video that's coming and the entire
0:53set of math skills too. But for now,
0:55that's it. Let's jump into it. So let's
0:57go through the math skills that are
0:59specifically highlighted within the
1:01theory section for the AS or paper one
1:04for A level biology. So here's the first
1:06math skill that comes up for topic one
1:08in the specification calculating pH from
1:11hydrogen ion concentration using the
1:14formula.
1:17Now this exact math skill has never
1:20actually come up yet but elements of it
1:22have. So principles around the log and
1:26also hydrogen ions but literally
1:28calculating pH hasn't come up but let's
1:30just go through how to calculate pH from
1:33the hydrogen ion concentration using the
1:35formula in case it does come up or
1:37principles of it do come up as well.
1:40First of all pH is a measure of the
1:42hydrogen ion concentration in a
1:44solution. The more hydrogen ions there
1:46are the lower the pH. And that's why
1:48acids which release hydrogen ions have
1:51pH values lower than 7. The formula that
1:55we use is pH= minus log to the base 10
1:59of the hydrogen ion concentration. The
2:02square brackets around H+ just mean
2:05concentration and it must be in moles
2:08per decime cubed which we can see here.
2:11Now the log function here is base 10 not
2:15natural log. So, make sure you're
2:17pressing the correct button on your
2:19calculator.
2:21The minus sign is there. So, that a high
2:24hydrogen ion concentration gives a low
2:27pH and a low hydrogen ion concentration
2:31gives a high pH. So, that's what we mean
2:33by it flips the sense. It basically
2:36shows you high hydrogen ion
2:38concentration means low pH.
2:42So if we were to go through then an
2:45example and imagine that the hydrogen
2:49ion concentration is 3.2 * 10 ^ of - 5
2:54moles per decime cubed. First you need
2:56to type that number into your calculator
2:59and then press the log button. You
3:02should get -4.4949.
3:07Remember the formula says minus log so
3:10that we can take the negative of this
3:12value which gives us positive 4.4949.
3:17Finally we would round our answer
3:20because the concentration was given to
3:22us in two significant figures. We give
3:25the pH to two decimal places and that
3:28makes the pH equal to 4.5.
3:32And here we can see it step by step
3:34everything that you should get. You can
3:37also work backwards if you know the pH.
3:39The hydrogen ion concentration is 10 to
3:41the power of minus pH. So for example,
3:45if pH is 7.35,
3:48the concentration works out as 4.47
3:52* 10us 8.
3:58Next then we've got plot data from
4:00enzyme practicals in an appropriate
4:02graphical format. So the sorts of things
4:06you would need to do then is create a
4:07graph using the data that you're given
4:09or from your actual experiments and you
4:11need to know what to plot. So your
4:13independent variable always goes on the
4:15x-axis and that's actually the case for
4:17any graph and the independent variable
4:19is what you are deliberately changing.
4:23Dependent variable always goes on the y
4:25ais and that is what you are measuring.
4:28You always need to make sure that you're
4:30using the correct units and you put the
4:32units on your axes. And for enzyme
4:35experiments, it should always be a line
4:37graph, not a bar chart. So here we can
4:40see an example of one. And here's some
4:42more just general graph rules that
4:44they'd be looking for if they were
4:45asking you to draw a graph in the exam.
4:48So they would give you a section of
4:50graph paper. And you need to make sure
4:52that you pick a scale so that you fill
4:55at least half of the graph paper,
4:57whether that's in an exam or in your
4:59required practical. You also need to
5:01make sure that your scale goes up by
5:03even increments. So you've got an evenly
5:05spaced scale. You should be plotting
5:08your graph with small crosses. So each
5:10data point you put a cross for that
5:12position. And we should have this smooth
5:14curve line of best fit. And that could
5:16be a curve like we can see here. Or you
5:18might actually get a straight line of
5:20best fit. unlikely in an enzyme
5:22practical because we do have this
5:23increase up to an optimum and then it
5:25either plateaus or it will decrease. And
5:28then lastly, we don't tend to join the
5:32line together dot to dot between those
5:34data points unless you have so few data
5:37points that you can't accurately predict
5:40the line of best fit. But in the exam,
5:42it's unlikely they would give that to
5:43you. That's more relevant if you did the
5:45experiment yourself and you only had
5:47three data points, for example.
