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Cambridge Pre-U

Biology

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Coursebook

Cambridge Elevate edition Original material ÂŠ Cambridge University Press 2016

Cambridge Pre-U Biology

S5: Cell replication

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describe and explain mitosis, with the aid of diagrams, in terms of chromosome, nuclear envelope and (where present) centriole behaviour with emphasis on the features of chromosome behaviour that contribute to the production of cells that are genetically identical to each other and to their predecessor outline the roles, in the control of the cell cycle, of checkpoints, extracellular growth factors and the cell-signalling pathway that activates protein kinases describe how telomere shortening determines the number of divisions of a cell by mitosis and the role of telomerase reverse transcriptase to reverse the telomere shrinkage in cells that must repeatedly divide throughout life (e.g. cells in the basal layer of skin, stem cells and some white blood cells) explain how some mutations can cause cancer, including those that cause proto-oncogenes to become oncogenes and those that reduce the activity of tumour-suppressor genes

S5.1

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Learning Outcomes

We shall see that there are two types of nuclear division, mitosis and meiosis, with different functions. The differences are due to differences in the behaviour of chromosomes. It helps to understand a little more about chromosomes before studying each type of nuclear division.

S5.2

question

5.1 How many telomeres are there in each chromosome in Figure 5.5 in the Coursebook?

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S5.3

The animation ‘DNA packaging’ at www.hhmi.org/biointeractive/dna-packaging deals with the structure of nucleosomes and chromatin and presents a short film of mitosis.

S5.4

question

5.2 Produce a large replica of Figure 5.6 in the Coursebook. Add poppit bead or pipe cleaner replicas of chromosomes and chromatids to each stage, assuming the cell has a diploid number of six (three pairs of chromosomes). Bead models of chromosomes can be made by joining together poppit beads to make a string. Chromosomes can be represented by two rows of beads, each row representing a chromatid, with some means of attachment to represent a centromere. A pipe cleaner is an alternative to a string of beads. Use two different colours for the two members of a homologous pair if possible. Make a long, medium and short pair.

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S5.5

During interphase, the cell grows to its normal size after cell division and carries out its normal functions, synthesising many substances, especially proteins, in the process. At some point during interphase, a signal may be received that the cell should divide again. If this happens, the DNA in the nucleus replicates so that each chromosome consists of two identical chromatids. The same signal also results in the replication of the centrioles, and hence the centrosome, so that a cell enters mitosis having replicated both its centrosome and its chromosomes. This coordination is important.

S5.6

S5.7 Control of the cell cycle

The term ‘chromosome’ was originally used for the double structures shown in Figure 5.2 in the Coursebook, consisting of two chromatids. We shall see that during mitosis the two chromatids separate. As soon as this happens, each chromatid should be called a chromosome. During the G1 phase there are 46 of these single chromosomes in humans – the diploid number, 2n. After the S phase there are 46 double structures.

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There must be strict control over when cells divide. The consequences of breakdown of this control in even a few cells is obvious in cancer, which is caused by uncontrolled cell division. For the cell cycle as a whole, correct timings, irreversibility and error-free completion are all factors that must be controlled. The importance of the control mechanisms is emphasised by the fact that they are basically the same in all eukaryotic cells and have been so well conserved since they first evolved over a billion years ago that many work perfectly when transferred between organisms as widely different as humans and yeast. Entry into a cell cycle is controlled by both stimulatory and inhibitory signals (‘accelerators’ and ‘brakes’). Regulation is carried out by a central control system of proteins, but overall control is by checkpoints. A checkpoint is a biochemical pathway that can delay or stop a cell cycle in response to a specific condition. Checkpoints exist to ensure that one cycle, or one part of the cycle, does not start before the previous one is successfully completed, and to ensure that a cycle starts only if conditions in the cell’s surroundings are suitable.

Protein kinases, cyclins and cyclin-dependent kinases

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The control system of proteins includes a group of proteins known as protein kinases. A kinase is an enzyme that catalyses the transfer of a phosphate group from one molecule (usually ATP) to another. Protein kinases add phosphate groups to large numbers of proteins, resulting either in their activation to carry out one of the key events in the cell cycle or their inhibition to prevent a cell cycle event occurring or being repeated. Each protein kinase is usually active at only one specific stage of the cycle. For example, the kinase that initiates mitosis phosphorylates proteins involved in the breakdown of the nuclear envelope and assembly of the mitotic spindle. Protein kinases themselves are controlled in various ways. The most important of these depends on the fact that for a cell cycle protein kinase to function, it must bind to a second protein called a cyclin. Hence the kinases are known as cyclin-dependent kinases (CDKs). There is great variety in the cyclins and their influence, which gives the opportunity for fine control. The overall level of cyclins increases during the cell cycle, starting at the beginning of interphase and ending with their breakdown at the end of nuclear division. Different cyclins appear at different times and in different places (for example, in the centrosome). For example in humans, cyclin D helps to move the cycle through the G1 to S phase, cyclins A and E are S-phase cyclins needed for DNA replication, and cyclins A and B are M-phase cyclins that initiate mitosis. Humans have at least 12 different cyclins and, since most CDKs can combine with more than one cyclin, a large number of combinations is possible. Removal of cyclins when their job is complete involves degradation by Original material © Cambridge University Press 2016

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proteasomes (see Chapter 1). This requires attachment of ubiquitin to cyclins via an enzyme cascade. Another way of regulating CDKs is by phosphorylation. There are phosphorylation sites on CDKs, some for inhibition and some for activation. Environmental conditions and checkpoints regulate the phosphorylation, an example of cell signalling (see Chapter 4 in the Coursebook). Some CDKs are also inhibited by certain proteins (cyclin kinase inhibitors, CKIs) or by being confined to specific locations until appropriate. Cyclin B1, for example, is needed in the nucleus for a cell to enter mitosis, but is found in the cytoplasm during interphase.

Quiescence and senescence

Not all cells are engaged in a cell cycle. When cell signalling indicates that division is not appropriate, cells can enter a state described as quiescent, or G0 (G zero). To re-enter a cell cycle, the cell must normally enter G1. As cells age, they typically lose the ability to divide, and enter the senescent state which leads to death. The role of telomeres in this process is discussed later. Question

5.3 Suggest three changes to the regulatory molecules described above that would be required to allow a cell to re-enter the cell cycle in the G1 stage.

Role of cell signalling

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When conditions are suitable, for example if sufficient nutrients are available, cells enter the cell cycle. They do so in response to certain signals, such as growth factors. Cell signalling involves molecules known as ligands, which bind to proteins. They are often extracellular and bind to and activate protein receptors in the cell surface membrane. This brings about a change in the target cell. Growth factors are extracellular ligands which are usually small proteins (peptides) that stimulate growth, and can include hormones. Binding of the growth factor triggers a sequence of biochemical reactions inside the cell involving second messengers (see Chapter 4). One consequence is that genes are switched on that bring about activation of CDKs. Cyclin D (a G1 cyclin) is one example. As noted, cyclin D is needed for the transition from the G1 to S phase.

Apoptosis and cancer

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There is a fundamental link between control of the cell cycle and the most common mechanism of programmed cell death known as apoptosis. Apoptosis occurs as a normal part of development to get rid of cells no longer needed, as when fingers and toes are sculpted out of a continuous web of tissue. Apoptosis not only plays other roles in the healthy body, for example limiting tissue expansion by killing some cells, but may also be involved in cancer. Cancer is a genetic disease caused by mutations in the genes that control the cell cycle. This results in uncontrolled cell division and consequent formation of undifferentiated masses of tissue known as tumours. If a cell becomes cancerous, it may be instructed to self-destruct by apoptosis. A common mechanism for doing this involves a protein that stimulates cell multiplication. If â&#x20AC;&#x2DC;survival factorsâ&#x20AC;&#x2122; such as growth factors are also present, the cells produced will survive and growth will occur. If survival factors are absent, the protein stimulates both cell multiplication and apoptosis, achieving a balance of no net growth. If apoptosis is blocked for some reason, cancer occurs.

S5.8 In early prophase, the two centrosomes are produced by replication of the centrosome during the S phase of interphase. Each contains two centrioles. From metaphase onwards, the separate chromatids should be referred to as chromosomes. Original material ÂŠ Cambridge University Press 2016

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S5.9

The poles of the spindle are where the centrosomes are located, one at each pole. As noted in Chapter 1 of the Coursebook, the centrosome is an organelle found in animal cells (not plant cells) that acts as the microtubule organising centre (MTOC) for construction of the spindle. Each centrosome replicates to form two centrosomes during the S phase of the cell cycle. Each centrosome consists of a pair of centrioles surrounded by a diffuse cloud of material which consists of a large number of proteins (the pericentriolar material). These proteins control production of the microtubules. Plant mitosis occurs without centrosomes.

S5.10 Telomeres

Role in DNA replication The problem

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DNA replication is described in Chapter 6. It will be useful to study this before reading the following. We have seen that during DNA replication, RNA primers must be added at the 5' ends of all the origins of replication on the leading and lagging strands before DNA polymerase can function. (Remember that the RNA primers are on the newly synthesised strands.) The question arises: what happens to these RNA primers? The answer is that they are removed by enzymes, but this leaves gaps in the newly synthesised DNA, particularly on the lagging strand with all its Okazaki fragments, each of which has an RNA primer. The gaps are sealed by the activity of DNA polymerase and ligase enzymes (see Chapter 6 for a reminder of the functions of these enzymes). However, a problem arises at each end of the chromosome/DNA. DNA polymerase will only add new DNA to a 3' end, so there must be an existing strand ‘upstream’ from it in the 5' direction. As Figure S5.1 shows, there is no more DNA in the 5' direction behind the final RNA primer, so DNA polymerase cannot replace the RNA with DNA once the RNA has been removed. This leaves unpaired overlapping 3' ends on the parent strand. direction of movement of replication fork during replication 5'

RNA primer

uncopied DNA

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DNA polymerase starts here

lagging strand

leading strand

parent strand

DNA polymerase starts here uncopied DNA

RNA primer is removed, leaving an overlapping single strand of parent DNA (blue strand)

Figure S5.1 Diagram explaining the need for telomeres. Red and blue strands are complementary.

