Please refer to the question papers which i hv handed out during our biology lesson. or download your papers at www.myschoolchildren.com Or click HERE
Answer to Section A:
* you can always get your school biology teachers to workout the answer as well, they hv enough experience to guide you through. You can also improve the answer below if you find them lacking/unfit.
Q 1
a) i) Y = carrier protein, Z= phospholipid bilayer
ii) Facilitated diffusion
iii) * water soluble, * large molecule
iv) *Higher concentration gradient,*more carrier proteins, * large surface area,* higher temperature
b)i) semi-permeable membrane is a membrane that allows certain substances to pass through it freely while others cannot.
ii)*volume of visking tubing decreases, * the surrounding solution is hypertonic compared to the solution in the visking tubing, *water diffuse out into the surrounding solution by osmosis
c) visking tubing doesn’t has carrier protein/ATP
a) i) Y = carrier protein, Z= phospholipid bilayer
ii) Facilitated diffusion
iii) * water soluble, * large molecule
iv) *Higher concentration gradient,*more carrier proteins, * large surface area,* higher temperature
b)i) semi-permeable membrane is a membrane that allows certain substances to pass through it freely while others cannot.
ii)*volume of visking tubing decreases, * the surrounding solution is hypertonic compared to the solution in the visking tubing, *water diffuse out into the surrounding solution by osmosis
c) visking tubing doesn’t has carrier protein/ATP
Q2
a) Producer: Hydrilla sp., Secondary consumer: duck/black beetle/small fish/…
b) Duck eats small fish in the pond.
c) *cellular respiration doesn’t occur, * therefore no ATP can be produced, *lotus plant cannot carry out active transport to pump mineral ions into the root.
d)*excessive fertilizer will be drained into the pond,*encourage rapid growth of algae,* eutrophication occurs,*algae covered the surface of pond, less light intensity can penetrate into the water to reach the submerged plant, *…..
e) Tadpole : 20,000kJ x 10% = 2,000kJ, Black beetle: 2,000kJ x 10% = 200kJ
a) Producer: Hydrilla sp., Secondary consumer: duck/black beetle/small fish/…
b) Duck eats small fish in the pond.
c) *cellular respiration doesn’t occur, * therefore no ATP can be produced, *lotus plant cannot carry out active transport to pump mineral ions into the root.
d)*excessive fertilizer will be drained into the pond,*encourage rapid growth of algae,* eutrophication occurs,*algae covered the surface of pond, less light intensity can penetrate into the water to reach the submerged plant, *…..
e) Tadpole : 20,000kJ x 10% = 2,000kJ, Black beetle: 2,000kJ x 10% = 200kJ
Q3
a) Cell X: Anaphase, Cell Y: Anaphase I
b) X: produce new cells for growth/repair dead tissues
Y: produce gametes
c) * Homologous chromosomes separate and move to opposite poles of the cell,
* Chromosome number are halved.
d) *8 daughter cells, *number of chromosome = 4
e) drawing: - we hv practised this many times, u can refer back to those drawings.
Guidelines:
*Cell X: draw a diploid cell that contains 4 chromosomes (2 big and 2 small unduplicated chromosomes)
*Cell Y: draw a haploid cell that contains 2 chromosomes (one big and one small chromosome, u can show crossing over if you like), chromosomes must be unduplicated and non-homologous)
f) *chromosomes cannot divide/separate correctly,*causing abnormalities of chromosome in daughter cells.
a) Cell X: Anaphase, Cell Y: Anaphase I
b) X: produce new cells for growth/repair dead tissues
Y: produce gametes
c) * Homologous chromosomes separate and move to opposite poles of the cell,
* Chromosome number are halved.
d) *8 daughter cells, *number of chromosome = 4
e) drawing: - we hv practised this many times, u can refer back to those drawings.
Guidelines:
*Cell X: draw a diploid cell that contains 4 chromosomes (2 big and 2 small unduplicated chromosomes)
*Cell Y: draw a haploid cell that contains 2 chromosomes (one big and one small chromosome, u can show crossing over if you like), chromosomes must be unduplicated and non-homologous)
f) *chromosomes cannot divide/separate correctly,*causing abnormalities of chromosome in daughter cells.
Q4
a)i) cakes/chocolates/….
ii) *constipation,*take more food containing fibre
iii) *build new cells for growth and to replace damaged tissues,*for synthesis of enzyme, antibodies,*……(refer to form 4, chapter 6 – assimilation)
b) P: 2,250kJ, Q: 800kJ, R:14,500kJ
ii) *Rice is the source of energy for carrying out activities while meat is mainly used for growth,*At that age, they are more active and therefore need more energy, while the growth process at this age has slowed down/almost complete.
c)i) *carbohydrate
ii) *eat more proteins,*proteins is used to produce new cells for growth of foetus.
a)i) cakes/chocolates/….
ii) *constipation,*take more food containing fibre
iii) *build new cells for growth and to replace damaged tissues,*for synthesis of enzyme, antibodies,*……(refer to form 4, chapter 6 – assimilation)
b) P: 2,250kJ, Q: 800kJ, R:14,500kJ
ii) *Rice is the source of energy for carrying out activities while meat is mainly used for growth,*At that age, they are more active and therefore need more energy, while the growth process at this age has slowed down/almost complete.
c)i) *carbohydrate
ii) *eat more proteins,*proteins is used to produce new cells for growth of foetus.
Q5
a) P: XaY, Q:XAXab) drawing of Punnet square – again we hv done this many times in class, pls refer to your previous answer.
* male gametes: Xa and Y
* female gametes : XA and Xacomplete the genotype and phenotype
c)*blood clot very slowly, * due to lack of blood clotting factors needed for blood clotting.
d) drawing of genetic cross diagram – we hv done this as well, pls check/refer to your chapter 6 notes and questions.
Parental phenotype: Blood group A (father) x Blood group B (mother)
Parental genotype: IA IA IB IO
meiosis:
gametes:
random fertilization:
F1 generation
genotype:
phenotype:
phenotypic ratio: 1 blood group A: 1 blood group AB
a) P: XaY, Q:XAXab) drawing of Punnet square – again we hv done this many times in class, pls refer to your previous answer.
* male gametes: Xa and Y
* female gametes : XA and Xacomplete the genotype and phenotype
c)*blood clot very slowly, * due to lack of blood clotting factors needed for blood clotting.
d) drawing of genetic cross diagram – we hv done this as well, pls check/refer to your chapter 6 notes and questions.
Parental phenotype: Blood group A (father) x Blood group B (mother)
Parental genotype: IA IA IB IO
meiosis:
gametes:
random fertilization:
F1 generation
genotype:
phenotype:
phenotypic ratio: 1 blood group A: 1 blood group AB
All the best!