5:50Next then is using a tangent to
5:52calculate the initial rate of reaction
5:54from a graph. So reaction rates and this
5:56is all linking to the enzyme topic for
5:58topic one. Reaction rates are often
6:01determined by analyzing experimental
6:04data represented on a curved graph like
6:07we saw here. And when a reaction
6:10progresses the rate is usually changing
6:13over time. So to measure the rate of
6:15reaction at a specific point, we have to
6:18use a tangent to the curve and calculate
6:21the gradient of the tangent rather than
6:24that line of best fit. So let's go
6:27through an example then. So step one is
6:29to draw an usual tangent and we have to
6:31identify the point of interest. So in
6:34this case it might be they want to know
6:36the rate of reaction at 6 minutes. So
6:39you need to find 6 minutes on your
6:41graph. Next then you need to use a ruler
6:45to draw a straight line which will be
6:47our tangent that touches the curve at
6:50exactly the point that you chose. And
6:53then either side of where it touches the
6:56gap between your tangent and the curve
7:00should have equal angles. We can see
7:02here the gap between those two that
7:03angle there and that angle there is
7:06equal. So that's how you know the angle
7:08or the position to place your ruler to
7:11then draw that tangent.
7:14Now we can find the gradient of the
7:16tangent. So choose two clear points on
7:19the tangent. It doesn't actually matter
7:21anywhere on the tangent where you do
7:23this cuz it's a straight line. So no
7:25matter where you do this on the tangent,
7:26you'll get the same answer. But what
7:28we're going to work out is the change in
7:31y. Which means if we did this point and
7:34this point, we'd be reading off the
7:36value here and reading off the value
7:38here on our y axis and working out
7:40what's the difference between those. And
7:42we divide that between the change in x.
7:45So the time here, which is zero, and the
7:48time here, which is 9.6 minutes. And in
7:51fact, we can see all of that written
7:53here and worked out. So we then do the
7:56change in y / the change in x which is
7:58our gradient comes to 11.458 and here
8:02are our units mg per decime cubed which
8:05came from the concentration divided by
8:08time which is why it's per minute. So we
8:11then go on to using base frequencies to
8:14calculate the proportion of other bases
8:16in complimentary DNA strands. This is
8:18still from topic one and it's within the
8:20DNA section. So some key things just to
8:23be aware of to start with DNA base
8:25pairing rules. Adinine is always going
8:28to bind with thymine. They're
8:29complimentary base pairs. Cytosine will
8:31always bind with guanine. So that means
8:34in a double stranded DNA molecule,
8:36whatever percentage of adinine bases you
8:39have, you will have the same percentage
8:41of thymine because they always have to
8:42pair opposite each other. And the same
8:44with cytosine. Whatever percentage of
8:47cytosine bases you have, that'll be the
8:49same as the percentage of guanine. And
8:52because those are the only four bases
8:53you can have, the total of all of those
8:56percentages has to equal 100. So for
8:59this one, your method would be identify
9:02the percentage given in the question. So
9:04which of these bases do you have the
9:06percentage for? use the base pairing
9:08rule to find out the complimentary base
9:11and then you would take that away from
9:13100 and then half the rest between the
9:16other bases. But let me just show you an
9:17example of that. So we've got a DNA
9:19molecule contains 30% adinine bases.
9:23What are the proportion of the other
9:24three bases? Well, if we've got 30%
9:26adinine will have 30% thymine.
