Sexual and Asexual Reproduction - Merit Tutors · 2018. 5. 25. · Question Number Answer Acceptable answers Mark 3(c) Any two from the following: • sexual reproduction involves

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Sexual and Asexual

Reproduction Mark Scheme

Level Edexcel

Subject Biology

Exam Board GCSE(9-1)

Topic Genetics

Sub Topic Sexual and Asexual Reproduction

Booklet Mark Scheme

Time Allowed: 57 minutes

Score: /47

Percentage: /100

Question Number

Answer Acceptable answers Mark

1(a)(i) 0.5 / 0.5 picogram Accept: 0.5 picograms

accept: the same (mass) as the sperm cell (1)

Question Number


1(a)(ii) C haploid (1)

Question Number


1(a)(iii) thymine with adenine, cytosine with guanine

(1)

Question Number


1(a)(iv) weak hydrogen bonds / hydrogen bonds / hydrogen (1)

H (bond) (1)

Question Number


1(b)(i) A description including three of the following points:

• cell divides / cell division /cell splits(1)

• two cells produced (1)

• (both) diploid (1)

• (both) cells are geneticallyidentical (1)

credit correct reference to stages of mitosis: DNA replication / chromosomes duplicate (1) Chromosomes line up along the equator / middle of the cell (1) chromosomes pulled to either end of cell (1) cytokinesis / cytoplasm splits (1) (3)


Question Number


1(b)(ii) A description including three of the following points:

• ref (to many) cell divisions/ eq (1)

• growth (1)

• ref to differentiation /specialisation (1)

• ref to stem cells (1)

accept: gets bigger / larger

accept: become specific cells

(3)


Question Number


2(a)(i) B courtship (1)

Question Number


2(b) An explanation linking

• protection (of femaleduring birth / of young) /concealment (1)

• from predators / untilstrong enough (to fend foritself) (1)

safer camouflaged

weather (2)

Question Number


2(a)(ii) avoid injury / do not waste energy

avoid a fight idea of dominance / submission feels threatened

Ignore : female will pick the biggest antlers / respect (1)

Question Number


2(c) (i) A description including two of the following

• can eat plants whichcontain tannins (1)

• larger food supply (1)

• plants not consumed byother herbivores / lesscompetition from otherherbivores / animals (1)

get more food / less likely to starve / won’t starve

(2)


Question Number


2(c) (ii) A description including

• (flower) attracts insects(1)

• which pollinate the flower(1)

• Idea that insect - flowerrelationship is specific (1)

attraction can be specific in terms of colour, size or scent or nectar or pollen

fertilise / reproduce for pollinate

e.g. bee and bee orchid(2)


Question Number


3(a) A description including four of the following points

• ref to meiosis (1)

• 4 cells produced (from oneparent cell) (1)

• haploid (cells) / cells havehalf the number ofchromosomes (1)

• cells are genetically different(1)

do not accept if there is a ‘t’

cells have one set of chromosomes / 23 chromosomes

(4)


Question Number

Indicative Content Mark

QWCQWC *3(b) A description including

• fertilisation of egg by sperm• ref to fusion of nuclei• forming diploid cell• ref to zygote

• (zygote) divides by mitosis• to form identical cells• several mitotic divisions• growth of foetus• examples of how fetus grows eg in height, mass

• stem cells in embryo• specialisation / differentiation of (stem) cells into

different cell types• examples of different cell types eg neurones, skin cells• development of fetus

(6)

Level

0 No rewardable content

1 1 - 2 • a limited description including 2 or more comments about oneprocess

• the answer communicates ideas using simple language and useslimited scientific terminology

• spelling, punctuation and grammar are used with limited accuracy2 3 - 4 • a simple description including 2 or more comments on 2 processes

• the answer communicates ideas showing some evidence of clarityand organisation and uses scientific terminology appropriately

• spelling, punctuation and grammar are used with some accuracy3 5 - 6 • a detailed description including 2 or more comments on all 3

processes• the answer communicates ideas clearly and coherently uses a

range of scientific terminology accurately• spelling, punctuation and grammar are used with few errors


Question Number


3(c) Any two from the following:

