Aqa a-level biology 7402 - paper 2 - unofficial mark scheme 13th June 2019

Original post by Zxr_

That hardy weinberg q- Me and my mates got 32% but people got like 46% or something?

I got 32

Reply 22

I think the dilution question was

0.4 and 39.6

Original post by RAKH000

Did you the other percentage question where you had to put it to 3sf?

Original post by 1Secord

I remember there was a question asking about photoionisation. ( 2 marks?)
Another question about how to estimate the population of sundews, a type of insect eating plant, in a marshy bog area. ( 5 marks )
The second part to this question was asking why it is beneficial to the sundews to eat insects when there is a low concentration of nutrients in the soil ( 2 marks ?)

There was another question about calculating the volume of distilled water and the concentrated IAA solution to make 40cm^3 of the solution with a concentration of 10 parts per million? ( I really can't remember the question that well, I just know I struggled with it )

Reply 23

z20010318

12

Ngl everyone’s talking about Hardy Weinberg when it’s 2 marks and you get a mark for just stating that 2pq is heterozygous- just move on lads and think about the *****er that is going to be paper3

Reply 24

Habib-Reh2

9

Succesion Question-
Pioneer species
Change in enviromental conditions as species dies and soil forms
Less hostile conditions as change in Abiotic factros
Change bio-div
Climax community established

For Quadrats Q-
Divide area/marsh/field into a grid
Random number generator
Select random coordinates
place Quadrats
Work out Percentage Cover

Photoionisation-
Light energy needed to oxidise Chlorophyll
Electrons move to the higher energy level move out of chlorophyll down ETC

lol that paper was awful only Qs i probably got right :frown:

Original post by RAKH000

Drop down any questions and answers from the test and I will produce an unofficial mark scheme based on most popular answers, thanks!

- Describe Succession-
How an ecosystem changes over time (change in species diversity and habitat diversity) – relies on environment being made less hostile by present species via death and decomposition leading to it being outcompeted and replaced by larger better adapted species.

- Percentage question (3sf)
- KJmy calculation question

-Hardy-Weinburg Calculation-
48%

(edited 4 years ago)

Reply 25

Luke Holland

6

0.64 is dominant homozygous and hetrozygous (if you have 1 dominant allele, it will still be expressed) so:
p^2+2pq=0.64.
q^2= 0.36 (1-0.64)
q= 0.6 (sqrt 0.36)
p= 0.4 (1-0.6)
2pq = 2*0.4*0.6= 0.48
=48%

Original post by namelessfeelsx

I got 32% as well... because dominant homozygous would be p^2 so p=0.8 (after square rooting 0.64), then 1-0.8=0.2, then 2pq= 2x0.8x0.2= 0.32

If it is wrong we would still get marks for multiplying 2 by two other numbers.

Reply 26

Jamil1

Original post by 1Secord

I also got 48%. I think the question told you the percentage of Grey body was 64%. That percentage included GG or Gg.
I did 1-0.64 = 0.36 - So 36% must be black body, gg.
Square root of 0.36 = 0.6.
1 - 0.6 = 0.4 ( Since 1 = p + q )
The question asked for heterozygous - 2*p*q
2 * 0.6 * 0.4 = 0.48 = 48%

Yeah thats what I did, GG and Gg can have grey bodies not just GG, or just Gg which i think some people did to get 32%

Reply 27

Luke Holland

6

yes i got 0.4 and 39.6 but it was a bit of an educated guess!

Original post by The Anarchist7

I think the dilution question was

0.4 and 39.6

Reply 28

Royalwave

4

Original post by The Anarchist7

0.36= q^2 or recessive
q= 0.6

P must be 0.4 so p^2 = 0.16 ( number of homozygous dominant)

Therefore, 2pq =
0.64 - 0.16 = 0.48

48 percent WAS the answer can’t see how 0.46 was right unless some one can explain otherwise

The equation is p^2+2pq+q^2=1 and if p^2 is 0.64, you cant assume q^2 is 0.36 because of the unknown 2pq.So you had to to use p+q=1 to find q.

Reply 29

Jamil1

Anyone get 1.62 micrometres for the length of the glucagon dna question? It said 10 base pairs so its 20 bases which is what I did, let me know

0.608 for Rf Value
177...something kJ for the bottom part of q1/2
48% for Hardy Weinberg

Reply 30

hannahmulholland

Original post by Royalwave

The equation is p^2+2pq+q^2=1 and if p^2 is 0.64, you cant assume q^2 is 0.36 because of the unknown 2pq.So you had to to use p+q=1 to find q.

But p^2 isn’t 0.64, p^2+2pq is

Reply 31

melonmelon

i got 3.24 as the bases are paired

Original post by Jamil1

Anyone get 1.62 micrometres for the length of the glucagon dna question? It said 10 base pairs so its 20 bases which is what I did, let me know

0.608 for Rf Value
177...something kJ for the bottom part of q1/2
48% for Hardy Weinberg

Reply 32

Interesting thought with base pairs but no. Base pairs are matched horizontally not vertically so 10 base pairs is the same height as 10 unpaired bases

Original post by Jamil1

Anyone get 1.62 micrometres for the length of the glucagon dna question? It said 10 base pairs so its 20 bases which is what I did, let me know

0.608 for Rf Value
177...something kJ for the bottom part of q1/2
48% for Hardy Weinberg

Reply 33

blueoreo

For haemoglboin I said chance to base sequence affects primary structure so different hydrogen bonds in the secondary structure, different overall tertiary structure for the beta chain, quaternary structure for haemoglobin different.

Formation of lactate reforms NAD which can be used in further glycolysis to produce ATP.

-The stem cells from siblings replace faulty stem cells, they can divide by mitosis and differentiate to form RBC, sustainable as body produces own cells instead of artificially synthesised ones, low chance of rejection as from siblings.
-Viral method stem cells more useful as involve own genes so no chance of rejection and no immunosuppresants much more useful long term; however use of virus may cause an inflammatory response.

-Fewer Ca2+ ions means tropomyosin doesn't have a confirmational chance to move and expose myosin binding site, fewer actinomyosin cross bridge formed, actin not pulled in sliding-filament, weaker muscle contraction, less ATP synthesis for this process as ATPase not activated by Ca2+ ions.

Reply 34

Gnuyheatmik

10

Original post by The Anarchist7

Interesting thought with base pairs but no. Base pairs are matched horizontally not vertically so 10 base pairs is the same height as 10 unpaired bases

No but it gave u the total number of bases not base pairs so u had to divide by two

Reply 35

I’m an idiot and did the correct working but I think I wrote 0.4 and 36.6

Original post by Luke Holland

yes i got 0.4 and 39.6 but it was a bit of an educated guess!

Reply 36

How does the myelin seath being damaged effect rhythms of the heart tho. Wording on this question is gonna be awful in the mark scheme

Reply 37

blueoreo

For gluconeogensis mention it's the synthesis of glucose from non-carbohydrate sources; glucagon assists in the process by activating enzymes that break down glycogen to glucose.

More insulin means more Glut4 proteins, absorb more glucose by active transport, reduces glucose concentration.

Inhibiting enzyme: cannot form cAMP from ATP, does not activate protein kinase a, does not convert glycogen to glucose, process inhibited glucose concentration remains low.

Why would you put in CSF? direct transmission to area of concern, reduces chance of patient choking on pills(?)

Reply 38

Jamil1