Complete Set of Bio AHL class notes

CLASS NOTES FOR THE ENTIRE IB BIOLOGY AHL SYLLABUS Topic 7: Nucleic acids and proteins 7.1 DNA structure 7.2 DNA replication 7.3 Transcription 7.4 Translation 7.5 Proteins 7.6 Enzymes Topic 8: Cell respiration and photosynthesis 8.1 Cell respiration 8.2 Photosynthesis Topic 9: Plant science [11h] 9.1 Plant structure and growth [4h] 9.2 Transport in angiospermophytes [4h] 9.3 Reproduction in angiospermophytes [3h] Topic 10: Genetics [6h] 10.1 Meiosis [2h] 10.2 Dihybrid crosses and gene linkage [3h] 10.3 Polygenic inheritance [1h] Topic 11: Human health and physiology [17h] 11.1 Defense against infectious disease [4h] 11.2 Muscles and movement [4h] 11.3 The kidney [4h] 11.4 Reproduction [5h]

TOPIC 7: NUCLEIC ACIDS AND PROTEINS Topic 7.1 –– DNA Structure  7.1.1 Describe the structure of DNA, including the antiparallel strands, 3’’––5’’ linkages and hydrogen bonding between purines and pyrimidines. The sides of the ladder of DNA consist of alternating phosphate groups and deoxyribose sugar. Since the sides are antiparallel, one side goes in the 3' to 5' direction, and the other goes in the 5' to 3' direction. There are two types of nucleotides. Pyrimidines [C/T] hydrogen bond to purines [A/G] to create the ““rungs”” of the DNA ladder. The pyramidines (with the bigger name) are smaller——with only a single carbon ring. Purines take up more space with two rings. Space filling models of the relative sizes of purines of pyramidines in base pairing were a critical piece of evidence when Watson and Crick originally worked out the structure of DNA. 7.1.2 Outline the structure of nucleosomes. A nucleosome consists of DNA wrapped around eight histone proteins and held together by another histone protein. 7.1.3 State that nucleosomes help to supercoil chromosomes and help to regulate transcription. 7.1.4 Distinguish between unique or single-copy genes are highly repetitive sequences in nuclear DNA. Only a small proportion of the DNA in the nucleus constitutes genes. The majority of DNA consists of repetitive sequences. Highly repetitive sequences (satellite DNA) constitutes 5––45% of the genome. The sequences are typically between 5 and 300 base pairs per repeat, and may be duplicated as many as 105 times per genome. Highly repetitive sequences were once classified as ““junk DNA””, showing a degree of confidence that it had no role. This addresses the question: To what extent do the labels and categories used in the pursuit of knowledge affect the knowledge we obtain? 7.1.5 State that eukaryotic genes can contain exons and introns. Topic 7.2 - DNA Replication 7.2.1 State that DNA replication occurs in a 5’’ ń 3’’ direction. The 5’’ end of the free DNA nucleotide is added to the 3’’ end of the chain of nucleotides that is already synthesized. 7.2.2 Explain the process of DNA replication in prokaryotes, including the role of enzymes (helicase, DNA polymerase, RNA primase and DNA ligase), Okazaki fragments and deoxynucleoside triphosphates. The process of replication begins at specific nucleotide sequences on the DNA strand called the origin. At these points that helicase splits the DNA into its two antiparallel strands. On the strand running in the 5' to 3' direction, DNA polymerase III latches on at one end of the opening, called the replication bubble, and begins to continuously lay a new DNA strand from free nucleotides in the nucleus. As always, an exact copy of the now-detached strand is formed from this template due to base-pairing rules. At the same time DNA polymerase III is laying new DNA, helicase is continuing to split the strands, thus allowing replication to continue uninterrupted. On the opposite strand running 3' to 5', replication is not so simple. Because new strands have to be laid in the 5' to 3' direction, DNA polymerase III cannot lay continuously as it can on the

other strand. Instead, RNA primase lays short segments of RNA primer nucleotides at many points along the strand. When one segment of primer comes in contact with another, DNA polymerase I attaches and replaces the RNA primer with DNA. These segments of DNA are called Okazaki fragments. Once these fragments have been laid, they are joined by yet another enzyme known as DNA ligase, which attaches DNA into the gaps between fragments and completes the new strand. The 3' to 5' strand with Okazaki fragments is called the lagging strand, while the leading strand is the continuously replicating one. 7.2.3 State that DNA replication is initiated at many points in eukaryotic chromosomes.

 Topic 7.3 –– Transcription 7.3.1 State that transcription is carried out in a 5’’ ń 3’’ direction. The 5’’ end of the free RNA nucleotide is added to the 3’’ end of the RNA molecule that is already synthesized. 7.3.2 Distinguish between the sense and antisense strands of DNA. The sense strand (coding strand) has the same base sequence as mRNA with uracil instead of thymine. The antisense (template) strand is transcribed. 7.3.3 Explain the process of transcription in prokaryotes, including the role of the promoter region, RNA polymerase, nucleoside triphosphates and the terminator. During transcription, a DNA sequence is read by RNA polymerase, which produces a complementary, antiparallel RNA strand. An RNA polymerase complex separates the two strands of the DNA and bonds the RNA nucleotides when they base-pair to the DNA template. RNA polymerase (together with several other proteins collectively known as called factors) bind to parts of the DNA called promoters in order for separation of DNA strands to occur. Transcription proceeds as nucleoside triphosphates (energized nucleotides——AUC and G) bind to the DNA template and are joined by RNA polymerase in the 5' to 3' direction. Transcription ends when RNA polymerase reaches a termination site on the DNA. When it reaches the terminator, the RNA polymerase releases the RNA strand. 7.3.4 State that eukaryotic RNA needs the removal of introns to form mature mRNA. Topic 7.4 –– Translation 7.4.1 Explain that each tRNA molecule is recognized by a tRNA-activating enzyme that binds a specific amino acid to the tRNA, using ATP for energy. Each amino acid has a specific tRNA-activating enzyme that helps tRNA combine with its complimentary mRNA codon. The enzyme has an active site that recognizes three things: a specific amino acid, ATP, and a specific tRNA. The enzyme attaches the amino acid to the 3' end of the tRNA. The amino acid attachment site is always the base triple CCA. Like the genetic code itself that uses 64 codons to code for 20 amino acids, there is degeneracy in tRNA functionality. To provide a one-to-one correspondence between tRNA molecules and codons that specify the 20 amino acids, 61 types of tRNA molecules would be required per cell (There are 3 stop codons). In practice, some tRNA molecules have anticodons that can match up with multiple codons - this phenomenon is known as ““wobble.”” In reality 31 different tRNA molecules are required to translate, unambiguously, all 61 sense codons of the standard genetic code.

