Overall

Paper 1

Paper 2

A* = 175-180

A* = 65/91

A* = 61/91

A = 153-160

A = 56/91

A = 52/91

B = 120

B = 47/91

B = 44/91

+Question

Debate

Marks

1.1 Figure 1 is an image of a neuromuscular junction.

Name the structures in the  neuromuscular junction labelled A-D

A-Mitochondria

B-Pre synaptic membrane

C-Vesicles containing neurotransmitter

D-synaptic cleftno

Mark schemes always accept mitochondria. It was pointing at ONE mitochondria nothing all of them.

For B can we say presynaptic membrane

I thought if it was knob they would use the big arrowthey did for D

The arrow was pointing at the membrane tho

Yh it is presynaptic membrane

Wasnt it pointing at the post synaptic membrane?or am i just a bit dumb

I’m like 99 percent sure it was presynaptic

Oh well i don't know if i got the marks now i was pretty sure i had them

It pointing at the presynaptic membrane oh then i only got i mark:( same :( whatever happens happe

 hi ns

Do you have to mention neurotransmitters for C, I put synaptic vesicle
You should be able to get a mark

That’s a relief thanks

2

1.2 something about ACH and the neuromuscular junction

(Do not mention the release of calcium ions from the sarcoplasmic reticulum)

• ACH diffuses across the synaptic cleft (via exocytosis). ACh binds to specific receptors on muscle membrane
• Causing Na+ channels to open
• Na+ diffuse into post-synaptic knob causing depolarisation
• If threshold is met, an action potential is initated

you dont get an action potential in the muscle, just a contraction.

You do get an action potentials in muscles that is propogates along the sarcolemma down T tubules.

Would it not be Calcium ion channels which open instead? No thats in the presynaptic membrane

Pretty sure it specifically asked you not to talk about Ca+ ion B s

4

1.3 Explain how inhibitory synapses causes hyperpolarisation of the membrane?

How it prevents another action potential from happening straight away?

• Inhibitory neurotransmitters hyperpolarise postsynaptic membrane
• Causes Cl- channels to open, so Cl- diffuse in & K+ channels open so K+ diffuse out
• More Na+ required for depolarisation, so reduces likelihood of threshold being met

CHLORIDE IONS ARE OUT OF SPEC

i pit hyperpolarisation so harder to reachon threshold for  depolarisati

I also said that yhhh input cl- is diffuses in hyperpolarizing it

chloride ions aren’t on spec so you can still get the marks without mentioning them

Just the last two points are required i’ve seen this exact question in a past paper and it was just more Na+ needed to reach threshold so less likely for action potential/depolarisation yeah ur right

i wrote about acytchloninesterase breaking down the ach 😪

• 2

1.4 Calculate the sarcomere when contracted by 20%

5.44x10^-2mm

Take total length of sarcomeres in myofibril (x34 sarcomeres) and times  noby 1.2 (20% change)

Then divide by 1000 to get mm

‘After contraction th  e wording of the question meaning the sarcomere was relaxed so surely it would be a 20% increase.

5.44x10^-2mm

I got 1.36 x10^-2

Oh bruhhhhhhhhhhhhh

^thats 20%, you needed 80%

It literally said each sarcomere was 2 micrometers so multiplied by 1.2 as it’s 20% increase for all and then multiplied by 36.

Anyone get 0.0864 so 8.64x10^-2

2

1.5 What happens when sarcomere contracts?

Tick box question

A band stays the same, I decrease, H decrease

3rd box was correct

hooray i put this at first then crossed it and put 2nd and then changed it back again😜 ur cooked enjoy clearing

1

02.3 explain why there is a higher glycogen concentration in fast muscles than slow muscle

Fast: rapid breakdown of glycogen for glucose for anaerobic respiration, provides ATP for energy as

short and intense activity.

Slow: long period of endurance activity so have lots of blood vessels for glucose supply to produce ATP by aerobic respiration for myofibril contraction

And prevent anaerobic respiration as well as lactic acid buildup

i also talked about fast bot containing much myoglobin

I thought the short twitch contained more myoglobin to store oxygen?

4

03.1 conclude and evaluate: how does absorption and colour affect rate of photosynthesis

Highest rate of photosynthesis at red and blue colours / Lowest rate of photosynthesis at green colour

Also higher absorbed ance higher rate of photosynthesis for the first part no?

I put that absorption and rate of photosynthesis directly proportional as when one increases the other i

ncreases

Yeah no shit

I wrote about chickens and hens wtf i thought it was acTUal hens  - u might b cooked brev

I lit started waffling abt cooking the hen and the absorbance

2

03.2 Name 3 factors were controlled in the experiment?

• CO2 conc
• Temperature
• Mineral ion conc in soil or water
• Light intensity
• Soil ph

Humidity? Yeah thought so

oxygen conc? Surely?!