• sexual reproduction involvestwo parents but asexualreproduction only involves one(organism / parent / cell) (1)

• sexual reproduction needsgametes / sex cells butasexual reproduction does not(1)

• sexual reproduction producesgenetically different organismsbut asexual reproductionproduces genetically identicaloffspring / clones (1)

ignore any reference to meiosis or mitosis

sexual reproduction results in variation but asexual reproduction does not

(2)



Question

Number


4(a) D haploid and haploid (1)

Question

Number


4 (b) A description linking three of the

following

(DNA is a) double helix (1)

the sides of DNA are made

from (alternating) sugars and phosphate (molecules) /

sugar phosphate backbone (1)

{paired / complementary} bases / A (joins to) T and C

(joins to) G (1)

(bases joined by/strands held

together by) hydrogen bonds (1)

Accept H bonds

Ignore h or H2 bonds

(3)


Question

Number


4(c) A description including four of

the following:

(the process is) translation (1)

(mRNA ) leaves the nucleus / enters the cytoplasm (1)

(mRNA joins to) ribosomes(1)

tRNA carries amino acids (1)

tRNA joins to mRNA / bases on tRNA matches bases on

mRNA (1)

(bases read as) {sets of three

/ triplets / idea of codons} (1)

(ribosome / mRNA holds tRNA so) amino acids are joined together / to make

polypeptides (1)

(4)

Total for Question 4 = 8 marks


Question number

Answer Mark

5(a) An explanation that combines identification – understanding (1 mark) and reasoning/justification – understanding (1 mark): • Mendel crossed homozygous tall and homozygous short

pea plants and produced all tall offspring (1)• therefore all the offspring had a heterozygous genotype

with one tall and one short allele showing that the tallallele is dominant (1) (2)

Question number

Answer Mark

5(b)(i) An explanation that combines identification – application of knowledge (1 mark) and reasoning/justification – application of understanding (1 mark): • asexual reproduction is a rapid reproduction technique

allowing the production of more plants• as there is no requirement for cross pollination/higher crop

yield/increased profit (2)

Question number

Answer Mark

5(b)(ii) An explanation that combines identification – application of knowledge (1 mark) and reasoning/justification – application of understanding (1 mark): • introduces variation into the population• which allows for natural selection of fitter plants/increased

chance of the population surviving (2)

Question number

Answer Mark

5(c)(i) C (1)

Question number

Answer Mark

5(c)(ii) An explanation that combines identification via a judgement (1 mark) to reach a conclusion via justification/reasoning (1 mark): • genotype is XDXd/she must have one dominant and one

recessive allele (1)• because her daughter must have received the recessive

allele and her son has inherited a dominant allele (1)

(2)


Inheritance Mark Scheme 1

Level Edexcel

Subject Biology


Topic Genetics

Sub Topic Inheritance

Booklet Mark Scheme 1


Score: /44

Percentage: /100

Question Number


1a(i) answers must be in this order.

dominant

HH

(2)

Question Number


1a(ii)H h

H HH Hh h Hh hh

1 mark for correct gametes 1 mark for correct offspring

If incorrect gametes allow 1 mark for correct Punnett square based on selected gametes

(2)

Question Number


1a(iii)

75% / ¾ / 0.75

accept error carried forward from their Punnett square

accept: 3 : 1

(1)

Question Number


1b(i) An explanation linking two of the following:

Huntington’s disease is caused by a dominant allele / CF is caused by a recessive allele (1)

only one allele for Huntington’s disease needs to be inherited to have the disease / would have the disease if heterozygous (or homozygous dominant)(1)

two alleles (recessive) need to be inherited to have CF / be homozygous recessive for CF (1)

Ignore refs to gene for allele against this marking point

Ignore refs to gene for allele against this marking point

(2)


Question Number


1b(ii) A mucus (1)

Question Number


1b(iii) An explanation linking two of the following:

(thick / sticky / more) mucus (1)

builds up in the tubes (of the reproductive system) (1)

(the mucus) blocks the flow of sperm (1)

Reject: mucus in

lungs/intestine

accept sperm duct / vas deferens

(2)