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These ends would be digested, and as a result the genetic information they hold would be lost, if a solution had not evolved. Every time a cell divided, a little more DNA would be lost from these ends. Eventually the loss of vital genes would result in cell death. (Incidentally, note that Figure S5.1 shows that the problem occurs at one end of each new DNA double helix. The next time these two DNA molecules replicated, the opposite ends would have the same problem. (Try this with more diagrams showing the 3' to 5' directions on all the DNA strands.) So after two cycles of division, both ends of every chromosome would need a remedy.

The solution

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The discovery of the enzyme telomerase revealed the solution to the problem outlined above. The ‘normal’ DNA molecule is made a bit longer by adding some more bases (nucleotides). These have no useful information but allow the DNA polymerase to start ‘upstream’ of the useful DNA. The enzyme that adds the extra bases is called telomerase, and the extra DNA is the telomere (see Figure 5.12 in the Coursebook). The main function of telomeres is therefore to prevent the loss of vital genetic information during cell division and to allow continued replication of a cell. Telomerase uses RNA as a template and makes a DNA copy of this template. It is therefore a type of reverse transcriptase. The enzyme is a ribonucleoprotein. This means – it has both RNA and protein component. Part of the RNA serves as a template for making complementary DNA. The 3' end of the parent strand is extended in the 5' to 3' direction by the addition of bases which are complementary to those in the short piece of RNA in the telomerase Figure S5.2. The process can be repeated many times. As a result, in human telomeres a six-base sequence, TTAGGG, is typically repeated up to several thousand times. Once the telomerase moves on, a DNA polymerase/primase and a ligase add the DNA that is complementary to (again in the 5' to 3' direction). To see an animation of telomerase in action visit www.garlandscience.com/ECB4-students and search for Chapter 6.6. Thus the telomere is a series of repeat sequences of DNA, rich in G and T bases with complementary C and A bases (Figure S5.2). When a cell divides, the only information lost is a small section of the telomere at each end of the chromosome, which has no harmful consequences. built-in RNA template of 11 bases (there are further non-template bases)

overhanging G-rich 3‘ end of parent DNA

5‘ T

A G G G

3‘ A

C 5‘

A G G G

A U

gap due to removal of RNA primer T

A G G G

3‘ A

A G G G

C 5‘

A G G G

3‘ A

A U

A G G G

A G G G C

A G G G

3‘ A

A G G G

A U C

A G

telomerase (made of protein and RNA) recognises the 3‘ overhanging telomere sequence at the end of the parent strand

3‘

A G 3‘

A U

A G G G

A U

A G G G C

A G G G

T A

A G G G

A U C 5‘

C 5‘

A G G G

A G 3‘

T A

3‘

A G 3‘

A G G G

A U

A U C

3‘

A G G G

This gap is later filled. An RNA primer is added and DNA polymerase copies the G-rich strand in the 5‘ to 3‘ direction further elongation. Synthesis ends with the sequence GGTTAG.

3‘ A

A U C 5‘

telomerase moves

5‘ T

3‘

further elongation – only 1 repeat per elongation

3‘ A

T A

A G G G

C 5‘

5‘ T

A U C

5‘

telomerase moves ready to add another repeat

5‘ T

3‘

C 5‘

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5‘

A G G G

G-rich strand is elongated in 5‘ to 3‘ direction (red letters) 5‘ T

3‘

5‘ T

A G G G 3‘

A G G G

C 5‘

5’

A G G G

A U

3‘

A G G G C

A G 3‘

A U C 5‘

Figure S5.2 Synthesis of a telomere by telomerase.

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Other functions of telomeres

telomere-binding proteins stabilise the chromosome

Chromosomes with free ends (ends without telomeres) have been shown to stick to one another. This is because free ends are recognised as breaks by the DNA repair machinery, which joins the free ends together. To prevent chromosomes with free ends sticking together, the telomere forms a loop at the end of the chromosome, rather like a knot. This is made possible by the fact that the end of the telomere becomes single-stranded, with the G–T-rich strand extending beyond the C–A-rich strand. This ‘tail’ bends around and pairs with the C–A-rich strand as shown in Figure S5.3. Telomeres have also been shown to play a role in the pairing of homologous chromosomes (synapsis) during meiosis I. triple-stranded structure (the D-loop or displacement loop)

T-G-rich strand (5‘ to 3‘)

5‘ 3‘

3‘

T T A G G G A A T C C C

3‘

T-loop (telomere loop) of several thousand bases. Ends in single-stranded DNA with a 3‘ end.

3‘ 5‘

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5‘

C-A-rich strand (3‘ to 5‘)

Figure S5.3 Sealing the end of a chromosome with a telomere loop.

Telomeres and ageing

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Since their discovery, telomeres have excited a lot of attention and research, because they are associated with both ageing and cancer. Telomeres get shorter by about 100 base pairs with every division unless replaced by telomerase. Most fully differentiated (specialised) human body cells (somatic cells) do not divide in this way, or do so only partially. When several thousand base pairs have been lost, cells stop dividing. If divisions continue to the point where vital DNA is lost, the cell dies. Evidence suggests this shortening of telomeres could be one of the mechanisms of ageing, whereby cells and organisms grow old and die. This, of course, suggests that by somehow preventing the loss of telomeres we might be able to slow down or even prevent ageing.

Stem cells

Stem cells retain the ability to divide an unlimited number of times. When a stem cell divides, each new cell has the potential to remain a stem cell or to develop into a specialised cell such as a blood cell or muscle cell.

Telomeres and cancer The great majority of cancer cells retain the ability to divide repeatedly by reactivating telomerase, eventually forming a tumour. It has been suggested that the presence of telomeres helps prevent cancers, because if a cell loses control of cell division and starts to divide repeatedly it will eventually use up

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its telomeres and die. Only those few cells that get around this problem by reactivating telomerase become cancerous. Research is being carried out into ways of inactivating telomerase in cancer cells.

S5.11

In the adult, stem cells are found throughout the body – for example, in the bone marrow, basal layer of skin, gut, heart, brain and hair follicles. Research into stem cells has opened up some exciting medical applications.

S5.12 Biological significance of meiosis

As a result of meiosis four new daughter nuclei are formed, each with half the number of chromosomes of the parent nucleus. The significance of meiosis is that it: • allows sexual reproduction to occur during a life cycle, because it prevents the doubling of the chromosome number with every generation • increases genetic variation in two ways, namely independent assortment of chromosomes and crossing over.

Independent assortment of chromosomes

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Figure S5.4 shows how a cell with a diploid number of six can produce eight genetically different types of gamete as a result of independent assortment of its chromosomes. The number of possible types of gamete produced as a result of independent assortment is 2n, where n is the haploid number. 7

Possible gametes as a result of independent assortment

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homologous pair of chromosomes

PARENT CELL 2n = 6

meiosis I

Figure S5.4 A cell with a diploid number of six can produce eight genetically different types of gamete as a result of independent assortment of chromosomes during meiosis.

question

5.4 a How many possible genetically different types of gamete can one human produce as a result of independent assortment of chromosomes?

b How many possible genetically different children can one human couple produce as a result of independent assortment of chromosomes and random fusion of gametes?

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Crossing over Crossing over during meiosis allows the recombination of genes (see Figure S5.5). GAMETES

PARENT CELL

50% AB

50% ab

A Aa a 25% AB B Bb b

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crossing over

25% ab

no crossing over

B Bb b

A Aa a

25% Ab

25% aB

Figure S5.5 An example of additional genetic variation in gametes as a result of crossing over during meiosis in a parent cell.

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Figure S5.5 shows the effect of crossing over on variation amongst gametes. The parent cell has two genes, A/a and B/b, that are situated far apart on the same chromosome and because of this the two genes are said to be linked. The top diagram shows what happens when there is no crossing over between the pair of homologous chromosomes. Half of the gametes have the allele combination AB and half will have ab. It is very unusual for meiosis to occur without crossing over between homologous chromosomes, but it does happen in male Drosophila (see ‘Autosomal linkage’ in Chapter 16). The lower diagram in Figure S5.5 shows the effect of crossing over between two non-sister chromatids in this homologous pair. When this happens half of the gametes have the allele combinations Ab and aB. In many parent cells crossing over may occur between both pairs of non-sister chromatids so that all of the gametes have the new combinations of alleles of these two genes (either Ab or ab). If gene loci are far apart like A/a and B/b in Figure S5.5 then it is highly likely that there will be a high proportion of gametes with allele combinations Ab and ab. If gene loci are close together on the same chromosome then there is a much smaller chance that crossing over will occur between them and far less chance of generating genetic diversity in the next generation. Crossing over is responsible for much genetic diversity (see ‘Crossing over’ in Chapter 16).