7.4.2 Outline the structure of ribosomes, including protein and RNA composition, large and small subunits, three tRNA binding sites and mRNA binding sites. A ribosome consists of two subunits: small and large. The two subunits separate when they are not in use for protein synthesis. In eukaryotes, the large subunit consists of three different molecules of rRNA (ribosomal RNA) and about 45 different protein molecules. A small subunit consists of one rRNA molecule and 33 different protein molecules. A ribosome has an mRNA binding groove and two tRNA binding sites. 80S and 70S is the ““Svedberg”” unit of ribosome size. It refers to rates of sedimentation in a laboratory centrifuge. 7.4.3 State that translation consists of initiation, elongation, translocation and termination. 7.4.4 State that translation occurs in a 5’’ ń 3’’ direction. During translation, the ribosome moves along the mRNA towards the 3’’ end. The start codon is nearer to the 5’’ end. 7.4.5 Draw and label a diagram showing the structure of a peptide bond between two amino acids. 7.4.6 Explain the process of translation, including ribosomes, polysomes, start codons and stop codons. Initiation Once the mRNA reaches the cytoplasm, the two halves of a ribosome attach themselves to it. The mRNA’’s 5' end is attached to the small subunit of the ribosome. AUG is the start codon on the mRNA. It forms hydrogen bonds with the anti-codon of the initiator tRNA molecule, which holds the amino acid methionine. The large ribosomal subunit has two tRNA binding sites: the A site and the P site. When initiation is done the initiator tRNA is in the P site. Elongation The next tRNA, carrying its specific amino acid, attaches itself to the next codon at the A site of the larger subunit. These two amino acids——methionine and the second amino acid——now make a peptide bond. The initiator tRNA breaks off and second tRNA carrying the newly formed dipeptide moves from the A site to the P site. Next, a third tRNA molecule attaches to the codon in the A site. The amino acid that is attached to the third tRNA makes a peptide bond with the amino acid from the second. The polypeptide chain is being elongated. The codon keeps moving through the ribosome in a 5' to 3' direction (from A to P). Translocation Translocation refers to the continuous process of ribosomes shifting along the mRNA chain, from the A site to the P site, like a conveyor belt in a factory; making the A site vacant for the next tRNA each time. Mass production of proteins is achieved by polysomes——multiple ribosomes lined up like identical pearls on a string each making the same protein in unison but each one step behind the neighbor ahead of it. Termination The three mRNA stop codons do not code for an amino acid. They terminate the translation process. The polypeptide is released and the mRNA fragments are recycled and used for RNA and DNA synthesis. tRNA is also recycled in the sense that it is continuously loaded, unloaded and reloaded.

7.4.7 State that free ribosomes synthesize proteins for use primarily within the cell, and that bound ribosomes synthesize proteins primarily for secretion or for lysosomes. Topic 7.5 - Proteins 7.5.1 Explain the four levels of protein structure, indicating the significance of each level. The four levels of protein structure are convenient ways of looking at proteins which are colossal biomolecules. The levels are obviously interdependent. Each level emphasizes a different aspect of biological significance. The primary structure is the basic order of amino acids in the polypeptide protein chain, before any folding or bonding between amino acids has occurred. The secondary structure refers to the repeated, regular structures that protein chains take on due to hydrogen bonding between amino acids. The secondary structure is usually in the form of twisting into an alpha helix or corrugations of a beta - pleated sheet. The tertiary structure is the complex, three-dimensional protein shape resulting from the folding of the polypeptide due to different types of bonds between the amino acids. These bonds include hydrogen bonds, disulphide linkages and electrovalent attractions. A disulphide linkage is depends on the presence of the amino acids cysteine which contains sulfur. An electrovalent bond can form between the negatively and positively charged molecules or amino acid groups along the chain. The tertiary level explains the specificity of three-dimensional shape of the active site. The quaternary structure is the hierarchical level for appreciating the largest and most complicated proteins, consisting of the two or more sub-units. Quaternary structure may also involve the binding of a prosthetic group (like iron in hemoglobin and Mg in chlorophyll) to form a conjugated protein. 7.5.2 Outline the difference between fibrous and globular proteins, with reference to two examples of each protein type. Fibrous proteins are best understood in terms of their stable, repetitive secondary structures, which could be in the alpha helix or beta pleated form. Two examples of fibrous proteins are keratin (in hair and skin) and collagen (in tendons, cartilage, and bones). Globular proteins are folded into three-dimensional shape. Enzymes and antibodies are globular proteins. 7.5.3 Explain the significance of polar and non-polar amino acids. Non-polar amino acids have non-polar (neutrally charged) R groups. Polar amino acids have R chains with polar groups (charged either positive or negative). Proteins with a lot of polar amino acids make the proteins hydrophilic and therefore able to interact with water. Proteins with many non-polar amino acids are more hydrophobic and are less soluble in water. With these characteristics, proteins tend to fold themselves so that the hydrophilic ones are on the inner side. This is important for protein channels running through membranes. Hydrophilic molecules and ions are to pass in and out of the cells easily through these hydrophilic channels. The hydrophobic edges of the channels align with the fatty acid chains of phospholipids which help control the stable positioning of the channels.

