生物讨论专区

Jec

这是essay题目。
Describe the main stages of glycolysis.

我会等半小时后才提供答案。

hellodizzy

这是你们trial？

Jec

为何这样问？你问我答：不是

hellodizzy

原帖由 Jec 于 16-9-2007 04:30 PM 发表
为何这样问？你问我答：不是

问而已。。。。。。。。

Jec

原帖由 Jec 于 16-9-2007 03:54 PM 发表
这是essay题目。
Describe the main stages of glycolysis.

我会等半小时后才提供答案。

答案:
Glycolysis can be divided into three main stages as follows:

·
Phosphorylation of glucose
Glucose is phosphorylated with a phosphate molecule, using up 2 ATP molecules and producing fructose phosphate.

·
Splitting sugar biphosphate
Further phosphorylation produces fructose biphosphate, which splits into 2 molecules of C3 sugar phosphate, called glyceraldehydes-3-phosphate (PGAL).

Both sugar phosphate molecules are isomers of each other and one of the molecules will then convert to become identical to each other.

·
Oxidation by dehydrogenation
Each PGAL molecule is further phosphorylated with inorganic phosphate and undergoes dehydrogenation to produce 1 molecule of glycerate-1,3-biphosphate and a pair of hydrogen atoms.

Each glycerate-1,3-biphosphate loses a phosphate molecule and produces 1 molecule of glycerate-3-phosphate.

Each dehydrogenation produces 2 ATP molecules, synthesized using the phosphate molecules that were released and 1 reduced NAD molecule carrying the removed hydrogen. It is then converted to another C3 compound called pyruvate.

As both C3 sugar phosphate molecules are converted to pyruvate, thus a total of 4 ATP molecules and 2 reduced NAD molecules are produced.

Since 2 ATP molecules are used during the first stage, the net gain of ATP is actually only 2 ATP molecules at the end of glycolysis.

Jec

下一道题目又来咯!
Essay question:
With the aid of a diagram, explain the light independent reaction or Calvin cycle.

Jec

解答:
The Calvin cycle takes place in the stroma of the chloroplast. ATP provides the energy and NADPH provides the reducing power needed for biosynthesis using carbon dioxide. Calvin cycle can be divided into three phases:

Carbon fixation
Catalysed by ribulose biphosphate (RuBP carboxylase), a C5 sugar named ribulose biphosphate (RuBP) takes up carbon dioxide to form the unstable C6 compound. RuBP is the carbon dioxide (CO2) acceptor. The C6 compound breaks down immediately to form 2 molecules of a C3 compound, glycerate-3-phosphate (GP) as the first product in photosynthesis.

Reduction
NADPH will then reduce 1,3-biphosphate (GP) by removing the oxygen from it and converting it to a C3 sugar, glyceraldehydes-3-phosphate (G3P), using ATP energy from the light dependent reaction.

For every three molecules of CO2, 6 molecules of 3GP are yielded. However, only 1 molecule of this C3 sugar can be counted as a net gain of carbohydrate. This molecule can be converted to glucose, sucrose, starch, amino acids and other products.

Regeneration of the CO2 acceptor (RuBP)
The other five molecules of G3P are converted into 3 molecules of RuBP to keep the cycle going. This process requires energy from ATP.

ynnee

谁又moe'CD,可以寄给我吗？

muse

感谢版主把帖子合并了
大家以后你们把题目放上来
我会把它们编辑进已分类好的#67－#91
想问版主能把它们移到第一面吗？
因为每次要按到第三面
有点不方便。。

muse

大家哪一课比较不会？
说出来，我尽量post练习上来
大家一起revise吗。。

muse

bio专区的帖竟然沉了
看来大家是忙着考试吧。。
我明天应该会上一些bio问题上来
大家加油！

Jec

大家好。我又回来了。首先，先上载一个Krebs Cycle 的video clip。
http://s222.photobucket.com/albums/dd145/cbjec/?action=view&current=krebs_cycle_reactions.flv

希望你们有所帮助。

[ 本帖最后由 Jec 于 21-9-2007 05:32 PM 编辑 ]

考试要到了...大家拼了!!
A cross of bean variety produce 9744 individuals with dominant tall genotype (T) compared with 256 individuals with recessive short genotype (t).
(a)By assuming that the crossing in the bean population above takes place in equilibrium , calculate the genotype frequency in the crossing .