Total for question 1 = 10 marks


Question Number


2(a)(i) XX ignore any superscript or

subscript letters/symbols

reject XY (1)

Question Number


2(a)(ii) An explanation linking two of the following

they did not inherit the (haemophilia) allele (1)

(allele is) located on X chromosome (1)

males receive X chromosome from their mother/Y chromosome from father (1)

B is homozygous dominant/ neither X chromosome from B has the allele for haemophilia (1)

ignore gene throughout

accept have the dominant/normal allele

accept disorder is located on the X chromosome

ignore mother is unaffected accept mother neither affected nor a carrier

accept mother for B and father for A

(2)


Question Number


2(a)(iii) HX Y

XH HXHX HYX Xh XHXh hYX

a Punnett square showing the gametes of individuals C and D (1)

a Punnett square showing the genotypes of the offspring (1)

25% / 0.25 / 1 in 4 probability of a child having haemophilia (1)

reject if allele shown on Y chromosome

50% of males have haemophilia

Punnett square must be interpreted correctly (3)


Question Number


2(b) An explanation linking the following

aseptic conditions (1)

prevent growth of unwanted organisms/prevent contamination (1)

OR

temperature /pH (1)

provide optimal conditions for growth /prevent enzymes denaturing (1)

OR

nutrient levels (1)

provide optimal conditions for growth (1)

OR

aeration/oxygen (1)

for aerobic respiration/ provide optimal conditions for growth / prevent anaerobic respiration (1)

ignore sterile

provide optimal conditions for enzymes/prevent micro-organisms being killed

(2)

(Total for question 2 = 8 marks)


Question Number


3a(i) homozygous recessive Accept in any order:

homozygous

recessive (alleles) (1)

Question Number


3(a)(ii) female gametes

e e

E Ee Ee

e ee ee

male gametes

correct gametes in male/female gametes headings (1)

correct offspring genotypes (1) (2)

Question Number


3a(iii) Any one of the following

• 1/2

• 0.50

• 2/4

• 50 %

• 1:1 / 2:2

Accept if 2 correct answers are given e.g. ½, 50%

evens chance

(1)


Question Number


3(a)(iv)A 0%

(1)

Question Number


3(b) A description including the following points

• reference to mucus (1)

• location described e.g.lungs / pancreas /reproductive system (1)

• consequence describede.g. breathing difficulty /infection / weight loss dueto blocking of enzymes /difficulty with digestion orabsorption / infertility (1)

Accept three symptoms described (3) Ignore: references to symptoms of sickle cell

Accept – airways for lungs

Accept fertility problems for infertility

Symptoms may include

diabetes (1) malnutrition (1) incontinence in females (1) sinusitis (1) nasal polyps (1) arthritis (1) (3)


Question Number


4(a)(i) D

(1)

Question Number


4(a)(ii) substitution (1) 27 ÷ 40

evaluation (1) 0.675 x 100 67.5 (%)

e.c.f from 3(a)(i)

accept 68(%) for 2 marks give full marks for correct answer, no working (2)

Question Number


4(b)(i)

b b

B Bb Bb

b bb bb

Female gametes

Male gametes

gametes in male/female gametes headings (1)

offspring genotypes (1) (2)

Question Number


4(b)(ii) 0.5 / 50% / 50/50 / ½ / 2/4 / 2:2 /even chance

evens

(1)


Question Number


4(b)(iii) homozygous

recessive

homozygous recessive

Accept any reasonable spelling of the term Reject heterozygous

(1)

Question Number


4 (c) an explanation linking three of the following

• speciation (1)

• different geographical areamay have differentselection pressures /environmental conditions(1)

• those individuals of aspecies suited /adapted /to this environment willsurvive and breed (1)

• adaptations/genes passeddown to the offspring

• new species unable tobreed with original (1)

named environmental conditions e.e.g. clima

Accept survival of the fittest

(3)


Question Number


5(a)(iii) An explanation linking two of the following points:

• two of the offspring fromgeneration II had CF (1)

• the children with cysticfibrosis must have inherited 1recessive allele from eachparent / children must have 2recessive alleles (1)

• both parents must have 1recessive allele / be carriersof the CF allele (1)