Topic 7.6 - Enzymes 7.6.1 State that metabolic pathways consist of chains and cycles of enzyme catalyzed reactions. 7.6.2 Describe the induced-fit model. The induced-fit model is an extension and refinement of the lock-and-key model. It accounts the ability of some enzymes to bind to several substrates should be mentioned. The German scientist Emil Fischer introduced the lock-and-key model for enzymes and their substrates in 1890. It was not until 1958 that Daniel Koshland in the United States suggested that the binding of the substrate to the active site caused a conformational change, hence the inducedfit model. This is an example of one model or theory, accepted for many years, being superseded by another that offers a fuller explanation of a process. 7.6.3 Explain that enzymes lower the activation energy of the chemical reactions that they catalyze. All reactions, either with or without enzymes, need collisions between molecules in order to occur. Many molecules have strong covalent bonds holding them together. They require powerful collisions at high speed in order to break these bonds. Increasing the rate of collision to a rate at which these reactions would occur normally would require prohibitive amounts of energy, usually in the form of heat. Enzymes compress substrate molecules slightly, selectively stressing certain of their bonds. Under these conditions much weaker collisions are required to break them. The activation energy——the amount of energy needed for a reaction to occur——is significantly reduced. 7.6.4 Explain the difference between competitive and non-competitive inhibition, with reference to one example of each. Competitive inhibition is when an inhibiting molecule that is structurally similar to the substrate molecule binds to the active site, preventing substrate binding. An example is the inhibition of the enzyme succinate dehydrogenase by malonate which is a product generated several steps later in the Krebs cycle. Non-competitive inhibition to an inhibitor binding to an enzyme (not to its active site) that causes a conformational change in its active site, resulting in a decrease in activity. An example is the action of the deadly poison Cyanide (CN) which inhibits cytochrome oxidase the enzyme that facilitates the use of oxygen at the very end of the electron transport chain in aerobic respiration 7.6.5 Explain the control of metabolic pathways by end-product inhibition, including the role of allosteric sites. Allostery is a form of non-competitive inhibition. The shape of allosteric enzymes can be altered by the binding of end products of an enzyme mediated reaction or pathway to an allosteric site; which is an area of the enzyme geographically separate from the active site. The binding at the distant allosteric site distorts the active site, thereby decreasing its activity. Accumulating metabolites——not required by the body for immediate use, can act as allosteric inhibitors of enzymes used much earlier in a metabolic pathway——effectively shutting down their own production. This is an example of homeostatic negative feedback on the molecular level. A good example is the ATP inhibition of phosphofructokinase in glycolysis. If usable energy is freely available the cell can afford to temporarily shut down anaerobic respiration.

TOPIC 8: CELL RESPIRATION AND PHOTOSYNTHESIS Topic 8.1 - Cell Respiration 8.1.1 State that oxidation involves the loss of electrons from an element, whereas reduction involves a gain of electrons; and that oxidation frequently involves gaining oxygen or losing hydrogen, whereas reduction frequently involves losing oxygen or gaining hydrogen. 8.1.2 Outline the process of glycolysis, including phosphorylation, lysis, oxidation and ATP formation. Glycolysis is the anaerobic splitting of sugar. In the cytoplasm, one hexose sugar is converted into two three-carbon atom compounds (pyruvate) with a net gain of two ATP and two NADH/H+. Glycolysis is common to aerobic and anaerobic respiration. Gylcolysis requires an initial investment of 2 molecules of ATP. This called phosphorylation. Adding the terminal phosphates of 2 ATP molecules destabilizes glucose and makes available energy locked up in its bonds. Two analogies that come to mind here are lighting a candle before it can give off heat and light; or kick starting a gasoline powered motorcycle. Lysis is the next step. The phosphorylated six-carbon sugar is split by enzymes into two threecarbon molecules of PGAL. Each PGAL is then simultaneously oxidized to PGA. Oxidation in this case is removal of hydrogen. A hydrogen ion is removed and added to the ion carrier NAD+ to form two molecules of NADH. Enough energy is released here to generate an ATP from each PGAL oxidation. Lastly each PGA is converted to pyruvic acid. This involves dephosphorylation. Each PGA donates its phosphate to an ADP to generate usable ATP. The net energy gain is 2 ATP. A total of 4 ATP are generated but 2 ATP were invested originally. 8.1.3 Draw and label a diagram showing the structure of a mitochondrion as seen in electron micrographs. 8.1.4 Explain aerobic respiration, including the link reaction, the Krebs cycle, the role of NADH + H +, the electron transport chain and the role of oxygen. LINK REACTION The link reaction is the first stage of aerobic respiration. This occurs in mitochondria in eukaryotes. The link reaction is relay between glycolysis and a series of reactions entailing a high yield of ATP called the Krebs cycle. The link reaction begins with the decarboxylation of pyruvate. A molecule of CO2 is removed. The product formed is a two-carbon acetyl group which reacts with coenzyme A (It is worth mentioning that this is a derivative of Vitamin B5, cysteine and ATP.) to form Acetyl Co-A. The decarboxylation process also involves oxidation. A hydrogen ion is removed, as in glycolysis, to form NADH/H+. In lipid metabolism the oxidation of the fatty acid chains also results in the formation of twocarbon atom (acetyl) fragments which then pass through the Krebs cycle. A fatty acid chain can produce multiple 2C acetyl groups. This is why fats can yield twice as much energy as can carbohydrates or proteins. Amino acids are first de-aminated, and then enter the Krebs cycle according to their various radicals. KREBS CYCLE In raw input for the Krebs cycle is the 2 carbon Acetyl group. The carbon balance is restored because at two stages in the cycle a molecule of CO2 is released. As in glycolysis oxidation is a key feature. During several steps along the way hydrogen ions are removed. As we shall see, the transport of hydrogen by carrier molecules is the key to producing high yields of ATP.