Where and when Meiosis occurs in: • animals – during gamete formation (gametogenesis) • plants – during spore formation, e.g. pollen sacs and embryo sacs in seed-bearing plants • fungi – during spore formation in sexual reproduction. Original material © Cambridge University Press 2016

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S5.13 Mitosis and cancer

Summary

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Cancers are thought to start when mutations occur in the genes that control cell division. These genes can code for a number of different proteins, such as growth factors and their receptors, as well as the signalling molecules inside the cells and the enzymes involved in their production. In theory, a mutation in any of these genes could result in uncontrollable cell division and, thus, cancer. Research has identified two groups of genes that are particularly important. One group, which is involved in normal cell growth and division, is the proto-oncogenes. A mutation in one of these genes could either result in its overexpression, i.e. more protein being produced, or its product being more biochemically active. These new (mutated) versions of proto-oncogenes are called oncogenes, because they are cancer-causing (‘onco’ comes from the Greek term for ‘bulk’ or ‘mass’, so ‘onco’ refers to tumours). Some of the changes are caused by simple point mutations within the genes themselves. Others result from the gene itself being copied multiple times or the movement of the genes throughout the genome and, possibly, in the region of the stronger promoter that will result in its overexpression. Proto-oncogenes have been seen to be involved in the normal stimulation of cell division. Additionally, certain other genes within cells inhibit cell division under normal circumstances; these genes are called tumour-supressor genes. Mutations in these genes result in either lower levels of expression or complete inactivation, leading to abnormally high levels of cell division and, thus, could cause cancer. Mutations are not unusual events, and most of the time they don’t lead to cancer. Most mutated cells are either affected in some way that results in their early death or are destroyed by the body’s immune system. Since most cells can be replaced, mutation usually has no harmful effect on the body. Cancerous cells, however, manage to escape both possible fates, so although the mutation may originally occur in only one cell, it is passed on to all that cell’s descendants. By the time it is detected, a typical tumour usually contains about 1000 million cancerous cells. Any agent that causes cancer is called a carcinogen and is described as carcinogenic.

Many enzymes are involved, but DNA polymerase is the enzyme responsible for making the new DNA strands.

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The cell cycle must be precisely regulated. This is achieved by cell signalling pathways that activate protein kinases, by extracellular growth factors and by biochemical pathways known as checkpoints that can delay or stop the cycle.

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Unless replaced by the activity of the enzyme telomerase, telomeres get shorter with every division until the vital DNA is unprotected. Cell death then results, either by an ageing (senescence) process or by a programmed death using the mechanism of apoptosis. Cells that replace telomeres can divide repeatedly, and include stem cells and cancer cells.

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S22: Nutrition

Learning Outcomes

compare and contrast the modes of nutrition, dentition and digestive systems of nonruminant herbivores and carnivores ■ recall the structure and function of the mammalian alimentary canal , including the histology of the stomach, ileum and pancreas ■ identify sites of production, activation and action of the following digestive enzymes in humans : amylase, maltase, endopeptidases (including pepsin and trypsin), exopeptidases and lipase ■ explain the parts played by bile, mucus and sodium hydrogencarbonate in digestion

S22.1: The evolution of the human diet

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What did our earliest human ancestors eat? Scientists think that the early human diet may have been omnivorous, similar to that of modern chimpanzees, which includes meat and insects as well as fruit and leaves. Microscopic analysis of early human teeth found in archaeological digs shows evidence of chewing tough material such as roots, nuts and seeds. In addition, butchery marks on Early Stone Age fossil animal bones suggests that these early humans had also begun eating meat and bone marrow. So which of our ancestors were the first butchers? There were several species of hominin living at that time, but only in species of the genus Homo have features linked to meat-eating been discovered. Such features are a decrease in tooth and gut size. These changes are seen especially in Homo erectus.

The increase in body size seen in the evolution of humans could be linked the increased importance of animal foods in the diet. This diet richer in proteins allowed an increase in body size.

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Why did their diet change? Meat and marrow are energy-rich foods packed with essential amino acids and micronutrients. Oily fish also offer resources rich in nutrients like omega-3 fatty acids needed for brain growth.

S22.1 Modes of nutrition in mammals: herbivores, carnivores and omnivores Organisms with specialised digestive systems are described as holozoic: food is taken into a gut, broken down into smaller molecules and absorbed into the body. In this chapter we will look particularly at how some mammals feed and absorb nutrients from their diet. Holozoic animals may be classified according to their dietary intake. • Herbivores feed exclusively on plant matter. • Carnivores feed on other animals. • Omnivores have a diet that includes both plants and animals. The diet of animals must provide metabolic fuel and also molecules for growth and repair. Metabolic fuel is normally taken in as carbohydrate and lipid, but after deamination protein may also be metabolised as fuel. Proteins and amino acids are required to build Original material © Cambridge University Press 2016

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and repair body proteins and may also contribute, in addition to sugars and phosphatecontaining food, to the synthesis of nucleic acids by providing a source of nitrogen.

Mammalian herbivore nutrition strategies

Mammalian herbivores are primary consumers. Their diet is low in protein but high in fibre. Their feeding is limited by time or resources. Those whose nutrition is time-limited use a feeding strategy of grazing and browsing. They feed for long periods during the day, moving through their habitat consuming whatever vegetation is available and not toxic or inedible. Some species of herbivorous mammal gather in herds for mutual grooming and protection. Where food resources are limited, some mammals may have to use a selective feeding strategy. It would expend too much energy to search out and consume specific nutritionallyrich foods. The size and metabolic rate of the organism will determine how much and how often it needs to feed. The white-headed capuchin, Cebus capucinus, for example, is an omnivore that adopts such a strategy. Fruits and insects form the most important parts of its diet. These capuchin monkeys feed at all levels in the forests Central America. They forage for insect food by stripping the bark from trees, breaking apart dead tree branches, rolling over stones and searching amongst leaf litter on the forest floor. They have even been seen to use stones to crack open hard fruits. Their favourite foods, such as ripe fruits, are not easy to find and capuchins are highly skilled in changing their feeding behaviour in order find them. Some herbivores, cattle and sheep, for example, are described as ruminants, as they have a specialised digestive tract that ferments food in a forestomach called the rumen.

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S22.2 Dentition

Mammals have four different types of tooth (Figure S22.1): incisors, canines, premolars and molars. Each type is adapted to specific functions. upper

adult 21-25 years old

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lower

permanent (adult) teeth

central incisor

second premolar

lateral incisor canine

first molar second molar

first premolar

third molar

Figure S22.1â&#x20AC;&#x201A; Human adult teeth.

Incisors are chisel-shaped and used for cutting and biting off pieces of food. Canines are long and sharp, to hold and tear at food. Premolars are used to crush and grind soft food. Molars are used for chewing and grinding hard food. As mammals have evolved to take advantage of food resources, the adaptation of their dentition can be a key means of identifying their usual diets (Figure S22.2). Herbivores

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that have to crop vegetation may use sharp-edged incisors, either in opposing rows top and bottom – for example, the horse – or lower incisors working against a horny pad on the upper jaw – for example, the sheep. Their canine teeth are absent or indistinguishable from incisors (some may have a vestigial canine tooth). Many also have a diastema (toothless space) between the incisors and premolars allowing the tongue to pull vegetation into the buccal cavity more efficiently. The grinding surfaces form interlocking W- and M-shaped grooves which wear down, exposing abrasive ridges of enamel between dentine. Their jaw movement is lateral as well as vertical, allowing sideways chewing movement along the grooves, grinding the rough plant material. As a consequence of the heavy grinding load that wears the teeth down, herbivore teeth grow continuously from their base.

Figure S22.2 The dentition of a horse (herbivore) compared with that of a dog (carnivore).

Question

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Carnivores also use their incisors to nip and nibble food, but the greatest contrast is seen in their canine teeth, which may have to grip moving prey and pierce tough hide. They are long and sharp, and in some cases interlocking even with the mouth slightly open. The premolars and molars are adapted as carnassials; these are blade-like teeth specialised for slicing rather than grinding meat. The carnivore jaw is restricted to a single plane of movement, opening and closing like a pair of scissors, with the upper and lower carnassial teeth slicing past one another rather than meeting on flattened surfaces.

22.1 M ake a table to summarise the differences between the jaw structure and dentition of the dog and the horse.

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S22.3 Human digestive system (omnivore) Once food has been ingested, mechanical and chemical digestion can begin. In many mammals, mechanical breakdown is achieved by chewing in the buccal cavity. This increases the surface area for enzyme action and eases swallowing. Chemical digestion may also begin if salivary amylase is secreted. The food can then pass through further stages in mechanical and chemical digestion, and absorption. The main regions of the alimentary canal (gut) are: mouth and buccal cavity; pharynx (throat); oesophagus; stomach; small intestine (duodenum, jejunum and ileum); large intestine (caecum and appendix, colon, rectum and anus). The other organs closely associated with the gut are the liver, gall bladder and pancreas (Figure S22.3).

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buccal cavity

salivary gland pharynx salivary gland

mouth

oesophagus

stomach

liver gall bladder

pancreas

colon small intestine caecum appendix

large intestine

rectum anus

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Figure S22.3 Diagram of human digestive system and associated organs.

The digestive tract is suspended from the dorsal wall of the abdominal cavity by connective tissue called mesentery. The mesentery brings nerve, lymph and blood supplies to the tract. mesentery

peritoneum

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duct

serosa

gland, e.g. liver and pancreas

the three layers of the mucosa

inner-circular outer-longitudinal

muscularis externa

submucosa mucosa

gut lumen

epithelium lamina propria muscularis mucosa

Figure S22.4 Transverse section (TS) of the mammalian alimentary canal showing the different tissue layers.

The lumen, through which the food travels, is surrounded by the gut wall (Figure S22.4). The gut wall consists of four distinct layers: the innermost mucosa, the submucosa, muscularis externa (external muscle layer composed of smooth muscle tissue) and serosa (outer layer). The thickness of each layer varies depending on the function of that particular section of the digestive tract. The muscularis externa consists of an inner circular muscle layer which narrows Original material ÂŠ Cambridge University Press 2016

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the lumen when it contracts, and an outer longitudinal layer which contracts parallel to the lumen. These antagonistic muscles enable peristalsis to move food along the lumen by waves of muscular contraction. The innermost mucosa also varies in the degree of infolding, which affects its surface area and also provides invaginations. Mucus and enzymes are secreted by the specialised epithelial cells that line the gut. Mucus protects the epithelial cells from chemical and mechanical damage. It also lubricates the movement of food especially in the oesophagus.

