β-Glucans and Pentosans and their Degradation Products in Commercial Beers

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生物专业英语第六课-5

生物专业英语第六课-5

作业
翻译句子
① ② ③ ④ ⑤
Lysosome contain…… Chlorophyll a molecule…… Meiosis is a special…… He also recorded …… P.A.levene discovered that……
课堂练习 判断
① Both prokaryotic and enkaryotic cells have a supporting cytoskeleton. ② The nuclear envolope is produced by the rough endoplasmic reticulum. ③ The Calvin-Benson cycle may take place either in light or in the dark. ④ Light-dependent reactions take place in the chloroplast stroma.
系统 进化分枝图
作业 复习前五课的重点词汇和句式 课堂练习 选择 ① Chromoplasts are a type of ( ). A.pigment B.storage bin C.nutrient tank D.plastid ② During protein synthesis a single mRNA molecule may be associated with several ribosomes to form ( ). A.a gene B.a lysosome C.a polysome D.smooth endoplasmic reticulum

③ In the first stage of photosynthesis water is ( ). A.oxidated B.phosphotylated C.reduced D.heated ④ In the ligjt-independent reactions of photosynthesis, the first electon acceptor for atmospheric CO2 is ( ) A.PGA. B. ATP C. RUBP D. a carbohydrate subunit ⑤ DNA is replicate during the ( ). A.S phase B. M phase C.G2 phase D. G1phase

生物专业英语 课文翻译

生物专业英语 课文翻译

教学内容:Cytoplasm: The Dynamic, Mobile Factory细胞质:动力工厂Most of the properties we associate with life are properties of the cytoplasm. Much of themass of a cell consists of this semifluid substance, which is bounded on the outside by theplasma membrane. Organelles are suspended within it, supported by the filamentous networkof the cytoskeleton. Dissolved in the cytoplasmic fluid are nutrients, ions, soluble proteins,and other materials needed for cell functioning.生命的大部分特征表现在细胞质的特征上。

细胞质大部分由半流体物质组成,并由细胞膜(原生质膜)包被。

细胞器悬浮在其中,并由丝状的细胞骨架支撑。

细胞质中溶解了大量的营养物质,离子,可溶蛋白以及维持细胞生理需求的其它物质。

The Nucleus: Information Central(细胞核:信息中心)The eukaryotic cell nucleus is the largest organelle and houses the genetic material (DNA) on chromosomes. (In prokaryotes the hereditary material is found in the nucleoid.) The nucleusalso contains one or two organelles-the nucleoli-that play a role in cell division. Apore-perforated sac called the nuclear envelope separates the nucleus and its contents from the cytoplasm. Small molecules can pass through the nuclear envelope, but larger molecules suchas mRNA and ribosomes must enter and exit via the pores.真核细胞的细胞核是最大的细胞器,细胞核对染色体组有保护作用(原核细胞的遗传物质存在于拟核中)。

分子生物学双语复习知识点

分子生物学双语复习知识点

生命基本特征(本质):生命是生物体所表现出来的复合现象,包括:自身繁殖,生长发育,新陈代谢,遗传变异,对刺激产生反应等生物学Biology:是一门研究生命的现象与本质及活动规律的科学。

它包揽了生命的各个方面,从生命的化学组成,细胞的结构与功能,个体生物学,生物的多样性,到生物的遗传、进化及生态等方面的完整知识体系。

分子生物学Molecular Biology:它是研究核酸、蛋白质等生物大分子的结构与功能,并从分子水平上阐述蛋白质与核酸、蛋白质与蛋白质之间相互作用的关系及其基因表达调控机制的学科,是人类从分子水平上真正揭开生物世界的奥秘,由被动地适应自然界转向主动地改造和重组自然界的基础学科。

Is a subject to understand the five basic cell behavior patterns (growth, division, specialization, movement, and interaction) in terms of the various molecules that are responsible for them.That is, molecular biology wants to generate a complete description of the structure, function, and interrelationships of the cell’s macromolecules, and thereby to understand why living cells behave the way they do.分子生物学的研究内容Research Contents of Molecular Biology :生物大分子的结构功能研究(结构部分,又称结构分子生物学):包括基因、基因组的结构;DNA 复制、转录、翻译(功能部分);基因表达调控研究(调控部分);DNA重组技术(又称基因工程)Structure and Function of Macromolecules (Structural Part, Also Known as Structural & Molecular Biology);DNA Replication, Transcription, Translation (Functional Part);Regulation of Gene Expression (Control Section);Recombinant DNA Technology (Genetics)F.Miescher就发现了核素(nuclein);Boyer 和Berg等发展了重组DNA技术,完成了第一个细菌基因的克隆;Sanger 等发明了DNA测序技术;Sanger、Maxam和Gilbert先后发明了三种DNA序列的快速测定法;Mullis等发明的聚合酶链式反应(PCR);90年代全自动核酸序列测定仪问世;生物芯片技术是生命科学研究中继基因克隆技术、PCR技术、基因自动测序技术后的又一次革命性技术突破;分子遗传学基本理论建立者Jacob和Monod最早提出的操纵元学说分子生物学的3条基本原理:构成生物体各类有机大分子的单体在不同生物中都是相同的;生物体内一切有机大分子的建成都遵循共同的规则;某一特定生物体所拥有的核酸及蛋白质分子决定了它的属性。

高三英语生物结构单选题50题

高三英语生物结构单选题50题

高三英语生物结构单选题50题1.The nucleus of a plant cell is responsible for_____.A.storing waterB.controlling cell activitiesC.producing energyD.breaking down waste答案:B。

本题考查植物细胞中细胞核的功能。

选项A,储存水是液泡的功能;选项C,产生能量主要是线粒体的功能;选项D,分解废物不是细胞核的主要功能。

而细胞核控制细胞的活动。

2.In animal cells, the organelle that is involved in packaging and transporting proteins is_____.A.lysosomeB.endoplasmic reticulumC.Golgi apparatusD.mitochondrion答案:C。