9:30Then we know that adinine and thymine
9:33are half of the bases. The other half is
9:35cytosine and guanine. So if 30 is
9:38adinine, 30 is thymine, then those
9:42together add up to 60. Take that away
9:44from 100 gives us 40% left. That has to
9:47be equally split between sides scene and
9:48guanine. So they must be 20% each. Next
9:52skill then is using the magnification
9:55formula. So magnification is image size
9:58divided by actual size. And you could
10:00have to rearrange that formula to work
10:02out the actual size of an object as
10:04well. unlikely they'd ask you to work
10:06out the image size because you would
10:08just look at the image and measure it
10:09with your ruler. So here's our formula
10:12again. And the size of the image is the
10:15size of the object in the actual
10:17microscope image. Whereas the size of
10:20the real object or actual size is the
10:24real life size of the structure you're
10:26looking at which is probably a cell or
10:28an organel in the cell. So we've got an
10:31example here. If the size of an image
10:34under the microscope is 200 micrometers,
10:38the actual size of the objects is 50
10:40micrometers and the magnification is
10:43this time we do 200 / 50 and that would
10:46tell us that the magnification is four
10:48times. So we can see just using that
10:50formula and the data that you were
10:52given. The units of measurements that
10:55you usually get are micrometers or
10:57nanometers for the actual size and it's
11:00normally actually micrometers because
11:03most of the organels in the cell are
11:06within the sizes that you would measure
11:09them in micrometers. We don't tend to
11:11use nanometers unless like it says here
11:14viruses are very very small. So you
11:16might have the units in nanometers. But
11:18when you're actually measuring it on
11:20your image size, you're likely going to
11:22be measuring in millimeters. But they
11:25usually ask you to give your final
11:27answer in micrometers. So you do have to
11:29do that conversion as well. And we'll
11:32see that in our worked example. So if we
11:34have a look at this one, we've got what
11:36is the magnification of the
11:39mitochondrian in your image? Show your
11:42working. So for this one, write out the
11:45formula. First of all, we've got
11:46magnification equals size of image
11:48divided by the size of the real object.
11:51The length of the image, if we were to
11:53measure it, it's been measured to be 84
11:56mm. And we've already been told that the
11:59actual length is 6 micrometers.
12:02So, we'd need to do 84 / 6. However,
12:06they have to be in the same units. and
12:0984 mm you can fit 1,000 microme into 1
12:15mm. So if we have 84 mm that means we
12:20have 84,000
12:22micrometers. So whenever you're
12:24converting your image size which will be
12:27in millime into micrometers you always
12:30times by a th00and as we then substitute
12:33in those values and we get 14,000. That
12:36is our magnification.
12:39The next math skill we've got is
12:41calculating the motic index from
12:43microscope slides. And here's the
12:45formula that you need to know for this
12:47one. The motic index is the number of
12:50cells that you can see in your
12:52microscope image in mitosis divided by
12:55the total number of cells. So what this
12:59calculation basically is is a measure of
13:01the proportion of cells undergoing
13:03mitosis and it's used to estimate how
13:06quickly a tissue is growing because if
13:08you have a higher motic index that
13:10indicates more cells are undergoing
13:11mitosis and therefore there must be more
13:13growth happening. So here's our formula.
13:16If we have a look at how you'd actually
13:18calculate this then you'd need to
13:20examine a microscope slide that shows
13:22dividing tissue. So for example, it
13:24might be an onion root tip field of view
13:27from your required practical two. You
13:30would then count the number of cells
13:32that you can see in any stage of
13:33mitosis. So whether it's prophase,
13:35metaphase, anaphase or telophase, count
13:38the total number of cells you can find
13:40in that field of view and then you'd use
13:43that formula number of cells in mitosis
13:45divided by a total number of cells. So,
13:47just to show you an example here, we've
13:49got a field of view that shows us some
13:53onion um root tip cells. And we can see
13:57if you to count them all, we've got 14
13:59cells. Four of them are in mitosis. So,
14:02we do 4 / 14. Our mitoic index is 0.29.
14:08Now, a few things just to point out.
14:09First of all, only ever count the whole
14:12cells. So, we can see here we've got a
14:14part of a cell. We've got a part of one
14:16there, a part there, there, there, and
14:19actually here as well. You wouldn't
14:21include those because if you can't see
14:23the whole cell, we can't actually see
14:25whether it's in mitosis or not. So only
14:27count on the whole cells. And the next
14:29thing is the way that we can tell if a
14:32cell is in mitosis or not is the
14:35chromosomes are visible. So this one
14:37here, we can see visible lines, which
14:39are the chromosomes. We can see visible
14:41lines in this one, this one, and this
14:42one. than in any particular
14:45organization. So it's probably prophase
14:48but we can see the chromosomes whereas
14:50here you don't see any lines. You don't
14:52see any chromosomes. Now the point I was
14:55making down here is this is just an
14:57image that I found um online but in the
15:00exam they would give you an image where
15:03it was really obvious where there was no
15:05debate about whether it is or is not in
15:09mitosis. It would be really obvious. You
15:11can see those lines. So just bear that
15:12in mind for the exam.