ORA if homozygous dominant then no CF offspring

Ignore: references to genes

ORA if homozygous recessive offspring would have CF

(2)

Question Number


5(a)(i)2 / two (offspring) 2 and 3 (1)

Question Number


5(a)(ii)D (1)


Question Number


5(c) An explanation linking two of the following:

• pedigree analysis will determinethe likelihood that their offspringcould inherit the CF allele(1)

• if heterozygous there is a 50%chance (that the CF allele) will bepassed on / if 2 heterozygousparents 25% chance the offspringwill have CF(1)

• if either parent is homozygousdominant there is 0% chance thattheir offspring could have thedisease(1)

Accept to see if they are a carrier of the CF allele

Accept ratios rather than percentages 2 in 4 chance

(2)

Question Number


5(b)correct gametes (1)

correct offspring (1)

B b

B BB Bb

b Bb bb

Accept bB instead of Bb

(2)




Level Edexcel

Subject Biology


Topic Genetics




Score: /50

Percentage: /100

Question number

Answer Mark

1(a) An explanation that combines identification – understanding (1 mark) and reasoning/justification – understanding (1 mark): • Mendel crossed homozygous tall and homozygous short

pea plants and produced all tall offspring (1)• therefore all the offspring had a heterozygous genotype

with one tall and one short allele showing that the tallallele is dominant (1) (2)

Question number

Answer Mark

1(b)(i) An explanation that combines identification – application of knowledge (1 mark) and reasoning/justification – application of understanding (1 mark): • asexual reproduction is a rapid reproduction technique

allowing the production of more plants• as there is no requirement for cross pollination/higher crop

yield/increased profit (2)

Question number

Answer Mark

1(b)(ii) An explanation that combines identification – application of knowledge (1 mark) and reasoning/justification – application of understanding (1 mark): • introduces variation into the population• which allows for natural selection of fitter plants/increased

chance of the population surviving (2)

Question number

Answer Mark

1(c)(i) C (1)

Question number

Answer Mark

1(c)(ii) An explanation that combines identification via a judgement (1 mark) to reach a conclusion via justification/reasoning (1 mark): • genotype is XDXd/she must have one dominant and one

recessive allele (1)• because her daughter must have received the recessive

allele and her son has inherited a dominant allele (1)

(2)


Question number

Answer Mark

2(a)(i) B (1)

Question number

Answer Mark

2(a)(ii) TACGTACATGGC (1)

Question number

Answer Additional guidance Mark

2(a)(iii) • 3.33 × 10−10 equals 0.33  nm(1)

• 0.33 × 250 = 82.5 (nm) (1)

maximum one mark if no conversion to nm

award full marks for correct numerical answer without working (2)

Question number


2(b)(i) • heterozygous accept alleles showing heterozygous genotype (1)

Question number

Answer Mark

2(b)(ii) • correct Punnett square (1)

• 75% normal fur pigmentation (1)

A a A AA Aa a Aa aa

(2)

Question number

Answer Mark

2(c) An explanation that combines identification – understanding (1 mark) and reasoning/justification – understanding (1 mark): • both parents must be heterozygous for the recessive

allele (1)• so the offspring must inherit the recessive allele from

each parent (1) (2)


Question Number


3(a)(i) D homozygous recessive (1)

Question Number


3(a)(ii) A description to include three of the following points:

tired / lethargic (1)

short of breath / reduced oxygen carrying capacity / problems exercising (1)

swelling of hands and feet (1)

painful / weak joints (1)

blocked blood vessels / blood clots(1)

Accept weak/fatigued/

Accept difficulty breathing

Accept reference to pain or painful episodes/sickle cell crisis

Ignore references to the shape of the red blood cell

Ignore references to mucus

(3)

Question Number


3(b)(i)

correct gametes (1)

correct offspring genotypes (1)

D d

D DD Dd

D DD Dd

Allow ECF for incorrect gametes

(2)


Question Number


3(b)(ii) 50(%) (1)

0(%) (1)

Answers must be in this order Possible ecf from the candidates Punnett square

Clip together with 1bi (2)



Question Number


4(a)(i) 75% (1)

Question Number


4(a)(ii) An explanation linking two of the following:

• Punnett square wouldpredict 50% normal 50%carrier (1)

• actual offspring are not50% carrier (1)

• the probability is applied toeach child not the overalloffspring (1)

accept ratios or probabilities instead of percentages

actual offspring are 75%

accept references to random assortment

(2)

Question Number


4(a)(iii) b b

B b Bb

bb b bb

Probability = 50% Ratio 2/4, 2:2, can be given for probability

1 mark for correct gametes

1 mark for completed Punnett square with correct probability

Accept reverse order for gametes Accept letters other than B/b (but alleles must be the same letter)

50% mark can only be given if 50% of the offspring are homozygous recessive

(2)

Question Number


4(a)(iv) A homozygous dominant (BB)

(1)



Question Number


QWCQWC *4(b) A explanation to include some of the following points:

Pedigree analysis would show the likelihood of their offspring inheriting the disorder Pedigree analysis should also be carried out on the partners of the third generation.

• X is not a carrier• X is homozygous dominant• and does not have sickle cell disease• The parents of X are heterozygous / his sister has sickle cell• so will not pass on the allele for the disease to offspring• if his partner is a carrier• there will be a 50% chance of the child being a carrier

• Y and Z are carriers of the sickle cell allele• Y and Z are heterozygous• The mother of Y has sickle cell / Y will inherit the sickle cell

allele• The parents of Z do not have sickle cell / mother is a

carrier/heterozygous• They have a 50% chance of passing the sickle cell allele

onto their potential offspring• If their partners were also carriers• There would be a 25% chance that the offspring will have

the sickle cell disease• There would be a 50% chance that the offspring would also

carry the allele for sickle cell diseaseExample Punnett square:

B bB BB Bbb B bb

(6)

0 Level No rewardable content 1 1 - 2 • a limited explanation the genetic profile of X,Y and Z or an

explanation of the use of pedigree analysis• the answer communicates ideas using simple language and uses

limited scientific terminology• spelling, punctuation and grammar are used with limited accuracy

2 3 - 4 • A simple explanation of the genetic profile of X, Y and Z and anexplanation of the use of pedigree analysis


• spelling, punctuation and grammar are used with some accuracy3 5 - 6 • a detailed explanation of the genetic profile of X, Y and Z and

explanation of the use of pedigree analysis plus either anexplanation of one genotype or a prediction of one of the offspringoutcomes

• the answer communicates ideas clearly and coherently uses a rangeof scientific terminology accurately

• spelling, punctuation and grammar are used with few errors


Question Number


5a (i) D - ff (1)

Question Number


5a (ii) An explanation linking two of the following:

• mucus blocks (pancreatic)duct /small intestine wall /digestive system(1)

• preventingenzymes/named enzymebeing released (into smallintestine) (1)

• less digestion of food (1)

• less absorption (1)(2)

Question Number


5b (i) an individual who has one recessive allele and one dominant allele (1)

2 different alleles

(1)

Question Number


5b (ii) An explanation to include two of the following:

• Rebecca is homozygousdominant(for CF)(1)

• Rebecca has (inherited)one dominant allele fromeach parent(1)

• cystic fibrosis involves theinheritance of tworecessive alleles (1)

Rebecca has (inherited) 2 dominant alleles

(2)


Question Number


QWC *5(c)

An explanation including the following points:

• A Punnett square or genetic diagram showing the followinggametes and offspring

D d

d D dd

d D dd

• mother gametes = d, d• father gametes = D, d• offspring = 50% Dd, 50% dd• probability of offspring with sickle cell disease = 50%• probability of carrier / heterozygous = 50%• both parents will give one allele to the possible offspring• the father can give either the dominant or recessive allele• the mother can only give a recessive allele• a dominant and recessive allele will result in heterozygous

offspring

(6)

evLevelL No rewardable content

1 - 2• Limited written explanation is provided by the candidate of the

inheritance and / or a genetic diagram/ Punnett square with onlycorrect gametes or offspring

• the answer communicates ideas using simple language and useslimited scientific terminology

