Carbohidratos y lípidos
Diploma del Bachillerato Internacional · IB Diploma · Biology · HL Tema 10 4:36 Narración en inglés · Subtítulos en inglés + 中文 quemados en pantalla
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Transcripción
Starch, a triglyceride and a protein can all contain carbon, hydrogen and oxygen, yet their structures and functions differ.
淀粉、甘油三酯和蛋白质都含有碳、氢和氧，但它们的结构和功能各不相同。
An element list alone does not identify a biomolecule.
仅凭元素列表无法鉴定一种生物分子。
Carbohydrates include monosaccharides and polymers joined by glycosidic bonds.
碳水化合物包括单糖和通过糖苷键连接的多聚体。
A triglyceride has glycerol joined to three fatty acids by ester bonds.
甘油三酯由甘油与三个脂肪酸通过酯键连接而成。
Proteins contain amino acid residues joined by peptide bonds, with side chains and folding contributing to their properties.
蛋白质含有通过肽键连接的氨基酸残基，侧链和折叠方式决定了其性质。
These structures support storage, membrane, or functional roles depending on the molecule.
根据分子不同，这些结构分别承担储存、膜或功能作用。
Peptide bond is a covalent linkage between amino acid residues.
肽键是氨基酸残基之间的共价连接。
Hydrolysis is breaking a linkage using water.
水解是利用水断裂化学键的过程。
Condensation forms a linkage with a small molecule, such as water, released in the simplified model.
缩合反应形成化学键，并在简化模型中释放水等小分子。
Hydrolysis uses water to break such a linkage.
水解反应则利用水来断裂此类化学键。
Polymer sequence, branching, polarity, and three-dimensional shape matter.
聚合物的序列、分支、极性和三维形状都很重要。
Saturated fatty acids have no carbon-carbon double bond, while unsaturated fatty acids have at least one. This difference can influence packing under stated conditions.
饱和脂肪酸不含碳碳双键，而不饱和脂肪酸至少含有一个；这一差异可能影响其在特定条件下的堆积方式。
In the lab: Use labelled molecular models and identify the actual linkage rather than memorizing shapes alone.
在实验室中，使用带标签的分子模型来识别实际的连接方式，而不仅仅是记忆形状。
Compare an attributed structure with the proposed function, mark polar and non-polar regions where justified, and distinguish a monomer from a residue within a polymer.
将指定的结构与提出的功能进行比较，在合理之处标记极性和非极性区域，并区分单体与聚合物中的残基。
Food tests provide evidence of chemical groups under specific conditions, not complete molecular structures.
食品测试仅在特定条件下提供化学基团的证据，而非完整的分子结构。
Known: a single unbranched peptide formed from 20 amino acids contains 19 peptide linkages.
已知：由20个氨基酸组成的单条无分支肽链包含19个肽键。
In a simplified condensation accounting model, forming these links releases 19 water molecules.
在简化的缩合反应计量模型中，形成这些键会释放19个水分子。
A triglyceride formed from one glycerol and three fatty acids has three ester linkages and releases three waters in the corresponding model.
由一个甘油和三个脂肪酸形成的甘油三酯含有3个酯键，并在相应模型中释放3个水分子。
These are linkage counts, not complete pathways in living cells.
这些仅为键的数量统计，并非活细胞内的完整代谢途径。
Your turn: A single unbranched peptide contains 35 amino-acid residues.
轮到你了：一条无分支肽链含有35个氨基酸残基。
How many peptide bonds join them?
连接它们的肽键有多少个？
Pause the video and work it out.
请暂停视频，自己算一算。
Check: one bond fewer than the residues, so 34 peptide bonds.
核对：肽键数比氨基酸残基少一个，即 34 个肽键。
Not every protein is an enzyme, and not every catalyst is a protein.
并非所有蛋白质都是酶，也并非所有催化剂都是蛋白质。
Lipids are not all polymers of repeating monomers.
脂质并不全是重复单体的聚合物。
A protein can lose function when its folding changes without every peptide bond being broken.
当蛋白质的折叠方式发生改变时，即使肽键未全部断裂，其功能也可能丧失。
Not true: Every protein is an enzyme and every lipid is a repeating-monomer polymer.
不正确。 并非所有蛋白质都是酶，也并非所有脂质都是由重复单体构成的聚合物。
Enzymes are one functional class of protein; structural and transport proteins are not enzymes.
酶只是蛋白质的一个功能类别，结构蛋白和运输蛋白不属于酶。
And lipids such as triglycerides are built from unlike parts, not repeating monomers, so they are not polymers.
此外，像甘油三酯这样的脂质是由不同组分构成的，而非重复单体，因此它们不是聚合物。
Carbohydrates include monosaccharides and polymers joined by glycosidic bonds.
碳水化合物包括单糖以及通过糖苷键连接的聚合物。
Condensation forms a linkage with a small molecule such as water released in the simplified model.
缩合反应形成化学键，在简化模型中会释放水等小分子。
Hydrolysis uses water to break such a linkage.
水解反应则利用水来断裂此类化学键。
Watch for: not every protein is an enzyme, and not every catalyst is a protein.
请注意：并非所有蛋白质都是酶，也并非所有催化剂都是蛋白质。
A single unbranched peptide has 28 amino acid residues.
一条无分支肽链含有28个氨基酸残基。
Find the number of peptide bonds.
求肽键的数量。
Condensation releases water in the simplified model.
在简化模型中，缩合反应会释放水分子。
The result is 27 peptide bonds.
结果是27个肽键。
Condensation joins two monomers and releases one water for each new bond.
缩合把两个单体连接起来，每形成一个键就释放一分子水。
Hydrolysis runs the other way: each bond broken uses one water.
水解方向相反：每断开一个键就消耗一分子水。
Two separate linear peptides each contain 10 residues.
两条独立的线性肽链各含十个残基。
How many peptide bonds exist before and after joining them end to end by one peptide linkage?
在通过一个肽键将它们首尾相连之前和之后，分别有多少个肽键？
State the model boundary.
请说明模型边界。
At first there are 2 times 9, which is 18 peptide bonds; the 20-residue chain has 19.
起初有 2 乘 9，即 18 个肽键；连成 20 个残基的链后有 19 个。
One new bond forms.
新形成了一个肽键。
This is linkage accounting, not a complete biosynthetic pathway.
这只是对化学键的计数，而不是完整的生物合成途径。
Pause and check yourself. Explain each point on screen in your own words, then change one number in the last example and say which assumption must still hold.
暂停一下，自我检查：用自己的话解释屏幕上的每个要点，再改动最后一个例子中的一个数字，说明哪个假设仍然必须成立。