动物细胞中,高尔基体负责包装和运输蛋白质。

选项A,溶酶体主要是分解废物;选项B,内质网参与蛋白质的合成等;选项D,线粒体产生能量。

3.Which of the following is NOT a part of the cell membrane structure?A.PhospholipidsB.ProteinsC.CarbohydratesD.Nucleic acids答案:D。

细胞膜的结构主要由磷脂、蛋白质和少量的糖类组成。

核酸不是细胞膜的组成部分。

4.The cell wall of a plant cell is mainly made up of_____.A.celluloseB.proteinC.lipidD.starch答案:A。

植物细胞的细胞壁主要由纤维素组成。

选项B,蛋白质不是细胞壁的主要成分;选项C,脂质不是细胞壁的成分;选项D,淀粉主要存在于细胞内储存能量。

5.The organelle that is known as the “powerhouse” of the cell is_____.A.nucleusB.chloroplastC.mitochondrionD.endoplasmic reticulum答案:C。

小学上册D卷英语第五单元暑期作业(含答案)

小学上册D卷英语第五单元暑期作业(含答案)

小学上册英语第五单元暑期作业(含答案)英语试题一、综合题(本题有100小题,每小题1分,共100分.每小题不选、错误,均不给分)1.Which animal is known for its ability to fly?A. FishB. DogC. BirdD. Cat答案: C2.The __________ is a major river that flows into the Gulf of Mexico. (密西西比河)3.The tree is _______ (full) of fruit.4.The hen lays _________. (蛋)5.What is the largest continent on Earth?A. AfricaB. AsiaC. EuropeD. North America答案: B6.ecological footprint) reflects human activity's impact. The ____7.What do we call the time of year when it is very cold and often snows?A. SummerB. WinterC. SpringD. Fall答案:B8.What is the term for a young seagull?A. ChickB. HatchlingC. CalfD. Kit答案:a9. A chemical reaction can occur in _____ solutions.10.Plants can be used for __________ (食物和药物).11.The bear catches fish in the cold ____.12.The sun is ___ in the afternoon. (setting)13.The firefly lights up the _______ (夜空).14.The _______ (青蛙) often sits on lily pads.15.I think friendship is one of the greatest gifts. Friends support each other through thick and thin. I’m grateful for my friend __________, who always knows how to cheer me up.16.My brother loves to explore __________ (新地方).17. A saturated solution holds the maximum amount of _____.18.In a chemical reaction, products are on the ______ side of the equation.19. A solution that can conduct electricity is called an ______ solution.20.What do we call the place where we learn?A. HomeB. SchoolC. ParkD. Office答案: B21.The crow is often seen in _________. (城市)22.They are _____ (walking) the dog.23. A ______ (水培) system allows plants to grow in water.24.My room is __________ than yours.25.The fall of the Roman Empire occurred in the _______ century.26.The puppy is ________ (顽皮).27.The __________ (维也纳会议) aimed to restore order after Napoleon.28.The ________ loves to jump and explore new places.29.What do we call a person who studies the structure of the Earth?A. GeologistB. BiologistC. ChemistD. Physicist答案: A30.In chemistry, a reagent is a substance used to cause a _____.31.My favorite animal to see at the zoo is the ________ (大象).32.The __________ (绿叶) provide oxygen.33.I love visiting the __________ because it has beautiful flowers and plants.34.I like _____ (to cook/to eat).35.The butterfly flaps its wings ______ (优雅).36.The parrot can be very ________________ (吵闹).37.What do you call a baby seal?A. PupB. CalfC. KittenD. Cub答案: A38.The ________ (向日葵) turns towards the sun.39.My pet ______ (狗) loves to go for walks.40.If I could have any superpower, I would choose __________ because __________.41.I like to help my mom ________ (种花) in the garden.42.The __________ (历史的双重性) showcases complexity.43.Rain helps the plants to ______ (生长).44. A ____ is a friendly pet that often wags its tail.45.She is _____ (baking) a pie.46.What do you call a person who studies insects?A. BiologistB. EntomologistC. ZoologistD. Botanist答案:B47. A flower can have many different ______ (结构).48.The ________ is very tall and strong.49.I like to ______ (参与) in charity events.50.The _____ (叶子) change color in autumn.51.In _____ (澳大利亚), you can find kangaroos.52.The __________ (小径) through the woods is peaceful.53.I like to ______ in the ocean. (swim)54.Napoleon Bonaparte was a military leader from __________. (法国)55. A playful ___ (小猫) pounces on a toy.56.The leaves change _______ in the fall, creating a lovely sight.57.I have a toy _______ that can jump high and far across the room.58. A __________ is formed by the accumulation of sediments in a coastal area.59. A chemical change alters the _____ of a substance.60.My cat likes to sleep on ____.61.I have a toy ______ (火车) that goes around the ______ (轨道).62.The teacher is _____ (kind/mean) to us.63.The monkey is _____ in the tree. (swinging)64. A ________ is a geographic region defined by its climate.65.The __________ is an important tool for environmental management.66.I love to ______ (观察) the stars at night.67.Some _______ can survive drought conditions.68.My _______ (小狗) loves to play fetch.69.I enjoy listening to audiobooks while __________.70.What do you call the fear of heights?A. AcrophobiaB. ClaustrophobiaC. AgoraphobiaD. Nyctophobia答案:A71.The chemical symbol for phosphorus is ______.72.I can make music with my ________ (玩具类型).73.Which planet is known as the Red Planet?A. VenusB. EarthC. MarsD. Mercury答案: C74.The ______ of a tree is often wider than its trunk. (树的冠层通常比树干更宽。