At the start of the Krebs cycle the 2C acetyl group joins the 4C molecule remaining at the end of the cycle to make 6C citric acid. The names of all intermediate compounds in the cycle are not required. C

2

+ C 4= C

6

ńC

5

ń C4 ń C4 ń C4 ń …… and back to the beginning again until death.

One turn of the Krebs cycle yields: 2 @ CO2 3 @ NADH/H+ (Each one can later yield 3 ATP) 1 @ FADH2 (This can later yield 2 ATP) 1 @ ATP directly by substrate level phosphorylation Remember that there are two turns of the Krebs cycle for each 6C glucose input. Also, when calculating the potential net ATP yield, two Link reactions and a Glycolysis must be taken into account. There is a potential yield of 38 molecules of ATP from the breakdown of one glucose molecule in aerobic respiration. If you truly understand aerobic respiration you should be able to locate all 38 ATP! The big picture understanding of aerobic respiration boils down to one simple idea. Glucose is a large stable molecule with lots of chemical energy trapped in its bonds. It is easy to release this energy explosively, say, by combustion; but that would damage the cell, and most of the energy would be lost as useless heat. The complex biochemistry of cell respiration has evolved to obtain——in a series of enzyme-controlled, gentle, incremental steps——as much chemical energy as possible to make ATP (from ADP and P). ELECTRON TRANSPORT CHAIN A further eking out of available potential energy to make ATP is made possible by the electron transport chain. The chain consists of a short series of coenzyme carriers, embedded in tandem in the crista of mitochondria that begins with NAD+ and ending in O2. It is only at this extreme terminus of the complex biochemistry that the oxygen that we breathe in is utilized. NAD+ ń FAD ń various cytochromes ń O2 Each time excited electrons are removed from their associated protons and passed down the chain enough potential energy is extracted to yield 3 ATP molecules. In the case of FAD, the second carrier in the chain and at a slightly lower energy level and a little more electronegative than NAD+, only 2 ATP are produced. 8.1.5 Explain oxidative phosphorylation in terms of chemiosmosis. Chemiosmosis explains precisely how ATP is produced from the electron transport chain. The enzymes controlling the steps in the chain, embedded in the crista membrane, are actually proton pumps. They use the energy from the passage of excited electrons to change their shape and pump disembodied H+ from the matrix of the mitochondria——which contains all the Krebs cycle machinery——into the inter-membrane space. Protons accumulate here and the pH is detectably lower than in the matrix. The accumulation of protons in the inter-membrane space creates a significant concentration gradient. The electron carrier molecules and their associated protein-pumping enzymes are coupled with ATP synthase in the inner membranes of mitochondrial crista. ATP synthase harnesses the concentration gradient of the protons. As protons whizz, from a high to a low concentration, through the open channels of ATP synthase, their energy is harnessed to make ATP from ADP and P. Only at this point do the generic protons rejoin generic electrons, previously transported by the chain, to form respiratory H2O with oxygen, the ultimate and most electronegative of the carriers.

8.1.6 Explain the relationship between the structure of the mitochondrion and its function. Cristae form a large surface area for the electron transport chain, the small space between inner and outer membranes for accumulation of protons, and the fluid matrix containing enzymes of the Krebs cycle. Topic 8.2 –– Photosynthesis 8.2.1 Draw and label a diagram showing the structure of a chloroplast as seen in electron micrographs. 8.2.2 State that photosynthesis consists of light-dependent and light-independent reactions. These should not be called ““light”” and ““dark”” reactions. 8.2.3 Explain the light-dependent reactions. Include the photo-activation of photosystem II, photolysis of water, electron transport, cyclic and non-cyclic photophosphorylation, photo-activation of photosystem I, and reduction of NADP+. Overview of the biochemistry of photosynthesis Essentially, photosynthesis is the biochemical transformation of light energy into chemical energy. There are family resemblances between the light dependent reactions of photosynthesis and oxidative phosphorylation in cell respiration. Both involve proton pumping and chemiosmosis; both pass electrons down an electron transport chain; both utilize a variant of the coenzyme NAD. Both use up, and generate, generous amounts of ATP. In respiration the energy source is the breaking of the chemical bonds of ““ready-made”” food molecules. In photosynthesis the energy source is simply photons of light from the sun. The devil is in the details. Cell respiration is decarboxylation (removal of CO2) and oxidation (the removal of hydrogen and electrons in an excited state) from food molecules. Photosynthesis does the opposite, but in a highly intricate and nuanced way. In photosynthesis, generic, low energy water molecules are the ultimate source of hydrogen and electrons. Photons are used to raise the energy levels of hydrogen (from the H2O) and electrons (also from H2O but via chlorophyll). H+ and e- are combined in the light independent reactions with generic CO2, from the air, to synthesize simple carbohydrates. Photosynthesis can be seen in terms of carboxylation (addition of CO2) and reduction (the addition of hydrogen and associated electrons, in an excited state). The big picture——but not the detailed biochemistry——is more or less the opposite to the decarboxylation and oxidative phosphorylation of cell respiration described above. Photo-activation of photosystem II The chlorophyll complexes of photosystem II and photosystem I are embedded in the thylakoid membrane of the chloroplast. Wedged between the two photosystems are the enzymes and carrier molecules of the electron transport chain. The light dependent reactions begin with a photon of light hitting a chlorophyll molecule in the photosystem II complex. This causes electrons to become excited and jump to a higher energy level.