22.2

QUeSTioNS a Describe how the muscles of the gut move a bolus of food by peristalsis. b

What part does peristalsis play in digestion?

The stomach

submucosa (contains submucosal plexus) muscularis externa (contains myenteric plexus)

(a)

oblique layer circular layer longitudinal layer serosa

mucosa

surface epithelium lamina propria muscularis mucosa

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The stomach is a chamber that is able to hold food for a period of digestion. The entry to and exit from the stomach are bounded by the cardiac and pyloric sphincters. These control the entry and exit of food, respectively. The stomach wall has an extra layer of oblique muscle in its muscularis externa that enables the stomach to churn its contents to mix them with gastric juices, which are secreted from the gastric pits in its wall (Figure S22.5).

stomach wall

gastric gland

gastric pit

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gastric pits

surface epithelium pepsinogen

pepsin

mitochondrion in parietal cell

parietal cell mucous neck cells

parietal cell

gastric glands

chief cell

chief cells

enteroendocrine cell

(b)

HCI

(c)

Figure S22.5 Diagrams showing the structure and histology of the human stomach.

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In the empty stomach, the mucosal layer has deep longitudinal folds called rugae. When the stomach is full, the rugae are much reduced in size because the tunica mucosa is stretched, or distended, to accommodate the large amount of food.

Histology of the stomach

Figure S22.6â&#x20AC;&#x201A; Photomicrograph of longitudinal section (LS) of human stomach (stained with haematoxylin and eosin): the mucosa is highly folded so it can expand when the stomach fills with food. Below the submucosa are three thick layers of smooth muscle that contract to mix food in the stomach.

Gastric juice

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The four types of cell found in the epithelium of the base of the gastric pits have specific roles which complement one another (Figure S22.7). The chief cells (also known as peptic or zymogenic cells) contain many ribosomes and secrete pepsinogen, an inactive precursor of pepsin. Pepsin is an endopeptidase that hydrolyses proteins at several points within their structure to produce many shorter chains. Parietal cells (also known as oxyntic cells) are packed with mitochondria and can secrete hydrochloric acid at pH 1. This acid has three key effects: dissolving minerals, particularly calcium salts; killing parasites; and activating pepsinogen to pepsin. Mucous neck cells secrete copious amounts of mucus, which protects the stomach epithelium from damage by the acid, pepsin and the movement of food inside the stomach. Enteroendocrine cells secrete the hormone gastrin, which regulates stomach movements and secretions.

Figure S22.7â&#x20AC;&#x201A; Photomicrograph of gastric glands of human stomach (stained with haematoxylin and eosin) showing chief cells (staining purple) and parietal cells (staining pink).

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Once thoroughly mixed with gastric juices, the semi-liquid food is called chyme. Very little of the products of protein digestion can be absorbed in the stomach: only some water, ions and also alcohol and some drugs, for example aspirin.

Small intestine

Once the chyme has spent a few hours in the stomach, it passes through the pyloric sphincter into the small intestine, which consists of the duodenum and ileum. The duodenum is a relatively short section of gut into which bile and pancreatic juices are secreted via the common bile duct. Bile is a yellowish green liquid produced in the liver of most vertebrates. It is a powerful emulsifying agent containing bile salts – steroid acids that have both hydrophilic and hydrophobic properties (see Chapter 2) – and bilirubin, which is a product of the breakdown of old red blood cells in the liver. In many mammals that have a fatty diet, the bile is stored in the gall bladder to be secreted in response to food entering the duodenum.

Pancreatic juice

The pancreas serves as both an endocrine and exocrine organ. Endocrine tissue is arranged in identifiable small patches called islets of Langerhans, which secrete two hormones into the bloodstream. Glucagon is secreted from the alpha cells, and insulin from the beta cells (Figure S22.8).

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Figure S22.8 Photomicrographs of human pancreas showing endocrine and exocrine tissue a shows an islet of Langerhans surrounded by acinar (exocrine) cells; b is an enlarged view of the larger of a smaller islet.

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The bulk of the cells of the pancreas are acinar cells, which form the exocrine tissue that secretes pancreatic juice into ducts leading to the duodenum (Figures S22.8 and S22.9). This secretion consists of sodium hydrogencarbonate ions – an alkali that neutralises the chyme – and various digestive enzymes, some of which are only activated by enzymes secreted by some epithelial cells on the villi.

Figure S22.9 Transmission electron micrograph of a pancreatic acinar cell. The cytoplasm of these exocrine cells has extensive rough endoplasmic reticulum (for protein synthesis), many mitochondria (to provide ATP) and many zymogen granules (for storage of inactive enzymes).

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Digestion of carbohydrates, proteins and lipids

The enzymes secreted by the digestive system are listed in Table S22.1, which shows the molecules they digest and the products of digestion.

Lipase

Site of production Salivary glands

Site of action Buccal cavity

Gastric glands

Stomach

Enzyme Salivary amylase Pepsin* Rennin*+

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Pancreatic amylase Trypsin* Chymotrypsin* Exopeptidase* (carboxypeptidase) Maltase Lactase Sucrase Exopeptidase (aminopeptidase) Dipeptidase

Gastric glands Pancreas

Pancreas

Epithelial cells lining small intestine

Stomach Small intestine

Small intestine

Surface of microvilli of small intestine

Substrate Starch Proteins Casein (milk protein) Lipids

Product Maltose Polypeptides Insoluble salt of casein Monoglycerides, fatty acids and glycerol

Starch Polypeptides Polypeptides Peptides

Maltose Peptides Peptides Dipeptides and amino acids

Maltose Lactose Sucrose Peptides

Glucose Glucose and galactose Glucose and fructose Dipeptides and amino acids

Dipeptides

Amino acids

* secreted in inactive form, e.g. pepsin as pepsinogen, trypsin as trypsinogen and chymotrypsin as chymotrypsinogen + see Chapter P1 for more information about rennin Table S22.1 The enzymes that catalyse the digestion of carbohydrates, proteins and lipids, their sites of production, the places where they are active, the substrates that they digest and the products of digestion.

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The main components of the human diet, carbohydrates (starch, glycogen, sucrose and lactose), proteins and lipids, are digested in stages in the alimentary canal. The three groups of hydrolytic enzymes (hydrolases) that catalyse the breakdown of these compounds are carbohydrases, proteases and lipases.

Digestion of carbohydrates

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Digestion of proteins

The digestion of the polysaccharides starch and glycogen begins in the buccal cavity and continues in the small intestine. The disaccharides sucrose and lactose are digested in the small intestine. Salivary amylase catalyses the hydrolysis of glycosidic bonds within molecules of starch and glycogen to form short chains of glucose and the disaccharide maltose. The optimum pH for salivary amylase is 6.5 to 7.5 and it is inactive at the pH of the stomach. This means that very little starch or glycogen is digested until food reaches the small intestine as food spends very little time in the buccal cavity. Acinar cells in the pancreas secrete pancreatic amylase which digests the remaining molecules of polysaccharide into maltose. Maltase is not secreted into the lumen of the small intestine. It is an enzyme located in the plasma membrane of the epithelial cells lining the small intestine with its active site facing the lumen (Figure S22.12). Maltase catalyses the hydrolysis of the glycosidic bond within maltose to release two molecules of glucose (Figure 2.5 in the Coursebook). Sucrase and lactase are two other carbohydrases that are situated in the cell surface membrane of epithelial cells in the small intestine. Sucrase catalyses the hydrolysis of the glycosidic bond in sucrose to release a molecule of glucose and a molecule of fructose. Lactase catalyses the hydrolysis of the glycosidic bond in sucrose to release a molecule of glucose and a molecule of galactose.

Proteins are broken down in several stages in the stomach and in the small intestine. Proteases act in three ways:

• endopeptidases catalyse the hydrolysis of peptide bonds within protein molecules to form shorter polypeptides • exopeptidases catalyse the terminal peptide bonds at the ends of polypeptides and shorter peptides to release amino acid molecules • dipeptidases catalyse the hydrolysis of the peptide bond between the amino acids in dipeptides (Figure S22.10).

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Saliva does not contain any proteases, so protein digestion begins in the stomach where cells in the gastric glands secrete the inactive enzyme pepsinogen. Pepsinogen is activated by hydrochloric acid to form pepsin by the removal of a chain of about 40 amino acids from one end of the molecule. The removal of this chain exposes the active site so activating the enzyme. Once activated, pepsin catalyses the removal of these short chains from other molecules of pepsinogen (Figure S22.5c). Pepsin is an endopeptidase that catalyses the hydrolysis of peptide bonds within polypeptide molecules. Pepsin cannot act on every peptide bond within a protein. The active site of pepsin has a shape that only accepts certain adjacent pairs of amino acids. The amino acid that contributes the N of the peptide bond (the ‘N’ of the C-N) has to be phenylalanine, leucine, tyrosine or tryptophan. This means that pepsin ‘cuts’ protein molecules to the left of these amino acids if the protein is shown with the N-terminal on the left as in Figure S22.10. The effect of pepsin is to break down polypeptides into shorter chains of amino acids, so giving many more ends that provide ‘targets’ for exopeptidases. Prorennin is secreted by some young mammals, e.g. calves and human infants, to convert the soluble milk protein casein into the insoluble calcium caseinate. Prorennin is converted into the active enzyme rennin when it is secreted into the stomach. Rennin helps retain milk in the stomach so that protein can be digested fully. Trypsin and chymotrypsin are two other endopeptidases. These enzymes are secreted in inactive form by acinar cells in the pancreas (Figure S22.8). Trypsin hydrolyses peptide Original material © Cambridge University Press 2016

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bonds by cleaving peptides on the right of the amino acids lysine and arginine (if the polypeptide is shown with the N-terminal on the left as in Figure S22.10). Similarly, chymotrypsin only acts on peptide bonds to the right of the amino acids phenylalanine, methionine, tyrosine and tryptophan, unless the adjoining amino acid is proline. Trypsinogen is activated by the enzyme enterokinase, which is secreted by epithelial cells of the small intestine. Enterokinase breaks a peptide bond releasing a short chain of six amino acids from the N-terminal of trypsinogen to form trypsin. Once this has happened, the trypsin molecule folds into a shape with an active site able to interact with its substrate. Trypsin can also catalyse the same reaction so activating more molecules of trypsinogen. It also activates chymotrypsinogen in the same way by removing a short chain of amino acids at the N-terminal. The digestion of proteins is completed by exopeptidases and dipeptidases. Exopeptidases catalyse the hydrolysis of peptide bonds at the ends of peptide chains made of three or more amino acids to form free amino acid molecules and dipeptides (Figure S22.10). There are two groups of exopeptidases:

• carboxypeptidases catalyse the hydrolysis of peptide bonds at the C-terminal of peptides, e.g. the tripeptides shown in Figure S22.10 • aminopeptidases catalyse the hydrolysis of the peptide bond at the N-terminal od peptides.