大学生物专业英语lesson_five

大学生物专业英语lesson_five
Chargaff´s(查伽夫,美国生物化学家) 法则:(1) 在DNA中腺嘌呤与胸腺嘧啶含量相等的,胞嘧啶与鸟嘌呤 含量相等;(2)腺嘌呤胸腺嘧啶比例和胞嘧啶鸟嘌呤比 例随物种不同而不同。
3 The Research race for
the Molecular structure of DNA
In the late 1940s and early 1950s, researchers looking for the structure of DNA drew upon Chargaff's insight, Levene´s ideas on DNA components, and two other lines of evidence.
Nuclei acid, originally isolated by Johann Miescher in 1871, was identified as prime constituent of chromosomes through the use of the red-staining method developed by Feulgen in the early 1900s.
他们的工作为其他研究者精确阐明 “酶是 如何影响复杂新陈代谢途径”铺平了道路。
粗糙脉孢菌作为木质纤维素降解真菌,不 仅具有完整的木质纤维素降解酶系,而且 还拥有全基因组基因敲除突变体库,是研 究丝状真菌纤维素酶表达分泌和木质纤维 素降解机制的优秀体系。
国内外利用粗糙脉孢菌系统,在木质纤维 素降解机制方面取得了显著进展,包括纤 维素酶信号传导、调控以及生物质降解后 糖的转运利用等。 2019/11/27
Thirty years later Beadle and Ephrussi showed a relationship between particular genes and biosynthetic reactions responsible for eye color in fruit flies.

生物专业英语期末打印版

生物专业英语期末打印版

GillCapHomozygous(纯合的):A diploid organism that has two identical alleles for particular characteristic.Heterozygous (杂合的):A diploid organism that has two different alletic forms of a particular gene.Phenotype(表型):The physical, chemical, and psychological expression of genes possessed by an organism.Genotype(基因型):The catalog of genes of an organism, whether or not these genes are expressed.Allele(等位基因):Alternative forms of a gene for a particular characteristic (e.g., attached earlobe genes and free earlobe genes are alternative alleles for ear shape).Monohybrid Crosses(单因子杂种):a hybrid produced by crossing parents that are homozygous except for a single gene locus that has two alleles.Dihybrid crosses(双因子杂种,双因子杂合子):A cross between individuals that differ with respect to two specified gene pairs.Law of independent assortment(独立分配定律,自由组合定律)::Members of one gene pair will separate from each otherindependently of the members of other gene pairs.Semiconservative replication(半保留复制):The method of replication of DNA in which the molecule divides longitudinally,each half being conserved and acting as a template for the formation of a new strand.Lagging strand(后随链):is synthesized in short stretches known as Okazaki fragments.DNA polymerase( DNA 聚合酶):An enzyme that brings new DNA triphosphate nucleotides into position for bonding onanother DNA molecule.Age structure(年龄结构):Of a population, the number of individuals in each of several or many age categories.Density-dependent factors(密度依赖因子,密度制约因子):population-limiting factors that become more effective as thesize of the population increases.1.As mitosis proceeds, the spindle microtubules play a crucial role in ensuring that both paired and separated chromatids move in the right directions at the proper times. Each half of the spindle forms as microtubules extend from each pole of a dividing cell to the region of the metaphase plate. During prophase, other microtubules, the centromeric fibers, extend outward from the spindle poles to structures on the chromosomes called kinetochores. During anaphase the fibers begin to shorten, and the chromatids begin to move apart.在有丝分裂过程中,是纺锤体微管确保了染色单体在适当时间以正确方向进行分离。

第七课 翻译

第七课    翻译

Characteristics of FungiThe approximately 175,000 species of fungi include some of the simplest multicellular organisms. Fungi have a variety of lifestyles. They may be saprobes that decompose dead organic matter; they may be parasites which obtain nutrients from living hosts; or they may live in symbiotic relationships with algae or with the roots of higher plants. In spite of these variations, however, all fungi carry out extracellular digestion: they secrete enzymes that digest organic matter, and then they absorb the resulting nutrients.大约175,000种真菌包括了一些极其简单的多细胞生物。

无论它们多么不一样,所有的真菌都进行细胞外消化:它们分泌酶来消化有机物,然后吸收由此产生的营养。

它们可以是分解死的有机物的腐生菌;也可以是从活的宿主获得营养的寄生者;或者它们可以和藻类或高等植物的根以共生的关系生活在一起。

Most fungi have the same basic body structure consisting of a main body or thallus composed of filaments called hyphae. In most species the walls of hyphal cells contain chitin. Hyphae in certain species may become specialized to form rhizoids, which serve as rootlike anchors, or they may become the feeding structures known as haustoria. Finally, hyphae may or may not be septate-having cross walls that segregate independent cells, each with at least one nucleus. Lower fungi are coenocytic; that is, they are one mass of cytoplasm that contains multiple nuclei.无论它们多么不一样,所有的真菌都进行细胞外消化:它们分泌酶来消化有机物,然后吸收由此产生的营养。

英语六级巅峰阅读附详解 第59期-生物知识

英语六级巅峰阅读附详解 第59期-生物知识

英语六级巅峰阅读附详解第59期:生物知识If you go down to the woods today, you may meet high-tech trees genetically modified to speed their growthor improve the quality of their wood. Genetically-engineered food crops have become increasingly common, albeit controversial. over the past ten years. But genetic engineering of trees has lagged behind.Part of the reason is technical. Understanding. and then altering, the genes of a big pine tree are more complex than creating a better tomato. While tomatoes sprout happily, and rapidly, in the laboratory, growing a whole tree from a single, genetically altered cell in a test tube is a tricky process that takes years, not months. Moreover. little is known about tree genes. Some trees, such as pine trees. have a lot of DNA-roughly ten times as much as human. And, whereas the Human Genome Project is more than half-way throughits task of isolating and sequencing the estimated 100,00 genes in human cells. similar efforts to analyzetree genes are still just saplings (幼苗).Given the large number of tree genes and the little that is known about them, tree engineers are starting with a searchfor genetic "markers". The first step is to isolate DNA from trees with desirable propertiessuch as insect resistance. The next step is to find stretches of DNA that show the presence of a particular gene. Then, when you mate two trees with different desirable properties, it is simple to check which offspring contain them all by looking for the genetic markers. Henry Amerson, at North Carolina State University, is using genetic markers to breed fungal resistance into southern pines. Billions of these are grown across America for pulp and paper, and outbreaks of disease are expensive. But not all individual trees are susceptible. Dr. Amerson’s group has found markers that distinguish fungus-resistant stock from disease-prone ing traditional breeding techniques, they are introducing the resistance genes into pines on test sites in America.Using generic markers speeds up old-fashioned breeding methods becauseyou no longer have to wait for the tree to grow up to see if it has the desiredtraits. But it is more a sophisticated form of selective breeding. Now. however.interest in genetic tinkering (基因修补) is also gaining ground. To this end, Dr.Amerson and his colleagues are taking part in the Pine Gene Discovery Project. an initiative to identify and sequence the50,000-odd genes in the pine tree's genome. Knowing which gene does what should make it easier to know what to alter.如果你今天走到森林里去,可能会看到高科技的树木——这些树通过基因改良促进了生长的速度或提高了木材的质量。