• spelling, punctuation and grammar are used with limited accuracy2 -4 • The genetic diagram/ Punnett square is correct for both gametes and

offspring with a simple explanation• the answer communicates ideas showing some evidence of clarity

and organisation and mostly uses scientific terminologyappropriately

• spelling, punctuation and grammar are used with some accuracy3 -6 • The Punnett Square/genetic diagram is complete and accurate for

both gametes and offspring plus percentage outcomes and a detailedexplanation of inheritance of CF

• the answer communicates ideas clearly and coherently uses a rangeof scientific terminology accurately

• spelling, punctuation and grammar are used with few errors

1

3

5

0




Level Edexcel

Subject Biology


Topic Genetics




Score: /12

Percentage: /100


Question Number


1(a) Genus;

Species;

Must be in the correct order (2)

Question Number


1(b) A suggestion including the following points:

rats with the mutationsurvive to reproduce (1)

pass on the allele whichmakes the offspringresistant to warfarin (1)

accept breed / produce offspring etc for reproduce

accept gene / mutation for allele

(2)

Question Number


1(c)

Correct gametes (1) Correct offspring (1)

If incorrect gametes are entered into the Punnett square but the offspring for those gametes are correct 1 mark can be awarded as an error carried forward

(2)


(Total for question 1 = 12 marks)

Question Number


QWCC 1(d) A explanation to include some of the following points

MRSA is a bacterial infection

number of cases increased from 1995 to 2006

MRSA is resistant to antibiotics so MRSA infection not easy to treat

number of cases were similar between 2005 and 2007

antiseptics killed the bacteria less bacteria were transferred from person to person

number of cases decreased from 2007

antiseptics kill bacteria on surfaces causing less infections from MRSA

(6)

LevelLev No rewardable content 1 –1 2 a limited explanation of the graph including correct data reading

or the use of antiseptics or antibiotics to kill bacteria/treat MRSA the answer communicates ideas using simple language and uses

limited scientific terminology spelling, punctuation and grammar are used with limited accuracy

2 –3 4 a simple explanation of one trend of the graph including correctdata reading and the effect of the use of antiseptics or antibioticsto kill bacteria/treat MRSA

the answer communicates ideas showing some evidence of clarityand organisation and uses scientific terminology appropriately

spelling, punctuation and grammar are used with some accuracy3 –5 6 a detailed explanation of at least two trends of the graph linking it

to antibiotic resistance and antiseptic programme the answer communicates ideas clearly and coherently uses a

range of scientific terminology accurately spelling, punctuation and grammar are used with few errors


DNA and

Protein Synthesis Mark Scheme 1

Level Edexcel

Subject Biology


Topic Genetics

Sub Topic DNA and Protein Synthesis



Score: /36

Percentage: /100

Question

Number


1(a) D haploid and haploid (1)

Question

Number


1 (b) A description linking three of the

following

(DNA is a) double helix (1)

the sides of DNA are made

from (alternating) sugars and phosphate (molecules) /

sugar phosphate backbone (1)

{paired / complementary} bases / A (joins to) T and C

(joins to) G (1)

(bases joined by/strands held

together by) hydrogen bonds (1)

Accept H bonds

Ignore h or H2 bonds

(3)


Question

Number


1(c) A description including four of

the following:

(the process is) translation (1)

(mRNA ) leaves the nucleus / enters the cytoplasm (1)

(mRNA joins to) ribosomes(1)

tRNA carries amino acids (1)

tRNA joins to mRNA / bases on tRNA matches bases on

mRNA (1)

(bases read as) {sets of three

/ triplets / idea of codons} (1)

(ribosome / mRNA holds tRNA so) amino acids are joined together / to make

polypeptides (1)

(4)



Question Number


2(a) A description that includes two of the following

• hydrogen bonds (1)

• between (complementary)base pairs (1)

H bonds accept singular

A and T, G and C but not the wrong pairings (2)

Question Number


2(b) • one bar the height of theguanine bar (34%) andone bar the height of thethymine bar (16%) (1)

• bars for cytosine andadenine shown the correctway round (1)

+/- 1 square (including sketches)

(2)

Question Number

Answer Mark

2(c)(i)

[all correct = 2 marks and 1 mistake = 1 mark]