15:15Next we've got is using a scale bar to
15:17calculate the actual cell sizes. So in
15:21an magnified image like we can see here,
15:24sometimes you actually get a scale bar
15:26and from that you have to work out the
15:28magnification and they don't give you
15:30the magnification. And in this one we've
15:32got a mitochondrian and we've been given
15:34this scale bar and we're told that scale
15:37bar is 200 nm. Now what that means is
15:41that line there represents the actual
15:45size of 200 nmters. Not that if you were
15:49to look at that that is literally 200 nm
15:51on the image that is the actual size. So
15:54what we would then need to do is get our
15:56ruler and measure what is the image size
15:59of that line because we know the actual
16:02size is representing 200 nm and
16:06measuring this 16 mm. cuz obviously it
16:09depends what device you're looking at,
16:11but let's say it's 16 mm. You could then
16:14use your image and your actual to work
16:17out the magnification.
16:19So 1 millm again, we need to convert our
16:21units here. 1 mm is 1 million nanome. So
16:27you would need to do time 1 million cuz
16:31we've got 16 mm and there are 1 million
16:34nanometers in 1 millm. So we actually
16:37have 16 million nanometers.
16:40So we do 16 million / 200. And then that
16:44tells us the magnification of this image
16:47is 80,000 times.
16:50The next math skill is plotting results
16:53from permeability or osmosis
16:55experiments. And this could come up in
16:58your required practical three and four,
17:00the ones where you're using potato
17:03plants for osmosis or the permeability
17:05of the beetroot membranes. So for the
17:07osmosis one, your experiment could be
17:10with plant tissue such as potato
17:12cylinders in different sucrose or salt
17:14concentrations where you're going to be
17:16measuring a change in mass or length of
17:19your plant tissue. Then if you're
17:21plotting your results, if we go back to
17:23one of our earlier slides, we were
17:25talking about you always have your
17:26independent on the x-axis, your
17:28dependent on the y-axis. And the
17:30independent is what you're deliberately
17:32changing. And in this case, we'd be
17:34deliberately changing the concentration
17:36of the sucrose solution. So that would
17:38go on your xaxis. And we are measuring
17:41our dependent variable, the percentage
17:43change in mass. Or we'd actually measure
17:45the change in mass. So the initial and
17:48final mass work out what the change in
17:50mass is and then you always plot it as a
17:52percentage so it's comparable taking
17:54into account the fact that they might
17:55not have been exactly the same mass at
17:58the start and that goes on your y-axis
18:01the permeability experiments. So this
18:03would be your beetroot discs, maybe at
18:06different temperatures or in different
18:08solvents or different concentrations of
18:10solvent. And what you'd be measuring is
18:13how much pigment is in the solution
18:16after a set amount of time. And you
18:17could do that using a calimeter. So you
18:20get a numerical value or quantitative
18:23rather than just subject subjective
18:25describing how dark the purple color of
18:29the pigment from beetroot is. So in
18:32terms of plotting the data, our
18:33independent variable depends what's
18:35actually been varied. It might be the
18:37temperature, it might be the type of
18:38solvent, it might be the concentration
18:40of the solvent and that goes on your
18:42xaxis.
18:43The dependent variable is what you're
18:45measuring and we are measuring the
18:47amount of light being absorbed in that
18:49colorimeter or sometimes it's measured
18:51as the amount of light that transmits
18:52through. I always go for absorbance.
18:55So thinking about our graph rules, it
18:58would be a line graph cuz it's
18:59continuous data. We have to label the
19:01axes and we have to give the units as
19:04well. You have to use a suitable scale
19:07meaning your graph fits at least half of
19:10the paper and you have to increase by
19:12equal amounts on your scale. And lastly
19:15plotting your data points with an X and
19:17then you draw a line of best fit which
19:19is likely to be a smooth curve. So
19:22here's some example data. We've got our
19:24sucrose concentration and we've got the
19:26percentage change in mass and here is
19:30our graph. So we've got percentage
19:31change in mass, sucrossse concentration
19:34and there's our curved line of best fit.