stiky ends名词解释

stiky ends名词解释

stiky ends名词解释sticky ends是一种在DNA分子的末端产生的粘性突出部分。

具体的英文解释是“the protruding ends of a DNA molecule that result from the action of restriction enzymes, often with a single-stranded overhang”。

sticky ends在DNA重组、连接和应用于基因工程方面起着重要的作用。

1. Restriction enzymes cleave DNA molecules, leaving sticky ends that can be used for DNA recombination.2. The sticky ends of the DNA fragments were ligated together using DNA ligase.3. The sticky ends of two complementary DNA molecules can be annealed together to form a double-stranded DNA molecule.4. The restriction enzyme EcoRI creates sticky ends witha single-stranded overhang of nucleotides.5. Sticky ends allow for precise and efficient DNA manipulation in genetic engineering.6. The sticky ends of DNA fragments can be labeled with fluorescent tags for visualization.7. The sticky ends of DNA can be used to insert aspecific gene sequence into a plasmid vector.8. The sticky ends of the cut DNA fragments can anneal to each other, forming recombinant DNA.9. The sticky ends of the DNA molecules facilitate the annealing process during PCR amplification.10. The sticky ends of a DNA molecule can be ligated to an adaptor sequence for next-generation sequencing applications.11. By using different restriction enzymes, sticky ends with different overhang sequences can be generated.12. The sticky ends of DNA fragments allow for the directional cloning of genes into expression vectors.13. The sticky ends of DNA molecules can be used to create chimeric proteins through gene fusion.14. The sticky ends of DNA are recognized by complementary base pairing with another DNA molecule.15. Sticky ends can be modified with specific functional groups for targeted DNA manipulation.16. The sticky ends of DNA can be blunted using a DNA polymerase to prevent self-ligation.17. The sticky ends of DNA fragments can be purified using gel electrophoresis.18. The sticky ends of DNA fragments provide a site for hybridization with complementary nucleic acid probes.19. The sticky ends of DNA fragments allow for the directional insertion of genes into expression vectors.20. The sticky ends of DNA can be annealed together in a hybridization reaction to form a DNA concatamer.。

高二英语询问植物界别单选题50题

高二英语询问植物界别单选题50题

高二英语询问植物界别单选题50题1. In the botanical garden, we saw a plant with seeds that are not enclosed in a fruit. Which type of plant is it?A. AngiospermB. GymnospermC. FernD. Moss答案:B。

解析:裸子植物(Gymnosperm)的种子是裸露的,不被果实包裹,这是裸子植物的重要特征。

被子植物(Angiosperm)的种子是被果实包裹的,蕨类植物(Fern)主要通过孢子繁殖,苔藓植物 Moss)也是通过孢子繁殖,所以A、C、D选项不符合。

2. A researcher in the plant laboratory is studying a plant that has flowers and produces seeds enclosed in a fruit. This plant is most likely to be a(n) _.A. GymnospermB. BryophyteC. AngiospermD. Pteridophyte答案:C。

解析:被子植物 Angiosperm)具有花朵,并且种子被包裹在果实内,这是被子植物区别于其他植物类群的典型特征。

裸子植物(Gymnosperm)种子裸露,苔藓植物(Bryophyte)没有种子,靠孢子繁殖,蕨类植物(Pteridophyte)也是靠孢子繁殖,所以A、B、D选项错误。

3. When we walked in the botanical garden, we noticed a plant without flowers but with cones. What kind of plant is it?A. AngiospermB. GymnospermC. AlgaeD. Fungi答案:B。

stiky ends名词解释

stiky ends名词解释

stiky ends名词解释Sticky ends (名词解释) :Sticky ends是DNA融合实验中常用的术语,用于描述由于酶切后DNA链断裂形成的具有突出单链DNA序列的末端。

这些具有互补的序列的单链DNA片段能够与其他DNA片段很好地结合,从而形成较稳定的DNA连接,常用于基因工程和重组DNA技术中。

Double-Stranded DNA sticky ends (中文:双链DNA黏性末端)是指由于酶切作用后两个DNA单链的部分被切除而留下的具有互补序列的末端。

这些末端的单链DNA片段具有黏性,可以与其他DNA序列的黏性末端互补结合,形成稳定的DNA连接。

以下是24个双语例句:1. The DNA fragments were ligated together using their sticky ends.DNA片段通过连接其黏性末端进行了连接。

2. Restriction enzymes can produce sticky ends on DNA fragments.限制性酶可以在DNA片段上产生黏性末端。

3. The sticky ends of the DNA can hybridize with complementary sequences.DNA的黏性末端可以与互补序列进行杂交。

4. The sticky ends allow for the formation of recombinant DNA molecules.黏性末端有助于形成重组DNA分子。

5. The DNA fragments were ligated at their sticky ends to create a longer sequence.DNA片段通过其黏性末端连接以创建更长的序列。

6. The sticky ends of the DNA were annealed to form a stable hybrid.DNA的黏性末端被退火形成一个稳定的杂交体。

高二英语生物分类单选题50题

高二英语生物分类单选题50题

高二英语生物分类单选题50题1. Which of the following belongs to the phylum Arthropoda?A. EarthwormB. StarfishC. ButterflyD. Sponge答案:C。