Carboxypeptidases are secreted by the pancreas in inactive form and they are activated by trypsin by removal of a short chain of amino acids. Aminopeptidases are secreted by the small intestine and act on the surface of epithelial cells. Endopeptidases and exopeptidases N-terminal

C-terminal

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digestion of a protein in the stomach

action of pepsin

action of trypsin

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digestion in the small intestine

action of carboxypeptidases

action of aminopeptidases and dipeptidases

digestion on the surface of epithelial cells

Key peptide bond broken by action of proteases

lysine

phenylalanine

any amino acid other than lysine or phenylalanine

Figure S22.10 Proteins are digested by endopeptidases, exopeptidases and dipeptidases. The products of protein digestion in the small intestine are free amino acids, tripeptides and dipeptides.

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cannot catalyse the hydrolysis of the peptide bond in dipeptide molecules. Dipeptidase enzymes situated on the surface of epithelial cells catalyses this reaction to release more amino acids and complete the digestion of protein.

Digestion of lipids

Lipids are not soluble in water so are much harder to digest than proteins and carbohydrates. Most of the lipids in the diet are triglycerides (Figure 2.12 in the Coursebook). The digestion of lipids occurs in two stages: • emulsification of lipid droplets into much smaller particles • hydrolysis of the ester bonds between glycerol and fatty acids.

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Lipase is the enzyme that catalyses the hydrolysis of ester bonds within each triglyceride molecule. Some lipase is produced by the stomach, but most digestion of lipids occurs in the small intestine. Lipid is not soluble in water so forms large globules to minimise the contact between triglycerides, which are hydrophobic, and water. Lipase is a globular protein and soluble in water. Therefore lipase can only act on the surface of these globules and as they are so large it would take a long time to complete digestion of triglycerides in the diet. Bile, which contains no enzymes, is secreted by the liver to reduce the particle size and decrease the time that fat needs for digestion. Bile salts are secreted by the liver to emulsify these globules of lipid. The hydrophobic portions of the bile salt molecules dissolve in lipid while their hydrophilic portions remain at the surface in contact with water. This reduces the surface tension so breaking the lipid globules into smaller particles, known as mixed micelles, with a much larger combined surface area. This effectively increases the concentration of substrate so increases the activity of lipase. Lipase is secreted in a form which is not particularly active. Acinar cells also secrete colipase, which is activated by trypsin when it reaches the duodenum. Colipase binds to lipase so stabilising the molecule in a form that can attach to emulsified lipid droplets and catalyse a hydrolysis reaction within the hydrophobic environment at the edge of the micelles as shown in Figure S22.11. Lipase can catalyse the hydrolysis of all three ester bonds between fatty acids and glycerol in a molecule of triglyceride. This releases free fatty acids and glycerol. However, in most cases lipase only acts on the outer two ester bonds forming two free fatty acid molecules and a monoglyceride from each triglyceride molecule. The chemical reactions of digestion occur within the food in the lumen as it passes through the buccal cavity, stomach and duodenum. Most digestion that occurs in the ileum occurs on the lining of the epithelial cells catalysed by enzymes that are very close to the transport proteins that absorb products of digestion (Figures S22.14 and S22.15). Pancreatic enzymes are adsorbed onto glycoproteins that form the glycocalyx on the surface of the cell surface membranes. Enzymes secreted by epithelial cells, such as maltase, sucrase and lactase, are embedded in the phospholipid bilayer with their active sites exposed to the outside (Figure S22.12). The crypts of Lieberkühn are the deep folds between the villi. The basal part of each crypt has multipotent stem cells which continually divide by mitosis to provide new cells to replace the cells of the epithelium that are worn away by the constant movement of food along the small intestine. The average life span of an epithelial cell of the small intestine is about 24 hours.

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1. large droplets of lipid enter the duodenum from the stomach

3. bile salts emulsify lipid droplets into much smaller droplets increasing surface area exposed to lipase

colipase

4. lipase and colipase enter duodenum from the pancreatic duct

2. micelles of bile salts enter the duodenum from the bile duct

lipase

5. colipase stabilises lipase and facilitates interaction with micelles

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fatty acid

6. hydrolysis of ester bonds of triglyceride catalysed by lipase

monoglyceride

12 fatty acid

7. products of hydrolysis diffuse from micelles and are absorbed by epithelial cells of small intestine

lumen

Figure S22.11â&#x20AC;&#x201A; The action of bile salts and lipase in digesting triglycerides.

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adsorbed enzyme e.g. amylase

sugar monomers in glycoprotein

glycocalyx

enzyme in plasma membrane e.g. maltase

plasma membrane of ileum epithelial cell

glycoprotein

phospholipid

Figure S22.12â&#x20AC;&#x201A; Digestive enzymes on the surface of epithelial cells that line the ileum. Some enzymes from pancreatic juice, such as amylase, become trapped within the glycocalyx of these cells. Other digestive enzymes, such as maltase, are made by the epithelial cells and are incorporated into the plasma membrane that faces the lumen.

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Villi

The surface area of the small intestine is greatly increased by inward-projecting villi (Figure S22.13). (The singular of villi is villus.) Each villus has a rich supply of blood and lymph vessels. Smooth muscle of the muscularis mucosa extends into each villus, agitating it as food passes along the gut by peristalsis. The epithelium consists of mucus-secreting goblet cells and columnar cells known as enterocytes, which have a brush border of microvilli to increase further the surface area for digestion and absorption.

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Figure S22.13 a is a photomicrograph of a cross-section of the ileum showing the mucosa folded into villi, the submucosa and the muscularis externa. b shows the epithelia of two adjacent villi. Goblet cells are stained dark purple. The brush border lining the columnar epithelial cells is also visible.

The products of digestion are absorbed by the columnar epithelial cells, which line the villi of the small intestine. These cells share many features in common with the cells that line the proximal convoluted tubules in the kidneys since both cells are involved in the movement of large quantities of monosaccharides and amino acids into the blood. In addition, enterocytes absorb and process the products of lipid digestion. These cells are specialised for absorption:

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• the luminal membrane forms a dense brush border giving a very large surface area for movement of products of digestion from the lumen into the cells • the luminal membrane has many carrier proteins for active uptake and facilitated diffusion • the basal and lateral membranes have many sodium-potassium pump proteins which help to maintain an electrochemical gradient for sodium ions which is used to drive the absorption of glucose, galactose and amino acids • many mitochondria to supply ATP required for sodium-potassium pumps and for the synthesis of triglycerides and proteins, including membrane proteins (enzymes, carrier proteins and sodium-potassium pumps).

Absorption of monosaccharides The products of carbohydrate digestion are glucose, galactose and fructose. Glucose and galactose are absorbed from the lumen across the luminal plasma membrane by cotransporter proteins known as sodium-dependent glucose transporters (SGLT1). Each of these carrier proteins transports a sodium ion together with either a molecule of glucose or galactose. The mechanism for this transport is the same as described for the proximal convoluted tubule and is sometimes called secondary active transport as it is driven by the electrochemical gradient for sodium ions (see ‘Reabsorption in the proximal Original material © Cambridge University Press 2016

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convoluted tubule’ in Chapter 14). The electrochemical gradient for sodium ions is also maintained by the high concentration of sodium ions in the lumen of the small intestine. Saliva, pancreatic juice and bile all contain high concentrations of sodium ions which are reabsorbed in the small intestine as well as in the large intestine just as they are in kidney tubules. Fructose diffuses from the lumen through a specific carrier protein (GLUT5) but without the co-transport of sodium ions so is an example of facilitated diffusion. While a meal is being absorbed, the concentrations of glucose, galactose and fructose increase inside the enterocyte so that all three pass through carrier proteins (GLUT2) in the basolateral membranes by facilitated diffusion into blood capillaries within the villi. Carrier proteins for glucose are described in ‘The control of blood glucose’ in Chapter 14.

luminal membrane

brush border

tight junction

GLUT5

SGLT1

ATP

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ADP +Pi ATP

lateral membrane

ADP +Pi

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GLUT2

infolding of basal membrane

monosaccharides and sodium ions enter blood capillaries

Key

fructose

glucose

sodium ions

potassium ions

sodium-potassium pump

Figure S22.14 The absorption of monosaccharides. Glucose and galactose are absorbed by secondary active transport, fructose is absorbed by facilitated diffusion.

