果蝇和哺乳动物的性别决定是都是XY型

Questions 85-92. Sex determination in fruit flies and mammals are both XY type, that is, XX leads to female and XY leads to male.
果蝇和哺乳动物的性别决定是都是XY型,即XX产生雌性,XY产生雄性
85. Some organisms have abnormal sex chromosomes such as XO (only have one X chromosome) or XXY (extra X chromosome). The most likely cause of the abnormal sex chromosome is: (1 point)
有些生物具有不正常的性染色体,如XO (只有一个X染色体)或XXY (额外的X染色体).不正常性染色体产生的原因是:
Error occurred in mitosis of fertilized egg.
受精卵的有丝分裂出现错误
Gene mutation
基因突变
Error occurred in meiosis in gamete formation. 在配子形成过程中减数分裂出现错误
Sex chromosomes in gametes are either lost or doubled in fertilization. 配子中的性染色体在受精过程中丢失或加倍
86. In organisms with XXY chromosome type, there is an extra X chromosome. How do you determine if this X chromosome is from sperm or egg practically (1 point)
在XXY型的生物体中,有一个额外的X染色体.你如何判定这个X染色体是来自精子还是卵子
Karyotype 核型分析
In situ hybridization 原位杂交
RFLP (restriction fragment length polymorphism) 多形性限制片段分析
DNA sequencing DNA定序
87. In mammals, XO leads to female and XXY leads to male. In fruit flies, XO leads to male and XXY leads to female. Which of the following is NOT correct (1 point)
在哺乳动物中,XO产生雌性,XXY产生雄性.在果蝇中,XO产生雄性,XXY产生雌性.下列哪个是不正确的 (1分)
Y chromosome in mammals is necessary for formation of a male organism.
哺乳动物的Y染色体是形成雄性动物所必需
The Y chromosome in mammals is required for sex organ to develop.
哺乳动物中的Y染色体是性器官发育所必需的
The Y chromosome in fruit flies is not functional.
果蝇中的Y染色体是无功能的
Number of X chromosome in fruit flies has impact on sex determination.
果蝇中的X染色体数目可影响性别决定
88. In mammals with abnormal sex chromosomes, the number of individuals with XO chromosome type is far fewer than the number of individuals with XXY chromosome type. It is therefore predicted that: (1 point)
在具有不正常性染色体的哺乳动物中,具有XO型染色体的个体数目远少於具有XXY型的个体数.因此可推测:
The individuals with XO chromosome are less capable of surviving than that with XXY chromosome.
与具有XXY染色体的个体相比,具有XO染色体的个体更不易存活
The individuals with XO chromosome are less capable of reproducing than that with XXY chromosome.
与具有XXY染色体的个体相比,具有XO染色体的个体更不易繁殖
The difference is related to gender of the individuals (XO leads to female and XXY leads to male).
差异与个体的性别相关(XO产生雌性,XXY产生雄性)
None of the above. 以上皆非
89. In both fruit flies and mammals, XX leads to female and XY leads to male. The gene products encoded by two X chromosomes of female individuals are nearly identical to those encoded by one X chromosome of male individuals. This is accomplished by gene dosage compensation. In mammals, it is accomplished by converting one X chromosome into Barr body (X inactivation). Which of the following about Barr body is/are correct (1 point)
在果蝇与哺乳动物中XX产生雌性,XY产生雄性.雌性两条X染色体编码的基因产物,与雄性一条X染色体编码的基因产物基本相同.这是通过基因剂量补偿实现的.在哺乳动物中,这是通过将一条X染色体转换成巴尔氏体(X失活)实现的.下列有关巴尔氏体的叙述,哪个(些)正确
Only normal female individuals have Barr bodies. 只有正常的雌性个体具有巴尔氏体
Only normal male individuals don't have Barr bodies. 只有正常的雄性个体不具有巴尔氏体
Barr body can always be used to determine sex of human beings. 巴尔氏体总是能用来确定人类的性别
The maximum number of Barr body is one 巴尔氏体最多只有一个
The number of Barr bodies equals the number of X chromosomes minus one.
巴尔氏体的数目等於X染色体的数目减一
A. 1, 3, 5
B. 2, 5
C. 4
D. 5
E. 1, 4, 5
90. No Barr body can be observed in normal female fruit flies because (1 point)
在正常的雌性果蝇中,为何观察不到巴尔体
X chromosome of fruit flies is too small
果蝇的X染色体太小
There is no mechanism of dosage compensation in fruit flies
在果蝇中没有剂量补偿的机制
There is no X inactivation in fruit flies
在果蝇中没有X失活
Heterochromatin is difficult to detect in fruit flies.
在果蝇中很难检测到异染色质
91. The fur color of cats is determined by genes on X chromosome. XA is the dominant allele for orange fur, while Xa is the recessive allele for black fur. Which of the following is true about the fur color of the offsprings from a XAXa female cat and XAYmale cat (1 point)
猫的皮毛颜色是由X染色体上的基因决定.XA是橙色皮毛的显性等位基因,而Xa是黑色皮毛的隐性等位基因.下列有关XAXa雌猫和XAY雄猫的后代皮毛颜色的0述,何者正确
They are all orange
它们都是橙色
All the female are orange and half the male are orange
所有雌猫都是橙色,雄猫有一半是橙色
Regardless of sex, half are orange, the other half have furs that are mosaic of orange and black.
若不论性别,有一半是橙色,另一半的皮毛则是橙色与黑色相互镶嵌
Those with mosaic furs are all female.
所有具有镶嵌皮毛颜色的全是雌猫
92. One of the genes controlling sweat gland in human is located on X chromosome. Two twin sisters show different phenotypes of the sweat gland. One has no sweat gland on her left arm while the other has on her left arm. Which of the following statements is/are true (1 point)
控制人类汗腺的一个基因位於X染色体上.一对双胞胎姐妹具有不同汗腺表型.一个在左臂上没有汗腺,另一个左臂上有汗腺.下列0述哪个(些)正确
The twin cannot be identical twin. 这对双胞胎不是同卵双胞胎
They both are heterozygotic of the gene. 她们都是此基因的异基因型
The reason for the different phenotype is random X inactivation.
不同表型的原因是随机的X失活
X inactivation must occur after first division of the zygote.