Electron transport in non-cyclic phosphorylation At this level, the electrons are not stable and a gradient of potential energy is established. Excited electrons pass down a concentration gradient through the electron transport chain where they meet photosystem I, situated at a slightly lower energy level. As in cell respiration, the enzymes controlling the steps in the chain are actually proton pumps. They use the energy from the passage of excited electrons to change their shape and pump disembodied H+ from the stroma of the chloroplast——which contains all the Calvin cycle (light independent reaction) machinery——into the thylakoid membrane space. Protons accumulate here and the pH is detectably lower than in the stroma. The accumulation of protons in the thylakoid membrane space creates a significant concentration gradient. The electron carrier molecules and their associated protein-pumping enzymes are coupled with ATP synthase in thylakoid membranes in close proximity to the photosystems. ATP synthase harnesses the concentration gradient of the protons. As protons whizz, from a high to a low concentration, through the open channels of ATP synthase, their energy is harnessed to make ATP from ADP and P. [It is worth comparing the language used here with the description in 8.1.5.] Photo-activation of photosystem I and the reduction of NADP The chlorophyll in Photosystem I also absorbs a photons. Electrons are boosted to a higher, unstable energy level just as in the case of photosystem II. The electrons are not stable so they move down——via more carriers——to a lower energy level. In tandem with a free proton (H+) from the thylakoid membrane space, these excited electrons reduce NADP to NADPH. NADPH* is shown with an asterix in the equation below to emphasize its extreme high energy and reducing propensity. NADP + e- + H+ ń NADPH* Photolysis of water Electrons lost from photosystem II are replaced by electrons from water as it splits by photolysis. The electrons lost from photosystem I are replaced by electrons coming down from the electron transport chain of photosystem II. Water is the proton source for chemiosmosis and the same protons replace those lost in the generation of NADPH*. Photolysis results in the O2 molecules released by photosynthesizing plants and cyanobacteria. Cyclic vs. non-cyclic photophosphorylation The Calvin cycle requires slightly more ATP than non-cyclic phosphorylation alone can provide. ATP production is boosted by cyclic phosphorylation. Cyclic phosphorylation recycles electrons around Photosystem I only. It involves no photolysis of water and does not use NADP as the final acceptor. No oxygen or NAPH* is produced. 8.2.4 Explain photophosphorylation in terms of chemiosmosis. Outlined under ““Electron transport in non-cyclic phosphorylation”” in 8.2.3 8.2.5 Explain the light-independent reactions. Include the roles of ribulose bisphosphate (RuBP) carboxylase, reduction of glycerate 3phosphate (GP) to triose phosphate (TP), NADPH + H +, ATP, regeneration of RuBP, and subsequent synthesis of more complex carbohydrates. Carboxylation The Calvin cycle, or light independent reactions, are well named. No light is required. A 1C molecule of CO2 is ““fixed”” for each for each turn of the cycle. Six turns generates a 6C glucose molecule. NADPH* from non-cyclic phosphorylation of the light dependent reactions is the potent hydrogen input that drives the cycle. The pivotal molecule is ribulose bisphosphate

(RuBP. The cycle starts with the carboxylation of 5C RuBP by CO2. The enzyme ribulose bisphosphate (RuBP) carboxylase (““Rubisco”” for short——the great carbon fixer) is thought to be the most abundant protein on earth! Reduction It is convenient to consider the six turns of the cycle all at once in order to keep track of the balance of carbon atoms. Six turns require 6CO2 and six NADPH*. The six CO2 molecules fixed by the six RUBP produces six unstable 6C intermediates that spontaneously break down to 12 molecules of GP (glycerate 3-phosphate). GP is not a sugar. Next, 12 molecules of GP are reduced to 12 molecules of TP (triose phosphate) by 12 NADPH* coupled with 12 ATP. 12GP + 12NADPH* + 12ATP ń 12TP + 12NADP+ + 12ADP + 12Pi Triose phosphate (TP), as the suffix of name confirms, is a true 3C sugar. Two of the 12 TP molecules combine to form a single 6C glucose——the most important end product of photosynthesis. The remaining 10 TP molecules are transformed in several stages——one of which involves energy input from ATP——to regenerate the 6 molecules of 5C RuBP needed to start the process all over again. Notice that the carbon balance is maintained; since 10 @ 3C (for TP) and 6 @ 5C (for RuBP) both add up to 30. The lollipop apparatus used to work out the biochemical details of the Calvin cycle shows considerable creativity. It won a Nobel Prize. 8.2.6 Explain the relationship between the structure of the chloroplast and its function. The chloroplast has an intricately folded inner membrane that provides a large surface area for embedded chlorophyll molecules. The folding of the thylakoid creates interconnected structures that look like stacks of coins, called grana. The tiny grana are able to orient themselves perpendicular to the sun. The exact mechanism for this is frontier cell biology. The thylakoids provide an inter-membrane space for the accumulation of protons to power ATP production by chemiosmosis. The fluid portion of the chloroplast, or stroma, contains the enzymes and various substrates of the Calvin cycle. 8.2.7 Explain the relationship between the action spectrum and the absorption spectrum of photosynthetic pigments in green plants. Plants are green because they reflect green light. They absorb light from other wavelengths. The absorption spectrum illustrates this. The action spectrum graph can be mapped onto the absorption spectrum almost exactly. The action spectrum measures directly the effectiveness of the various wavelengths of light in actually driving photosynthesis.