Absorption of amino acids and peptides Amino acids are absorbed across the luminal membranes in much the same way as glucose by sodium-dependent carrier proteins (Figure S22.15). However, amino acids differ from each other as they have different R groups (see Appendix 1 in the Coursebook). These

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ATP

PEPT1

ADP +Pi

ATP ADP +Pi

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sodium-dependent amino acid carrier protein

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R groups give amino acids different properties thus one type of carrier protein cannot transport all 20 of the different amino acids needed to make proteins. There are four different types of carrier proteins each specific for some of the amino acids. The basal and lateral membranes have carrier proteins for facilitated diffusion of amino acids out of the enterocytes into the blood. As shown in Figure S22.10 dipeptides and tripeptides are also produced by protein digestion in the lumen of the small intestine. Enterocytes absorb small peptides up to four amino acids in length. These peptides are co-transported with hydrogen ions by peptide transporter proteins (PEPT1). Once inside the cytosol, peptides are digested by dipeptidases and aminopeptidases to maintain the concentration gradient for peptides to continually enter the cell. The brush borders of enterocytes have a low pH and therefore a high concentration of hydrogen ions. This high of hydrogen ions concentration is maintained by a sodium-H+ exchange protein that moves sodium ions into the enterocyte in exchange for hydrogen ions that are pumped out. This is another example of the electrochemical gradient for sodium ions being put to work â&#x20AC;&#x201C; in this case for movement of peptides into the cells and hydrogen ions out of the cells.

amino acids and sodium ions enter blood capillaries in the villi

Key amino acids

sodium ions

potassium ions

hydrogen ions

action of dipeptidase

Figure S22.15â&#x20AC;&#x201A; Absorption of the products of protein digestion: amino acids, dipeptides and tripeptides.

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Some peptides are also absorbed by endocytosis. This method is also used for absorption of much larger protein molecules in young infants for the first few days after the birth. Undigested antibody molecules in mother’s milk are absorbed in this way to give infants passive immunity against any infectious diseases that their mothers have experienced (see ‘Active and passive immunity’ in Chapter 11).

Absorption of fatty acids and monoglycerides

movement of fatty acids, glycerol and monoglycerides to SER

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synthesis of triglycerides in membranes of SER

lipid particle

diffusion of fatty acids, glycerol and monoglycerides through phospholipid bilayer

diffusion of fatty acids through transport proteins

The micelles formed during the digestion of lipids move close to the cell surface membranes of enterocytes. These micelles contain undigested triglycerides, monoglycerides, fatty acids and glycerol. All of these can diffuse through phospholipid bilayers into these cells without the need for any specific protein carrier. However, enterocytes also have some transporter proteins specific to fatty acids. These must have hydrophobic cores to allow fatty acids to pass from micelles into the cytoplasm.

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Golgi apparatus

proteins from RER form a coat around lipid particle

chylomicron released by exocytosis

chylomicrons enter lacteals in villi and then flow into the lymphatic vessels

Figure S22.16 Absorption of the products of lipid digestion.

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Once inside the cytoplasm of the enterocytes, fatty acids, glycerol and monoglycerides enter the smooth endoplasmic reticulum where they are reformed into triglycerides and phospholipid molecules. Rough endoplasmic reticulum in enterocytes makes several different forms of protein that move to the SER to coat lipid particles that contain triglycerides and other lipids, especially cholesterol that is also absorbed from the lumen of the small intestine. The protein-coated lipid particles are transported in vesicles from the SER to the Golgi apparatus where another protein molecule is added to make mature particles, known as chylomicrons. Many of these particles are packaged into vesicles to travel to the basal and lateral membranes where they are secreted by exocytosis. Some chylomicrons enter capillaries, but the majority enter lacteals and are transported in lymphatic vessels to enter the blood over a longer period of time than would happen if they all entered capillaries in the small intestine. Chylomicrons deliver triglycerides and cholesterol to tissues, especially adipose (fat-storage) tissue, muscle fibres (as a source of energy) and liver cells. Bile salts are absorbed together with the products of lipid digestion and recycled through the blood to the liver. Questions

22.3 E xplain the difference between the action of endopeptidases and exopeptidases and their roles in protein digestion.

Large intestine

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22.4 How are the different products of digestion absorbed?

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The large intestine consists of the caecum and appendix, colon and rectum. In humans, the caecum and appendix have no significant function, compared with that in herbivores, which is considered later in this chapter. The appendix may have an additional role associated with the immune system, as it has lymph nodes. In humans and carnivores, the semi-digested material that reaches the colon contains digestive secretions, undigested cellulose, dead mucosal cells and the microorganisms that inhabit the gut. The majority of these microorganisms are Escherichia coli, the bacterium that has an important role in synthesising vitamins B and K. Water, mineral salts and vitamins are absorbed in the colon. The residue, which is about 50% bacteria, is termed faeces. Its colour derives mostly from the decomposing bile secretions. The final section of the gut is the rectum, which holds the faeces prior to defecation via the anus. The anus consists of two sphincter muscles; the inner ring of smooth muscle is under autonomic control, and the outer is made of striated muscle under voluntary control.

Questions

22.5 a Describe the role of bile in digestion and absorption.

b Describe the change in pH of the digestive tract along its length and explain its significance and how it takes place.

S22.4 Specialisation of non-ruminant herbivore digestive system Mammals that feed mainly on grass gain most of their energy from cellulose and other polysaccharides that form plant cell walls. Cellulose is a polymer of β-glucose

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(see ‘Cellulose’ in Chapter 2 of the Coursebook). Animals do not make enzymes that can digest the glycosidic bonds between the β-glucose monomers in cellulose. To survive on a diet of grasses and similar plants, herbivorous mammals live in symbiosis with microorganisms that make the enzyme cellulase to digest cellulose, metabolise β-glucose and release short-chain fatty acids that are absorbed by the host animal’s blood system. Conditions inside the gut of all animals are anaerobic as any oxygen is quickly absorbed by cells lining the gut or by any symbionts and parasites living within the gut. Food rich in cellulose needs to remain within the host’s digestive system to give time for microbial breakdown and fermentation. This requires organs with a large storage capacity. Ruminants, such as cattle, sheep and goats, have three extra chambers at the front of their gut between the oesophagus and the stomach. These extra chambers provide space for a community of many species of microorganisms, including bacteria, protoctists and fungi. Horses and rabbits are non-ruminant herbivores. The first part of their alimentary canals is much the same as it is in humans and carnivores. The oesophagus empties into a stomach with a fairly small capacity adapted for the digestion of protein. The small intestine is adapted for digestion and absorption of starch, proteins and fats just as described for the human digestive system above. Ruminants tend to eat fairly quickly and swallow food into the first of their extra chambers before resting somewhere, regurgitating the food that they have swallowed and chewing it again (‘chewing the cud’). The microbes secrete cellulase, digest cellulose and absorb the products of digestion. They grow in number and in so doing produce a large quantity of protein. From the rumen food passes through the rest of the intestine and the microbial protein is digested and amino acids are absorbed. Non-ruminants do not have extra chambers at the front of their intestine, instead microbial digestion of cellulose occurs in the large intestine within the caecum and the colon, both of which are much enlarged compared to other mammals. Unlike ruminants, much of the microbial protein is not digested and absorbed because protein digestion occurs in the small intestine. Rabbits practice coprophagy, eating the soft faecal pellets that they produce at night so that much of their food goes through the gut twice so that microbial proteins are digested and absorbed in the small intestine on the second time through the gut. Horses do not do this. Their natural feeding strategy is to eat small meals quite frequently during the day. Kept under natural conditions, horses graze for up to 17 hours a day to gain sufficient energy and nutrients, especially proteins. After passing through the small intestine food passes into the caecum through a narrow opening. The caecum is a blind-ending sac with muscle fibres to mix the food and pass it onto the colon through another small opening that is very close to the entrance to the caecum. The colon is huge and is folded on itself several times (Figure S22.17). Conditions within the large intestine are anaerobic. The symbiotic microorganisms use the products of cellulose breakdown in their respiration. Pyruvate produced at the end of glycolysis is converted into a variety of short-chain fatty acids, such as acetate (ethanoate), propionate and butyrate (butanate) which are excreted by the bacteria. These fatty acids are absorbed by the mucosa of the large intestine and metabolised by the horse as a source of energy. Maintaining the community of microorganisms in the large intestine requires maintenance of constant conditions. The temperature is maintained near 37 oC and as horses tend to eat frequently the microorganisms are supplied with suitable substrates to digest and metabolise. However, the production of fatty acids tends to decrease the pH although constant absorption into the blood maintains a constant concentration of fatty acids and hence a constant pH. As in other mammals secretions of the pancreas and small intestine contain sodium hydrogencarbonate, which buffers the contents of the large intestine. If the pH decreases too much some of the microorganisms cannot survive, the composition of the community changes and the supply of energy to the horse may decrease and the production of gases within the large intestine increase, causing pain and discomfort.

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stomach

large colon

small intestine

rectum oesophagus

small colon

caecum

Figure S22.17 A diagram of the digestive tract of the horse.

S22.5 Specialisation of the carnivore digestive system

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The dog is a pure carnivore. The overall length of its digestive tract is rather short: about five to six times the length of the animal’s body (Figure S22.18). The salivary glands serve merely to lubricate, and do not have an important digestive function. Food is rarely chewed into small portions, but ‘wolfed’ down whole. Although relatively small, the dog’s stomach is all that is needed, as the food of a carnivore, wholly meat and fat, is nutrient dense, allowing one small meal to suffice for many hours. The stomach and small intestine function in the same way as the human gut. As there can be large amounts of fat in the diet, bile is important. As with humans, there is no enzyme in the carnivore gut capable of digesting cellulose. It is interesting to note that domestic dog species produce much more pancreatic amylase than their wild counterpart, the wolf. This may be evidence of their adaptation to sharing the human diet, as dogs became domesticated and used by early humans as hunting companions. The small intestine does not join the large intestine in a straight line, but at a right angle. At this point is a small appendage, 5 cm in length, called the caecum. This has no functional use in a carnivore and is merely vestigial, unlike the caecum of the herbivore, which is essential.