X失活一定在受精卵第一次分裂后发生
1, 2, 3, 4
1
2, 3
3
2, 3, 4
93. Mycorrhizae are symbiotic associations of fungi and plant roots.Which of the following is/are true about mycorrhizae (1point)
菌根是真菌和植物根的共生体.下列有关菌根的0述,哪个(些)正确
They are often harmful to plant roots while beneficial to fungi.
它们通常对植物的根有害,而对真菌有益
They are often beneficial to plants but harmful to fungi
它们通常对植物有益,对真菌有害
They are helpful for plants to absorb water and minerals.
它们帮助植物吸收水和矿物质
They could even help the older root region above the root hair area to supply minerals to plants.
它们能帮助根毛区上方较老根区为植物提供矿物质
A. 1, 3, 4
B. 2, 3, 4
C. 3, 4
D. 3
94. Stomata of a plant open when guard cells (1point)
当保卫细胞怎麽样时,植物的气孔开放
accumulate water by active transport.
通过主动运输聚集水
B. sense an increase in CO2 in the air spaces of the leaf.
感受到叶的气隙中CO2的含量增加
C. become more turgid because of an influx of K+, followed by the osmotic entry of water.
变得更加肿胀,因为K+流入,随后水渗透进入
D. sense that water content of whole plant is low.
感到整株植物水含量低时
95. Which of the following processes of plants could be regulated by phytochrome (1point)
植物下列的过程中,何者能受光敏色素的调节
(1) seed germination 种子萌发
(2) flowering 开花
(3) shoot elongation 茎的延伸
(4) open and closure of stomata 气孔的开闭
A. 1, 2, 3, 4
B. 1, 2, 3
C. 1, 2
D. 1
96. If N represents population size, r represents the difference in per capita birth rates and death rates, K represents the carrying capacity, t represents time, which of the following equations best describes logistic growth of the population (1point)
如果N代表人口数,r代表人均出生率和死亡率的差异,K代表最大负荷量,t代表时间,下列何者最能描述人口的逻辑增长
D
97. Which of the following is usually the limiting proce
98. Which of the following ecosystems has the lowest primary production per square meter (1point)
下列哪个生态系统中具有最低的每平方米初级生产量
A. a salt marsh 一块盐沼
B. an open ocean 一片开阔的海洋
C. a grassland 一片草原
D. a tropical rain forest 一个热带雨林
99. Which of the following is/are true about Archaea and Eubacteria (1point)
下列有关古细菌与真细菌的0述,哪个(些)正确
They don't have nuclear envelope
它们不具有核膜
They both have branched chain in membrane lipids
在它们的膜脂质中都具有分支链
They have one kind of RNA polymerase
它们具有一种RNA聚合酶
They have circular chromosomes.
它们具有环状的染色体
A. 1, 2, 4
B. 1, 4
C. 2, 3
D. 1, 2, 3
100. Four major groups of fungi are recognized. They are chytrids, zygote fungi, sac
101. Chlorophyll a is involved in both light energy absorption and initial electron transfer of photosynthesis. Which of the following are true about the chlorophyll a (1point)
叶绿素a参与光能的吸收及光合作用的起始电子传递.下列有关叶绿素a的0述,哪些正确 (1分)
(1) The position of chlorophyll a in photosystems has a strong influence on the function of chlorophyll a.
叶绿素a在光合系统中的位置,对叶绿素a的功能有极大影响
(2) Chlorophyll a in photosynthetic reaction center is chemically modified so that it performs initial electron transfer.
叶绿素a在光合作用反应中心经化学修饰,以便使它进行起始电子传递
(3) Part of chlorophyll a is structurally related to haeme group of haemoglobin.
叶绿素a的部分结构与血红素中的血质基相关
(4) Part of chlorophyll a is structurally related to carotenoids. 叶绿素a的部分结构与类胡萝卜素相关
A. 1, 2, 3, 4
B. 1, 3, 4
C. 3, 4
D. 1, 2
102. In measurement of photosynthetic electron transfer, intact chloroplasts are isolated and used for electron transfer rates under different conditions. Which of the following is correct (1point)
在测定光合作用电子传递时,叶绿体可被完整的分离出来,用以测定不同环境下的电子传递速率.下列叙述何者正确
A. Addition of an uncoupler leads to an increased rate of electron transfer. 加入解偶联剂,导致电子传递速率增加
B. Cyclic electron transfer starts only when linear electron transfer is inhibited. 只有当线性电子传递被抑制时,循环电子传递才开始
C. ATP synthesis could only be observed with continuous light illumination. 只有连续性光照时,才有ATP的合成
D. Oxygen evolution by chloroplast suspension is absolutely dependent upon the presence of CO2.
叶绿素悬浮液的放氧速率完全依赖CO2的存在
103. The figure shown below is a diagram of evolutionary tree. Which of the following statements about evolution are true and deducible from the figure (2 points)
下图所示为一演化树,下列有关演化的0述,哪些正确并可从图中推断 (2分)
All eucaryotic cells contain mitochondria.
所有真核细胞都含有线粒体
(2) Symbiosis of the eucaryotic ancestor with autotrophic cell precede the symbiosis with the cell taking advantage of the oxidative metabolism.
真核细胞祖先与自营细胞的共生,早於其与利用氧化代谢细胞的共生
(3) There is common ancestor of eubacteria and eukaryota, archaebacteria are a group with unique and independent origin.
真细菌与真核生物具有共同的祖先,古细菌是一类具有独特的,独立起源的生物群
(4) Ancestral eukaryote was anaerobic.
真核细胞祖先是厌氧的
(5) None of the recent photosynthetic bacteria are related to the chloroplasts.
最近的光合细菌中,没有一个与叶绿体相关
(6) Mitochondria and chloroplasts has similar genomes. 粒线体与叶绿体具有相似的基因组
(7) Mitochondria are present in the cells of the plants, animals and fungi.
植物,动物及真菌的细胞具有线粒体
(8) Fungi lost chloroplasts during evolution.
在进化过程中,真菌丢失了叶绿体
(9) Bacteria are highly homogenous group of organisms which quickly diversified genomes and metabolisms during the last billion years.
细菌是同源高的一群生物,其基因组及代谢途径在近十亿年中发生很快的分化
(10) Chloroplasts and mitochondria are results of independent endosymbiotic events.