G G C T A G T T G

C C GG A G C A A C

(2)

Question Number


2(c)(ii) three / 3 Reject any other numbers given (1)

Question Number


2(d) ribosome(s) / polysome(s) Ignore cytoplasm Reject any other structure given

(1)


Question Number


3 (a)(i) mitosis reasonableitos phoneticspelling provided there is a ‘t’ ignore asexual reproduction (1)

Question Number


3 (a)(ii) Any two from the following:

• same characteristics in offspringas parent plant /bestcharacteristics inherited / clonesproduced / identical (1)

• easier to generate newplants/propagate (1)

• quicker to produce new plants(1)

• cheap /idea that the plants willnot run out / no need to buy newplants / seeds (1)

Accept same as parent plant

(2)

Question Number


3(b) Stage 1

• to break open cells/release cellcontents / release DNA /dissolveproteins (1)

Stage 3

• to precipitate DNA from thesolution/to separate DNA (fromother components)/ (1)

Accept break down cell membrane / cell wall

Accept to make DNA visible

ignore refs to freezing the DNA (2)

Question Number


3 (c)(i) C 4(1)


Question Number


3 (c)(ii) • location drawn anywhere in

cytoplasm (1)• correct name - nucleus (1)

chloroplast / mitochondria

NB these are stand alone mark points (2)


Question Number


4(a) A description including the following linked points

• ref to a gene (coding forprotein)(1)

• sequence of basesdetermines sequence ofamino acids (1)

• idea of one code / triplet /codon / 3 bases (for oneamino acid) (1)

• several amino acids makeup a protein /(poly)peptide (1)

• transcription / detail oftranscription (1)

• translation / detail oftranslation (1)

Accept on either DNA or RNA base pairs

Accept a chain of amino acids

eg mRNA made

eg mRNA attached to ribosome (4)


Question Number


QWCQWC *4(b) A description including some of the following points in a logical sequence

Points relating to DNA structural features: • two strands• double helix• (contains) bases• A, T, C, G• adenine / A paired with thymine / T• guanine / G paired with cytosine / C• hydrogen / H bonds joining bases

Contributions from Scientists:

• X-ray (crystallography) being used• to show helical structure• to show diameter of molecule

• how base pairs are arranged was shown• how strands are arranged was shown• modelling

• reference to using other people’s ideas(6)

Level 0 No rewardable content 1 1 - 2 • a limited description that includes either:

at least three DNA features OR one contribution• the answer communicates ideas using simple language and uses

limited scientific terminology• spelling, punctuation and grammar are used with limited

accuracy2 3 - 4 • a simple description that includes at least three features of DNA

and at least one contribution OR two features of DNA and twocontributions.


• spelling, punctuation and grammar are used with some accuracy3 5 - 6 • a detailed description of the structure of DNA that includes at

least three features and two contributions.• the answer communicates ideas clearly and coherently uses a

range of scientific terminology accurately• spelling, punctuation and grammar are used with few errors


Question Number


4(c) An explanation to include two of the following points linked together

• genes / base sequence (onhuman chromosome)identified (1)

• identification of faulty /mutated genes (1)

• people can be tested for agenetic disorder (1)

• ref to development of genetherapy (1)

• idea that appropriate/early /personalised /genomic medication /counselling can be given(1)

Accept base pair sequence gene map

Accept idea that genes can be linked to disease

Accept diagnosis of cancer

Accept a description of gene therapy

(2)



DNA and

Protein Synthesis Mark Scheme 2

Level Edexcel

Subject Biology


Topic Genetics

Sub Topic DNA and Protein Synthesis



Score: /29

Percentage: /100


Question Number


1(a)(ii)

translation Accept spellings such as transation, transalation

reject: transcription (1)

Question Number


1(a)(iii) D ribosome (1)

Question Number


1(b) An explanation including two of the following points:

• ref to specific shape (1)

• to bind to substrate / formenzyme substrate complex(1)

• for reaction to take place /catalysed(1)

• joining together{substrates / molecules} /break down {substrates /molecules} (1)

• ref to lock and keymechanism / hypothesis(1)

(2)

Question Number


1(a)(i) met, val, lys

met – val – lys

MET VAL LYS Accept mix of upper and lower case

Accept: metvallys / metvalys

Not necessary to separate the words out.