TOPIC 9: PLANT SCIENCE Topic 9.1 –– Plant structure and growth 9.1.1 Draw and label plan diagrams to show the distribution of tissues in the stem and leaf of a dicotyledonous plant. Note that plan diagrams show distribution of tissues (for example, xylem and phloem) and do not show individual cells. They are sometimes called ““low power”” diagrams. 9.1.2 Outline three differences between the structures of dicotyledonous and monocotyledonous plants. Most familiar plants are ““dicots”” with branched veins in their leaves and seeds that are divided into two halves. The ““monocots”” are also very familiar as crop staples. Monocots include the grasses, corn and rice. Monocot leaves have parallel veins and their seeds are in one piece. Other differences include the distribution of vascular tissue in stems, two or one cotyledon, floral organs in multiples of 3 in monocotyledonous versus 4 or 5 in dicotyledonous, fibrous adventitious roots in monocotyledonous versus tap root with lateral branches in dicotyledonous. 9.1.3 Explain the relationship between the distribution of tissues in the leaf and the functions of these tissues. This should be restricted to dicotyledonous plants. The functions should include: absorption of light, gas exchange, support, water conservation, and the transport of water and products of photosynthesis. Leaves are the organs of photosynthesis. They are very thin and present a large surface area for the absorption of light energy. The internal structure is sandwiched between an upper and a lower epidermis. The upper epidermis is transparent and covered with a waxy cuticle to prevent water loss. Just below the upper epidermis is a dense layer of palisade cells. Palisade cells are the main photosynthetic cells. Below the palisade layer is a sparse layer of spongy mesophyll cells that assist photosynthesis and also scaffold air spaces. CO2 and O2 and H2O vapor circulate freely here. Mostly in the lower epidermis, are pairs of guard cells that surround pores called stoma. During the day stomata allow Co2 to enter the leaf and O2 to exit. Stomata also control the rate of transpiration flow——the net loss of H2O vapor through the leaves. Vascular bundles containing phloem and xylem permeate the leaf in order to facilitate the gross transport of water and minerals (up from the leaf) and the translocation of sugars and other end-products of photosynthesis (back to the roots and up to the meristems for growth and to any flowers for nectar, pollen or——later——fruit production). 9.1.4 Identify modifications of roots, stems and leaves for different functions: bulbs, stem tubers, storage roots and tendrils. Sugars are translocated in the phloem mostly as sucrose. Sugars are stored as starch in plants. Some plants have modified storage organs such as the thick roots of carrots and the stem tubers of potatoes. 9.1.5 State that dicotyledonous plants have apical and lateral meristems. Apical meristems are sometimes referred to as primary meristems, and lateral meristems as cambium. Meristems generate new cells for growth of the plant. 9.1.6 Compare growth due to apical and lateral meristems in dicotyledonous plants. Apical meristems as the name suggests are embryonic tissues at the leaf and flower buds, and at the bud and root tips. Apical meristems allow the plant to grow in length. Lateral meristems

are found wedged between xylem and phloem in vascular bundles and as a cylinders of dividing, undifferentiated cambium cells that run the length of the plant. Lateral meristems allow the plant to get thicker and more robust. This is known as secondary growth. Plant meristems are good places to observe the stages of mitosis. 9.1.7 Explain the role of auxin in phototropism as an example of the control of plant growth. Positive phototropism is the growth of a plant towards light. Sunflowers famously do this. Auxin is a plant hormone (a chemical messenger——just as hormones are in animals) that promotes cell in the tips of shoots. Classic controlled experiments involving chopping off the shoot tip, shielding the tip from light with a silver paper cap and placing agar between the shoot and its severed tip; prove that auxin is located in the tip and activated by light. Auxin travels downwards on the opposite side to ambient sunlight and stimulates cell elongation on the ““darker side”” by the uptake of H2O. The asymmetrical elongation makes the tip to bend towards the light. Topic 9.2 –– Transport in Angiospermophytes 9.2.1 Outline how the root system provides a large surface area for mineral ion and water uptake by means of branching and root hairs. 9.2.2 List ways in which mineral ions in the soil move to the root. There are three processes: diffusion of mineral ions, fungal hyphae (mutualism), and mass flow of water in the soil carrying ions. 9.2.3 Explain the process of mineral ion absorption from the soil into roots by active transport. Plants take up vital mineral ions against a concentration gradient by active transport through gated channels. This, of course, requires energy in the form of ATP. As in animal membranes, H2O cannot be pumped directly by active transport. However H2O can be moved locally as needed by actively pumping K+ ions to increase a solute concentration gradient so that a controlled amount of H2O follows by passive osmosis. Auxin works in this way. 9.2.4 State that terrestrial plants support themselves by means of thickened cellulose, cell turgor and lignified xylem. 9.2.5 Define transpiration. Transpiration is the loss of water vapor from the leaves and stems of plants.

 9.2.6 Explain how water is carried by the transpiration stream, including the structure of xylem vessels, transpiration pull, cohesion, adhesion and evaporation. Xylem tubes are dead, lignin (woody) tubes that have sieve-like ends to allow the free flow of water. Xylem vessels are derived from a line of single cells whose outer cellulose cell walls become lignified and fused. This results in a tiny lumen, much smaller than the lumen of an animal blood capillary. Water moves through xylem because it is pulled from above as water evaporates through the stomata on the undersides of leaves because of transpiration. Confined inside the xylem tubes, the cohesive and adhesive properties of water, caused by hydrogen bonding between the polar molecules, come dramatically into play. The tiny cylinders of liquid water travelling through xylem vessels have the tensile strength of steel wires! Osmosis and capillary action also play a role in the transpiration stream.