叶绿体和粒线体是两次独立内共生事件的结果
A. 1, 2, 5
B. 3, 4, 7
C. 4, 7, 10
D. 6, 8, 10
E. 4, 9, 10
104. The figure shown below is an imag
105. Siamese cat is an example of the animals with melanin synthesized in both sexes mostly at the body extremities. That makes snout, ears, tail and feet much darker than the rest of the body. Explanation of this type of the body coloration is that: (1 point)
暹罗猫是一个很好的动物例子,在雌,雄猫的身体末端处都合成黑色素.所以在它们的鼻子,耳朵,尾巴和足部比身体的其他部分更黑.下列何项叙述可以说明这种类型身体著色的现象 (1分)
Only at the body extremities the enzyme tyrosinase (responsible for the synthesis of the melanin) is synthesized.
酪氨酸酶(负责黑色素的合成)只在身体末端处合成.
The only places where one of the X chromosomes that have dominant allele of the tyrosinase is NOT inactivated
只发生在一条具有酪氨酸酶显性等位基因的X染色体不失活的部位
Melanin is synthesized only in the colder parts of the body because Siamese cat bears temperature sensitive allele for enzyme producing melanin.
因为暹罗猫产生黑色素的酵素之等位基因为温度敏感性,所以黑色素只在身体较冷的部位合成.
Melanocytes are localized only at the snout, ears, tail and feet 0 the rest of body is without melanocytes.
黑色素细胞只位於鼻子,耳朵,尾巴和足部,身体的其他部分则没有黑色素细胞
The body extremities are more exposed to the UV-radiation which stimulates production of the melanin.
由於身体末端暴露於紫外线照射较多,因而激发黑色素的产生
106. Retinoblastoma protein (Rbp) and p53 are examples of the anti-oncogens (tumor suppressor). Which of the following statements about these proteins is true (1 point)
视网膜胚细胞瘤蛋白(Rbp)和p53是抗致癌基因(即瘤抑制子).下列0述,何者正确 (1分)
A. Mutation in the p53 gene (when p53 lost its regulatory function) can stop the cell cycle.
p53基因发生突变(使p53失去它的调控功能),会使细胞周期停止
B. Overproduction of the Rbp in the retina can cause cancer.
在视网膜中产生过量的Rbp,会产生癌症
C. Cells with mutated p53 are predisposed to malignancy.
p53发生突变的细胞倾向成为恶性肿瘤
D. Cells with mutated Rbp are resistant to malignancy.
Rbp突变的细胞抵抗恶性肿瘤的产生
E. Various viruses incorporated homologs of the p53 and Rb into their genomes and use these proteins for the transformation of the host cell.
许多病毒会将p53和Rb的同源基因插入它们的基因组后,他们会表现这些蛋白,使宿主细胞发生转形
107. Extracellular matrix is responsible for the mechanoelastical properties of the tissues. Which of the following molecules is NOT a component of the extracellular matrix is: (1 point)
细胞外基质负责组织的机械及弹性(mechanoelastical)性质.下列何者不是细胞外基质的组成分子
A. elastin 弹性蛋白
B. cytokeratin 细胞角质蛋白
C. laminin 层粘连蛋白
D. collagen 胶原蛋白
E. chondroitin sulphate 硫酸软骨素
108. Prions are unique infectious agents formed only from protein called PrP. W
109. Algae were supplied with a radioactive isotope of Carbon, 14C, and allowed to photosynthesis. After a period of time, the light was switched off and the algae were left in the dark. The graph shows the relative amount of some radioactive labelled compounds over the period of the experiment. (1 point)
提供藻放射性碳同位素14C后,进行光合作用.一段时间后,关闭灯光,并将藻置於黑暗中.下列曲线图显示在实验期间,一些放射性标记的化合物的相对量. (1分)
Which line is representing the amount of glycerate 3-phosphate (3GP), ribulose biphosphate (RuBP) and sucrose (1 point)
Fill out the correct number of the line in the correct box.
哪条曲线代表甘油酸-3-磷酸(G-3-P),二磷酸核酮糖(RuBP)和蔗糖 请在下列表格中,填写正确线条的字母.(1分)

Compound
化合物
Line
曲线
(1) 3GP
A
(2) RuBP
C
(3) Sucrose
蔗糖
B
110. Methylene blue acts as a hydrogen acceptor. It is blue in oxidised state, but goes colourless when it is reduced by accepting hydrogen atoms. (1 point)
亚甲基蓝作为一个氢受体.在氧化状态下,它是蓝色的;当它接受氢原子而被还原时,它会变成无色.(1分)
Methylene blue + hydrogen reduced methylene blue
(blue) (colorless)
亚甲基蓝(蓝色)+氢 ( 还原型亚甲基蓝(无色)
A student likes to investigate this reaction. He prepares four test tubes as shown below
一个学生喜欢研究这个反应.他准备了四个试管,内含各种反应物质,如下表所示.
Tube A
Tube B
Tube C
Tube D
Distilled water
蒸馏水
-
2 ml
2 ml
2 ml
Glucose solution
葡萄糖溶液
2 ml
2 ml
-
2 ml
Methylene blue solution
亚甲基蓝溶液
1 ml
1 ml
1 ml
-
Yeast solution
酵母溶液
2 ml
-
2 ml
2 ml
All tubes were incubated at a temperature of 30 °C. The colour was recorded at the start and after intervals of 5 and 15 minutes. The results are shown in the table.
所有的试管都培养在30 °C.分别在培养后 5分钟和15分钟时纪录颜色.结果如下表所示.