解析:节肢动物门(Arthropoda)的典型特征包括具有分节的附肢等。

蝴蝶(Butterfly)属于节肢动物门。

蚯蚓(Earthworm)属于环节动物门 Annelida)。

海星 Starfish)属于棘皮动物门Echinodermata)。

海绵 Sponge)属于多孔动物门 Porifera)。

2. The organism which is classified in the class Mammalia should have the following feature:A. Gills for breathingB. Feathers on the bodyC. Hair or fur and produce milk to feed their youngD. Scales on the body答案:C。

解析:哺乳纲(Mammalia)的生物具有毛发或皮毛并且能够产奶哺育幼崽。

用鳃呼吸(Gills for breathing)是鱼类等水生生物的特征,它们属于鱼纲等,不属于哺乳纲。

身上有羽毛(Feathers on the body)是鸟类的特征,鸟类属于鸟纲(Aves)。

身上有鳞片(Scales on the body)是爬行动物等的特征,爬行动物属于爬行纲(Reptilia)。

3. Which kingdom does the mushroom belong to?A. AnimaliaB. PlantaeC. FungiD. Protista答案:C。

解析:蘑菇属于真菌界(Fungi)。

动物界(Animalia)的生物具有能运动、异养等特点。

(英汉对照)分子生物学-- 名词解释

(英汉对照)分子生物学-- 名词解释

α helix α螺旋A helical secondary structure in proteins.Pl. α helices. 蛋白质中一种螺旋形的二级结构。

复数:α helices。

α-amanitin α鹅膏蕈碱A toxin that inhibits the three eukaryotic RNA polymerases to different extents. Name derives from mushroom of genus Amanita in which toxin is found. 一种能不同程度地抑制三种真核生物RNA聚合酶的毒素。

名称来自于产生此毒素的Amanita属蘑菇。

β-galactosidase β-半乳糖苷酶Enzyme that cleaves lactose into galactose and glucose. Name origin: the bond cut by this enzyme is called a β-galactosidic bond. 将乳糖分解为半乳糖和葡萄糖的酶。

名称来源:该酶切割的键称为β-半乳糖苷键。

β sheet β折叠A secondary structure in proteins, relatively flat and formed hydrogen bonding between two parallel or anti-parallel stretches of polypeptide. 蛋白质的一种二级结构,相对平坦,在两条平行的或反向平行的肽段之间形成氢键。

σ subunit σ亚基Component of prokaryotic RNA polymerase holoenzyme. Required for recognition of promoters. 原核生物RNA聚合酶全酶的组成成分。

专英三课文译文

专英三课文译文

How Biologists Define a SpeciesModern biology generally define a species as group of actually or potentially interbreeding populations that is reproductively isolated from the such groups. Members of a species can interbreed with each other, but they cannot breed with organisms belonging to another species. One advantage of the standard of reproductive isolation is that it is very precise. Notice, however, that it can only be applied to organisms that reproduce sexually. Asexual reproducers, including most prokaryotes, many plants, and some animals, must be classified into species on the basis of physical (biochemical or morphological) traits.现代生物学大体将物种定义为实际上的一群或潜在的杂交群体,即从这个群体隔离繁殖的后代。

种族中个体可以互相交配,但种间不能。

这种繁殖隔离的一个优势是很精确。

但仅适用于性繁殖的生物。

非性繁殖,包括大多数原核生物,许多植物,某些动物,需要通过自然特征进行归类。

Preventing Gene ExchangeTwo general types of mechanisms operate to block the exchange of genes between individuals of related groups. The first general type is made up of prezygotic isolating mechanisms that prevent the formation of zygotes. Prezygotic isolation falls into two categories: ecological and behavioral. In the first case, two related group may become adapted to slightly different environments-perhaps varying soil types or food sources. Over time, these genetic differences become so great that successful cross-fertilization can no longer take place. In behavioral isolation, related groups evolve differing behaviors such as specific mating rituals-that restrict the exchange of genes to members of the same group.两种基因型机制阻碍了相关群体中个体的基因交换。

分子生物学复习部分资料中英文题目

分子生物学复习部分资料中英文题目

分子生物学复习部分资料中英文题目名词解释(probe)探针:分子杂交中和待测核苷酸链碱基互补的具有特定序列的被标记的核苷酸链,可用于检测核酸样品中存在的特定基因。

(molecular hybridization)分子杂交:是利用DNA变性与复性这一基本性质来进行DNA或RNA定性或定量分析的一项技术。

(gene chip)基因芯片:指单位面积有规律地紧密排列的特定的DNA片段的支持物。

(gene library)基因文库:是指一个包含了某一生物体全部DNA序列的克隆群体。

()cDNA文库:是包含某一组织细胞在一定条件下所表达的全部mRNA经逆转录而合成的cDNA序列的克隆群体,它以cDNA片段的形式贮存着该组织细胞的基因表达信息。

(genomic DNA library)基因组DNA文库:是指生物的基因组DNA的信息(包括所有的编码区和非编码区)以DNA片段形式贮存的克隆群体。

(transgenic technology)转基因技术:采用基因转移技术使目的基因整合入受精卵细胞或胚胎干细胞,然后将细胞导入动物子宫,使之发育成个体的技术。

(transgenosis)转基因: 转基因技术中被导入的目的基因(transgenic animal)转基因动物:转基因技术中目的基因的受体动物(Somatic cell nuclear transfer)核转移技术:将动物的一个体细胞核全部导入另一个体的去胞核的的激活的卵细胞内,使之发育成个体,即克隆(clone)。