 9.2.7 State that guard cells can regulate transpiration by opening and closing stomata.

9.2.8 State that the plant hormone abscisic acid causes the closing of stomata.

 9.2.9 Explain how the abiotic factors light, temperature, wind and humidity, affect the rate of transpiration in a typical terrestrial plant. As with wet clothes hanging on a washing line, high temperature, low humidity and plenty of wind increase the rate of water loss. Air just below a stoma tends to be at around 100% humidity. Wind constantly introduces new air at a lower humidity, relentlessly replenishing the concentration gradient. Plants tend to open their stomata during the light of day in order to allow in CO2, which is a limiting factor for photosynthesis. For this reason light tends to increase the rate of transpiration; although many plants close their stomata around mid-day to limit excessive H2O loss. 9.2.10 Outline four adaptations of xerophytes that help to reduce transpiration. Xerophytes are plants adapted to desert conditions. Adaptations to an extremely dry environment include: reduced leaves, rolled leaves, spines, deep roots, thickened waxy cuticle, reduced number of stomata, stomata in pits surrounded by hairs, water storage tissue, low growth form, CAM (crassulacean acid metabolism) and C4 physiology. CAM plants open their stomata and absorb carbon dioxide only at night when cooler temperatures limit evaporation. They fix CO2 with an enzyme called PEP carboxylase, which is extremely efficient. The CO2 is converted to organic acids which accumulate at night and are used to power photosynthesis during the daytime when sunlight is available. In CAM plants the storage of CO2 as organic acids and their subsequent use is separated temporally. C4 plants can open their stomata in the day. Instead of temporally separating the reactions, they have a distinct leaf anatomy that separates the two reaction stages spatially. In C4 plants, CO2 is also fixed by PEP carboxylase to make 4C oxaloacetate. This occurs in rosettes of mesophyll cells, which surround bundle sheath cells, where the regular Calvin cycle takes place. This peculiar layout of photosynthesizing cells in C4 leaves is called kranz anatomy, after the German word for wreath. 9.2.11 Outline the role of phloem in active translocation of sugars (sucrose) and amino acids from source (photosynthetic tissue and storage organs) to sink (fruits, seeds, roots). Unlike xylem, phloem is very much a living tissue with cytoplasmic cells arranged into tubes that distribute sucrose, amino acids, and other organic nutrients throughout the plant. Overall phloem translocates food made in the leaves down to the roots and other non-photosynthetic parts of the plant, from source to sink. Active transport often occurs locally in phloem tissue. Topic 9.3 –– Reproduction in Angiospermophytes 9.3.1 Draw and label a diagram showing the structure of a dicotyledonous animalpollinated flower. Limit the diagram to sepal, petal, anther, filament, stigma, style and ovary. 9.3.2 Distinguish between pollination, fertilization and seed dispersal. Pollination is the prerequisite to fertilization. Pollination is the transfer of pollen onto the stigma of a carpel by wind or animal carriers. After recognizing a compatible stigma, the pollen grain hatches and extrudes a cytoplasmic pollen tube which extends down through the entire

length of the stigma until it finds the ovule which is surrounded by an ovary. At the end of its journey, the pollen tube transfers a pair of haploid, male nuclei into the ovule where fertilization——the fusion of gametes——takes place. The seed is derived from the ovule. Fruit are derived from the ovary. The fruit is responsible for seed dispersal ——which ensures that new plants do not compete with the established parent plants for light, H2O and other essential nutrients. Dispersal may be by animal, wind or water. 9.3.3 Draw and label a diagram showing the external and internal structure of a named dicotyledonous seed. The named seed should be non-endospermic. The structure in the diagram should be limited to testa, micropyle, embryo root, embryo shoot and cotyledons. 9.3.4 Explain the conditions needed for the germination of a typical seed. Seeds are resistant and dormant which help the survival of the plant because seeds can wait for their optimal environment to grow. When they are provided with the right conditions their dormancy breaks and they germinate. In order to germinate plants generally need water, a specific temperature range and a good supply of O2. Seeds vary in their light requirements. Some seeds need to be passed through an animal digestive system before becoming viable. Others need chilling by frost or even a toasting in a forest fire! 9.3.5 Outline the metabolic processes during germination of a starchy seed. Absorption of water precedes the formation of gibberellin in the embryo’’s cotyledon. This stimulates the production of amylase, which catalyzes the breakdown of starch to maltose. This subsequently diffuses to the embryo for energy release and growth. 9.3.6 Explain how flowering is controlled in long-day and short-day plants, including the role of phytochrome. Limit this to the conversion of Pr (red absorbing) to Pfr (far-red absorbing) in red or white light, the gradual reversion of Pfr to Pr in darkness, and the action of Pfr as a promoter of flowering in long-day plants and an inhibitor of flowering in short-day plants.

TOPIC 10: GENETICS Topic 10.1 - Meiosis 10.1.1 Describe the behavior of the chromosomes in the phases of meiosis. Overview Meiosis is a special variant of cell division that produces genetically diverse haploid gametes. Meiosis follows the same strategy as mitosis which produces diploid clones. Meiosis consists of two successive divisions producing 4 haploid daughter cells. Everything hinges on homologous chromosomes lining up at Prophase I. Prophase I As in mitosis the replicated chromosomes become supercoiled and are visible as two sister chromatids attached at the centromere. Homologous chromosomes align. Non-sister chromosomes can randomly touch and form chiasmata. Sections of homologous chromosomes may be exchanged at these points. This process, also called recombination, partially explains the genetic uniqueness of gametes produced by meiosis.

Meanwhile, the nucleus breaks down and dissolves and microfilaments form that will later pull the homologous away from each other towards the poles of the cell. Prophase I represents the longest of the stages, and may occupy 90% of the time spent in meiosis. Metaphase I The chromosomes line up along the equator of the cell in this stage. They line up randomly. This innocuous fact is the second vital factor that contributes to the vast genetic variation in gametes. At the two poles of the cell are located the two centrioles. Spindle fibers are attached to the centromere of each chromosome. Anaphase I The chromosomes are pulled by the spindle fibers of each centriole on the opposite poles. During the first meiotic division sister chromatids remain attached at their centromeres. Telophase I The two groups of homologous chromosome pairs move farther away from each other and the cytoplasm constricts between two groups of chromosomes. This results in the formation of two cells each with a single complete set of homologous chromosomes. . Prophase and Metaphase II The cell move into the second meiotic division immediately. The chromosomes remain visible as sister chromatids. They do not line up at the equator in any special way. Anaphase II The spindle fibers pull the individual chromatids apart. This results in the splitting of the centromere and the final separation of the sister chromatids. Telophase II Each group of chromosomes becomes surrounded by a nuclear envelope and the chromosomes become thin and diffuse again. The nucleolus reappears and the spindle fibers disappear. 10.1.2 Outline the formation of chiasmata in the process of crossing over. The separated sister chromatids produced at the end of Anaphase II are by no means genetically identical if crossing over has occurred. An allele combination can change from PQ/PQ and pq/pq to PQ/Pq and pQ/pq. 10.1.3 Explain how meiosis results in an effectively infinite genetic variety in gametes through crossing over in prophase I and random orientation in metaphase I. A single pair of homologous chromosomes can only line up in one way. Two pairs can line up two ways; three pairs in eight different ways: ABC, ABc, AbC, Abc, aBC, aBc, abC, abc Eight permutations are 23, or 2n where n is the haploid number of chromosomes. Humans have n = 23 chromosomes. This provides 223 combinations, or 8,388,608 different combinations. When crossover is also taken into account, it is easy to see how a single ejaculation may contain at least 100.000.000 genetically unique sperm. 10.1.4 State Mendel’’s law of independent assortment. ““Each pair of alleles segregates into gametes independently””

