Form6 Biology

蝎杰

１想请问这题问题如何解答
THE PERCENTAGE OF BASE IN ONE CHAIN OF THE DOUBLE HELIX DNA IS AS FOLLOWED
T＝４０％   C＝２２％ .WHAT IS THE PERCENTAGE OF A＋G BASES（COMBINES）IN THE SAME DNA STRAND
２.WHAT IS THE PERCENTAGE OF A＋G（COMBINES）IN THE COMPLEMENTRY DNA

雨天之魂

蝎杰 于 5-1-2005 22:46  说 :
１想请问这题问题如何解答
THE PERCENTAGE OF BASE IN ONE CHAIN OF THE DOUBLE HELIX DNA IS AS FOLLOWED
T＝４０％   C＝２２％ .WHAT IS THE PERCENTAGE OF A＋G BASES（COMBINES）IN THE SAME DNA STRAND
２.WHAT IS THE PERCENTAGE OF A＋G（COMBINES）IN THE COMPLEMENTRY DNA  

首先要知道的是在 DNA 的 base pairing 里, A 配 T, G 配 C 噢.
然后在 DNA 里也只会有 A,G,T,C 四种 base.
在同一条 strand 里, 如果说我已经知道 T =40%, C =22%,
那么剩下的就是 A+G 咯, 对不对?
所以 A+G = 100-(40+22).
         = 38%

反过来说, 如果我知道这边的 T 是 40%, 那么对面的 A 也应该 =40%.
知道了 C =22%, 也就是说对面的 G =22%.
那么对面那排的 A+G 就 =40% + 22%
                      = 62%

有问题欢迎提问. 我把帖子合并到 form 6 biology 了噢.

蝎杰

谢谢替我解答。我还想问关于protein synthesis的process.尤其是transcription。mRNA detaches from gene从这里开始不大明白。

蝎杰 于 6-1-2005 02:41 PM  说 :
谢谢替我解答。我还想问关于protein synthesis的process.尤其是transcription。mRNA detaches from gene从这里开始不大明白。

首先,你知道mRNA是從TRANSCRIPTION來的對不對

在之前  必須由RNA POLYMERASE 來transcribe DNA sequence

就出現了pre-mRNA

再經過modification後  才得到mRNA

以上都在NUCLEUS內進行

-----------------------------------------------------------------

mRNA後來會從NUCLEUS出來  到RIBOSOME(一大一小兩個PROTEIN SUBUNIT)進行TRANSLATION

TRANSLATION的MECHANISM是怎樣呢?  很難用文字說明白   上網或書上看看圖就了解了

其實PROTEIN SYNTHESIS跟當初mRNA的TRANSCIPTION有點像

都是拿周圍的材料(只是由mRNA的nucleutides換成是PROTEIN的Amino acids)

一直接一直接  接到做出成品為止

當TRANSLATION完成PROTEIN後  mRNA就會被分解成nucleutides 以便下次TRANSCIPTION再用

雨天之魂

RealBallaz 于 6-1-2005 15:14  说 :



首先,你知道mRNA是從TRANSCRIPTION來的對不對

在之前  必須由RNA POLYMERASE 來transcribe DNA sequence

就出現了pre-mRNA

再經過modification後  才得到mRNA

以上都在NUCLEUS內進行 ...

差不多是这样, tRNA 和 rRNA 的部分有待补充.
如果可以的话, 请再给我几天的时间. 第三课的笔记要出炉了.
迫不及待也无所谓, 我直接解说也可以. 但总没有笔记里的来得全面和完整.

雨天之魂

Enzyme
-biological catalyst produced by lviing cells.
-lower the activation energy.
-speed up the rate of biochemical reactions in cell but remain unchanged at the end of reactions.
-most are globular protein (不是全部)

Classification of enzyme
oxidoreductase -involve in redox reaction(reduce/oxidise by transfer e-, H+, O)
transferase -transfer a group
hydrolase -breaking by water
liase -breaking without using water
isomerase -rearrange the structure of a molecule to form isomers
ligase -joining of molecules.
例子在以后会慢慢看到, 现在不需要背. 大概知道它们的分别就好.