gene knockout基因剔除:建立在同源重组基础上的有目的去除动物体内某种基因的技术。

(functional cloning)功能克隆:通过对一种致病基因功能的了解来克隆该致病基因。

(positional cloning)定位克隆:从一种致病基因的染色体定位出发逐步缩小范围,最后克隆该基因。

(genetic diagnosis)基因诊断:直接检测基因结构及其表达水平是否正常,从而对疾病作出诊断的方法。

托福考试阅读题练习

托福考试阅读题练习

托福考试阅读题练习Passage 1Insect Wings文章解析给了一幅翅膀图,各种专业名词,顿时懵圈。

第一段:引入话题,翅膀化石fossil,引发很多猜测speculation。

第二段:讲了原始翅膀protowings,提出了一个theory,讲了翅膀的作用。

第三段:说第一个理论存在问题,protowing需要和身体以某种方式连接hinged,而且需要有肌肉控制翅膀的拍动,所以提出了另外一个理论解释翅膀的起源。

其中很多专业名词,但是题目特别简单(问什么导致了insect最终能飞行,定位在最后一句话finally,larger muscler让昆虫飞起来。

)第四段:支持第二个理论,反对第一种理论的证据是化学传感器chemical-sensing organs不在背上,在legs上。

[第二三段中有EXCEPT题,问第一个理论的problems是什么,两段的开头句逐个对应即可。

]第五段:提出第三个理论,protowings来自于水中的一种昆虫insect的腮gill。

问句:翅膀是怎么从水中到飞行的?答案是通过翅膀产生推动力浮在水面上,当muscle越来越强大,滑的越来越快,最后一飞冲天[有事实信息题]。

第六段:讲飞行的作用:虽然上面的理论还有争议,但是wing的作用是传播dispersal或avoidpredator,尤其考虑到predators的`数量,躲避predator才是最重要的目的[有句子简化题,注意转折关系]。

躲避predator的机制是一种预警机制warning system,通过sense wind or vibration[有事实信息题,定位后能看懂wind即可选出。

]解析:第一篇文章有关动物的描写是三篇中最难的,难点在于:1. 很多专业术语和专有名词,处理时采用首字母缩写,化零为整的方法,关注句子整体意思和句间关系;2. 长难句遍布,很多倒装句和修饰成分,处理时找句子的主干,课下多做长难句的分析练习。

生物专业英语第三版蒋悟生编__课文翻译

生物专业英语第三版蒋悟生编__课文翻译

Inside the Living Cell: Structure andFunction of Internal Cell PartsCytoplasm: The Dynamic, Mobile Factory细胞质:动力工厂Most of the properties we associate with life are properties of the cytoplasm. Much of the mass of a cell consists of this semifluid substance, which is bounded on the outside by the plasma membrane. Organelles are suspended within it, supported by the filamentous network of the cytoskeleton. Dissolved in the cytoplasmic fluid are nutrients, ions, soluble proteins, and other materials needed for cell functioning.生命的大部分特征表现在细胞质的特征上。

细胞质大部分由半流体物质组成,并由细胞膜(原生质膜)包被。

细胞器悬浮在其中,并由丝状的细胞骨架支撑。

细胞质中溶解了大量的营养物质,离子,可溶蛋白以及维持细胞生理需求的其它物质。

The Nucleus: Information Central(细胞核:信息中心)The eukaryotic cell nucleus is the largest organelle and houses the genetic material (DNA) on chromosomes. (In prokaryotes the hereditary material is found in the nucleoid.) The nucleus also contains one or two organelles-the nucleoli-that play a role in cell division. A pore-perforated sac called the nuclear envelope separates the nucleus and its contents from the cytoplasm. Small molecules can pass through the nuclear envelope, but larger molecules such as mRNA and ribosomes must enter and exit via the pores.真核细胞的细胞核是最大的细胞器,细胞核对染色体组有保护作用(原核细胞的遗传物质存在于拟核中)。

清华大学微生物(Microoganism)复习题库

清华大学微生物(Microoganism)复习题库

微生物题库及答案中文版2004-2005秋季学期说明:1.以网上流传的“微生物答案1[1][1].2版”为基本框架,题目和顺序均来自该版。

2.答案主要参考材料为科学出版社的中译版《微生物生物学》和老师的讲义。

3.绝大多数答案为中文,个别用中文难以表达或繁琐的部分保留了英文叙述。

4.由于现有图片都不甚清晰,涉及图片的部分均未加入,建议打印后徒手绘之。

(文档中相应部分均留有空间)。

1. Introduction to Microbiology1.what is Microbiology?研究微生物的科学,微生物是一大群种类各异的微小生物体,他们以单细胞或细胞群体存在,包括病毒。

2.what are Microorganisms?微生物是一大群种类各异的微小生物体,他们以单细胞或细胞群体存在,也包括病毒。

例如:细菌,古细菌,藻类,真菌,原生动物等。

3.4.粘菌。

5.a)且往往是的DNAb)c)d)e)f)谢以及复制功能。

6.a)b)7.a)Small volume, large surface areab)Fast absorption and conversionc)Rapid duplication and growthd)Strong adaptabilitye)8.Populations, communities and ecosystems?裂。

9.Brief history of microbiology?f)1684 Antonie van Leeuwenhoek (discovery of bacteria) 列文虎克发现细菌g)1857-1864 Louis Pasteur (lactic acid fermentation, yeast alcohol fermentation,spontaneous generation theory) 巴斯德和自然发生说的破灭(发现乳酸发酵和酒精发酵)h)1881-1884 Robert Koch (pure culture, cause of tuberculosis, Koch’spostulates, cause of (霍乱) cholera)科赫的疾病细菌说,纯培养;Koch假设,结核和霍乱病茵的发现。

