2009 STPM BIOLOGY PAPER 2 (ESSAY QUESTIONS)

星星之玲

5 (a) (i) Distinguish glycosidic bonds found in starch from those found in cellulose. [4 m]

        (ii)Explain why these differences are of biological importance.[6 m]

   (b) Explain the role of carbohydrate in membrane. [5 m]


6 . With the aid of a labeled diagram, describe the process of transcription[15 m ]


7 (a) Explain the role of liver in the metabolism of protein. [10 m]

   (b) Describe the Cori cycle and its function in the metabolism of carbohydrate.[5 m]


8 (a)Explain the interaction of phytochromes in flowering.[10 m]

   (b) Describe the involvement of hormones in an apical dominance. [5 m]


9. (a) Describe the assumption that maintain the Hardy-Weinberg equilibrium.[8 m]

    (b) Phenylketonuria is a genetic disorder due to homozygosity of a certain recessive allele which occurs
in one out of every 15000 individuals. Determine the number if individuals in a population which are carrier of the disorder. [7 m]

10 (a) (i) Describe the procedure of quadrat sampling.[3 m]

          (ii) State 3 condition for a reliable estimate of a population. [3 m]

     (b) (i) Explain the capture-recapture technique.[4 m]

           (ii)State the assumption that need to be considered for the application of the capture-recapture technique.    [ 5 m]




can anyone give some point...some question is really dun know how 2 do...

[ 本帖最后由 星星之玲 于 19-11-2009 06:34 PM 编辑 ]

darksider

原帖由 星星之玲 于 2009/11/19 06:28 PM 发表
5 (a) (i) Distinguish glycosidic bonds found in starch from those found in cellulose. [4 m]

        (ii)Explain why these differences are of biological importance.[6 m]

   (b) Explain the role o ...

5(a)(i)
-cellulose ; beta-1,4-glycosidic bond ; each beta glucose is connected to adjacent glucose which is inverted via this bond
-starch ; amylose : alpha-1,4-glycosidic bond , amylopectin : alpha-1,4-glycosidic bonds and alpha-1,6-glycosidic bonds

5(a)(ii)
-cellulose ; beta-1,4-glycosidic bonds allow formation of hydrogen bonds between adjacent cellulose polymers ; high tensile - strength and large space in between the microfibril (thus allowing water and other substance to pass through ; hence, can carry out structural role, eg, in cell wall.

-starch ;
-alpha-1,4-glycosidic bonds cause amylose to coil into a compact molecule ; together with its osmotically inactive nature, it is a good storage molecule.

-alpha-1,6-glycosidic bonds cause amylopectin to branch ; short chain means "more ends" so difficult to break down ; osmotically inactive ; good storage molecule

5(a)(iii) glycocalyx attached to extrinsic protein ; cell-to-cell recognition , cell adhesion , cushion membrane , receptor , immunity ( can be recognised by lymphocyte).

891108

今年蛮难下

purple_pig_33

题目真的很奇怪
尤其是structure
有两题的分数特别怪
单单那个添mRNA得就可以8分？
我都不知道我到底有填错吗

darksider

原帖由 darksider 于 2009/11/19 06:52 PM 发表


5(a)(i)
-cellulose ; beta-1,4-glycosidic bond ; each beta glucose is connected to adjacent glucose which is inverted via this bond
-starch ; amylose : alpha-1,4-glycosidic bond , amylopectin :  ...

9. (a) Describe the assumption that maintain the Hardy-Weinberg equilibrium.[8 m]

    (b) Phenylketonuria is a genetic disorder due to homozygosity of a certain recessive allele which occurs
in one out of every 15000 individuals. Determine the number if individuals in a population which are carrier of the disorder. [7 m
a)
-no migration ; if there is immigration ; foreign alleles are introduced into the gene pool ; if there is emigration , it will result in losing alleles which means gene pool will be affected. ; if there is no migration, all this would not happen and the gene pool is likely to be maintained

-random mating ; will ensure that no alleles are favoured ; thus won't affect allele frequency in the population

-no mutation ; will not affect the allele frequencies ; if mutation occurs , some organisms in the population will die and result in changes in allele frequencies ; thus affect the gene pool

-large population ; small changes in the genotypes of individuals are unlikely to impose much effect on the allele frequencies of the population and it would affect much if the population is small.

-no natural selection : (not in syllabus so dun mention better)

These build up the basis of hardy-weinberg law which states allele frequency of a population would remain constant from generation to generation provided that above factors are considered.

9(b) Let recessive allele for the disorder be r and the dominant allele for the disorder be R.
p + q = 1 , p^2 + q^2 + 2pq = 1

rr = q^2 = 1/15000
q = sqrt (1/15000)

p = 1-q = 1- sqrt (1/15000)

carrier's alleles / genotype = Rr and rR = 2pq

number of carriers in the population = 2pq x 15000 = 242.9xxx = approx 243

work on it... final answer = 243 carriers.

[ 本帖最后由 darksider 于 19-11-2009 08:46 PM 编辑 ]

darksider

原帖由 purple_pig_33 于 2009/11/19 08:40 PM 发表
题目真的很奇怪
尤其是structure
有两题的分数特别怪
单单那个添mRNA得就可以8分？
我都不知道我到底有填错吗

If tRNA's anticodon is 5' AUA 3', then its corresponding codon in mRNA is 3' UAU 5'...