-Reactans have to reach a high energy intermediate state(transitional state) before the products are formed. Activation energies are required for reactants to reach the state.
(关于 activation 的 fact)

-Enzyme brings the products to close proximity, correct orientation, and stressing/distorting chemical bonds. The shape of the substance is slightly changed, bonds are broken, and new ones are formed.
(关于 how enzyme lowers the a. energy)

Rate of reaction affected by
-[enzyme], [substrate], pH(optimal), temperature(optimal).

Cofactors: non-protein substances requrie by certain enzyme.
-not denatured by high temp.
-not destroyed at the end of reaction (may be regenerated)
-can be organic/inorganic compound.

a)activator -inorganic ions, attach to enzyme/substrate to help in the formation of enzyme-substrate complex. eg. Cl- in amylase.

b)Coenzyme -organic compound, doesn't bind to the enzyme closely, can exist on its own and reacts freely with other compound. eg. NAD with dehydrogenase enzyme.
依赖性较低, 可自己行动.
c)Prosthetic group -organic compound, attach to enzyme by covalent bond (form conjugated protein) eg. FAD with dehydrogenase enzyme, Haem in haemoglobin.
依赖性高, 会紧紧合并.

Mechanism:
-Lock and key hypothesis(Emil Fischer).
active site and substrate are exacly complementary. The shape of the substrate(key) fits into the active site of the enzyme(lock) rigidly, forming an enzyme-substrate complex.两个的形状是固定的, 完美配合.

-Induced-fit hypothesis(Daniel Kahsland)
active site is felxible, not exactly complementary to the shape of the substrate. Binding induces a slight change in the shape of enzyme to enclose the substrate, making the fit more precise.
enzyme的形状并不固定, 可稍微改变来配合substrate.

Michaelis-Menten constant:

V= Vmax[S]/ (Km + [S]) , Km = [S] at 1/2 Vmax.
low Km = high enzyme affinity /efficiency.
记得 Km 最基本的 formula, 越低的 Km 就表示 enzyme 越有效率(好).

Inhibitor:
原本 inhibitor 可依照 competitive/non-competitive 和 reversible/ irreversible 分类的, 可是中六课程直接把 competitive + reversible 和 non-competitive + irreversible 放在一组. 知道实际上不然就可以了, 答题的时候还是照着课程吧.

Competitive
-similar structure to substrate molecule, bind to active site.
-eg. malonic acid inhibits succinic acid dehydrogenase(跟succinic acid抢)
-reversible bonding, can be neutralised by increasing [substrate]
-affects the Km but not Vmax. (只有 enzyme 的效率降低, 效应的最高速度不会改变)

Non-competitive
-no structural similarity with the substrate, wouldn't compete with substrate for the active site.
-attach to allosteric site /prosthetic group/ active site (自由度比较高)-eg. heavy metallic ions (Ag+, Hg+)
-irreversible bonding, effect cannot be neutralised.
-Vmax is affected but not the Km. (只要 enzyme 被合并到, 就直接废掉, 余剩之 enzyme 个别的效率并不会被影响到. 效应整体的速度却因为可用的 enzyme 越来越少而减低了.)

记得以上两个部分的最后一句, 在 Lineweaver-Burk plots 里就不成问题了.
competitive 会让 -1/Km 越来越往外出去, 1/Vmax 没有改变.
non-competitive 会让 1/Vmax 越来越高上去, -1/Km 则没有改变.

End product inhibition
-biochemical pathway in which the final end product act as an reversible non-competitive inhibitor for the 1st step in the metabolic pathway. (一种 negative feedback 以控制体内各种物质的份量)

Enzyme Immobilisation
-entrapment in gel, in microcapsule, by covalent bonding, adsorption, cross-linking of enzymes to other molecules.(固定enzyme的方法)

Uses
-produce high fructose syrup -glucose isomerase
-produce glucose dipstick -glucose oxidase/peroxidase
-lactose free milk -lactose
-semisynthetic penicillins

Advantages
-do not contaminate end product, no purification is needed, enzyme can be reused, cheaper cost, stabilise enzyme.