Wind speed?

2

03.3 How are products of light dependent reactions used in light-independent reaction.

Do not include details about the activity of RuBP.

• GP is reduced into Triose Phosphate (TP) by NADPH (which is oxidised and turns into NADP)

• This process requires energy that is provided for by ATP
• ATP provides energy for the regeneration of RuBP from Triose Phosphate (TP)

Do you have to mention GP is glycerate-3-phosphate?

Dont think aqa ever wanted the actual names

NO I EMAILLED THEM THEY SAID THY WANTED IT

GP can be abbreviated but TP omust be Triose Phosphate

Didnt it say only to make triose phosphate, not the regeneration,

it’s talking about the activity of RuBP, regeneration is fine

3

04.1 how to tell if grey fly heterozygous

Cross with black fly

Observe phenotypic ratio of

Offspring

e.g.

If any black offspring produced, fly heterozygous.

If all grey offspring, fly must be homoz

Or look at parents genotype

“Test Cross” was in a mark scheme before.

Would you be allowed gene sequencing?

I drew two monohybrid crosses for this, is that ok?

2

04.2 why is it not x-linked

• If x-linked, any female offspring inherits dominant allele for grey body from father, and will be grey-bodied, but black-bodied female offspring present

Were the marks for 04.1 and 04.2 not the other way round, nope this question was 1 mark and 4.1 was 2 marks

1

04.3 hardy weinberg

18%

19% grey bodied, 81% black bodied,

0.9 g, heterozygous = 2*0.1*0.9

48% :( is 48 = 1 mark?

I think u get 48-49 if you assumed 0.19 was homozygous grey-bodied only, but 0.19 is for any grey-bodied

i spent like 15 minutes on this goddamn question because i forgot the hardy weinberg equation

— btw you always get 1 mark for identifying heterozygous is 2pq

is the is not 9????

IF U PUT 0.18 DO U GET A MARK ?

2

04.4 cross

RrGG x Rrgg

Cross table stuff

RrGg, rrGg, (RRGg would not be alive, they wanted valid births 👍)

Curly grey, normal grey,

2:1 depends on how they want the ratio give it

Isn’t it 3:1 cuz RR die it didn’t ask ratio of surviving offspring just offspring in general

Take a look at the table

It did it said all RR dont make it past embryo stage

Im a bit confused 12 survive and 4 die so 3:1 right?

No - it asked for ratio of flies surviving into adulthood so 2:1

Was it not 2:1 ratio because the lethal gene kills homozygous dominant- yes i think it was 2:1

Yeah idk how they want the answer so i just put that

My last “phenotype” i just wrote dead im assuming they can read

Hey e

Wait i thought it was 3:1 coz there were ? i thought this too

⅔ would be awarded for either ratio due to ecf as per older markschemes

Not 3 to 1? Why is the table 4 by 4. I’m pretty sure it’s 2 to 1.

Didn’t the qs ask for the phenotypes of surviving adults so 2:1 so don’t include RR cos it dies

3

05.1how can sampling be used to estimate the number of fat hen plants in the whole field

Need a method of random sampling e.g. making axis with tape measures and using random number generators to make coordinates to place quadrats and count numbers of plants in each one

Repeat until you have enough so is representative of area (aqa usually say at least 10 i think) and calculate mean number of plants per quadrat

Convert this to number of plants per m2 and multiply by area of field

(or mean number per quadrat * i number of quadrats?)

P

5

05.2 why decrease potato production

Interspecific competition

Nitrate ion conc decreases as wild hen plant use most of it

Less light for photosynthesis

Fat hen plant is 2m tall so blocks sunlight

This causes the wild hen plants to outcompete the potato cells decreasing the production/population size of potato plants

3

05.3

6.69kg

550 * 0.77*10^-6 * 20000 * 79%

think this is the first maths bio q i have it ever done right

2

06.1 describe and explain relationship between IAA concentration growth rate and cell plasticity

Higher concentration, greater growth and plasticity

Maybe mention increase in conc from 10^-7 to 10^-6 (middle bit) had greater effect on increase than the other concentrations

It was explain so would have to say higher IAA conc → more cell elongation → higher plasticity etc

Could also mention that plasticity started to decrease at one point which is very likely due to the cell wall being at the point of bursting open

Was it not explain?

no it was evaluate the curve and fin d tangent no it wasn’t evaluate

Did you guys extrapolate?

Theye was no tangent?

You had to draw the tangent

No

2

06?.2 percentage increase

570%

0.6 to 4 (scale was on thie right hand side of tabl is e), 2sf

TWO SIGNIFICANT FIGURES

480%? No

670%? No

Why not 670%????

