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a-level生物难吗(干货整理A-level生物A)

a-level生物难吗(干货整理A-level生物A)C = 3 = triose 三糖(CH2O)n where n is a number between 3 and 9. They are classified according to the number of carbon atoms. The monosaccharides you will have to know fall into these categories (CH2O)n 其中n是3到9之间的数字。它们是根据碳原子的数量来分类的。你必须知道的单糖属于这些类别。:Carbohydrates contain 3 elements 碳水化合物包含3种元素:Carbohydrates are found in one of three forms 碳水化合物以三种形式存在:General formula 通式:

A-Level Biology(生物)越来越受欢迎,但据说学习内容太多,让不少小伙伴望而生畏。Alevel生物的覆盖面比较广,知识点比较多,涵盖了动物,植物,微生物三个内容,而且从宏观(比如生物圈)到微观(比如涉及到细胞结构,生物反应等),具有背诵和记忆的特点,所以需要学生具有扎实的基础,对各个章节的知识点都牢固地掌握。如果知识基础不牢固的话,接下来的知识的应用,做题也增加了一定难度。

另外A-Level生物的深度比较深,就是一个知识点不仅仅需要你记住,还要在记住的基础上学会运用,学会解决实际问题。比如光合作用photosynthesis,呼吸作用respiration,我们不仅要记住它们的反应过程,还要学会怎样测量反应速率,怎样运用影响因素来实现生产过程中的最大化。

a-level生物难吗(干货整理A-level生物A)(1)

今天给大家带来的笔记是:生物分子和酶--碳水化合物

Biological Molecules and Enzymes——Carbohydrates

Carbohydrates contain 3 elements 碳水化合物包含3种元素:

  1. Carbon 碳 (C)
  2. Hydrogen 氢(H)
  3. Oxygen 氧(O)

Carbohydrates are found in one of three forms 碳水化合物以三种形式存在:

  1. Monosaccharides 单糖
  2. Disaccharides (both sugars) 双糖
  3. Polysaccharides 多糖

a-level生物难吗(干货整理A-level生物A)(2)

Monosaccharides 单糖

General formula 通式:

(CH2O)n where n is a number between 3 and 9. They are classified according to the number of carbon atoms. The monosaccharides you will have to know fall into these categories (CH2O)n 其中n是3到9之间的数字。它们是根据碳原子的数量来分类的。你必须知道的单糖属于这些类别。:

C = 3 = triose 三糖

C = 4 = tetrose 四糖

C = 5 = pentose 戊糖

C = 6 = hexose 六糖

Trioses: (e.g. glyceraldehydes) intermediates in respiration and photosynthesis.

Tetroses: rare.

Pentoses: (e.g. ribose ribulose) used in the synthesis of nucleic acids (RNA and DNA) co-enzymes (NAD NADP FAD) and ATP.

Hexoses: (e.g. glucose fructose) used as a source of energy in respiration and as building blocks for larger molecules.

All but one carbon atom have an alcohol (OH) group attached. The remaining carbon atom has an aldehyde or ketone group attached.

三糖:(如甘油醛),呼吸作用和光合作用的中间产物。

四糖:罕见。

五糖:(如核糖、核酮糖),用于合成核酸(RNA和DNA)、辅酶(NAD、NADP、FAD)和ATP。

六糖:(如葡萄糖、果糖),在呼吸作用中作为能量来源,并作为较大分子的构建块。

除一个碳原子外,所有的碳原子都有一个酒精(OH)基团相连。剩余的碳原子连接有一个醛或酮基。

Chain form 链条形式:

a-level生物难吗(干货整理A-level生物A)(3)

Ring form 圆环形式:

Due to the bond angles between the carbon atoms it is possible for pentoses and hexoses to form stable ring structures. The carbon atoms are numbered 1 to 5 in pentoses and 1 to 6 in hexoses.

Depending on the orientation of the OH group on carbon 1 the monosaccharide can have either α or β configurations.

由于碳原子之间的键角,五糖和六糖有可能形成稳定的环状结构。五糖中的碳原子编号为1至5,六糖中的碳原子编号为1至6。

根据碳1上OH基的方向,单糖可以有α或β构型。

a-level生物难吗(干货整理A-level生物A)(4)

Disaccharides and glycosidic bonds 双糖和糖苷键

These are formed when two monosaccharides are condensed together. One monosaccharide loses an H atom from carbon atom number 1 and the other loses an OH group from carbon 4 to form the bond.

The reaction which is called a condensation reaction involves the loss of water (H2O) and the formation of an 1 4-glycosidic bond. Depending on the monosaccharides used this can be an α-1 4-glycosidic bond or a β-1 4-glycosidic bond.

这是在两个单糖凝结在一起时形成的。一个单糖从1号碳原子上失去一个H原子,另一个从4号碳原子上失去一个OH基团以形成该键。

这个反应被称为缩合反应,涉及到水(H2O)的损失和1 4-糖苷键的形成。根据所使用的单糖,这可以是一个α-1 4-糖苷键或一个β-1 4-糖苷键。

a-level生物难吗(干货整理A-level生物A)(5)

The reverse of this reaction the formation of two monosaccharides from one disaccharide is called a hydrolysis reaction and requires one water molecule to supply the H and OH to the sugars formed.

Examples of Disaccharides

Sucrose: glucose fructose

Lactose: glucose galactose

Maltose: glucose glucose.

Maltose: glucose glucose.