很多人没填 3' ... 5'... 如果没有，八分就 bye bye.

organelle 1 = chloroplast (很多人写 mesophyll cell)
organelle 2 = vacuole

Why in the absence of oxygen , E would not occur?
-oxygen is the final electron accepor in E. In the absence of oxygen, there is no way to get rid of NADPH and FADH2 produced so the organism will resort to either B or C (anaerobiosis) *becuz can get rid of NADH which is used to reduce pyruvate to lactate or ethanal.

What is chloride shift ?
-it refers to movement of chloride ions into the blood in order to balance the potential gradient (any other that refers to charge difference) created when HCO3- diffuse into blood plasma

What is the advantage of CAM plant?
-can preserve water efficiently

What is the example of CAM plant ?
-cactus

Differences btw CAM and C4 plant ?
CAM
-photosynthesis involves mesophyll cells only
-stomate close at day and open at night
-non-dimorphic
-does not show kranz anatomy(if u dun use dimorphic / mesophyll cell)

C4
-photosynthesis involves mesophyll cells and bundle sheath cells
-stomata open at day and close at night
-dimorphic
-show kranz anatomy

How many ATP molecules are produced ?
22 ( 6 NADH = 18 ATP , 2 FADH2 = 4 ATP )

Which codon are degenerate?
Find two codons which have their similar first and second nucleotide but their third nucleotides are different.
- reason : They code for the same amino acid

[ 本帖最后由 darksider 于 19-11-2009 08:56 PM 编辑 ]

bell_25

原帖由 purple_pig_33 于 19-11-2009 08:40 PM 发表
题目真的很奇怪
尤其是structure
有两题的分数特别怪
单单那个添mRNA得就可以8分？
我都不知道我到底有填错吗

糟糕。。我填错了。。
我朋友说要reverse那个sequence。。from3‘ to 5’
我的8分飞走了。。

生物难题

原帖由 darksider 于 19-11-2009 08:48 PM 发表


If tRNA's anticodon is 5' AUA 3', then its corresponding codon in mRNA is 3' UAU 5'...

很多人没填 3' ... 5'... 如果没有，八分就 bye bye.

organelle 1 = chloroplast (很多人写 mesophyll cell)
...

如果anticodon is 5' AUA 3'， mRNA 不可以是 5' UAU 3'吗？
mRNA是READ从5' TO 3' 的。
不是3' TO 5'。

bell_25

对。。我就是那个笨蛋。。
那两个我通通错到完。。我真糊涂。。
那么请问differences of C4 and CAM plants你写什么?
还有advantages呢？
谢谢

bell_25

原帖由 生物难题 于 19-11-2009 08:52 PM 发表


如果anticodon is 5' AUA 3'， mRNA 不可以是 5' UAU 3'吗？
mRNA是READ从5' TO 3' 的。
不是3' TO 5'。

那么如果anticodon是 5' AUG 3'呢？

生物难题

原帖由 bell_25 于 19-11-2009 08:54 PM 发表


那么如果anticodon是 5' AUG 3'呢？

如果ANTICODON是 5' AUG 3'，
那mRNA CODON 答案应该是 5' CAU 3'。
不是这样吗？

darksider

原帖由 生物难题 于 2009/11/19 08:52 PM 发表


如果anticodon is 5' AUA 3'， mRNA 不可以是 5' UAU 3'吗？
mRNA是READ从5' TO 3' 的。
不是3' TO 5'。

anticodon and codon are anti parallel

http://profiles.nlm.nih.gov/SC/B/C/B/S/_/scbcbs.pdf

page 544

http://books.google.com/books?id=T2CZWJy3LZkC&pg=PA437&lpg=PA437&dq=codon+and+anticodon+antiparallel&source=bl&ots=RI7FVmr4YX&sig=UpJghUtEsfQ6Iyur9vZE_CAsKh0&hl=zh-CN&ei=yEEFS_fFO5HM6wO52aDBCg&sa=X&oi=book_result&ct=result&resnum=5&ved=0CCkQ6AEwBA#

[ 本帖最后由 darksider 于 19-11-2009 09:11 PM 编辑 ]

bell_25

不知道喔。。有些人写5' CAU 3'。。可是有些人写 3’ UAC 5'
可是铁定的是我的8分不见了。。我真的太太粗心了。。

purple_pig_33

不是5到3么？？？


我找过书本了
tRNA是从5到3的
而且amino acid是attach在3那边的
然后mRNA也是5到3

所以是5到3

darksider

原帖由 purple_pig_33 于 2009/11/19 09:09 PM 发表
不是5到3么？？？


我找过书本了
tRNA是从5到3的
而且amino acid是attach在3那边的
然后mRNA也是5到3

所以是5到3

purple_pig_33

好的
我错了
http://waynesword.palomar.edu/codons.htm
我要去撞墙了

hung90

应该是同样是5' 到 3'. 但里面的 sequence 需要reverse. 我就知道没这么容易.

darksider

今年的题目看似容易，但其实很多都很tricky.

purple_pig_33

没有A了啦
我就很奇怪为什么8分那么容易拿

我哭死了

faun

work on it... final answer = 243 carriers.

我的答案是240. 可以接受吗?