Colour of content
溶液的颜色
Tube A
Tube B
Tube C
Tube D
At start
在开始时
Blue
蓝
Blue
蓝
Blue
蓝
Colourless
无色
After 5 minutes
5分钟后
Colourless 无色
Blue
蓝
Blue
蓝
Colourless
无色
After 15 minutes
15分钟后
Colourless
无色
Blue
蓝
Pale blue 浅蓝
Colourless
无色
Which test tube can be characterized as a control in this investigation and which test tube is useless (1 point)
Fill out the correct letter
哪个试管在这个实验中可以作为对照组,哪个试管是没用的 在下列表格中,填写正确的字母(1分)
Tube
(1) Control 对照
B
(2) Useless 无用的
D
111. Morgan crossed Drosophila of two known genotypes, BbVv x bbvv, where B, the wild-type (grey) body, is dominant over b (black body) and V (wild-type wing) is dominant over v (vestigial, a very small wing). Morgan expected to see four phenotypes in a ratio 1:1:1:1. But he observed:
Morgan将两种已知基因型的果蝇进行杂交,即BbVv x bbvv;其中,B是野生型(灰色身体),相对於b(黑色身体)是显性的;V(野生型翅膀)相对於v(发育不全,一个非常小的翅膀)是显性的.莫干(Morgan)期待看到四种表型的比例是1:1:1:1.但是他看到的如下:
Wild type 野生型: 965
Black vestigial 黑色,发育不全: 944
Grey vestigial 灰色,发育不全: 206
Black normal 黑色,正常: 185
These results were explained in assuming linkage of alleles together with genetic recombination (crossing over).
这些结果可以用假定的等位基因的连锁和遗传重组(互换)来解释.
In this particular example the recombinant frequency (defined as the ratio of recombinants in relation to the total offspring) is: (1 point)
这些结果可以用假定的等位基因的连锁和遗传重组(互换)来解释.
在这个特例中,互换率(定义为重组子代相较於全部后代的比值)是多少
A. 0.205
B. 0.170
C. 0.108
D. 0.900
E. 0.080
112. 70% of the population of Beijing is able to taste phenylthiocarbamide. The ability to taste (T, taster) is dominant over the inability to taste (t, non-taster).
What percentage of the offspring of 'tasters' will be non-tasters (2 points)
北京70%的人口能够辨别苯基硫脲的味道.辨味的能力(T,taster)相对於不能辨味者(t,non-taster)是显性的.试问有辨味能力者(taster)的后代中为不能辨味(non-taster)的比例是多少 (2分)
25%
15%
13%
20%
7.5%
Questions 113-116. Wild type individuals of Drosophila have red eyes and straw-coloured bodies. A recessive allele of a single gene in Drosophila causes glass eye and a recessive allele of a different gene causes ebony body.
果蝇的野生型个体具有红色的眼睛和淡黄色的身体.果蝇中一个单基因的隐性等位基因会产生玻璃眼(glass eye)的表型,而另一个基因的隐性等位基因则产生乌黑身体(ebony body)的表型.
A student crosses pure breeding wild type flies with pure breeding flies having glass eye and ebony body and the resulting F1 flies showed all the wild type phenotype for both features. On crossing the F1 flies among themselves the student expect a 9:3:3:1 ratio but the results are not like that. The actual offspring showed:
一个学生将纯品系的野生型果蝇与另一个具有玻璃眼和乌黑身体的纯品系果蝇进行杂交,产生的第一代(F1)果蝇全部是野生型的表型.以第一代(F1)果蝇进行杂交,此学生期待一个9:3:3:1的比例;然而结果并不是这样.后代实际的表现情形如下表所示:
Eye
眼
Body
身体
Number of flies in F2
F2代中果蝇的数目
Wild
野生型
Wild
野生型
164
Wild
野生型
Ebony
乌黑
37
Glass
玻璃
Wild
野生型
59
Glass
玻璃
Ebony
乌黑
28
There are two possibilities:
- The differences from 9:3:3:1 are coincidental (null hypothesis accepted).
The differences do not occur by coincidence (null hypothesis rejected).
有两种可能性:
(1) 比例不同於9:3:3:1是巧合(接受虚无假设).
(2) 比例的不同并不是巧合发生的(反对虚无假设).
You are required to check this applying the 2 (chi square) test.
你需要应用卡方 (chi 的平方) 分析法进行核对.
For this situation, e.g. degree of freedom, the following diagram with 2 values should be used: 在这种情况下,比如自由度,你应该用下面2数值的图:
Question 113. The calculated 2 is (3 point)
计算得到的2是多少 (3分)
A. 10.11
B. 2.84
C. 14.33
D. 11.40
Question 114. Indicate the degree of freedom (df) for this test: (1 point)
指出该实验的自由度(df)是多少 (1分)
A. 2
B. 3
C. 4

Question 115. Determine the probability that the deviation of the observed results from expected results is due to chance alone. (1 point)
决定观察的结果偏离预期的结果是起因於巧合的可能性是多少
A. About 1%
B. About 2%
C. About 5%
D. About 8%
Question 116. To explain the observed deviation of the 9:3:3:1 ratio the student suggested some possibilities.
由於观察到的结果偏离9:3:3:1比率,该学生提出了几种可能的解释
(1) linkage of both the alleles两个等位基因连锁
(2) crossing over 互换
(3) incomplete dominance不完全显性
Which combination of suggestions is the correct explanation (1 point)
下列何者是正确的解释
A. 1, 2
B. 1, 3
C. 2, 3
D. 1, 2, 3
117. Which of the following diagram offers a correct representation of the urea urine of a person in hunger strike, who finally died. (1 point)
因饥荒而最终饿死的人,其尿中尿素含量的变化曲线图为下列何者 G
118. Wilhelm von Osten gave performances with his horse called smart Hans. He stated that he taught his horse to make calculations. But in fact this isn't true at all. He had taught the horse to mind his hidden but trigging indications. As a result the horse made the desired movements: swinging the correct number of times with his foreleg. After that the horse got some reward.
Wilhelm von Osten宣称他的马匹smart Hans经过训练后可运算数学,事实上并不正确,他只是教会他的马匹辨识他隐藏起来的讯号,当正确答案出现时,马匹便会挥动前脚并获得奖赏.