For this question the increments were 0.6 and 4, you had to do 4.0-0.6 and then divide it by the initial value which was 0.6 and then multiply it by 100 to get the percentage. Did it ask for 2sf i got 569.9

2

06?.3to

Dilute by factor of 10 with nine parts water, mix well, then repeat

• 99cm3 water to 1cm3 of 10^-1 solution
• Or 1 part solution to 9 parts water twice
• Sometimes ask to mention stir or mix at the end

 Can you just add series of dilution

I put 1cm3 solution, 99cm3 water

You have to say diluted twice by 10, not once by 100

You can put either since there was nothing in the question suggesting there was an intermediate of 10^-2 moldm-3

Oh it’s fine it’s only 1 mark

lol

1

06?.4 experimental design for concentration and stem

 growth

1 dish with distilled water and 5 with different concentrations

Same length of stem segments to ensure control of Surface Area.

Same growth/development stage of stem

Same volume of the GA concentration solutions

Same temperature, carbon dioxide oxyge  n glucose supply, same light intensity

Same time for them to have growth

Same plant species

Measure stem segment before and after the experiment to see the growth.

Do many per petri dish and calculate mean growth rate per GA concentration, plot a graph of mean rates vs conc to see relationship.

Also put lid on petri dishes to prevent evaporation of water that will affect GA concentration (saw similar thing in past paper ms so thought it could be right)

This answer won’t score full marks

Different initial lengths so calculate percentage change instead of change in length

yea i wrote that

Did anyone say cut the stems same length for a control variable? <- i did yes same and its correct

Surely it needs to be longer thaln 20 mins?? I put 2 weeks i put 24 hours. ← SAME

 I put 48-72 hours yes

Maybe also sterile equipment no c ml ompetition with bacteria e I’m tc

i thought the petri dishes were in there to throw us off i’m so cooked

Guys does anyone know how the mark allocation for these types of questions are - do u only get like a certain amount of marks for ur controls and the rest for the actual method?? Like is it capped at 2 or 3 for controls and the rest for method or something?

5

Genes part 1

3 bands on left, 5 bands on right. Only 1 band (the 3rd band) should be at equal places on experiment 1 and 2.

(as first and last section is made smaller when cut more so they travel further)

—--  

         —---

—--   —---

—--

       —-----

       —----

       —----

hi

(allow me this is very ro it ugh)

This is what people in my class did i thought the 3 for a were bigger so wouldnt be at same distance

2

7.2 How would you determine the size of each fragment using gel electrophoresis?

• Use a “ladder” aka a DNA fragment with known lengths sizes / number of base pairs at known positions.
• The bands separate out, heaviest first, so can compare to “ladder” row to see which bands are similar/different

ALTERNATIVELY

• Run a standard with DNA fragments under the same conditions
• Compare to position of unknown DNA fragments to estimate their size
• Shorter DNA fragments travel further

2

How many fragments would a plasmid have instead?

4

But it said enzyme cuts linear DNA and plasmid has circular DNA, not linear, so wouldn’t it be 0?

Nuh uh it still cuts cause it targets a specific bases

1

Insert gene to plasmid

• Restriction enzymes cut vector DNA
• DNA ligase joins DNA fragments to ve ctory DNA forming phosphodiester binds  

ound!!!!

can you say to anneal the two DNAs together or do you have to say phosphodiester bonds? No I don’t think you do

Is it incorrect to say restriction enzyme cuts plasmid?

2

why applicable to type II and not type I

Type 1 incurable caused by inability to release sufficient amounts of insulin?

Obesity risk factor for type II

Type2 caused by high bc levels leading to inse bc it nsitivity of insulin receptors

Type 2 still releases insulin

can I say

Type 1 is insulin dependent so weight loss would work because won’t make insulin

Type 2 could be caused by obesity so weight loss ids an effective treatment but type 1 not caused by obsesity

2

Idk how to insert another table bit but why was the experiment valid

Large same size to ensure representative sample.

Wide range of ages to sample representative

Computer generated list so bias avoided in selecting patients- random sampling

Control group used to can compare effects of weight loss to no weight loss

Done over 2 years so+ long term effects known - reliable

3

Calculate difference

71  the

190*(45.2% - 4.7%)

45.2% calculates to 76. 42.5% calculates to 71. Which one was it?

2

Evaluate the effectiveness of weight loss treatment

Higher % of volunteers went into glucose remission in Group P compared to Group q

Higher % of volunteers lost a weight equal to or more than 15kg in Group P compared to Group q

However

a higher % of people that lost a weight equal or more than 15kg went into glucose remission in group Q compared to group P (do  I es this imply that there was more weight loss in group Q for those who actually bothered to lose weight?)

No stats test. Questions which have statement using all information in the mark scheme say ignore no stats test! Girl be quiet u can hav stat test

Don’t know exactly how much weight was lost, only if it was more than 15kg

Don’t know exactly how much weight was gained (last row)

1.9% gained weight in both group P and group Q.