Individuals with dominant tall genotype (TT or Tt) : 9744
Individuals with recessive short genotype (tt):256
Total:10000

According to Hardy-Wernberg equation ,
p²+2pq+q² =1
and

p+q=1

where,
p²=frequency of genotype of homozygous dominant tall (TT)
2pq=frequency of genotype of heterozygous (Tt)
q²=frequency of genotype of homozygous recessive short (tt)
p=frequency of dominant allele (T)
q=frequency of recessive allele (t)



frequency of genotype of homozygous recessive short (q²)
256/10000=0.03

Frequency of recessive allele (q)
Square root of q² =0.17

Frequency of dominant allele (p)
p+q = 1
p = 0.83

frequency of genotype of homozygous dominant tall (p²)
(0.83)²=0.69

Frequency of genotype of heterozygous (2pq)
2(0.83)(0.17)=0.28

(b) State the necessary assumptions in the above crossing .
-population size is large enough .
-no mutation
-no immigration or emigration occurred. That’s mean no gene flow in and out from the population.
-no natural selection .
-mating is random .

(c) Give the meaning of alleles frequency in a population .
-allele frequency is the ratio of the sum of a particular allele to the sum of the all of the allele in a population .

[ 本帖最后由 庄Ch'ng 于 21-9-2007 05:46 PM 编辑 ]

Jec

接下来，我scan了２００６年生物考卷１。请试试看。我还会提供答案的，放心！But, later.

Jec

Jec

Jec

各位网友，是时候对答案啦。。。
1. B     
--- DNA is deoxyribose, so the molecule does not contain oxygen.
2. B
--- A wrong because nucleotides consist of nitrogenous base, pentose sugar (either purine or pyrimidine) and phosphate. C also wrong because instead of covalent bonds, they are hydrogen bonds which bind nucleotides and bases together. D also wrong because it is not hexose, it is pentose sugar.
3. B
--- In the thylakoid membrane, light dependent reaction (or known as noncyclic photophosphorylation) takes place whereas in stroma, light independent reaction (or known as cyclic photophosphorylation) takes place. As a result, X is granum and its function is to split water molecules, i.e. photolysis of water; Y is stroma, its relavant function is to fix carbon dioxide from the atmosphere, hence, the process is known as CO2 fixation.


待续。。。

muse

jec你好厉害
简直就可以做老师了吗。。

Jec

原帖由 muse 于 22-9-2007 09:45 AM 发表
jec你好厉害
简直就可以做老师了吗。。

Paiseh...
在下有个小小梦想， 希望华人子弟能score好STPM，BIO考得顶瓜瓜。。。

[ 本帖最后由 Jec 于 22-9-2007 10:47 AM 编辑 ]

Jec

延续答案checking吧。
现在来到第4题。
4. B
--- Function for Golgi Body (or apparatus) is to modify proteins and form lysosomes, secretory vesicles and peroxisomes. However, the synthesis of lipid is the function for smooth endoplasmic reticulum.

5. C
--- In prokaryotes, the nuclei do not separate from cytoplasm because there are no nuclear membranes.

6. D
--- Try to decipher codon by finding their respective complementary anticodon. Please be careful because in the case where the direction of codon is 5' to 3', then in tRNA the direction of anticodon is vice versa, i.e. 3' to 5'.
Example:
Codon             5'-CGA-3'      5'-AUG-3'      5'-UUC-3'      5'-UAU-3'
Anticodon         3'-CGU-5'      3'-UAC-5'      3'-AAG-5'      3'-AUA-5'

7. C
--- Inorganic catalyst can withstand high heat exposure.Option II and option III are the best answers although option I cannot be denied its accurate statement, but there is apparently no such combination choice in the question.

待续。。。（我的中文输入和打字速度很慢）