19:37And this actually links to the next math
19:39skill determining the water potential
19:42from the intercept on a graph. So using
19:45that same concept we've got our sucrose
19:47concentration percentage change in mass
19:50and we've got our data plotted. the
19:52intercept. So this is where if we were
19:54to draw a dash line at where we have
19:57zero change in mass on our y axis that
20:00would be the water potential inside of
20:03the plant tissue which in this case is
20:05potato. And the reason for that is if
20:08you don't have any change in mass in an
20:10osmosis experiment that means there's
20:12been no net movement of water into the
20:16potato plant or out of the potato plant.
20:19And if there's no net movement of water
20:21basmosis, that tells us that the water
20:24potential inside of the potato tissue
20:27must be the same as the water potential
20:29of the sucrose solution. And that then
20:32indicates to us that at whatever point
20:36um the concentration is that we
20:38intercept at zero, that must mean that's
20:41the same concentration inside of the
20:44potato. instance, it tells us that the
20:47potato tissue must have a sucrose
20:49concentration of 0.4 cuz that is where
20:52the concentration is. Now, if you wanted
20:54to know what that was as a water
20:56potential, it's not literally just 0.4
20:59cuz that's a secret concentration, not a
21:01water potential concentration. So,
21:03sometimes there's an extra mark for
21:05saying you then need to look up in a
21:07table to see what water potential that
21:09concentration correlated to.
21:12The next skill, calculate the surface
21:15area to volume ratios of simple objects.
21:17So for example, cubes. So why does this
21:20matter? This is now topic three. Small
21:24organisms and cells exchange substances
21:27faster. So it links to the topic to do
21:29with gas exchange and absorption.
21:33Surf area to volume ratio shows how
21:36efficient an exchange surface is. So if
21:39you have a high surface area compared to
21:42the volume or surface area to volume
21:43ratio, that means you'll have faster
21:46diffusion or faster exchange relative to
21:49the size. And that could be diffusion of
21:51gases, absorption of substances, it
21:53could even be loss of heat across the
21:55surface. And if you have a lower surfer
21:58to volume ratio, you have a slower
22:00exchange system. And that is then when
22:02you start to see organisms have evolved
22:04over time and have these adaptations to
22:07increase exchange even though the entire
22:09organism has a low surfer to volume
22:12ratio. So for example gas exchange in
22:14the alvoli in the lungs or gas exchange
22:16in gills. So the formula for a cube
22:19would be working out the surface area
22:22which is 6 times because there's six
22:24faces of a cube the length squared
22:27because the area of one side of the cube
22:30is the length time the length and then
22:33there's six faces* 6. The volume of a
22:37cube would be the length cubed or in
22:39other words the height time the width
22:41time the length but because it's a cube
22:44those will all be the same dimension.
22:46And then it's whatever the surface area
22:48was divided by the volume. And we always
22:51present that as whatever the answer is
22:53to that calculation. We have that value
22:56to 1. So it's a ratio still. So we've
22:59got an example here. Cube with um
23:02dimensions of 1 cm. So 1 cm in height,
23:05length, and width. If we were to do our
23:07surface area, the length squared is
23:10still 1 * 6 cuz there's six faces to
23:14that cube. So that gives us 6 cm
23:16squared. The volume is the length cubed.
23:19So 1 * 1 * 1 which is 1. So 1 cm cubed.
23:24So that means our surface area divided
23:25by volume is 6 / 1. So we have a ratio
23:28of 6 to 1.
23:31Next then we have rearrange and use the
23:33formula for pulmonary ventilation rate
23:36which is pulmonary ventilation rate or
23:38PVR is tidal volume times breathing
23:40rate. So the PVR the pulmonary vol
23:43ventilation rate that is the volume of
23:46air ventilated per minute and that would
23:48typically be in decimeters cubed is our
23:51unit for volume per minute. Tidal volume
23:55is the volume of air per breath and
23:57because it's volume it's going to be in
23:58decubed and then breathing rate is
24:01number of breaths per minute. So that
24:04would be our formula. To rearrange it,
24:06if you wanted to find the tidal volume,
24:09it'd be your pulmonary ventilation rate
24:11divided by breathing rate. To work out
24:13the breathing rate, it would be
24:15pulmonary ventilation rate divided by
24:17the tidal volume. So, we've got a few
24:19examples. Then, a person has a tidal
24:22volume of 0.5 decimeters cubed and a
24:25breathing rate of 12 breaths per minute.
