Download Key Area 4-6 HOMEWORK Marking Scheme

Key Area 4-6 HOMEWORK Marking Scheme
Key Area 4 (total 17)
1
2
3
4
5
6a
6b
7
8
B
A
A
B
A
The sequence / order of amino acids is changed (from that point). OR
Different amino acids (in enzyme). 1 mark
The shape/active site of the enzyme will change. 1 mark
250
They result in a short/incomplete protein/polypeptide
OR
The mRNA cannot bind to the ribosome
OR
They prevent translation/tRNA molecules binding to mRNA
1 gene mutations alter the base/nucleotide type, sequence/order of DNA 1
2 they include substitution, insertion and deletion
All 3 for 1
3 missense, nonsense and splice site are all substitution mutations 1
4 description of a gene mutation including reference to bases/ nucleotide
Any two descriptions 1
OR diagrams with bases labelled
5 remaining one description 1
6 substitution affect only one/two triplets/few bases/are point mutations 1
7 and so only slightly alter/alter few amino acids in the amino acid sequence
of the protein 1
8 insertion/deletion affect many triplets/all codons after the mutation/are
frame-shift mutations 1
9 and so affect many amino acids in a protein/all amino acids after the
mutation 1
Maximum 6
10 mutagenic agents cause/induce/increase the
rate/frequency/chance/likelihood of mutation 1
11 they include (ir)radiation/examples/chemical (agent)s/examples 1
Not – structure of
enzyme changes.
Changes the protein
produced
Changes mRNA
sequence
Key Area 5 (total 17)
1
2
3
4
5ai
5aii
B
C/3
D
D
The 3' end of the strand contains deoxyribose/sugar/an OH group and the 5'
end contains phosphate.
The DNA is heated.
5aiii
5aiv
Primer.
Name – (DNA) polymerase. 1 mark
Role – Adds / attaches / joins nucleotides. 1 mark
5b
5c
Amplification.
Variable numbers of repetitive sequences / tandem repeats
OR
Repeat sequences of bases / nucleotides
20.5
Amplification / (mass) replication / (mass) copying of DNA
Separation of the (DNA) strands / breaking hydrogen bonds between strands /
denaturing (of the DNA)
They bind / anneal / join to (the ends of the) target / complimentary
sequences (of DNA being copied).
This is closer to the optimum temperature for DNA / taq polymerase.
DNA profiling / gel electrophoresis
6a
6b
6c
6d
6e
7
If temperature is
specified it must be
90oC plus.
Not - RNA polymerase
Taq polymerase – also
correct
Genetic screening /
genetic fingerprinting
Key Area 6 Task 1 (total 29)
1
2
3
4
5
6ai
6aii
6b
7a
7b
C
A
B
B
B
pH = 8
temperature = 38
pH = 6·6 to 8·8 or 2.2 and
temperature = 20 to 54ºC or 34ºC
1. The substrate attaches to the active site (on the enzyme).
2. The active site orientates / lines up the molecules.
3. This causes an induced fit/the enzyme to change shape slightly so that
the active site fits better.
4. This lowers the activation energy (required for the reaction to occur).
5. Products are released as they have a low affinity for / attraction to the
active site.
Any 3 from 5
Lactose can be broken down into two sugars/monosaccharides/ glucose
and galactose OR
Lactose is built up from two sugars/monosaccharides/glucose and galactose
1. volume of milk
2. volume of enzyme/lactase
3. concentration of enzyme/lactase
4. temperature of the milk/solution
5. age of milk
7c
Correct scales and labels on axes – 1 mark
Points correctly plotted and lines drawn going to zero – 1 mark
Lines distinguished from each other (eg key given) – 1 mark
At least 1 zero needs to be marked on the axes
7d
7e
7f
8ai
8aii
8b
Human milk contains more lactose (than cow’s milk)
Lactose/substrate is used up/starts to limit the rate of reaction
Repeat experiment and calculate an average
22 and 28
39% and 18%
one mark for correct choice of axes with units
one mark for points correctly plotted with line
one mark deducted for using less than 50% of graph paper
If error is made in (a) but correctly plotted – mark is given
Pepsin works best at a pH 2/optimum pH for pepsin is 2
0
Because the starting lengths are different
With egg white but no pepsin/pepsin replaced with water
temperature,
duration,
width of tube,
concentration/volume of enzyme,
type of pepsin,
egg boiled for same time,
type of egg (white)
(any two for one mark)
Repeat investigation at each pH value
8ci
8cii
8ciii
8civ
8v
8vi
ºC essential at least once
Products/compounds/things/
carbohydrates
Type of milk
Concentration of milk
Mass of lactase
Time
Temperature on its own
Room temperature
-1 for using less than 50% of
either
axis
-1 for transposing axes
-1 for drawing a bar graph
Pepsin works best at low pH
It’s easier to compare results
A tube with distilled water
Size of tube/beaker
Time
Key Area 6 Task 2 (total 32)
1
2
3
4
5
6
7a
7b
7c
7d
7ei
7eii
7f
8a
8b
8c
D
C
A
D
A
A
Oxygen is produced (becoming trapped in the filter paper causing it to float).
