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首页 >  备考指南 > 10月爱德思A-Level生物U1-U2考情回顾


来源:      浏览:      发布日期:2019-12-18 14:51





Unit 1、Unit 2、Unit 3是AS的考试,卷面满分分别为80、80、50,最后折算为120、120、60满分的相应成绩计入最后的总分之中;

Unit 4、Unit 5、Unit 6则是A2的考试,卷面满分分别为90、90、50,也一样最后折算为120、120、60满分的相应成绩计入总分之中。

最终的总分为600分满分,达到480分为A,420为B,以此类推。而A*的要求为在达到A(480分)基础之上,A2的3个单元,也就是Unit 4、Unit 5、Unit 6的总分要达到270分以上。


A-Level生物 Unit 1 试卷分析

1. Risk factors of CVD and experiment analysis

(a) Describe the diagram

(i) General Trend: Smokers>non-smokers

The risk of CVD increases as age increase,

Detailed description: smokers will increase the Risk of CVD especially after 45, by 2.6%

(ii) Other Risk factors:

Obesity / unhealthy diet and lack of excercise can raise the risk for non smokers and non-smokers

Exercise and healthy diet can reduce the risk

(b) Calculation

2. Genetic analysis

(a) Genetic diagram Punnett square or pedigree


(a) Function of thromboplastin/fibrin/thrombin and their solubility

Thromboplastin: catalyze the reaction of Prothrombin to thrombin, soluble

Thrombin: catalyse the reaction from fibrinogen to fibrin, soluble

Fibrin: form a mesh, trap the platelets and red blood cells to form a clot, stop the blood flow, insoluble

(b) Sex-related inheritance

The gene is located on sex chromosome X

The dominant allele code for normal proteins and recessive allele lead to blood clotting disorder

 When the female individual has the disorder, she must a homozygote for recessive allele, XaXa

XAXa is not affected

Male have only one X chromosome, therefore the heterozygote of Xa will directly lead to the  disorder(XaY)

4. Carbohydrate and experimental design

(i) B

(ii) bond to join monosaccharide to disaccharide: glycosidic bond

(iii) describe the result

General trend+ extreme value + use of figure

(iv) experimental design :

Control the variables (EX. the concentration )+ digitalize the sweetness( use a figure from 0-5 to record the sweetness)+wide sample( have more than 100 volunteer)+ proper intervals between tasting 2 samples( prevent the lack of sensitivity)

(c) Similarity and difference between amylose and amylopectin


D:Branches/bond/terminals/rate of hydrolysis

5. Transport across the membrane

Active transport/Facilitated diffusion

Similarity:Use protein, transport polar molecules

Difference:Direction and the energy use

(a) the difference of active transport and Diffusion

L: diffusion K:Active transport

(i)Temperature can change the Rate of diffusion

(C) (i)Fluidity of membrane and phospholipid bilayer

(ii) Surface area to volume ratio and the rate of diffusion

6. DNA and genetic code

(a) Check the code table and line the amino acid according to the code

(b) Substitution/deletion/addition

(ii)Degenerated code+ meaningful mutation: change the Amino Acid chain/stop the chain early——change the primary structure of protein

7. Blood pressure and artery structure

(a) (i) calculation of the difference of Diastolic pressure and Systolic pressure

(ii)Calculation of MAP :pay attention to the indication of elastic artery

(iii) Artery pressure too low—— lead to the low force and the low pressure in artery—— slower blood flow—— rate of Gas exchange(O2)/CO2) ——Tissue exhauseted for anaerobic respiration

(b) (i) Calculation with reduction of a fraction

(i) C

(ii) Smooth endothelium to reduce the friction

(c) (i) Thick wall to withstand the pressure and elastic fibre to stretch and recoil

(ii) CVD and the elastic fibre_atherosclerosis

8. The amino acid cycle

(a) The structure of amino acid

(b) Specificity of enzyme

(c) (i) According to the cycle , citrulline will be less, ornithine will be excessive, and carbamyl phosphate will be excessive

(ii) the vesicle fuse with then cell and transmit the mRNA that code for OTC to the cell ,attaching  to the ribosome and transcribed into Enzyme.

A-Level生物 Unit 2 考试分析

1. 主考点:Biological molecules

(1) Polysaccharide in plant: cellulose+starch

(2) Amyloplast:Cell organism that store the starch

Comparing and  contrast cellulose/starch

2. 主考点:Gametes and genetic material

(a) Calculation + transform the figure into standard form + retain the correct significant figures : (b) genetic material of gametes and body cells

3. 主考点:Cell structure and orgaelles

(a) (i) cell structure in a light microscope

(ii) nucleus in the electron microscope

(b) magnification and resolution


(ii)Calculation on magnification

Image size(in length)/ actual size(in length)=magnification

(iii) the function of centrioles

Release spindle fibres, forming spindle

Spindle fibre consists of microtubules, so does centrioles (microtubule organizing center)

Replicate at interphase and pull apart at (late) prophase

4.classification + fertilisation in plant (in a research)


(ii) evidence of classification of 3 domains

molecular evidence: DNA electrophoresis and the protein coded. Build a phylogenetic tree based on those evidence


Cell division : for archaea, the binary fusion is controled by cell cycle

Archaea and eukaryotes are more closely related

Been supported by former research and been repeated verifiedby other scientists of the science community