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oesophagus

stomach

caecum

colon rectum

small intestine

Figure S22.18 The digestive system of the dog. The total length of the digestive system is about 9 metres, two thirds of which are the small stomach and small intestine.

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Question 22.6 C ompare how herbivore and carnivore digestive tracts are adapted to their respective diets.

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The digestive systems of all mammals consist of a mouth, pharynx, oesophagus, stomach, duodenum, ileum, caecum, appendix, colon, rectum and anus. The liver and pancreas have important associated roles in bile production and enzyme secretion, respectively. The herbivore diet is rich in cellulose, which is broken down in non-ruminants by microorganisms in their caecum and appendix. Herbivore dentition, particularly their continuously growing molar teeth, is also adapted to grinding plant material down. Carnivore dentition is adapted to catching, piercing and slicing animal prey. The digestive tracts of carnivores are comparatively shorter than those of herbivores. Mechanical breakdown occurs through chewing in the mouth and churning in the stomach. Salivary amylase digests starch while it is in the mouth. Inactive pepsinogen is secreted by chief cells in the gastric mucosa. It is activated to pepsin by hydrochloric acid secreted by parietal cells. Pepsin is an endopeptidase, hydrolysing proteins into many shorter polypeptides. Mucus is secreted to protect the stomach wall. The pancreas’s exocrine function is to secrete amylase, lipase, trypsinogen and sodium hydrogencarbonate, which neutralises the acid chyme. Trypsin is also an endopeptidase. Other enzymes, called exopeptidases, hydrolyse the ends of peptidase chains producing dipeptides. Hepatocytes in the liver secrete bile, which contains bilirubin derived from haemoglobin from damaged red blood cells. Bile is secreted into the duodenum and emulsifies fats, thereby speeding the activity of lipase in its digestion of fats to fatty acids and glycerol. Disaccharides and dipeptides are hydrolysed by enzymes in the membrane of the columnar epithelia of the small intestine, which have microvilli. The monosaccharides and amino acids are transported into the capillaries within the villi. Fatty acids and glycerol are packaged by the Golgi bodies within the cells to be transported as chylomicrons into the lacteals within the villi.

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Summary

Original material © Cambridge University Press 2016

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Chapter title

Section title

The cell

This section looks at life in terms of the cell. Knowledge of eukaryotic and prokaryotic cellular structure and the processes which take place in all living cells is fundamental to explaining how life ‘works’. Cell biology has medical and commercial applications and these are also considered in this section.

1.1 Eukaryotic cell structure • microscopy • cell membranes • organelles: structure and function Candidates should be able to:

(c) discuss the importance of cell membranes in delimiting and compartmentalising cells, in selecting and moving substances across membranes and in providing surfaces for reactions (d) describe and explain the fluid mosaic model (e) discuss the roles of membrane proteins including transporters (channels and carriers [including CFTR]), receptors and antigens

Section number

Title

3–4

6–12

S1.6

Relative advantages of light and electron microscopes

Cell structure

Electron microscopy

S1.6

Relative advantages of light and electron microscopes

Cell membranes and transport

Structure of membranes, Movement of substances into and out of cells

74–77

79–88

S4.1

Introduction

Cell membranes and transport

Structure of membranes

Cell membranes and transport

Structure of membranes, Cell signalling

S4.4

More about facilitated diffusion

Cell membranes and transport

Structure of membranes

Cell membranes and transport

Movement of substances into and out of cells

S4.4

More about facilitated diffusion

S4.5

More about osmosis in plant cells

S4.6

Selective transport across membranes

S4.7

Comparing channel and carrier proteins

S4.8

Note about active transport

FO R

(g) explain how and why different substances move across membranes including simple and facilitated diffusion, osmosis, active transport, endocytosis (phagocytosis and pinocytosis), exocytosis (secretory pathway)

Add to which chapter of CB?

Cell biology and microscopy

(f) describe the factors affecting the permeability and fluidity of membranes

End page

Cell structure

EV IE

(b) explain and distinguish between resolution and magnification with reference to light microscopy and electron microscopy

Content

(a) discuss the relative advantages of light and electron microscopes

Start page

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Syllabus content

Cambridge International Pre-University Biology

Original material © Cambridge University Press 2016

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(C)oursebook or (S)upplement or (B)oth

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Chapter title

Section title

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Cell structure

Ultrastructure of an animal cell, Ultrastructure of a plant cell

Start page

End page

(h) recognise the following cell organelles and describe their functions

Location:

Syllabus content

Cambridge International Pre-University Biology

• nucleus • nuclear envelope • nucleolus • rough and smooth endoplasmic reticulum

EV IE

• ribosomes • Golgi apparatus • lysosomes • secretory vesicles • proteasomes • mitochondria • chloroplasts • vacuoles • cell walls

• cilia and flagella. Practical Candidates should be able to:

• centrioles

Cell structure

Worked example 1

(ii) measure the size of objects viewed with a light microscope and calculate the magnification of an image or drawing

Cell structure

Worked examples 1–4

(iii) produce drawings of an organism, a section through a small organism and a part of an organism as seen under the light microscope

(iv) produce correctly labelled and annotated drawings of cells from microscopic examination and from electron micrographs

(v) recognise organelles in a variety of cells from across the four eukaryotic kingdoms

Cell structure

Ultrastructure of an animal cell, Ultrastructure of a plant cell

FO R

(i) use a light microscope, stage micrometer scale and eyepiece graticule

Original material © Cambridge University Press 2016

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Section number

Title

S1.3

More about differences between plant and animal cells

S1.8

More about the nucleus

S1.12

Vacuoles

S1.13

More about centrioles and centrosomes

S1.14

Cilia and flagella

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Cambridge International Pre-University Biology

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Title

Cell membranes and transport

Box 4.1, Box 4.2, Box 4.3, Box 4.4

81, 86

Box S4.1

Investigating osmosis in animal cells

(vii) estimate the water potential of a plant tissue by investigating the change in length or mass of suitable plant tissue

Cell membranes and transport

Box 4.4

S4.5

More about osmosis in plant cells

(viii) estimate the solute potential of plant cells using percentage plasmolysis of suitable plant epidermal cells

Cell membranes and transport

S4.6

More about osmosis in plant cells

(ix) investigate the effect of temperature and different solvents on the permeability of membranes

Cell membranes and transport

(x) investigate endocytosis and intracellular digestion in a protoctist, such as Paramecium or Vorticella, or using yeast stained with neutral red.

Box S4.4

Observing endocytosis and intracellular digestion in a protoctist

• antibiotics • reproduction Candidates should be able to: (a) outline key structural features of prokaryotic cells including: unicellular, 1–5 µm diameter, peptidoglycan cell walls, lack of membranebound organelles, naked circular DNA, 70S ribosomes (b) outline the structure of the cell walls of Gram-positive and Gramnegative bacteria and the significance of the structure for the use of antibiotics

Box 4.2

S25

S25.3

Structure of prokaryotic cells

S25

S25.3

Structure of prokaryotic cells

S25

S25.5

An example of a bacterial pathogen: Agrobacterium tumefaciens

EV IE

Content

• pathogenic bacteria

Box 4.4

1.2 Prokaryotic cells • structure of prokaryotic cells

(vi) investigate the movement of materials through cell membranes, for example by diffusion, osmosis and active transport

S25

S25.3

Structure of prokaryotic cells

(e) outline the mechanism of asexual reproduction by binary fission in a typical prokaryote.

S25

S25.4

Asexual reproduction in prokaryotes

(i) investigate Gram staining of bacterial cell walls

S25

Box S25.1

Investigating Gram staining of bacterial cell walls

(ii) investigate the effect of penicillin or other antibiotics on bacterial growth (e.g. by use of antibiotic susceptibility discs).

S25

Box S25.2

Investigating susceptibility of bacteria to antibiotics

Practical

FO R

(d) explain the mode of action of penicillin on bacteria (as an example of an antibiotic) and explain why penicillin does not affect viruses

Original material © Cambridge University Press 2016

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Nucleic acids and protein synthesis

Section title

Start page

1.3 Cell replication Content

• mitosis • meiosis Candidates should be able to:

• DNA replication

(b) outline the cell cycle including growth, DNA replication, mitosis and cytokinesis

(c) describe and explain mitosis, with the aid of diagrams, in terms of chromosome, nuclear envelope and (where present) centriole behaviour with emphasis on the features of chromosome behaviour that contribute to the production of cells that are genetically identical to each other and to their predecessor

(d) outline the roles, in the control of the cell cycle, of checkpoints, extracellular growth factors and the cell-signalling pathway that activates protein kinases

(e) describe how telomere shortening determines the number of divisions of a cell by mitosis and the role of telomerase reverse transcriptase to reverse the telomere shrinkage in cells that must repeatedly divide throughout life (e.g. cells in the basal layer of skin, stem cells and some white blood cells)

The mitotic cell cycle

Fig 5.12

102

(f) describe meiosis, with the aid of diagrams, in terms of chromosome, nuclear envelope and (where present) centriole behaviour with emphasis on the features of chromosome behaviour that contribute to reductional division

Inherited change

Meiosis

368

Inherited change

Meiosis

368

(i) sequence images of eukaryotic cells undergoing mitosis

The mitotic cell cycle

Mitosis

(ii) prepare and view slides of root tip squashes or other material showing mitosis

The mitotic cell cycle

Box 5.1

102

Add to which chapter of CB?

Section number

Title

113

117

The mitotic cell cycle

Mitosis

S5.4–8

Cell replication

The mitotic cell cycle

Mitosis

101

S5.7

Control of the cell cycle

S5.7

Control of the cell cycle

S5.10

Telomeres

373

S5.12

Biological significance of meiosis

369

S5.12

Biological significance of meiosis

EV IE

FO R

Candidates should be able to:

(g) explain how independent assortment and crossing over can contribute to genetic variation (details of stages within prophase I are not required).