What kind of learning behaviour is this (1 point)
这属於何种学习行为
A. adaptation 适应
B. conditioning 条件化作用
C. habituation 习惯化作用
D. imitation 初始作用
E. imprinting 印痕
F. insight 顿悟
G. Fixed action pattern(FAP)固定动作模式
119. A snail crawling across a board will withdraw into its shell when you drop a marble on the board. Repetition of dropping marble will lead to weaker withdraw action and in the end the snail will ignore the marble dropping. Which of the following terms do apply for the disappearance of the withdraw action (1 point)
在爬行中的蜗牛身旁投下一颗小石头,会促使蜗牛把身体缩进壳中,重复投下石头,其收缩反应会越来越小,最后甚至会完全忽略不再反应.下列何者可解释此种现象
Adaptation 适应
Conditioning 条件化
Habituation 习惯化作用
Imprinting 印痕
Insight 顿悟
learned behaviour 学习行为
1
2,4
3,6
4, 5
5, 6
120. Bonsai trees need water with very low lime content. Which types of water could be used to water them (1 point)
Bonsai树需要含有少量石灰的水分方可存活,下列何者是合用於灌溉
Carbonated mineral water 碳酸矿泉水
Rain water 雨水
Tap water with high water hardness 硬水
Tap water with high water hardness treated by leaving it over night with a mix of peat and crushed stones and filtrating it before use 放置过夜再利用泥碳过滤后的硬水
Molten snow 溶雪
1, 5
2, 5
1, 3
4, 5
E. 2, 4, 5
121. Observe the diagrams 1 to 4 representing cross sections of the ovaries of different flowers.
图1到图4,描绘了不同花的子房横截面.
Match the numbers in front of the placentation type (A-D) with the corresponding diagram.
请将胎座式类型(A-D)与相应的图匹配起来.
Axile placentation. 中轴胎座
B. free central placentation. 自由中枢胎座式,或称中央独立胎座
C. Marginal placentation. 边缘胎座式,即边缘胎座.
D. Parietal placentation. 侧膜胎座
Match the number with correct placenta type. (1 point)
将图上编码和正确的胎座式类型(A-D)匹配起来(1分)
Type 类型
Answer答案
1
B
2
A
3
D
4
C
122. Which curve shows the correct time course of the production of saliva in a human after the intake of citric acid (1 point)
以下哪一条曲线正确的表示了人服用柠檬酸后,唾液生成的时间过程 B
B
Questions 123-125. The behavior of eight Humboldt penguins (Spheniscus humboldti) is investigated in a larger group of penguins in a zoo enclosure. The animals can be distinguished by marks or their individual pattern of black dots on their white thorax. To document the relationship of the penguins, their nearest neighbor (closest animal in the enclosure) was recorded in short time intervals during day time in a period of several weeks. The table shows the relatively stable mean values for the frequency of neighbors for the four male (M1 - M4) and four female (F1 - F4) penguins.
某人调查某动物园围栏0的一大群洪保德企鹅中八只企鹅的行为,利用戴标记或根据各企鹅白胸上黑点的不同样式来辨识.为了证明企鹅彼此的关系,在长达数周的时间中,於白天时段多次记录(各有短时间间隔)最靠近它们的企鹅.下表为四只雄企鹅(M1 0 M4)和四只雌企鹅(F1 0 F4)邻居的相对频率平均值.
M1
M2
M3
M4
F1
F2
F3
F4
M1
2
5
1
0
3
7
77
95
M2
2
0
9
9
75
1
2
98
M3
5
0
0
0
0
78
6
89
M4
1
9
0
80
8
0
0
98
F1
0
9
0
80
7
0
0
96
F2
3
75
0
8
7
0
0
93
F3
7
1
78
0
0
0
7
93
F4
77
2
6
0
0
0
7
92
95
98
89
98
96
93
93
92
Several months later the same animals were observed again yielding the following values.
几个月后同样又观察这些相同的动物,得出了下面的数值.
M1
M2
M3
M4
F1
F2
F3
F4
M1
4
8
2
1
4
11
60
90
M2
4
0
12
12
65
1
5
99
M3
8
0
0
0
1
62
9
80
M4
2
12
0
70
14
0
1
99
F1
1
12
0
70
10
0
1
94
F2
4
65
1
14
10
0
3
97
F3
11
1
62
0
0
0
10
84
F4
60
5
9
1
1
3
10
89
90
99
80
99
94
97
84
89
During the following years the tendency of these values remained the same.
在以后的几年0这些数值的趋势保持相同.
123. Analyze the tables and determine the mating system of the Humbodt penguins. (1 point) 分析以上表格,以确定洪保德企鹅的配对系统是
A. promiscuity 乱交的
B. polyandry 一雌多雄的
C. polygyny 一雄多雌的
D. monogamy 一雌一雄的
124. Which is the most common polygamous relationship (1 point)
哪一个是最常见的多配偶关系
A. promiscuity 乱交的
B. polyandry 一雌多雄的
C. polygyny 一雄多雌的
D. monogamy 一雌一雄的
125. Which group of animals do the penguins belong to (1 point)
企鹅属於哪一种类群的动物
Ratitae (birds with flat breast and weak breast muscles)
平胸总目(平胸且胸肌较弱的鸟)
Carinatae (birds with strong breast muscles) 突胸总目(胸肌强壮的鸟)
Neither, they are not birds 都不是,它们不是鸟
126. Substrate(s) of RUBISCO is (are): (1 point) 酵素RUBISCO的受质是:
Phosphoenolpuruvate (PEP) 磷酸烯醇丙酮酸
Ribulose-bis-phosphate (RuBP) 二磷酸核酮糖
Oxaloacetic acid (OAA) 草醋酸
Phosphoglyceric acid (PGA) 磷酸甘油酸
Carbon dioxide (CO2) 二氧化碳
Phosphoglyceraldehyde (GAP) 磷酸甘油醛
Oxygen (O2) 氧气
1, 2, 5
1, 5
2, 5
1, 2, 6
2, 5, 7
127. The diagram shows a section through a mammalian ovary. The numbers indicate different develop stages. (1 point)
下列为哺乳类卵巢构造示意图,数字代表不同的发育阶段
Choose the correct sequence of numbers in which the structures develop.
请选择正确的发育过程

A. 1, 2, 3, 4, 5
B. 5, 4, 3, 2, 1
C. 5, 2, 4, 1, 3
D. 5, 2, 4, 3, 1
E. 2, 4, 1, 3, 5
Questions 128-131. PKU and albinism are two autosomal recessive disorders, unlinked in human. If a normal couple produced a boy with both disorder, they want to have the second child:
苯丙酮尿症与白化症为两种隐性染色体遗传疾病,假如一对正常的夫妇生下一个同时具有两种疾病的男孩,他们想再次生育.