(c) Fertilisation in plants

The pollen tube nucleus:controls the growth of the pollen tube ;

controlling the production of {enzymes / protein} ;

enzymes being {produced / released / secreted  

enzymes being used to {digest / break down (tissues of style ;

idea of forming a pathway through style to embryo sac

For sperm to reach the egg cell and fertilaize the egg and the polar bodies

(d) Describe the outcome of a research

Describe Bar chart:

(1) general trend ( negative correlation but not linear )

(2) extreme value

(3) Use data

(ii) Describe the main composition of the middle lamella

The Verapamil will inhibbit the uptake of Ca2+

Pectin + Ca2+→Calcium Pectate in middle lamella

Calcium Pectate bind to cellulose of primary cell wall , stablizing the cell wall

Verapamil will weaken the cell wall by reducing the Calcium Pectate forming

Separate adjacent cells

5. 主考点:Medicine testing and Minerals in plant

(a)Nitrogen in plant:DNA synthesis(base)/Protein synthesis(amino group in amino acid)

(b)secretory protein and involved organelles

Golgi apparatus:

This enzyme is secretory protein

Modify/add(reduce)carbohydrate chain(group)to protein/enzyme

Transport protein (enzyme)/form/vesicle to Membrane

(c)Drug test :

Test on healthy animal(rabbits or mice)/animal infected with the diseases

Phase 1 Test on healthy Volunteers : check on The Harmful side effects and safe dosage for human

Phase 2 : Test on a small group of healthy volunteers or volunteers with disease, covering  different gender and age: check on the effectiveness

Phase 3 : Test on a small group of healthy volunteers or volunteers with disease, covering  different gender and age: check on the effectiveness

6. Evolution and adaption

(a) The definition of endemism(B)

(b) The difference of anatomical,physiological and behavioral adaption

Anatomical/physiological /anatomical adaption

(c) Discuss the advantage of adaption according to the given material

Repel the predator by producing poison

threaten the predator by bright color and spot on the skin

Adhesive disc on toes to facilitate moving and climbing

(d) The natural selection and Hardy-Weinbergg equasion

With the selection pressure(predator hunting)

New mutation coming out

Individuals with advanced Mutation (producing poison)survive/Individuals without this advantage are hunted

Individuals with advanced Mutation have the opportunity to give rise to offsprings

The gene pool changes , the frequency of the mutated gene increases, and that of gene without mutation decrease

More frog can produce poison

7. 主考点:Plant based material

(a) Properties of Lignin

(i)Waterproof give tensile strength

(ii)Similarities and differences of sclerenchyma and xylem


Dead cell without cytoplasm

Secondary cell wall with elastic fibre


Sclerenchyma: thicker wall, support the stems and leaves

Xylem: supply water /mineral ions and support the leaves and stems

(b) Sustainabity

Renewable material can be can be regrown in short time

Can be broken down biologically

Cause no pollution


(i) Calculation based on a constant increasing rate

(ii) Describe and analyze the given diagram

Possible outcome


Those plants can be regrown fast, renewble

After being burned, only CO2 is released, causes no pollution

And the biofuel can be biologically broken down

Some biofuel can release less CO2 producing the same amount of energy( hemp:12)


Biofuel (especially biodiesel) can release more CO2

CO2 release: Soybean(49)>Sugar beet(40)> rapeseed(37)

The production of the material can consume water , fertilisers, pesticides

Fertiliser and pesticide are not environmentally

Can cause pollution to the soil and water

Water: producing Biodiesel can use more water Rapseed, soybean > hemp and sugar beet

Fertilizer:Sugar beet>Rapeseed>soybean >hemp

Pestiside:Sugar beet,Rapeseed,soybean>hemp(with figure)

Conclusion:hemp consumes less resource to produce,release less CO2 when burning

           Soybean consumes more resource to produce, release more CO2

8. 主考点:Gene expression

(a) Structure of the sperm

C (not asking where the acrosomal enzyme stores)

(b) Embryo development

Morula( formed after cell division of zygote, not differentiated)

(c) Differentiation

(i) Epigenetic modification

Histone methylation and acetylation/DNA methylation and demethylation/RNA splicing/transcriptional factors

(ii) Switching on/off the gene

Through Histone/DNA methylation,some gene is switched off/active,and can not be expressed。

Through Histone acetylation/DNA demethylation, some gene is switched on/inactive, and can be expressed.

Transcription factors regulation:bind to the enhancer can increase the rate of expression

Transcription and translation into protein (enzyme) can be regulated

Cell can have different structure,can carry out different function(e.g. red blood cell :haemoglobin,carry oxygen)

Permanently modifies cell

The effect of environment on gene expression

Alleles : C,Cch,Ch,ca

C is dominant to Cch, ca

Ch is dominant to  ca

 C code for normal tyrosinase, can catalyse the forming of melanin

Ch  is a recessive mutation of C,  code for a mutation of tyrosinase, which can only be active in low temperature

Skin on nose , ears legs and claws are in low temperature because they emit the heat by producing liquid and increase of surface

The a mutation of tyrosinase is active in these area,  so the skin can produce the melanin, the fur is black, in other area, the tyrosinase denatures, the fur is white.

ca code for inactive tyrosinase, can not catalyse the formation of melanin

The genotype and environment can both determine the phenotype(the fur color)



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