24:28What is the PVR? So this time it' be 0.5
24:31* 12 and that tells us the PVR is 6
24:34decimeters cubed per minute. Another
24:37example we've got a person has a PVR of
24:407.2 decime cubed per minute and a
24:43breathing rate of 18 breaths per minute.
24:45What's the tidal volume? So this time be
24:477.2 so PVR divided by the breaths per
24:51minute 18. So the volume the tidal
24:54volume is 0.4 decimeters cubed. Last
24:58option then a person has a tidal volume
25:00of 0.6 decimeters cubed and a PVR of 9
25:04decimeters cubed per minute. The
25:06breathing rate would be 9 for PVR
25:10divided by the tidal volume which is
25:120.6. So they have 15 breaths per minute.
25:18Math skill 12 on this list is from topic
25:21four using 2 to the^ of n to calculate
25:25the number of chromosome combinations
25:27and this links to meiosis where n is the
25:30number of pairs of chromosomes and the
25:32number of homologous pairs. So the
25:34principle here is that during meiosis
25:37chromosomes line up at the equator in
25:40independent segregation and then
25:42separate apart and which side of the
25:45equator the maternal and paternal
25:47chromosome align is random. So each
25:50homologous pair can arrange in two
25:52different ways either on the left or the
25:54right of the equator. So if we want to
25:57know how many possible different
25:58combinations there are that we could get
26:00in the gametes, it's two because we have
26:03pairs of chromosomes to the power of n
26:07where n is the number of pairs you have
26:08the number of homologous pairs. And in
26:11humans we have 23 homologous pairs of
26:14chromosomes. So that means 2 to the^ of
26:1723 which comes to over 8 million
26:20possible gameamt combinations that we
26:23can create before you even take into
26:26account crossing over and random
26:29fertilization of these gameamtes. So
26:31it's another way just to demonstrate how
26:34large the variation genetic variation is
26:37that is introduced from meiosis with a
26:39mathematical formula.
26:42Next then we've got interpret data using
26:44logarithmic scales and this often links
26:46to aseptic technique and growing
26:49bacteria because we use a log scale when
26:51you have a really large range of data.
26:55So many biological processes involve
26:58exponential growth and decay. So for
27:00example bacteria replication but we can
27:03also actually all see this in the growth
27:05of cancer cells. It comes up quite a bit
27:07in paper two the same skill. So
27:09logarithms help compare values that have
27:12a very large range. So comparing very
27:16small values to very large ones. And we
27:19can have logarithm to the base 10 or log
27:2310. And that is used for bacterial
27:25growth and dilution calculations.
27:28Natural logarithm or ln. That's mainly
27:30what we see. But you can get log e as
27:32well is applied in growth rate
27:34equations. But they'd normally give you
27:36an equation that you'd need to use if
27:38you had to do that. So just a bit more
27:41information on this. Here we can see
27:43some original values and this could be
27:47number of bacteria in a solution after a
27:49certain amount of time. And we can see
27:51here we've got data that is very very
27:54small going up to very very large. And
27:56it' be difficult to plot this on a graph
28:00so that you can actually get a scale
28:01that you could fit all of this on. And
28:03actually this probably isn't data from
28:05the number of bacteria because that's
28:06not a whole number and nor is that and
28:08you'd have to have a whole number for
28:09number of bacteria but it's
28:11demonstrating that concept. You've got a
28:13really really large range. So instead
28:16you'd have to convert your raw original
28:18values into log values log 10. So we can
28:22see here we've converted to log 10. And
28:25what that means is if you present your
28:26original value um in the order of
28:28magnitude value instead and this then
28:31gives us a much smaller range and we can
28:34then plot that on a graph. Now you can
28:38actually also get graph paper which is
28:42logarithmic graph paper and we can see
28:44here in this graph that's what we have.