1. size/surface area/diameter/mass (weight ok)/thickness/type
of filter paper/disc
2. concentration of hydrogen peroxide
3. height/depth/volume of hydrogen peroxide or
height/depth/volume/size/shape/dimensions of beaker
4. soaking time
5. temperature of the solution
6. pH
Any 3 points for 2 marks, 2 points needed for 1 mark
Ten discs were used at each concentration or
The experiment was repeated at each concentration or
An average time was taken for each concentration
Use discs soaked in water (added to hydrogen peroxide) or
Use discs containing no catalase (added to hydrogen peroxide)
Correct scales, labels and units on axes (average time(s) is
okay)
Points correctly plotted and line drawn
1. As (catalase) concentration increases reaction rate/rate of
hydrogen peroxide breakdown increases.
2. At higher concentrations/above 1% the reaction rate levels
off.
Award 1 mark only if both answers mention time for disc to
rise instead of reaction rate
Inhibitor binds to the enzyme/catalase altering the shape of
the active site
or
Inhibitor binds to/blocks the active site of the enzyme/catalase
or
Inhibitor competes with the substrate/hydrogen peroxide for
the active site.
Trypsin / the enzyme digests / breaks down gelatine / protein and releases
the (dark) chemicals.
Temperature of solution / trypsin
pH
Volume / depth of solution / trypsin
Size / length / area of film
Age / type/ thickness of film / thickness of gelatin
Age of trypsin
(any 2)
Repeat the procedure at each concentration (and then calculate an average).
Length/shape of filter
paper
Volume of catalase
Time (on own)
amount
Dry disc
Disc by itself
-1 for using less than
50% of either
axis
-1 for transposing axes
-1 for drawing a bar
graph
-1 for extrapolating to
zero
Any explanations eg
limiting
factors/optimum
conditions
Denatured
Trypsin digests the
colour
Temperature of
room/test-tube
Test-tube dimensions
Mass of film
Volume of gelatine
Source of trypsin
Amount
Repeat the
investigation
Repeat with different
8d
Axes correctly drawn and labelled 1 mark
Must have trypsin conc (%) and time for film to clear (s)
Points correctly plotted and line drawn 1 mark
8e
There is more trypsin / enzyme (molecules) / active sites to react with the
gelatine / protein / substrate.
Surface area of film / size of film / thickness of gelatine is limiting the rate of
reaction
OR
The size of the film / gelatine is too small to allow all enzyme molecules to
react with it
OR
The reaction / clearing the film requires a minimum time to occur / is going as
fast as it can
1. Enzymes are catalysts/speed up metabolism/chemical reaction/lower
activation energy
2. Temperature: enzymes have an optimum temperature/temperature at
which they work best/work best at 37oC (or labelled graph to illustrate)
3. pH: all enzymes have an optimum pH/pH at which they work best (or
labelled graph to illustrate)
4. Denaturing: a change occurs in the structure/shape/active site of the
enzyme at high temperatures/when the pH changes
5. Inhibitors: slow up/stop enzyme activity
6. Competitive inhibitors: attach to/block the active site so keeping out the
substrate molecule OR inhibitor competes with substrate for active site
7. Non-competitive inhibitors: attach to another part of an enzyme and
changing the shape of the active site/enzyme (so the substrate molecule does
not fit)
8. Substrate concentration: increasing substrate concentration increases
activity until a point when activity levels off (or labelled graph to illustrate)
9. Explanation that activity levels off when all enzyme active sites are reacting
with substrate molecules/enzymes are working at fastest rate possible
10. Enzyme concentration: increasing enzyme concentration increases the
rate of reaction (or labelled graph to illustrate)
11. Explanation that activity increases due to more active sites being added
(12. Vitamins/minerals/cofactors/coenzymes/other enzymes activate
enzymes)
8f
9
solutions
Remove one mark for
bar graph or for using
less than half of the
graph paper
Enzyme is no longer
limiting the reaction
Substrate
concentration is
limiting the reaction
Other factors are
limiting the reaction