Disadv.
-active site may be covered, not free moving, less chance for substrate to enter active site.

Biosenser
immobilised enzyme -> transducer -> amplifier -> microelectronic producer -> numerical reading. (生物感应器的流程, 从尾端的 enzyme 那里被感应到, 再转换成数码资料)

用处:
-in manufacturing industry -[ ] of raw material/finish product.
-agricultural -[ ] of mineral/element of soil.
-environmental study -monitor pollution.
-medical field -[ ] of chemicals, blood sugar lvl.

DNA Experiment.

Griffith -proved heated dead bacteria could transform another living bacteria.

R type- rough colony /non-virulent Pneumococcus bacteria.
S type- smooth colony /virulent Pneumococcus bacteria.

1. living R. + mouse = mouse remains healthy.
2. living S. + mouse = mouse gets Pneumonia, S. is obtained from dead mouse.
3. heat killed S. + mouse = mouse remains healthy.
4. living R + heat killed S + mouse = mouse gets Pneumonia,
                                      S. is obtained from dead mouse

Avery at el.(证实 transform living bacteria 的是 DNA)
-follow Griffith's experiment by using Pneumococcus.
-S. type bacteria were heated, cooled, and homoginised.
-Carbohydrate, lipid, protein, RNA, DNA were extracted/purified.
-each + living R-type bacteria individually.
-DNA transforms R-type bacteria into S-type bacteria.

Beadle and Tatum
-Neurospora(spores) were irradiated by X-ray to induce mutation.
-cultured in complete medium.
-mycelium cross with wild type of opposite strain.
-sexual reproduction take place. Asci each with 8 ascuspores were produced.
-Ascuspore were dissected, separately cultured in complete medium.
-Growth = no mutation, thus discarded.
-No growth = mutation, further test by adding 20 different amino acid individually into 20 test tubes.
-The adding of lost amino acid will induce growth in that particular test tube.
-One gene has been mutated, one enzyme related to the synthesis of that amino acid is lost.
-Later, one gene-one enzyme -> one gene one protein -> one gene-one polypeptide (since some protein was controlled by 2 genes, eg. haemoglobin)
(Neurospora 的苞子被用 X-ray 来促成突变, 然后在完整的环境里培养. 用正常的(相反性别的) mycelium 来跟它们交配, asci(被繁殖出来的部分) 里有 8 个 ascuspores(小苞子)被分解后再用完整的环境来培养它们. 有发育的表示正常, 于是被淘汰. 没有发育的表示有突变发生, 被继续用在接下来的实验. 把那些有突变的分别放入 20 个试管里, 每个试管里有一种特定的 amino acid. 一个 gene 的突变会令它们缺乏一种特定的 enzyme 来制造 a.acid. 只有正好补足所缺乏那种 a.acid 的试管会有发育.)

Mehelson and Stahl
-prove the semi conservative DNA replication.
-E.coli bacteria cultured in 15N(一般 nitrogen 14 的 isotope), produce heavy DNA, which move to lower part of centrifugation tube(因为比较重, 所以沉淀的位置比较低).
-They were moved to normal N medium.
-after 25 min., DNA is extrated by ultracentrifugation.
-results shows hybrid type DNA.(double strands 里一条是特别重的(with 15N), 一条是普通的)
(DNA 的 double helix 构造里, 其中一条是续承自母体的, 另外一条是新制造的)

DNA structure-antiparallel, complementary base pairing, double alpha-helix, nucleotide consists of A,T,G,C base, deoxyribose and phosphate as back bone.
得大概会画几个nucleotide(包括bases, deoxyribose, phosphate group) 连接在一起的形状.