(1)


Question Number


1(c) A description including three of the following points:

• a mutation is a change in agene (sequence) / base pairs(1)

• (change in DNA) causes achange in amino acid(s)/order of amino acids (1)

• change in shape of {activesite / protein / enzyme} (1)

• prevent / reduce bindingto substrate (1)

• enzyme can no longerfunction / reduced function(1)

• enzyme could be moreeffective (1)

accept change enzyme function / stops the function of the enzyme

ignore: refs to denaturation (3)


Question number

Answer Mark

2(a)(i) B (1)

Question number

Answer Mark

2(a)(ii) TACGTACATGGC (1)

Question number


2(a)(iii) • 3.33 × 10−10 equals 0.33  nm(1)

• 0.33 × 250 = 82.5 (nm) (1)

maximum one mark if no conversion to nm

award full marks for correct numerical answer without working (2)

Question number


2(b)(i) • heterozygous accept alleles showing heterozygous genotype (1)

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Answer Mark

2(b)(ii) • correct Punnett square (1)

• 75% normal fur pigmentation (1)

A a A AA Aa a Aa aa

(2)

Question number

Answer Mark

2(c) An explanation that combines identification – understanding (1 mark) and reasoning/justification – understanding (1 mark): • both parents must be heterozygous for the recessive

allele (1)• so the offspring must inherit the recessive allele from

each parent (1) (2)


Question number

Answer Mark

3(a) An explanation that combines identification – application of knowledge (1 mark) and reasoning/justification – application of understanding (1 mark): • the CD4+ count is significantly below the normal range

because the HIV has destroyed the {white blood cells/CD4+cells} (1)

• so the person is more susceptible to opportunistic infectionsand classified as having AIDS (1) (2)


Question number

Indicative content Mark

*3(b) Answers will be credited according to candidate's deployment of knowledge and understanding of the material in relation to the qualities and skills outlined in the generic mark scheme.

The indicative content below is not prescriptive and candidates are not required to include all the material which is indicated as relevant. Additional content included in the response must be scientific and relevant.

AO2 (6 marks) • isolate an antigen from the pathogen which causes the

STI• inject the antigen into a mouse/rodent• collect lymphocytes producing an antibody to the STI

antigen• fuse the B-lymphocyte with a myeloma cell• production of a hybridoma• hybridoma produces a monoclonal antibody against the

antigen of the STI• attach the monoclonal antibody to coloured bead/indicator• incorporate into a test strip. (6)

Level Mark Descriptor

0 No awardable content Level 1 1-2 • The explanation attempts to link and apply knowledge and

understanding of scientific enquiry, techniques and procedures,flawed or simplistic connections made between elements in thecontext of the question. (AO2)

• Lines of reasoning are unsupported or unclear. (AO2)Level 2 3-4 • The explanation is mostly supported through linkage and

application of knowledge and understanding of scientific enquiry,techniques and procedures, some logical connections madebetween elements in the context of the question. (AO2)

• Lines of reasoning mostly supported through the application ofrelevant evidence. (AO2)

Level 3 5-6 • The explanation is supported throughout by linkage andapplication of knowledge and understanding of scientific enquiry,techniques and procedures, logical connections made betweenelements in the context of the question. (AO2)

• Lines of reasoning are supported by sustained application ofrelevant evidence. (AO2)


Question number

Answer Additional guidance

Mark

3(c) An explanation that combines identification – understanding (1 mark) andreasoning/justification – understanding (3marks):• a single strand of messenger RNA is

transcribed from the gene in thenucleus (1)

• messenger RNA molecule binds tothe ribosome (1)

• the triplet code from the mRNA ismatched by a complementary tRNAanticodon at the ribosome (1)

• tRNA transfers amino acids to thepolypeptide chain in a specific order(1)

to gain maximum marks the process must be in a logical sequence

(4)

Documents

Sexual and Asexual Reproduction - Merit Tutors · 2018. 5. 25. · Question Number Answer Acceptable answers Mark 3(c) Any two from the following: • sexual reproduction involves