Don’t know exact blood glucose concentration so unable to tell if it was close to / well beyond threshold

Don’t know if they came out of remission after

Losing weight reduces the effect of type II diabetes as less glucose concentration in the blood

i

Could you talk about other factors ie , diet smoking etc

Other factors sometimes do show up in these mark schemes

It was specifically asking you about weight loss, they didn’t provide any data about any other factors  

It was asking with regards to diabetes not just losing weight

5

SAN and heart contraction

Myogenic - initial stimulus for contraction initiated in the heart itself

1. SAN sends electrical impulse causes atria to contract at the same time

2. Impulse travels across atria, causing atria to contract to AVN. Non conductive tissue prevents the impulse from spreading to ventricles.

3. There is a delay before the AVN sends out an impulse to allow ventricles to fill up before they contract

4. Conducts through the bundle of His which branches into Purkyne tissue.

5. Wave of excitation causes ventricles to contract at the same time from the apex up

5

Explain results (p values)

Below 0.05 means significant increased risk of disease with factor So more than 0.95 probability that results were significant and not due to chance

Hyperthyroidism, age and the high blood pressure were significant impact.

Can you say no critical value?

3

10.1 causes of phenotypic diversity

Mutations resulting in different alleles

Crossing over during meiosis resulting in new combinations of alleles

Independent Segregation

Named environmental factor (e.g. sunlight)

I put changes to genotype and environmental factors because phenotype is genotype + environment

Yeah can you put env factors  cus it specifically asked that so like temp cus that affects size of smth?

Can we put Genetic and environmental factors? Can you talk about epigenetic variability?

2

10.2 Why is it advantageous to be able to camouflage based on surroundings?

Camouflage prevents predation

More likely to survive, reproduce fertile offspring with advantageous allele.

since it was two marks can we mention it increase their attractiveness for successful mating?

2

10.3 Disruptive selection with sympatric speciation

EVALUATE

Same habitat/geographical location

Mutation causes dark/light colour or nocturnal/diurnal behaviour

So the two groups don’t interact and no gene flow

So offspring inherit different alleles

Allele frequency of the two groups change

Reproductive isolation

Over many generations the two groups can no longer interbreed to produce fertile offspring (In the text it said that the geckos aren't different species? (use of mitochondrial dna may be inaccurate though)

Define disruptive selection?

Dark geckos able to hide from predators during the day, while pale geckos would be seen so they are active at night to avoid being seen during the day

Evaluate points against:

Whether they could interbreed to produce fertile offspring was not tested

Only mitochondrial genes anaylsed, should also analyse chromosomal genes

Pale geckos on walls and dark geckos on trees so different habitats in same ecosystem = geographic isolation so possibly allopatric speciation

The evaluate wording threw me off but i jsut wrote the normal  sympatric speciation answer

I put a definition of disruptive section and then just described sympatric speciation

I DID DIRECTIIONAL SELECTION THIS IS CORRECT

I said that as a counter point asw

Why is the first point same habitat when it said the geckos lived in different areas

It said the geckos both live in in an isolated habitat, so the same habitat no?

Yeah but wouldnt they be geographically isolated since pale live on walls and dark live on trees or smth

Maybe but question said sympatric speciation so i never thought abt it could probably get you a mark tho idk, this question was weird asf

True

5

10.4 Mitochondrial genes are analysed and have found that the colour-changing gecko is a distinct creature.

How has new technology allowed for rapid comparison of genes?

Gene machine. Mitochondria genes can be identified by analysing amino acid sequence or protein. DNA code universal so can find DNA sequence. Since you work backwards from mRNA, there are no introns so its faster. The closer the similarities between a DNA sequence, the more closely related the species are.

DNA base sequencing,

Specific type of geckos all have same mutation in mitochondrial DNA(2 marks)

Sequencing is now automated so genome s can be compared rapidly (1 mark)

Id assume sth like

Gene machine and PCR for rapid production of primers and amplification of DNA

Low genetic diversity/variation in nocturnal pop than diurnal pop, and diurnal pop has all the alleles in nocturnal pop, this means nocturnal pop split off from parental diurnal pop and there is limited g ene flow between the two

DNA probing, gel electrophoresis

Similar base sequence indicates genetic similarity

3

10.5 Geckos - capture and recapture precautions and formula

Dont include details of population change

Precaution - marking doesn’t cause harm (harm only not enough for mark - needs survival)  to geckos or affect chance of reproduction

marking also shouldn't impede movement to ensure random mixing

Ensure marking does not come off so results are accurate

Ensure geckos have to time to fully (uniformly) disperse before second sample taken.

No deaths or births

No immigration/ emigration

Population size= (number of geckos in sample 1 x number in sample 2)/ number marked in sample 2

Id assume they were talking about not increasing predation and not decreasing.      reproduction

3