DNA replication
1.Helicase unwinds parental double helix.
一开始, enzyme helicase 把 double helix 打开来.
2.Single-strand binding protein stabilises the unwound parental DMA.
一个名叫 ssbp 的 protein 出现来 stabilise 那些被分开来的两条 strands.
3.The leading strand is synthesised continuously from 5'-3' direction by DNA polymerase.
在其中一条母体DNA上, DNA polymerase 把 DNA nucleotide 一粒接一粒, 不中断的制造出新的另外一条来., 这条叫 leading strand
4.The lagging strand is synthesised discontinuously. Primase synthesises a short RNA primer, which is extended by DNA polymerase to form Okazaki fragments.
在另外一条母体上, Primase 制造了一小段的RNA primer, 暂时配对在另外一条母体DNA上. 这条RNA primer 被DNA polymerase 以 DNA nucleotides 继续延长., 形成一个个的缺口, 叫 O.fragments.
5.Another DNA polymerase replaces the RNA primer with DNA.
(那一小段的 RNA 被另外一段 DNA 取代, 现在全部已经是DNA了)
6.DNA ligase joins the Okazaki fragments.
DNA ligase (enzyme ) 把那些缺口全部合并.
7.Finally 2 identical strands of DNA are formed.
新的两串DNA被形成.



注: 红色的是母体DNA, 新形成的有一半还是母体DNA.

Protein synthesis

Transcription
-H-H bonds are broken by helicase, double helix splits.
-strand with cistron(sequence of base codes for polypeptide) =sense strand.
-strand without cistron =nonsense strand.
-RNA polymerase attach to the promoter site next to the cistron, initiates transcription in 5'-3' direction.
-nucleotides from cytoplasm with base complementary to the base on DNA are added one by one to elongate the strand.
-RNA polymerase detaches on 'termination codon', mRNA moves away from DNA.
2 strands of DNA are re-joined by H-H bonds.
-The pre-mRNA in eukaryotes contains exons and introns.
Spliceosome splits of the introns. Exons are joined to produce shorter, mature mRNA.
(Prokaryotes' RNA have no introns)
(DNA 被分开, 带有可用资料的部分就是cistron. 有cistron的那条是 sense strand. RNA  polymerase 附到 cistron 稍前的 promoter site 开始'复制'资料. 一粒一粒能够配对 DNA base(A->U, T->A, G<->C) 的 nucleotides 被配对上去, 然后连接起来形成一条 mRNA 一直到停止讯号. DNA 被重新合并. 被制成的 pre-mRNA 里 intron/exon 的部分. intron 被 spliceosome(enzyme) 去除掉, 剩下的 exon 被合并起来形成成熟的 mRNA.)



注:蓝色部分是 intron, 红色的是 exon. 蓝色的被去除了, 剩下的红色就合并起来.

-rRNA + protein = large/small ribosomal subunits = ribosomes
(in nucleolus)

-tRNA being folded into a 3D clover-leaf shape held by H-H bond. Aminoacyl-tRNA synthetase activates amino acids (use ATP) to form aminoacyl-tRNA complex, which act as anticodon, carrying an amino acid.
(课程里没有另外提起 rRNA 和 tRNA 怎么来, 就直接假设都是透过 transcription 来的)

Translation
-mRNA binds to ribosome, anticodon of tRNA carrying animo acid to mRNA.
-The tRNA complex translate the codon of mRNA, and leave an amino acid.
-Everytime the ribosome moves by one codon in a 5'-3' direction, a new aminoacyl-tRNA with a complementary anticodon arrives and binds with the new codon on mRNA, until the ribosome encounters a stop codon(UAA, UAG, UGA)
-An enzyme catalyses the formation of a peptide bond between 2 a.a.
-Hydrolysis of the bond between the tRNA to the last a.a in the polypeptide chain occurs.
-Synthesised p.p chain may then coil/fold/chemically modified to form a functional protein.
(之前一个过程中形成的 mRNA 带着密码(codon), 附到 ribosome, 以 tRNA complex 的 anticodon 来解读. ribosome 逐渐从 5'往 3' 的方向移动, 每一次的解读会让一颗新的 amino acid 被留住. 一颗接一颗会被连住形成一条 polypeptide chain(primary structure) 一直到最后的停止讯号. 到达停止讯号时, 最后一颗会和连接住的 tRNA 分解开. 形成的 p.p chain 会被卷,折成 secondary/tertiary/quaternary structure, 形成一个有特别功能的 protein.)