28:46you don't have even
28:49subdivisions
28:51on this graph. And it's important that
28:54you know what each of those subdivisions
28:56means and how to read it off. So we can
28:59see here that where we've got 10^ the 2
29:03and that is the first line at 10^ the 2.
29:06So we've then got 1 2 3 4 5 and it
29:09actually goes up 10 divisions. The first
29:12line, the first point on 10^ the 2. What
29:14that means is 10 2 * 1 which is 100.
29:19This here is now the third line up
29:21because we've got that was line one,
29:23line two, line three. So that means it's
29:2610^ the 2 * 3 which is 300. This one
29:31here we've got 10 4. That's the first
29:33line. Second, third, fourth, fifth. So
29:37that's 10 4 * 5. And so on and so on. So
29:40that's how you read off on the y- axis
29:42on one of these graphs. Um so each of
29:45those lines is representing times
29:48whatever number line that is.
29:52Number 14 on our list is use and
29:55interpret the index of diversity formula
29:57which is from topic 4. So here is the
30:00formula. They usually give this to you
30:02in the exam but they don't often tell
30:03you what each component means. Sometimes
30:05they do but they don't always. So D is
30:08index of diversity and this is a measure
30:10of biodiversity taken into account
30:13species richness and also the number of
30:15individuals within the each of the
30:17populations of species. Capital N is the
30:21total number of organisms. So every
30:23living thing in that community
30:26lowerase N is the number of organisms of
30:29just one species. And that's why we have
30:32this symbol here, sum of, because this
30:35part you have to do for every species in
30:37your community. And then you add up all
30:40of those for all of your species. And
30:42that's normally where people go wrong on
30:44this formula. But if we go through um
30:46some examples,
30:48first of all, why do we actually do
30:50this? As I said, it's a measure of
30:51biodiversity in the community. The
30:53higher the index of diversity or D, that
30:56tells us we've got a greater
30:57biodiversity, which means we've got more
31:00species and more balanced population
31:02sizes in the species, meaning each
31:05species probably has a large population.
31:08A lower index of diversity means lower
31:11biodiversity. So, you might have fewer
31:13species or you might just have one
31:15species that dominates. So you might
31:17have 10 species present, but you've got
31:19a million individuals in one of those
31:23species and only five or two or very low
31:26number in all of the others and
31:27therefore they're at risk of going
31:28extinct. So it's not actually very high
31:31biodiversity.
31:33So here's an example. They wouldn't give
31:35you this column in the exam. They don't
31:37make it that easy. They tend to say
31:38here's your species. Here's the
31:40population size. So capital n is the
31:43total number of individuals in this
31:46community. So we'd need to add up that
31:48column and that comes to 50. For this
31:51bit we need to do the sum of lowerase n
31:54* lowerase n minus one. So we have to do
31:57that for all of the species present. And
32:00as I said they don't give you that
32:01column. So I just tend to add it on in
32:03the exam and draw it on myself. So this
32:06would be 20
32:09* 20 - 1. So 20 * 19. This would be 15 *
32:1414.
32:16This is 10 * 9. And this is 5 * 4.
32:20You're always doing n * n - 1. And then
32:23it's the sum of that column. And that
32:27comes to 700. So this numerator here
32:30then is the total number of individuals
32:33which was 50 * 50 - 1. So that is 50 *
32:3749 which is 2450
32:40divided by our denominator which was the
32:42sum of lowerase n * lowerase n minus one
32:47and it was a sum of all of those which
32:48was 700. So we've got an index of
32:51diversity overall value of 3.5. Now one
32:54thing just to try and gauge whether
32:56you've done this correctly using this
32:59exact formula you usually get an answer
33:02between 1 and 10. So if it's drastically
33:06over 10, you've gone wrong somewhere.
33:08And if it's lower than one, you've
33:10probably gone wrong. If it's lower than
33:12zero, you've definitely gone wrong.
33:14Next, then calculating mean values for a
33:17data set. So this one's quite
33:18straightforward. Mean is a measure of
33:20val of an average. So it's the sum of
33:22all the values divided by the number of
33:24values you have. So add up all of your
33:26repeats, divide by the number of repeats
33:28you had. So we've got an example here.