DNA replication

(a) outline the semi-conservative replication of DNA

Practical

End page

Syllabus content

Cambridge International Pre-University Biology

(iii) investigate meiosis using prepared slides and photomicrographs of plant or animal tissues. [cf Section 3.6 Practical (i)]

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Section number

Title

S3.1

Why enzymes are essential to life

S3.1

Why enzymes are essential to life

S3.2

Features of the active site

Syllabus content

Cambridge International Pre-University Biology

1.4 Enzymes Content • structure and function of enzymes

• enzyme kinetics • commercial uses of enzymes

(a) discuss why enzymes are essential to life

(b) describe the structure and properties of enzymes including their role as catalysts in catabolic and anabolic reactions (both intracellular and extracellular)

(e) describe end product inhibition and allosteric regulation (including phosphofructokinase and ATP) (f) explain the impact of deficiency of phenylalanine hydroxylase (PAH) in phenylketonuria (PKU) as an example of an inherited error of metabolism (g) describe and explain the effect of competitive and non-competitive inhibitors on enzyme activity

Enzymes

Mode of action of enzymes

Factors that affect enzyme action

S3.3

Allosteric regulation

S3.5

Phenylketonuria (PKU)

Enzymes

Enzyme inhibition

Enzymes

Immobilizing enzymes

(i) outline commercial applications of using immobilised enzymes, including the production of lactose-free milk

Enzymes

Immobilizing enzymes

S3.6

Some commercially important enzymes

(j) explain the principles of operation of dip sticks containing glucose oxidase enzymes, and biosensors that can be used for quantitative measurement of glucose.

Homeostasis

Urine analysis

319

320

S14.5

More about biosensors

Practical

FO R

(h) explain the advantages of enzyme immobilisation

Enzymes

EV IE

(d) describe, explain and investigate factors affecting enzyme kinetics including the effect of temperature, pH, substrate and enzyme concentration in terms of activation energy, kinetic energy, successful collisions, complementary shape and fit, as well as active site/substrate interactions including Vmax

Candidates should be able to:

(i) carry out investigations into the properties of a variety of enzymes in relation to the effect of temperature, pH, inhibitors and concentrations of enzyme and substrate

Enzymes

Box 3.1

(ii) investigate the effect of immobilisation of enzymes on re-use of enzymes, ease of removal of enzyme from product and thermostability of enzymes.

Enzymes

Box 3.2

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Chapter title

Section title

Start page

End page

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Section number

Title

S12.5

Chemiosmosis

S12.7

More about respiration without oxygen

S12.6

More about respiration

S12.7

More about respiration without oxygen

Syllabus content

Cambridge International Pre-University Biology

1.5 Respiration Content • ATP

• chemiosmosis • glycolysis • anaerobic respiration

Candidates should be able to: Y

(b) outline chemiosmosis as a system in prokaryotes and eukaryotes in which:

• electrons may gain energy from oxidation of chemical substrates and that this energy may be used to do work • energetic electrons pass through the electron transport system to release energy

Energy and respiration

ATP

270

272

Energy and respiration

Respiration

272

275

EV IE

(a) explain the need to release energy to drive metabolic reactions and the role of ATP

• reactions within mitochondria

• the released energy is used to transfer protons out through membranes

• as these protons diffuse back through the membrane, their kinetic energy is used in membrane-associated ATP synthase to add phosphate to ADP, forming ATP (c) outline glycolysis (phosphorylation to fructose 1,6-bisphosphate, hydrolysis to triose phosphate, oxidation and dephosphorylation to pyruvate)

Energy and respiration

Respiration

272

273

Energy and respiration

Respiration

273

(d) outline the link reaction and Krebs cycle within the mitochondrion, general principles of dehydrogenation and decarboxylation to produce ATP, and reduced NAD and FAD

Energy and respiration

Respiration without oxygen

277

(f) compare and contrast the energy released per molecule of glucose substrate in aerobic and anaerobic conditions and explain the reasons for the difference.

Energy and respiration

Respiration

274

FO R

(e) outline anaerobic respiration in animals limited to the oxidation of reduced NAD to regenerate NAD and conversion of pyruvate to lactate and, at the same level of detail, compare and contrast this with anaerobic respiration in yeast and plants

275

Present?

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Chapter title

Section title

Practical

investigate the effect of temperature on the rate of respiration using simple respirometers.

End page

Candidates should be able to: (i) investigate the rate of glucose respiration by yeast in aerobic and anaerobic conditions

Start page

Syllabus content

Cambridge International Pre-University Biology

Add to which chapter of CB?

Section number

Title

Box S12.1

Investigating the rate of glucose respiration by yeast

Box S12.1

Investigating the rate of glucose respiration by yeast

S6.8

RNA

S16.1

Note for Figure 16.9

S6.1

Designing genomes from scratch

S6.11

More about triplet codes

S6.13

Protein synthesis

1.6 Genes and protein synthesis

Content • the gene and genetic code • protein synthesis

• inheritance and Mendelian genetics • mutations • genetic conditions Candidates should be able to: (a) define a gene as a unit of inheritance or as an ordered sequence of nucleotides located at a particular locus on a particular chromosome, which codes for a particular protein or, in certain cases, a functional or structural RNA molecule and discuss the limitations of these definitions with reference to introns, exons and promoters

EV IE

• control of gene expression

(b) describe the genetic code and discuss the extent to which it is true that the code is universal to all organisms

Nucleic acids and protein synthesis

The structure of DNA and RNA

111

(d) describe, in outline, eukaryotic introns, exons and the splicing of mRNA

S6.15

Post-transcriptional changes

(e) define the term proteomics and outline its importance to biomedicine (limited to diagnosis and drug design)

S6.15

Post-transcriptional changes

(f) describe the control of gene expression in prokaryotes, including the roles of a regulator gene, repressor, promoter and operator in the transcription of structural genes

Gene control in prokaryotes

118

113

Inherited change

DNA, RNA and protein synthesis

374

Nucleic acids and protein synthesis

Genetics

(c) explain protein synthesis in terms of transcription and translation, including the roles of DNA, mRNA, tRNA and ribosomes

FO R

Inherited change

389

122

391

Present?

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Chapter title

Section title

Supplement

(g) state, with examples, the differences between continuous and discontinuous variation (limited to relative range of phenotypes, relative number of genes and alleles involved, and relative impact of the environment)

Selection and evolution

Variation

(h) define and use the terms allele, locus, phenotype, genotype, dominant, recessive and codominant

Inherited change

(i) use genetic diagrams to solve dihybrid crosses, including those involving sex linkage, autosomal linkage, epistasis, codominance and multiple alleles

Inherited change

(j) use and interpret the chi-squared test to test the significance of the difference between observed and expected results

(l) explain how some mutations can cause cancer, including those that cause proto-oncogenes to become oncogenes and those that reduce the activity of tumour-suppressor genes (m) describe gene mutations, limited to substitution, deletion and insertion (n) explain, with reference to sickle cell anaemia and cystic fibrosis, how gene mutations might affect expression of a protein and thus affect phenotype (issues related to genetic conditions need to be handled with sensitivity). Practical Candidates should be able to:

375

377

Sex linkage, Dihybrid crosses, Interactions between loci, Autosomal linkage, Crossing over

379

386

Inherited change

The chi squared test

386

387

Inherited change

Mutations

387

389

Inheriting genes

The mitotic cell cycle

Cancer

103

105

Inherited change

Mutations

387

389

Inherited change

Mutations

387

389

(i) investigate genetics using locally available materials (e.g. locally available plants), germinating seedlings (e.g. rapid-cycling Brassica), Drosophila, fungi (such as Sordaria fimicola), ‘genetic tomatoes’, prepared materials such as ‘genetic corn-cobs’ and any other materials that yield suitable numerical information

401

(k) describe the effects of ionising radiation on living cells including DNA damage which is repaired, DNA damage that leads to apoptosis, and DNA damage causing mutations

FO R

(ii) investigate continuous and discontinuous variation with any available materials (e.g. people, plants with suitable single-gene and polygenic characteristics, polymorphic snails, etc.). 1.7 Applications of cell biology Content

End page

398

EV IE

Note: Candidates are not expected to recall the equations or symbols for the chi-squared test.

Start page

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Section number

Title

S16.5

Epistasis

S5.13

Mitosis and cancer

S6.9

Genetic mutations

S16.7

Cystic fibrosis

Syllabus content

Cambridge International Pre-University Biology

• principles of genetic engineering • isolating genes • cloning DNA

• vectors and insertion into host cells

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Chapter title

Supplement

Genetic technology

Section title

• identifying and cloning transformed cells • gene therapy and genetic profiling (DNA fingerprinting)

End page

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Section number

Title

• stem cells – isolation and uses • ethical issues surrounding genetic engineering and the use of stem cells

(a) discuss the potential and actual advantages and disadvantages of transferring genetic material by genetic engineering compared to selective breeding

(b) explain strategies that are available to isolate the desired gene from the genome of the gene donor including:

• use of restriction endonucleases to fragment the genome, and use of electrophoresis and complementary gene probes to identify relevant fragments from the gene (c) outline the principles of PCR as used to clone and amplify DNA and discuss the source and importance of Taq polymerase (d) explain strategies that are available to insert DNA into host cells including:

• inserting the DNA into a plasmid vector using restriction enzymes and DNA ligase, followed by uptake of the recombinant plasmid by a host cell

Tools for the gene technologist

464

466

Genetic technology

Polymerase chain reaction

471

473

Genetic technology

Tools for the gene technologist

464

466

Genetic technology and agriculture

481

483

• use of Agrobacterium tumefaciens in inserting DNA into plant cells (e.g. in creating Golden RiceTM and Golden Rice 2)

Genetic technology

EV IE

• use of mRNA and reverse transcriptase

Candidates should be able to:

S19.2