128. What is the chance of the second child with PKU (1 point)
第二个小孩得到苯丙酮尿症的机率为何
A.1/2
B. 1/4
C. 2/3
D. 1/16
129. What is the chance of the second child with both traits (1 point)
第二个小孩同时得到苯丙酮尿症和白化症的机率为何
A. 1/2
B. 1/4
C. 1/8
D. 1/16
130. What is the chance of the second child with either PKU or albinism (1point)
第二个小孩得到苯丙酮尿症或白化症中任何一种的机率为何
A. 1/2
B. 3/4
C. 3/8
D. 3/16
131. What is the chance for them to have a normal child (1 point)
第二个小孩为正常的机率为何
A. 1/16
B. 4/9
C. 9/16
D. 6/16
132. How is the trait inherited (1 point)
133. If D= dominant, d = re(1 point)
134. What is the genoty
135. If IV-2 married to a man from an unrelated family,
For the alleles D and d, which individua
137. If this trait is instead quite common in the population,
138. There are several types of human blood cells such as erythrocytes and monocytes. They all come from stem cells. Which of the following is/are correct about the stem cells of blood cells (1 point)
人体具有红血球及单核球等多种血球细胞,他们全来自干细胞,下列有关血球干细胞的叙述何者正确
(1) B cells come from lymphoid stem cells. B细胞源自淋巴干细胞
(2) T cells come from lymphoid stem cells. T细胞源自淋巴干细胞
(3) Erythropoietin stimulates production of erythrocytes from myeloid stem cells
红血球生成素可刺激骨髓干细胞产生红血球
(4) Neutrophils and basophils have same stem cells.
嗜中性球和嗜碱性球源自相同干细胞
(5) Lymphoid stem cells come from myeloid stem cells.
淋巴干细胞源自骨髓干细胞
1, 2, 3, 4, 5
1, 2, 3, 4
1, 3
1, 2, 4
139. Which of the following role(s) does platelets play in clotting process (1 point)
下列何者为血小板在凝血过程中所扮演的角色
They help to form plug for protection against blood loss.
协助形成血栓以减少血液丧失
They release chemical signals for fibrin formation.
释放化学讯号促使血纤维蛋白形成
(3) They release chemical signals for reducing blood pressure.
释放化学讯号以降低血压
A. 1, 2
B. 1, 2, 3
C. 2, 3
D. 1, 3
140. Which of the following is NOT involved in allergic response in human (1 point)
下列何者与人体的过敏反应无关
Histamine. 组织胺
Mast cell. 肥大细胞
Plasma cell 浆细胞
Platelets血小板
141. There are several sensory receptors in human skin. Which of the following is located in deepest position of the skin (1 point)
人体皮肤中具有多种受器,下列何种受器位於皮肤最深层的构造中
A. Sensory receptor for pain. 痛觉受器
B. Sensory receptor for cold. 冷觉受器
C. Sensory receptor for heat. 热觉受器
D. Sensory receptor for strong pressure. 强力压觉受器
142. One of the mutant zebra fish has a reduced number of hair cells in neuromast of its lateral line system. Which of the following will happen (1 point)
某一突变种斑马鱼,其测线系统中神经瘤上的毛细胞大量减少,下列何者将会发生
The mutant fish will not be able to detect depth of water.
此变种鱼无法侦测水的深度
The mutant fish swims slowly.
此变种鱼的游泳速率变慢
The mutant fish could not detect sound of its prey.
此变种鱼无法侦测猎物的声音
The mutant fish could be impaired in detecting water
movement around its body.
此变种鱼侦测身旁水流运动的能力受损
A. 1, 2
B. 3, 4
C. 4
D. 2, 4
143. Hemoglobin is responsible for transporting oxygen from lung to tissues. Bohr shift is one of the most important properties of hemoglobin. Which of the following is NOT true about Bohr shift (1 point)
血红素可将氧气由肺运至组织细胞,波耳效应(Bohr shift)是红血球的一项重要特徵,有关波耳效应下列何者错误
Additional oxygen is bound by hemoglobin in lung when pH decreases.
当血液pH值下降时,红血球可在肺部结合更多的氧气
Additional oxygen is released from hemoglobin at a lower pH.
当血液pH值下降时,红血球可释出更多氧气
CO2 is involved in Bohr shift.
二氧化碳与波耳效应有关
Bohr shift helps tissues to obtain more oxygen in exercise.
波耳效应可帮助组织在运动时获得更多氧气
144. Which of the following is/are NOT true about the difference in digestive tracts of carnivores and herbivores (1 point)
下列有关肉食性及草食性动物消化道构造差异的叙述,何者错误
Carnivores usually have a bigger stomach.
肉食性动物常具有较大的胃
Carnivores usually have a shorter colon.
肉食性动物常具有较短的结肠
Herbivores usually have a longer cecum.
草食性动物肠常有较长的盲肠
1, 2
1
2, 3
3
Questions 145-148. Hemophilia and color blindness are X-linked recessive traits. When a color-blind woman married to a hemophiliac man,
血友病和色盲是X-性联遗传隐性性状.当一位色盲的女性和一位患有血友病的男性结婚.
145. What is the chance for them to have a normal son (1 point)
他们生出一个正常儿子的机率为何
50%
0%, all their sons will suffer from color-blind
他们的儿子都将患有色盲
0%, all their sons will suffer from hemophilia
他们的儿子都将患有血友病
It depends on the recombinant frequency.
他们的儿子是否患病完全由重组机率所决定
146. If their son was married to a woman whose mother was color-blind, and whose father was not color-blind what is the chance for them to produce a normal daughter (1 point)
如果他们的儿子和一名其母亲是色盲的女性结婚,该女性的父亲并无色盲,他们生出一名正常的女儿的机率是多少
0%
50%
75%
100%
147. If their daughter was married to a n
148. If they have a color-blind daughter, (1 point)
Questions 149-152. Huntington disease is a rare fatal disease. People with this disease start to show symptoms in their 40's. Peter's father (John) has Huntington disease. John's father (Peter's grandfather), who also had this disease, had 11 children (5 sons and 6 daughters). Among them, 6 (3 sons and 3 daughters) of them had got the disease and five died from it.
亨丁顿症是罕见的致命疾病.患有这种疾病的人在40岁后开始显现病症.彼得的爸爸(约翰)患有亨丁顿病.约翰的爸爸(彼得的爷爷)也有这种病,并且生了11个孩子(5个儿子和6个女儿).他们中有6个人(三个儿子和三个女儿)患有此病,五个人死於此病.