33:30So here's all the repeats. Add them all
33:33up and there were five repeats. So we
33:35divide by five. We've got a mean of
33:37nine. Now this goes hand inhand with
33:40standard deviation. And although it says
33:43on the spec calculate standard
33:44deviation, you might be asked to do that
33:46in your required practicals, but they
33:49won't ask you to do it in the exam
33:50because they do state it take up too
33:52many marks because of the time you'd
33:55need for it for the number of maths
33:57marks you're allowed. But you do
33:59interpret standard deviation and that
34:01comes up every single year. So when
34:03we're doing an experiment and we've
34:04collected the data, we usually calculate
34:06a mean. But it's important to do more
34:09than just calculate your average or
34:11mean. It's better to also do the
34:13standard deviation because that tells us
34:16how spread out all of those results are
34:19compared to your mean. So it gives us a
34:22better idea of the reliability, how
34:23spread out all of those data points are.
34:26So standard deviation compared to the
34:28range of data is better as well because
34:30the range just tells you the highest and
34:32the lowest. Standard deviation is
34:34considering all of your repeats. How
34:37spread out are they compared to the
34:38mean. So in calculating standard
34:40deviation, one standard deviation is
34:43taken into account 68% of the data and
34:47two times whatever your standard
34:49deviation is includes 95% of the data.
34:52And they normally give you that as a
34:54statement in the exam. They'll say
34:57standard deviation has been included and
34:59it's two times the standard deviation
35:00which includes 95% of the data. Now we
35:04don't need to get into the nitty-gritty
35:06of the maths behind that. What you need
35:08to know is that means for you that if
35:11when you take into account your standard
35:13deviations
35:15against the mean, if those values
35:17overlap, it tells us you do not have a
35:20significant difference between those two
35:22means. So that is the point of telling
35:25you that fact. So just a few other
35:27points that we've got here. The standard
35:29deviation shows the spread of data
35:30around the mean like we said whereas the
35:32range only shows you the highest and the
35:34lowest value. So it's not as useful. The
35:37standard deviation reduces the effect of
35:40anomalies because it's taking into
35:42account all of the data. Whereas the
35:44range includes anomalies. It's just
35:46showing you the highest and the lowest.
35:48So it can be quite skewed. And the
35:50standard deviation, this is the big
35:52thing that it's used for. It's used to
35:54indicate whether a difference between
35:55mean sets of data sets is significant or
35:58not. That's the main thing that is
36:01assessed on in the exam every year. So
36:03let's just focus on that and have a
36:05look. We've got an example of some data
36:07here. Two different types of treatment.
36:09Mean population size. Here's our mean.
36:11And we're told in brackets this is 2 *
36:13the standard deviation. So what you
36:16would need to do is from our highest
36:19value, highest mean 7.3. We'd need to
36:23minus 0.8.
36:25From our lower, which is 5.6, we need to
36:27add on 0.7. And then C. 5.6 + 0.7. Does
36:33that take the mean higher than 7.3 minus
36:370.8? And if it does, that means the
36:41standard deviations overlap. And if it
36:44doesn't, then that means they don't
36:46overlap. If you have an overlap, that
36:50would mean there is no significant
36:52difference between the means. Even
36:53though those numbers are different, 5.6
36:556 and 7.3 might be different. When you
36:58take into the stand take into account
36:59the standard deviations, they're not
37:01significantly different. You could also
37:04get it on a graph, which is easier to
37:06interpret cuz you can then just visibly
37:07see does this standard deviation line
37:11cross the same point on the y-axis as
37:13this one. And we can see here it
37:15doesn't. So that means we can visibly
37:17see those means are different. The
37:19standard deviation bars also don't
37:21overlap. So it tells us that there is a
37:24significant difference between those
37:26means. So hopefully you found this
37:29helpful. If you did find this helpful,
37:31then in my AQA math skill workbook, I
37:34have got every single math skill that is
37:37listed in the math skills section of the
37:39specification. In this workbook, you
37:42have an explanation of every math skill,
37:44a modeled example, practice questions,
37:46and all of the answers. And I'll link
37:48that in the description. But that is it
37:50for today's video. Look out for the
37:52paper 2 version and the entire set of
37:55skills still to come.

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