注: 红色部分是 aminoacyl-tRNA complex 的形成,
    绿色部分是 rRNA 形成 ribosome 的过程,
    最后蓝色的部分就是 translation.
    摘自 longman biology.

[size=-3]

[ Last edited by 雨天之魂 on 7-1-2005 at 06:12 PM ]

蝎杰

不好意思，又是我，还有问题不懂...
1.if there were34amino acids and DNA only contained 2 types of nitrogenous bases, what would be the minimum number pf bases per codon that could for protein?
2.in the sequence for sickle cell anaemia,adenine replaces thymine in a triplet, forming the triplet.during translation of the mutantm,the amino acid valine is incroporated into the haemoglobin molecule instead of glutamic acic.
what is the anticodon in the transfer molecule carrying this valine?

雨天之魂

1.if there were34amino acids and DNA only contained 2 types of nitrogenous bases, what would be the minimum number of bases per codon that could for protein?

没有接触过这类的题目, 可以提供答案吗?
triplet codon 都是 3 个 bases per codon 的呀.

2.in the sequence for sickle cell anaemia,adenine replaces thymine in a triplet, forming the triplet.during translation of the mutantm,the amino acid valine is incroporated into the haemoglobin molecule instead of glutamic acic.
what is the anticodon in the transfer molecule carrying this valine?

         normal    mutant
DNA      -CTT-     -CAT-
mRNA     -GAA-     -GUA-
a.acid  glutamic   valine

codon of mutant = GUA
anticodon       = CAU

蝎杰

选择它有答案，如下:
A ３ B ４ C ６ D８

铁蛋

蝎杰 于 7-1-2005 06:09 PM  说 :
不好意思，又是我，还有问题不懂...
1.if there were34amino acids and DNA only contained 2 types of nitrogenous bases, what would be the minimum number pf bases per codon that could for protein?
2.i ...

1. 答案是 6. 因为 2^6 = 64, 一共有 64  codon vs 34 个 amino acid, 剩下的 30 个 codon 使到这个 genetic code degenerate. 了解的方法和 4 个 base 一样. 4^3 = 64 codon vs 20 amino acid, 所以出现每个amino acid 超过 1 个 codon 代表 (degeneracy of the genetic code). 当然，2^8 也是可能的，可是问题要 minimum number.

creme

有谁做了plant project?
怎么知道名字??我看到头都大了.....
很多花都差不多嘛....

别问

不好意思，有些问题想请教。。。
关于糖尿病的

这是我在某网站找到的
Persistent hyperglycemia can draw water into the eyes and cause visual blurring. This may persist for several weeks after correcting the hyperglycemia because of the slow diffusion of glucose out of the eyes.

而我学校老师给的笔记是这样
symptoms of daibetes melliltus
# poor vision due to osmotic loss of water from the eye lens

是哪个才对呢？
（还是我搞错它的意思？）

浮萍

Blurred vision : when the blood glucose is high, there is also a higher glucose content in the lens because the extra blood glucose will gets into the lens of the eye， 那lens 里的水就少了（osmotic loss of water from the lens ). This causes the lens to absorb extra water (osmosis), in an attempt to dilute the glucose in the lens. When extra water enters the lens, it alters the shape of it. The lens loses its ability to focus light for the eye, which you perceive as blurred vision.
应该是酱吧???

别问

呵呵，还是不大懂
是不是这两句其实都是对的？
首先是失去水分，接下来就向外拿水？

还有osmotic loss of water 是什么意思？

不好意思。。。

[ Last edited by 别问 on 26-1-2005 at 09:04 PM ]

别问

浮萍 于 26-1-2005 04:36 PM  说 :
the extra blood glucose will gets into the lens of the eye， 那lens 里的水就少了（osmo ...

是不是其实不是水少了，而是成 hypertonic？

还有我想问，当level of blood glucose is high,其实那glucose会diffuse 进cell of the body 的吗？
毕竟有gradiant of cincentration

可以的话，为什么会发生muscle wasting 及tiredness 呢？
不行的话，又是为什么?