149. How is the trait inherited (1 point) 遗传特徵为何
autosomal recessive体染色体隐性
autosomal dominant体染色体显性
sex-linked recessive性染色体隐性
sex-linked dominant性染色体显性
150. What is the possibility that Peter will also develop the disease (1 point)
彼得患有此病的机率为何
50%
25%
75%
67%
151. Peter is married to a normal woman, what is the possibility that their first child will eventually develop the disease (1 point)
彼得和一个正常的女性结婚,他们的第一个孩子患有此病的机率为何
50%
25%
75%
67%
0
152. If Peter's mother-in-law died from the same disease, what is the possibility that their first child will eventually develop the disease (1 point)
如果彼得的岳母死於同一种病,他们的第一个孩子患有此病的机率为何
3/16
4/16
7/16
9/16
12/16
153. Trophic levels are indicated below with numbered lines in the flowchart. Write the appropriate tropic level name in the space provided next to its number. Write ONLY the letter of the tropic characteristic. (1 point)
下列流程图为不同营养层及能量流向的示意图,各营养层及能量流向以数字标示,请将代表正确叙述的英文代号填入答案栏中,注意第一题无须填写答案
NOTE: Left-hand circle in flowchart is Heat; right-hand circle in flowchart is To detritivores.
流程图中左方圆形代表热能,右方图形代表吃碎屑动物
1. no answer required
2. _C________
3. __A________
4. ___B_______
5. __D________
6. __F________
7. ___E_______
A. energy used in cellular respiration
用於细胞呼吸作用的能量
B. secondary consumers次级消费者
C. tertiary consumers三级消费者
D. energy in wastes耗损掉的能量
E. primary producers初级生产者
F. primary consumers初级消费者
【C A B D F E】
154. Match the biome in the figure below with the appropriate plotted area (a, b, c, d, e, and f) in the climograph. (1 point)请将下图中的生物群落区和气象图中小区块(a, b, c, d, e, 和f)相配对(1分)
1. ___F____ arctic and alpine tundra极地和高山冻原
2. ___E____ coniferous forest针叶林
3. ___A____ desert沙漠
4. ___B____ grassland 草原
5. ____C____ temperate forest 温带森林
6. ___D_____ tropical forest热带森林
横轴:年平均降雨量(cm)
纵轴:年平均气温(℃) 【E F A B C D】
155. Referring to the action potential graph below, write the letter (from the graph) that corresponds with the appropriate action potential action on the right of what is occurring at that stage of the action potential. (1 point) Note, there could be more than one choice for each question.
请看以下动作电位变化图,图右为发生动作电位各个阶段的电位变化,请将图中的字母写到相应的电位变化描述栏.(1分)注意,每一个问题可能有多个答案.
【E DF AE B】
1.___E____ The membrane is unable to respond to any further stimulation regardless of intensity
无论刺激强度多大,膜都不能产生反应
2.__ _DE_____ Sodium gates close, and potassium gates re-open
钠离子通道关闭,钾离子通道门再度开启
3.___AF_ ___ Both sodium and potassium voltage-gated channels are closed钠离子和钾离子电压敏感型通道关闭
4.____B____ Stimulus opening of some sodium channels促使部分钠离子通道开启
156. Molting is a process observed in insects. Which of the following statements is/are true (1 point)
昆虫的蜕皮过程中,下列何者正确 (1分)
The exoskeleton of insects is largely made of protein and chitin.
昆虫的外骨骼大部分是由蛋白质和几丁质组成的.
The structure of chitin is similar to that of bacterial cell wall peptidoglycan.
几丁质的结构和细菌细胞壁肽聚糖的结构相似.
No enzyme has been found to digest chitin.
还没有发现可以消化几丁质的酶.
Molting can be observed in all arthropods.
所有的节肢动物都可以观察到蜕皮.
The only place that is not covered by exoskeleton is the joints between the body and walking legs.
唯一未覆盖著外骨骼处是身体和腿间的连结点.
1, 2, 4, 5
1, 4
1, 3,4, 5
1, 5
157. The mechanism of molting has largely been revealed. The figure below is a diagram of such a process. Boxes A, B and C represent 3 different growth hormones and molting hormones. Fill in the answer boxes by choosing correct letter. (1 point)
下图为蜕皮过程的示意图,方格中A,B与C分别代表三种荷尔蒙或与蜕皮作用有关的荷尔蒙,请於正确叙述旁的答案栏中填入正确的英文代号.

Answer: A-C
brain hormone (BH)
脑激素
A
juvenile hormone (JH)
青春期激素
B
molting hormone MH)
蜕皮激素
C
158. The figure below shows 4 different circulation systems of vertebrates. From left to right, these are the circulation systems of (1 point)
下图为四种脊椎动物循环系统的示意图,由左至右分别为哪些动物的循环系统
mammals, reptiles, amphibians, and fish, respectively.
哺乳类,爬虫类,两生类,鱼类
fish, amphibians, reptiles, and mammals, respectively.
鱼类,两生类,爬虫类,哺乳类
mammals, amphibians, reptiles, and fish, respectively.
哺乳类,两生类,爬虫类,鱼类
mammals, amphibians, fish, and reptiles, respectively.
`哺乳类,两生类,鱼类,爬虫类
159. Match the numbers shown below with correct structures in the figure in question above (question158). (1 point)
请将上图中用英文字母所标示的构造,与下表中的名称进行配对
Questions 160-162. The structure of a mammalian kidney is shown below.
下图为哺乳类肾脏构造的示意图
160. Match the following terms with correct structures shown in the figure. (1 point)
请将上图中用英文字母(A~G)所标示的构造,与下列名称进行配对
①collecting duct 集尿管
②glomerulus 肾小球
③distal tubule 远端肾小管
④Bowman`s capsule 鲍氏囊
⑤proximal tubule近端肾小管
⑥ureter 输尿管
⑦afferent arteriole 入球小动脉
Answer
A-G
①
F
②
B
③
C
④
D
⑤
E
⑥
A
⑦
G
161. The substances that be reabsorbed by proximal tubule is/are: (1 point)
下列哪些物质可由近端肾小管所吸收
(1) Na+
(2) Cl-
(3) Water水
(4) Glucose葡萄糖
(5) Amino acids胺基酸
(6) Urea尿素
1, 2, 3
6,
1, 2, 4, 5,
1, 2, 3, 4, 5
4, 5
162. In the kidney, ultra-filtration occurs in which of the following structures (1 point) 在肾脏中超滤过作用发生在下列哪些构造
Nephrons 肾元
Bowman's capsule 鲍氏囊
Proximal tubule 近端肾小管
Distal tubule 远端肾小管
Collecting duct 集尿管
1, 2
2, 3, 4, 5
3, 4, 5
2, 3, 4
1, 2, 3, 4
Questions 163-166. The sensory transduction by a taste receptor is shown the figure below. Structure A, B, and C different channels, The sequential events of the transduction is labeled by number 1 through number 7. A portion of the cell is magnified.
