AQA A-level Biology 7402 - Paper 1 - 6th June 2019

Reply 80

tsunderhi_

17

Original post by Kayyy30

can someone PLEASE explain how to do this :frown:

The answer is 6 x 10^6

Lmao I need help with it too

Reply 81

LYLN

12

Original post by Kayyy30

can someone PLEASE explain how to do this :frown:

The answer is 6 x 10^6

The initial no. of bacteria is 300. Then u do 300 x 10^4 and divide by 0.5

Reply 82

momotori_xx

6

0.5m^3 / 1x10^-4 = 5000m^3
There are 300 bacteria in the diluted culture
so 300/5000 = 0.06 then you convert to cm^3 by doing 0.06*1x10^8 = 6000000 = 6x10^6

Original post by Kayyy30

can someone PLEASE explain how to do this :frown:

The answer is 6 x 10^6

Reply 83

Kayyy30

12

but how would you get to 6000000

Original post by LYLN

The initial no. of bacteria is 300. Then u do 300 x 10^4 and divide by 0.5

Reply 84

Kayyy30

12

but dont you convert m3 to cm3 by x10^6 , not 10^8??

Original post by momotori_xx

0.5m^3 / 1x10^-4 = 5000m^3
There are 300 bacteria in the diluted culture
so 300/5000 = 0.06 then you convert to cm^3 by doing 0.06*1x10^8 = 6000000 = 6x10^6

Reply 85

queenxluna

6

okay we are told that the dilution factor is 1x104 we know 0.5cm3 contains 300 bacteriaso 1 cm3 = 600 bacteriathe question says BEFORE BEING DILUTEd so it must be 1000 times more stronger, so you do 600x104 and get 6000000

the m3 should be cm3

How are you guys all feeling for tomorrow?

Reply 88

hxrrisonjames

6

Original post by Evil Homer

How are you guys all feeling for tomorrow?

A bit nervous, never know what's going to come up. Hope the maths / stats questions are nice

Reply 89

Kayyy30

12

OOHHHH i see. So you bass just do 300 divided by 0.5 , which gives you 600. then multiply by 10^4?
THanks! :smile:

Original post by queenxluna

okay we are told that the dilution factor is 1x104 we know 0.5cm3 contains 300 bacteriaso 1 cm3 = 600 bacteriathe question says BEFORE BEING DILUTEd so it must be 1000 times more stronger, so you do 600x104 and get 6000000

Reply 90

LYLN

12

Original post by Kayyy30

but how would you get to 6000000

I think there’s a typo in your copy... it should say 0.5cm^3

Reply 91

RUhanne

4

deffo a typo done that question it should be 0.5cm^3

Reply 92

tsunderhi_

17

Original post by queenxluna

okay we are told that the dilution factor is 1x104 we know 0.5cm3 contains 300 bacteriaso 1 cm3 = 600 bacteriathe question says BEFORE BEING DILUTEd so it must be 1000 times more stronger, so you do 600x104 and get 6000000

Perfect explanation tysm x

Reply 93

Mcpatel24

8

Are stats likely to come up in this paper. If so what areas is it likely to come up in?

Reply 94

username3275206

5

What practicals and statistics come up in paper 1? Does anyone know?

Reply 95

LifeIsSoHard

12

So the m3 was a typo, it’s supposed to say cm3

Original post by queenxluna

okay we are told that the dilution factor is 1x104 we know 0.5cm3 contains 300 bacteriaso 1 cm3 = 600 bacteriathe question says BEFORE BEING DILUTEd so it must be 1000 times more stronger, so you do 600x104 and get 6000000

Reply 96

medic2001

11

Original post by mehdi_bioace

What practicals and statistics come up in paper 1? Does anyone know?

I'm not sure but these are all the year 1 practicals:
1)Investigate The Effect Of Enzyme or Substrate Concentration On The Initial Rate Of Reaction
2)Using A Light Microscope/How To Perform A Root Tip Squash/Slide Preparation/Mitotic Index
3)Identifying The Water Potential Of Plant Tissue/Colorimetry & Calibration Curves/Serial Dilutions/The Potometer
4)Factors Affecting Membrane Permeability/ Investigate The Effect Of Temperature or Alcohol Concentration On Membrane Permeability
5)Dissection
6)Investigate The Effect Of Antimicrobial Substances On Microbial Growth/Aseptic Technique

Reply 97

GCSEStudent903

13

https://www.youtube.com/watch?v=5-9mn4ZWZ-Y

Videos for the AS required practicals - others on the same channel.

Reply 98

GCSEStudent903

13

Transcripion is started by a "transcription factor" which is a DNA Helicase-RNA Polymerase complex.

DNA Helicase breaks hydrogen bonds between complementary base pairs (exposing the strands).
RNA Polymerase adds RNA nucleotides, forming phosphodiester bonds.

Original post by 0rg4n1c

during transcription, is it RNA polymerase that breaks the hydrogen bonds between dna strands or is it DNA helicase? I've read different things in different textbooks and online