（希望我的问题不会太白痴。。）

[ Last edited by 别问 on 26-1-2005 at 09:24 PM ]

浮萍

不晓得你的老师所说的osmotic loss of water from the lens 是否就是指lens hypertonic， 你还是去请教他吧。

For blood glucose to get into the cells 需要insulin， 不是因为conc gradient

When we eat, the pancreas is supposed to automatically produce the right amount of insulin to move glucose from blood into our cells. In people with diabetes, however, the pancreas either produces little or no insulin, or the cells do not respond appropriately to the insulin that is produced. Glucose builds up in the blood, overflows into the urine, and passes out of the body. Thus, the body loses its main source of fuel even though the blood contains large amounts of glucose.

雨天之魂

别问 于 26-1-2005 20:54  说 :

呵呵，还是不大懂
是不是这两句其实都是对的？
首先是失去水分，接下来就向外拿水？

还有osmotic loss of water 是什么意思？

不好意思。。。

glucose 水平提高会令水的纯度下降. 形成 hypertonic.
旁边细胞的水份(hypotonic)就会涌入来化淡它
.
osmotic loss of water 应该是旁边的细胞失水.

如果真的 glucose 是进到 eye lens 里的话, (还没明白为什么会进去)
eye lens 本身会被水涌入. 不会失水.

我不清楚它的构造, 没有时间去做研究.
不过关于 osmosis 的部分确实是我说的那样.

雨天之魂

还有我想问，当level of blood glucose is high,其实那glucose会diffuse 进cell of the body 的吗？
毕竟有gradiant of cincentration

transport of glucose across membrane 的方式是 facilitated diffusion.
而相关的 channel protein 是一个 uniport system.

Molecules may cross the membrane in either direction, depending only on the direction of the gradient. ...
..
Examples of passive mediated transport.
Glucose transport in many cells (by a uniport system)
....
Uniport: a single molecule is moved in one direction across the membrane.

糖会 diffuse 入细胞里., 可是更大部分的是细胞的水份会涌出来尝试化淡它.
举例来说, 高血糖的人容易头晕, 因为脑部细胞的水份会被扯出来化淡血管里的高糖份.

可以的话，为什么会发生muscle wasting 及tiredness 呢？
不行的话，又是为什么?

muscle wasting 跟高血糖没有关系.
它也叫 muscle atrophy, 或肌肉退化. 主要是因为少用到肌肉才会导致它退化.

The majority of muscle atrophy in the general population results from disuse

muscle tiredness 也没什么关系. 它是因为 oxygen debt.
关于 oxygen debt 的, 请看下文:

Physiological state produced by vigorous exercise,
in which the lungs cannot supply all the oxygen that the muscles need.
In other words, the lungs and bloodstream, pumped by the heart,
cannot supply sufficient oxygen for aerobic respiration in the muscles

bluedog

想知道lethal genes 的example 有一个是
fur colour in mice
wild mice have grey-coloured fur,a condition know as agouti.some mice have yellow fur,cross breeding yellow mice produces offsrring in the ratio 2 yellow fur :1 agouti fur(不是3:1因为yellow is dominant to agouti and that all the yellow coat mice are heterozygous.the stypical Mendelian ratio is explained by the fetal death of homozygous yellow coat mice.Examination of the uteri of pregnant yellow mice from the above cross revealed dead yellow fetuses.similiar examination of the uteri of crosses between yllow fur and agouti fur mice revealed no dead yellow fetuses.The explanation is that this cross would not produce homozygous yellow (YY)mice!)

想知道为什么YY的fetuses会死，还有，when哪个fetus 死在妈妈的uterus后，哪个妈妈还能活吗？哪个fetus的尸体有从妈妈的uterus出来吗，还是流在uterus里？
thanks u!!!

[ Last edited by bluedog on 27-1-2005 at 02:14 PM ]

别问

浮萍 于 26-1-2005 10:16 PM  说 :
不晓得你的老师所说的osmotic loss of water from the lens 是否就是指lens hypertonic， 你还是去请教他吧。

For blood glucose to get into the cells 需要insulin， 不是因为conc gradient

When we eat ...

谢谢你哦
不过我一直单纯的以为，glucose可以facilitated diffusion 进cell。。。
所以。。。