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β-Glucans and Pentosans and their Degradation Products in Commercial Beers Makoto Kanauchi1, Wakana Ishikura1 and Charles W. Bamforth2,*ABSTRACTJ. Inst. Brew. 117(1), 120–124, 2011Several commercial beers have been analyzed for their content of β-glucans, pentosans and their degradation products using high performance liquid chromatography, thin layer chromatog-raphy and chemical and enzymic analysis procedures. The beers tested contained high levels of residual high molecular weight pentosan, but much less high molecular weight β-glucan. The beers also contained sizeable levels of oligosaccharides, espe-cially trisaccharides, reflecting the incomplete degradation of polymeric materials in malting and mashing and the inability of yeast to ferment them. There is substantially more β-linked glu-cosyl material in beer than pentosyl substances, although the higher molecular weight of the latter probably makes it more likely to represent soluble fibre. In respect of fibre claims for the beers examined, even for the beer containing the least pentosan, it seems that less than a litre of the product would afford suffi-cient material.Key words: beer, beta-glucan, carbohydrate, oligosaccharides, pentosan.INTRODUCTIONBrewers have long been concerned about the mixed linkage β-glucans deriving from the cell walls of the starchy endosperm of barley because of the problems that they can cause in brewing1,3,20. They have paid rather more attention to these viscous polysaccharides than they have to the arabinoxylans, which are also found in the walls but in lesser quantities1,4. However it is now recognized that these too can be problematic as they have similarly high viscosities and, furthermore, are not degraded as exten-sively as the glucans during malting17. The aim, then, has invariably been one of minimizing the quantity of such molecules in wort and beer. However, when approached from a food perspective, complex cereal polysaccharides are championed for their potential benefits in the human diet7,9,11,19,23. On this basis, might incompletely degraded cell wall polysaccharides emerging through the brewing process represent soluble fibre if present in beer12? Fur-thermore, any β-glucan degradation products larger than glucose are not assimilable by the body and may con-stitute prebiotic material6. Such may also be the case for pentosan degradation products10,16.Much attention has been paid to the enzymic mecha-nisms by which the cell wall components are de-graded5,18,21. However there has been a dearth of studies in which comparison has been made of levels of these mole-cules and their digestion products in different types of beer. This is the topic addressed in the present paper.METHODS AND MATERIALSBeersThe beers were purchased locally (Miyagi) and are de-scribed in Table I. These are individual examples of the beers concerned. All were analyzed in duplicate. Carbohydrate analysis by high performanceliquid chromatography (HPLC)Beers were diluted five-fold with deionised water and filtered (Dismic 13HP, Cat. #13HPO45AN, Millipore Co. Millipore Billerica, MA, USA). Carbohydrates were ana-lyzed by HPLC (Shodex GPC – 101, Showa denko K.K.) using an Asahipak GS-220 HQ column (7.5 mm ID × 300 mm) at 50°C. Samples (10 µL) were eluted by water at a flow rate of 0.6 mL per min with detection by refractive index. Quantification was by measuring peak area in com-parison to reference standards.Carbohydrate retention times were determined using standards in the range 25–1,250 mg/L: D(+)-glucose (Cat. #049-00591, Wako, Wako Pure Chemicals Industries Ltd. Osaka, Japan); D(-)-Arabinose (Cat. #013-04572, Wako); D(+)xylose (Cat. #244-00302, Wako); maltose (Cat. #138-00611, Wako); maltotriose (Cat.#137-06541, Wako); D(+)-cellobiose (Cat. #036-07401, Wako); D(+)-cel-lotriose (Cat. #C1167, Sigma-Aldrich Corporation, St. Louis, MO, USA); cellotetraose (Cat. #400402, Seika-gaku Biobusiness Corporation, Tokyo Japan); cellopen-1 Department of Food Management, Miyagi University, 2-2-1 Hata-tate Taihaku-ku Sendai Miyagi, 982-0215 Japan.2Department of Food Science & Technology, University of Califor-nia, Davis, CA 95616-8598, USA.* Corresponding author. E-mail: cwbamforth@ Publication no. G-2011-0316-1112© 2011 The Institute of Brewing & Distilling Table I. Products investigated.Beer DescriptionA American lager, rice adjunct, low bitternessB American lager, corn syrup adjunct, low bitternessC Japanese dry beerD Japaneseall-maltbeerE IrishstoutF German Oktoberfest beerG Japanese high malt beer120JOURNAL OF THE INSTITUTE OF BREWINGVOL. 117, NO. 1, 2011 121taose (Cat. #400404, Seikagaku); barley β-glucan (Cat. #G6513, Sigma-Aldrich); oat spelt xylan (Cat. #X0627, Sigma-Aldrich). The samples were filtered as above. The retention times for standards are shown in Fig. 1. It was observed that the β-glucan was retained on the column to a significant extent, whereas xylan was eluted in the void volume. This could be because the xylan is of a substan-tially higher molecular weight, or because of interactions between the glucan and the column, leading to delayed elution from the column.Carbohydrate analysis by thin layer chromatography (TLC)Samples (2.5 μL) of beer and standards of D(+)-glu-cose, D(-)-arabinose, D(+)-xylose, maltose, maltotriose, D(+)-cellobiose, D(+)-cellotriose, cellotetraose, cellopen-taose, β-glucan or xylan (each solution being 5.0 mg/mL) were spotted onto a TLC plate (silica gel 70 plate, 5 cm × 10 cm, Cat. #193-08381, Wako). The plate was developed with an ethyl acetate, methanol, acetic acid and watermixture (60:20:15:5) before detection of migration usingFig. 