Grade 10 life sciences - cdn.24.co.za

[Pages:10]MARKS: 150

NATIONAL SENIOR CERTIFICATE

GRADE 12

LIFE SCIENCES P1 NOVEMBER 2009 MEMORANDUM

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Life Sciences/P1

2 NSC ? Memorandum

DoE/November 2009

SECTION A

QUESTION 1

1.1

1.1.1

1.1.2

1.1.3

1.1.4

1.1.5

1.2

1.2.1

1.2.2

1.2.3

1.2.4

1.2.5

1.2.6

1.3

1.3.1

1.3.2

1.3.3

1.3.4

1.3.5

1.3.6

1.4

1.4.1

1.4.2

1.4.3

1.4.4

D33 B33 C33 C33 A33

(5 x 2) (10)

Fallopian tube3/oviduct

Graafian follicle3

Umbilical vein3

Oxytocin3

Seed3

Cervix3

(6)

E3

G3

F3

I3

A3

D3

(6)

Transcription3

(1)

53

(1)

GCU3 ? CAU3 ? UGG3

(3)

(a) The sequence of the amino acids will change3/the actual

amino acids could change and a new/different protein could

form3

(2)

(b) Mutation3

(1)

(8)

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1.5

1.5.1

Anaphase II3

(1)

1.5.2

Chromatids3 are pulled towards the poles3

(2)

1.5.3 A Spindle fibre3

B Cell membrane3

(2)

1.5.4 (a) 83

(1)

(b) 43

(1)

1.5.5 Ovary3

(1)

1.5.6 No3

(1)

1.5.7

Humans would have 233 chromosomes/46 chromatids in this

phase. In this diagram only 4 chromosomes3/8 chromatids are

shown/incorrect3 number 3of chromosomes

(2)

1.5.8

- Reduction/halving of chromosome number3/keep chromosome

number constant from generation to generation/prevents

doubling of chromosome number at fertilisation

- Promotes/contributes to genetic variation3

Formation of gametes/cells containing one allele of a gene pair3

(Mark first TWO only)

(2)

(13)

1.6

1.6.1

The flowers with petals attracted more insects3 for pollination3

than the flowers without petals

(2)

OR

The flowers without petals may not have attracted insects3

therefore less pollination3

1.6.2

Some of the pollen tubes that developed were from the same

flower 3/self-pollination occurred and only make little growth into

the style3 /not all pollen grains make it to the ovary/does not

fertilise the ovule

(2)

1.6.3

Repeat the investigation and use the average3

Increasing the size of the sample3

Use the same size flowers3

Use the same colour flowers3

Use the flowers of the same apple tree3

Ensure that all the flowers are pollen-free at the beginning of

the investigation3

Use the same number of flowers3

The same number of days for pollination3/prevention of pollination/

for fertilisation to take place

any (3)

(Mark first THREE only)

(7)

TOTAL SECTION A: 50

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SECTION B

QUESTION 2

2.1

2.1.1

2.1.2

2.1.3 2.1.4

2.1.5 2.1.6

2.2

2.2.1

2.2.2

2.2.3

2.2.4

A - prostate gland3

B - vas deferens / sperm duct3

E - urethra3

G - nucleus3

(4)

C - Stores sperms temporarily3/sperms mature here (Mark first ONE only)

F - Contain enzymes to break down the cell membrane of the egg

cell3

(Mark first ONE only)

(2)

D3 testis 3/seminiferous tubules

(2)

To keep the testes at a temperature that is (about 3 oC) lower than

body temperature3

A lower temperature is necessary for the production of healthy

sperm3/so that healthy sperms can survive

(2)

(a) Interstitial cells3/Cells of Leydig

(1)

(b) Testosterone3

(1)

(a) Severing of the vas deferens3

Will not allow sperms to pass to urethra and into the female3

and hence no fertilisation results3

any (2)

(b) Yes3

(1)

(c) HI virus is carried in body fluids3/ seminal fluids/saliva/blood

Can infect a person through open wounds3/blood transfusion/

sexual intercourse

Therefore vasectomy does not stop the transmission of HIV

(2)

(17)

Accept day 14 or day 153

(1)

Days 0 - 73

(1)

- Causes the follicle to burst open3/stimulates ovulation

- Stimulates the formation of the corpus luteum3

(Mark first ONE only)

(1)

- LH levels remain low up to day 12/13 3

- Then it increases sharply up to day 143

- After which it decreases and remains low3

(3)

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2.2.5 As the oestrogen level increases 3

the thickness of the endometrium also increases3

(2)

2.2.6 Maintain the increase in the thickness of the endometrium3

for greater chance of implantation3

(2)

2.2.7 No3

(1)

2.2.8 The progesterone level3has dropped3/not maintained/corpus

luteum has started to degenerate

(2)

(13)

[30]

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QUESTION 3

3.1

3.1.1

(a) IAIB 33

AB33

(2)

(b) IAi33/ IAIo IBi 33/ IBIo ii33/ IoIo AO33 BO33 OO33

(6)

3.1.2 It is a sex-linked3disease

caused by a recessive allele3

carried on the X3 chromosome

Males need only one recessive allele3to have the disease because

they have XY combination,

while females have to have both recessive alleles3 to have haemophilia

because they have an XX combination

any (4)

(12)

3.2

3.2.1 Normal female: Chromosome pair 23 = XX 3

Female with Turner's syndrome: Only one X3 chromosome

(2)

3.2.2 She will not be able to have children3 since her sex organs will

not develop3/no menstrual cycle because there are underdeveloped

gonads and therefore no hormones

(2)

(Mark first ONE only)

(4)

3.3

3.3.1 102 3 x 100 %

120

1

= 853%

(2)

3.3.2 Equal number of boys and girls3

Take a much larger sample3/repeat samples in another school/

another population

(2)

(Mark first TWO only)

(4)

3.4 P1

phenotype Black x Brown3

genotype

Bb x bb3

Meiosis

G

B, b x b 3

Fertilisation

OR

F1

genotype Bb and bb3

phenotype Black and brown3

gametes b B Bb b bb

1 mark for correct gametes 1 mark for correct genotypes

1 mark for stating P1 and F1 1 mark for stating meiosis and fertilisation

any (6)

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3.5 - Although contraceptives are easily available nowadays, many teenagers

are not well informed about them3

- Some people feel that morality has decreased significantly3

- Families, nowadays, are less likely to provide teenagers with care

and discipline3

- Teenagers are more exposed to sex in the media in these days3

- Teenagers are increasingly able to make their own decisions3

- Abortions are now legal and easily available3

(Mark first FOUR only)

any (4)

30

TOTAL SECTION B: 60

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SECTION C QUESTION 4 4.1

4.1.1 1 : 2 : 13 3

4.1.2

The proportions of different genotypes for coat colour of cattle

WW RR

DoE/November 2009

(2)

RW

Rubric for the mark allocation of the pie chart

Correct type of graph Caption Correct proportions of slices

Label / key for each slice

1 1 1: 1 correct slice 2: 2 to 3 correct slices 1 mark for each label

NOTE: If the wrong type of graph is drawn: marks will be lost for

"correct type of graph" as well as for drawing the slices in correct

proportions.

(7)

4.1.3 Both alleles3 for fur colour are equally dominant3 and therefore both are expressed in the phenotype3

OR

Neither of the alleles3 for red or white colour are dominant over

each other3 and therefore no one colour alone is expressed/

masked in the phenotype3

(3)

(12)

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