下图为味觉受器感觉讯息传导的示意图,构造A,B,C分别为三种不同的离子通道,其讯息传导过程依序以数目字1~7标示,部分的细胞构造并已放大.
Structure A which is responsible for event 3 is a (1 point)
与传导过程3有关的A构造为:
Potassium channel 钾离子通道
Calcium channel 钙离子通道
Sodium channel 钠离子通道
Neurotransmitter channel神经传导物通道
Glycine channel 甘胺酸通道
164. Structure C which is responsible for event 5 is a (1 point)
与传导过程5有关的C构造为:
Potassium channel钾离子通道
Calcium channel 钙离子通道
Sodium channel钠离子通道
Neurotransmitter channel神经传导物通道
Glycine channel甘氨酸通道
165. Event 4 by structure B (1 point)
由构造B负责的传导过程是为:
depolarizes membrane potential. 去极化膜电位
increases membrane permeability 增加膜的通透性
transports more sugars molecules into the cell.
传送更多糖分子进入细胞膜内
transports signal molecules into the cell so that the cell starts to synthesize neurotransmitters.
传送讯息分子进入细胞膜内,增加神经传导物质的合成
transports precursor molecules of neurotransmitters into the cell so that the cell can synthesize neurotransmitters.
传送前驱分子进入细胞膜内,增加神经传导物质的合成
166. Which of the following statements is/are true about the action potentials shown as D and E in the figure (1 point)
有关动作电位(图D与E)的叙述何者正确
(1) They were recorded after and before sugar molecules were present, respectively.
图D与E分别於糖分子出现后和出现前纪录到的动作电位
(2) They were recorded before and after sugar molecules were present, respectively.
图D与E分别於糖分子出现前和出现后纪录到的动作电位
(3) The action potential observed after sugar reception is triggered by an increase of calcium ions which stimulate neurotransmitter release.
检测到糖后观察到的动作电位,是由钙离子的增加引起的,钙离子刺激释放神经传导物
(4) The a ction potential observed after sugar reception is triggered by an increase of Potassium ions which stimulate neurotransmitter release.
检测到糖后观察到的动作电位,是由钾离子的增加引起的,钾离子刺激释放神经传导物
(5) The action potential is recorded from taste sensory receptor cells.
动作电位是记录自受体感觉细胞
2, 3
1, 3
2, 4
2, 5
2, 4, 5
167. Wh定位
168. Both snake and weasel hibernate. Which of the following is correct (1 point) 蛇和黄鼠狼都会冬眠,下列叙述何者正确 (1分)
A. They will die when temperature decreases below the critical temperature.温度若降至临界低温以下,它们均将死亡.
B. Weasel will die when temperature decreases below the critical temperature. 温度若降至临界的低温以下,黄鼠狼将死亡.
C. Snake will die and weasel will wake up when the temperature decreases below the critical temperature.温度若降至临界低温以下,蛇会死而黄鼠狼会醒来.
D. Weasel keeps low body temperature and slow heart rate during whole period of hibernation. 黄鼠狼在冬眠中始终保持低的体温和慢的心跳速率.
169. It is possible to predict bird diversity based on forest types.
170. Four quantity pyramids are shown below. Which is representative for plant-aphid-ladybug (1 point)
下列4个数量金字塔中,何者代表植物-蚜虫-瓢虫


A B C D
171. Which of the following ecosystems has the highest net primary productivity (1 point) 下面哪个生态系统具有最高的净初级生产力 (1分)
A.Tropical rain forest 热带雨林
B.Open ocean 公海
C.Northern coniferous forest 北方针叶林
D.Farm lands 农场
172. The Figure below shows vertical distribution of some parameters (Chlorophyll, Phosphate, Primary production and Temperature) of North Pacific during summer. 下图为夏季北太平洋中一些参数(包括叶绿素,磷酸盐,初级生产和温度)的垂直分布.
From left to right, letter a through letter d represent: (1 point)
由左至右(字母a至字母d)分别代表:(1分)
A. Temperature, phosphate, chlorophyll and primary production
温度,磷酸盐,叶绿素和初级生产
B. Chlorophyll, phosphate, temperature and primary production
叶绿素,磷酸盐,温度和初级生产
C. Primary production, phosphate, temperature, chlorophyll
初级生产,磷酸盐,温度和叶绿素
D. Phosphate, temperature, primary production and chlorophyll.
磷酸盐,温度,初级生产和叶绿素.
173. The length of a food chain in a food web is often quite short. Usually, the length is shorter than 5 links. Which is mostly likely reason for the shortness of the food chain (1 point)
食物网中一个食物链的长度经常很短,通常其长度少於5个联结.下面何者是食物链如此之短的最可能原因 (1分)
A. The population of final predator is often too large.
最终捕食者的族群经常都太大.
B. The primary producers can sometimes be indigestible.
初级生产者有时不能被消化.
C. Only about 10% of energy in on link can be converted to organic matters in next trophic level.
一个联结中只有大约10%的能量可以被转化至下一个营养级的有机物.
D. Wintertime is too long and low temperature limits primary productivity.
冬天太长,低温限制了初级生产力.
真菌
动物
植物
光合细菌
其他细菌
原核生物0先
粒线体
叶绿体
古细菌
真细菌
无氧呼吸的真核生物0先
1 2
3 4
A
D
E
B
C
Answer
A-G
1 Sinus venosus 静脉窦
A
2. Atrium心房
G
3 Pulmonary vein肺静脉
F
4 Pulmonary artery肺动脉
E
5 Conus arteriosus动脉锥
C
6 right Atrium右心房
G
7 Left ventricle左心室
D