1.Calibration trace for HPLC of carbohydrates.Fig. 2. HPLC analysis of beer carbohydrates.an anisaldehyde reagent spray [26 mL ethanol, 1.5 mL anisaldehyde, 1.5 mL sulphuric acid (98%) and 0.5 mL acetic acid] with heating at 100°C for 5 min in an oven. Assay of total β-glucan and total pentosan Total β-glucan was assayed by the method of Bamforth2. Pentosan was assayed as described by Douglas13.RESULTS AND DISCUSSION The carbohydrate spectrum of a range of beer styles was analyzed by HPLC (Fig. 2). All of the beers appeared to contain substantial levels of residual pentosan, as indi-cated by the large peak eluting essentially in the void vol-ume of the column. [Note that very high molecular weight materials other than pentosan will also elute in the void volume. These would include any unconverted starch.] By contrast the peak for high molecular weight β-glucan was much smaller for each of the beers, consistent with previ-ous claims that there is much more extensive degradation of β-glucan than pentosan in malting and brewing17.There were a number of peaks in the various beers that did not coincide with the standard sugars that were run in the study. These eluted before the sugars and are pre-sumed to be oligosaccharides, likely both α- and β-linked, arising from the degradation of β-glucan, pentosan and starch. Table II displays the levels of oligosaccharides and monosaccharides determined from the HPLC analysis. Although the values for the β-glucan degradation products are all cited on the basis of the calibration performed with cellulosic oligosaccharides, it is possible (though not yet verified) that there may be a contribution from oligosac-charides that include a single β1→3 bond: the primary endo-β-glucanases in malt catalyze the hydrolysis of a β1→4 bond on the reducing side of a β1→3 linkage, thereby primarily producing oligosaccharides with a sin-gle β1→3 linkage on the reducing end and varying num-bers of β1→4 linked glucosyl units25. It is presumed thatthese oligosaccharides migrate on HPLC in a similar mode to the purely cellulosic oligosaccharides with an identical degree of polymerization.Perusal of Fig. 1 and of Tables II and III reveals that for all of the beers tested there was much more β-glucan than pentosan, however the molecular mass of the glucan was much lower. A total β-glucan level of the order of 4–6 g/L would at face value mean that 0.2 L of beer or less would satisfy a dietary fibre claim minimum of >0.75 g per serving24 . However, as reviewed by Wood26, it appears only to be high molecular weight β-glucan that has benefit as fibre. Accordingly, it would seem that it is the arabi-noxylan fraction that represents any soluble fibre present in the beer. Assuming a similar minimum level of 0.75 g to allow claims to be made, then at the levels measured here, beer F would deliver a sufficiency when consumed at approximately a third of a litre, whereas for beer E ap-proximately twice this consumption would be needed.Lower molecular weight carbohydrates were also ex-amined by thin layer chromatography (Fig. 3). For each of the beers, the predominant species was a trisaccharide – either α- or β-linked. There were much smaller quantities of disaccharide and arabinose in each of the beers. This is in accord with the data from the HPLC analysis.The absence of xylose from the beers is indicative of an absence of xylan degradation in malting and mashing (c.f. also the abundance of high molecular weight pento-san suggested in Fig. 2). This may relate to the impact of a xylanase inhibitor15. Free arabinose was detected (Table II) showing that some arabinofuranosidase activity was occurring22.The absence of glucose was consistent with the prefer-ential metabolism of this sugar by brewing yeast. With the exception of Beer B, the maltose levels in the beers exam-ined were within the range generally reported4,8 and were vastly less than what is typically found in wort (typically approximately 60 mg/mL for a wort of 10°Plato) (Table II). Similarly, there are measureable quantities of malto-triose in all but one of the beers, such levels again being comparable with those previously reported4,8 and veryTable II. Lower molecular weight carbohydrates in beer. Data represents the mean of duplicate determinations.Carbohydrate (μg/mL)Beer A Beer B Beer C Beer D Beer E Beer F Beer G Cellopentaose 1,401 (±9) 508 (±28) 1,350 (±160) 701 (±1) 585 (±503) 1,462 (±230) 1,804 (±6)Cellotetraose -a1,904 (±7) - 414 (±0) - 342 (±17) -Cellotriose 4,245 (±147) - 3,305 (±53) - 4,589 (±1,923) - 4,391 (±55)Maltotriose 1,902 (±84) 5,414 (±207) 1,349 (±21) 5,387 (±58) - 5,838 (±53) 2,317 (±162)Cellobiose-813(±406)- - - -109(±0) Maltose 1,124 (±68) 6,795 (±99) 1,509 (±20) 1,946 (±196) 928 (±584) 2,953 (±51) 2,559 (±25)Glucose - - - - - - - Arabinose 105 (±43) 493 (±37) 163 (±96) 444 (±57) 102 (±0) 208 (±13) 413 (±69)Xylose - - - - - - -a Not detected.Table III. Total β-glucan and pentosan in beer. Data represents the mean of duplicate determinations.Carbohydrate (μg/mL)Beer A Beer B Beer C Beer D Beer E Beer F Beer Gβ-Glucan 4,109 (±26) 6,042 (±155) 4,099 (±121) 5,369 (±80) 4,182 (±20) 4,011 (±46) 5,081 (±203)Pentosan 882 (±133) 1,115 (±172) 1,004 (±70) 1,805 (±116) 866 (±107) 1,788 (±165) 1,360 (±85)122JOURNAL OF THE INSTITUTE OF BREWINGVOL. 117, NO. 1, 2011 123much less than in conventional worts (approx. 20 mg/mL for wort of 10°Plato).Several factors will impinge on the spectrum of carbo-hydrates found in beer. These will include inter alia grist materials (the malt and its degree of modification, ad-juncts), brewhouse conditions and fermentation condi-tions. Accordingly, it is not possible to simply interpret from the data generated here the mode of production of the beers, which is of course commercially sensitive. It might additionally be noted that the history of the beers, being purchased off the shelf as they were, is unknown and so no consideration of any changes that may have occurred in the beer during storage can be made. However gross changes in the levels of the carbohydrates would be unexpected, especially as there was no visible evidence in the beers of any biological or non-biological breakdown. Furthermore, it must be emphasized that we were not at-tempting to make any definitive comparisons between the beers tested for their absolute content of the various car-bohydrates, recognizing that there may be batch-to-batch variations in the precise levels. Simply, we observed broad similarities in composition from beer to beer in terms of the nature of the carbohydrates that were present.CONCLUSIONSBeer contains high levels of residual high molecular weight pentosan, but much less high molecular weight β-glucan, consistent with the latter being the more exten-sively degraded component from the starchy endosperm cell walls of barley. Beer also contains sizeable levels of oligosaccharides, especially trisaccharides, reflecting theincomplete degradation of polymeric materials in malting and mashing and the inability of yeast to ferment them. There is substantially more β-linked glucosyl material in beer than pentosyl substances, although the higher mo-lecular weight of the latter probably makes it more likely to represent soluble fibre. In respect of fibre claims for the beers examined, even for the beer containing the least pentosan, it appears that less than a litre of the product would afford sufficient material.REFERENCES1. 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