Sucrose is used in many plants for transporting food reserves often from the leaves to other parts of the plant. Lactose is the sugar found in the milk of mammals and maltose is the first product of Starch digestion and is further broken down to glucose before absorption in the human gut.

这个反应的反面,即从一个二糖形成两个单糖,被称为水解反应,需要一个水分子向形成的糖提供H和OH。

二糖的例子

蔗糖:葡萄糖 果糖。

乳糖:葡萄糖 半乳糖。

麦芽糖:葡萄糖 葡萄糖。

麦芽糖:葡萄糖 葡萄糖。

蔗糖在许多植物中用于运输食物储备,通常从叶子到植物的其他部分。乳糖是哺乳动物乳汁中的糖,麦芽糖是淀粉消化的第一个产物,在人体肠道吸收之前进一步分解为葡萄糖。

a-level生物难吗(干货整理A-level生物A)(6)

Biochemical tests 生物化学测试

All monosaccharides and some disaccharides including maltose and lactose are reducing sugars. These can be tested for by adding Benedict's reagent to the sugar and heating in a water bath. If a reducing sugar is present the solution turns green then yellow and finally produces a brick red precipitate. Non-reducing sugars can also be tested for using Benedict's reagent but first require addition of an acid and heating to hydrolyse (break apart) the sugar. The acid must then be neutralised using an alkali like sodium hydroxide before carrying out the test as described above.

所有的单糖和一些双糖包括麦芽糖和乳糖都是还原糖。可以通过在糖中加入本尼迪克特试剂并在水浴中加热来检测。如果存在还原糖,溶液会变成绿色,然后变成黄色,最后产生砖红色沉淀。非还原性糖也可以用本尼迪克特试剂进行检测,但首先需要加入酸并加热以水解(分解)糖。然后必须用氢氧化钠等碱性物质中和酸,再进行上述测试。

Polysaccharides 多醣体

a-level生物难吗(干货整理A-level生物A)(7)

Polysaccharide

多糖:

Function

职能:

Structure

结构:

Relationship of structure to function

结构与功能的关系:

starch

淀粉

Main storage polysaccharide in plants.

植物中的主要储藏多糖。

Made of 2 polymers - amylose and amylopectin.

Amylose: a polymer of glucoses joined by α-1 4-glycosidic bonds. Forms a helix with 6 glucose molecules per turn and about 300 per helix.

Amylopectin: a polymer of glucoses joined by α-1 4-glycosidic bonds but with branches of α-1 6-glycosidic bonds. This causes the molecule to be branched rather than helical

由两种聚合物组成--直链淀粉和直链淀粉。

直链淀粉:由α-1 4-糖苷键连接的葡萄糖的聚合物。形成一个螺旋,每圈有6个葡萄糖分子,每个螺旋约有300个葡萄糖分子。

直链淀粉:由α-1 4-糖苷键连接的葡萄糖的聚合物,但有α-1 6-糖苷键的分支。这导致分子是支链的,而不是螺旋的。

Insoluble therefore good for storage.

Helix is compact.

The branches mean that the compound can easily hydrolysed to release the glucose monomers.

不溶于水,因此利于储存。

螺旋状结构紧凑。

分支意味着该化合物很容易水解以释放葡萄糖单体。

glycogen

糖原

Main storage polysaccharide in animals and fungi

动物和真菌的主要储存多糖

Similar to amylopectin but with many more branches which are also shorter.

与直链淀粉相似,但有更多的分支,也更短。

The number and length of the branches means that it is extremely compact and very fast hydrolysis.

枝条的数量和长度意味着它非常紧凑,水解速度非常快。

Cellulose

纤维素

Main structural constituent of plant cell walls

植物细胞壁的主要结构成分

Adjacent chains of long unbranched polymers of glucose joined by β-1 4-glycosidic bonds hydrogen bond with each other to form microfibrils.

由β-1 4-糖苷键连接的葡萄糖长而不分支的聚合物的相邻链互相氢键,形成微纤维。

The microfibrils are strong and so are structurally important in plant cell walls.

微纤维很强,因此在植物细胞壁中结构上很重要。

a-level生物难吗(干货整理A-level生物A)(8)

Functions of carbohydrates 碳水化合物的功能
  1. Substrate for respiration (glucose is essential for cardiac tissues).
  2. Intermediate in respiration (e.g. glyceraldehydes).
  3. Energy stores (e.g. starch glycogen).
  4. Structural (e.g. cellulose chitin in arthropod exoskeletons and fungal walls).
  5. Transport (e.g. sucrose is transported in the phloem of a plant).
  6. Recognition of molecules outside a cell (e.g. attached to proteins or lipids on cell surface membrane).

呼吸作用的底物(葡萄糖对心脏组织是必不可少的)。

呼吸作用的中间物(如甘油醛)。

能量储存(如淀粉、糖原)。

结构(如纤维素、节肢动物外骨骼和真菌壁中的几丁质)。

运输(如蔗糖在植物的韧皮部被运输)。

识别细胞外的分子(如附着在细胞表面膜上的蛋白质或脂质)。

a-level生物难吗(干货整理A-level生物A)(9)

Biochemical test 生物化学试验

Iodine solution or potassium iodide solution can be used to test for the presence of starch. A positive result changes the solution from an orange-brown to a blue-black colour. - refer to gcse and title biochemical test for carboydrate.

碘溶液或碘化钾溶液可以用来测试淀粉的存在。阳性结果使溶液从橙褐色变为蓝黑色。- 参考GCS和职称生化测试中的卡博德酸盐。

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