The law of independent assortment was soon found to have exceptions when pairs of genes are linked on a chromosome; but the law that Mendel discovered in the 19th century does operate for the majority of pairs of genes. 10.1.5 Explain the relationship between Mendel’’s law of independent assortment and meiosis. Mendel's law of independent assortment goes hand in hand with meiosis. The law of independent assortment is a description of the empirical fact that chromosomes separate into cells in meiosis independently of one another; that is, the distribution of genes for one trait does not affect the distribution of genes for any other trait. The commonsense exceptions, of course, are ““linked genes”” that happen to be located on the same chromosome.

Topic 10.2 - Dihybrid Crosses and gene linkage 10.2.1 Calculate and predict the genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes. In monohybrid crosses the F2 generation produces 3:1 phenotypic ratios for heterozygotes; and 1:1 phenotypic ratios for the test cross. In dihybrid crosses the analogous ratios are 9:3:3:1 and 1:1:1:1. Visualizing the F2 generations of heterozygote dihybrid crosses requires monster 4 x 4 Punnett squares. The haploid gametes for the diploid genotype RrSs are RS, Rs, rS and rs. 10.2.2 Distinguish between autosomes and sex chromosomes. There is an important exception to the rule of homologous chromosomes for human somatic cells. Human females have a homologous pair of X chromosomes (XX) and males have one X and one Y chromosome (XY). All other chromosomes——the non-sex determining chromosomes—— are called autosomes, 10.2.3 Explain how crossing over between non-sister chromatids of a homologous pair in prophase I can result in an exchange of alleles. Refer to 10.1.2 ““Outline the formation of chiasmata in the process of crossing over.”” 10.2.4 Define linkage group. A linkage group is a group of genes whose loci are on the same chromosome 10.2.5 Explain an example of a cross between two linked genes. Alleles are usually shown side by side in dihybrid crosses, for example, TtBb. In representing crosses involving linkage, it is more common to show them as vertical pairs, for example:

This format will be used in examination papers, or students will be given sufficient information to allow them to deduce which alleles are linked.

10.2.6 Identify which of the offspring are recombinants in a dihybrid cross involving linked genes. Real data is messy but should follow the ratios within statistical limits. If the monohybrid ratios 3:1 or 1:1 occur for known dihybrid crosses where you were expecting 9:3:3:1 and 1:1:1:1, you can be quite sure that the genes are ““linked”” on the same chromosome. If known dihybrid crosses conform to the monohybrids ratios mostly, but with small numbers of aberrant, phenotypes arising, crossover is the likely explanation.



 



Topic 10.3 - Polygenic Inheritance 10.3.1 Define polygenic inheritance. Polygenic inheritance is an additive effect of two or more genes on a single phenotypic character. These characters are usually those that have more than just two phenotypes, or even an apparently continuous variation of phenotypes, such as human skin color. 10.3.2 Explain that polygenic inheritance can contribute to continuous variation using two examples, one of which must be human skin color. The selective advantage of dark skin to protect against ultraviolet light and light skin to allow vitamin D production could be mentioned. The correlation between skin color and intensity of sunlight is clear, though the selective advantages of particular skin colors can now be overcome by the use of sun-block creams and vitamin D supplements. Eye color, height, intelligence and countless other human characteristics are at least partially explained by polygenic inheritance.



TOPIC 11: HUMAN HEALTH AND PHYSIOLOGY Topic 11.1 - Defense against infectious disease 11.1.1 Describe the process of blood clotting. Clotting factors are released from platelets (incomplete fragments of cells without a nucleus) and damaged cells nearby the injury or opening, resulting in the formation of the enzyme thrombin. Thrombin catalyzes the conversion of plasma-soluble fibrinogen, which is always present in the bloodstream, into the fibrous protein fibrin which captures red blood cells and immobilizes the fluid portion of blood. Platelets enter this fibrous mass and send out sticky extensions to one another. Platelets eventually contract, forcing out liquid, and scabbing over the wound.

It is critical that blood clotting does not occur spontaneously, hence a complicated cascade of enzyme and clotting factor mediated reactions. One clotting factor, Factor 8, is the missing molecule in hemophilia sufferers. It is interesting to note that blood clots even faster in the presence of Epinephrine (adrenaline). 11.1.2 Outline the principle of challenge and response, clonal selection and memory cells as the basis of immunity. This is intended to be a simple introduction to the complex topic of immunity. The idea of a polyclonal response can be introduced here. During the lifetime of each individual an internal evolution or ““clonal selection”” occurs in response to pathogen exposure. Each person has a slightly different immune response as a result. A wide variety of B cells inhabit the bone marrow. Antigens are foreign proteins. Antigens are produced by pathogens. They alert and are recognized by immune cells to their presence in the body. Antigens have specific shapes (tertiary structures) that come into contact with the specific receptors of the appropriate B cells. The B cells whose receptors bind with antigens are selected and made in multiple copies. These clones of B cells then divide further into plasma cells, which produce antibodies targeted to the pathogen. Antibodies are Y-shaped proteins that are secreted into the blood stream. Memory cells are produced as a response to encountering a specific pathogen. They persist for a long time and are ready to destroy the antigens as part of a powerful secondary response if they are encountered again. 11.1.3 Define active and passive immunity. Active immunity is immunity due to the production of antibodies by the organism itself after the body’’s defense mechanisms have been stimulated by antigens. Passive immunity is immunity due to the acquisition of antibodies from another organism in which active immunity has been stimulated, including via the placenta, colostrum, or by injection of antibodies. 11.1.4 Explain antibody production. Limit the explanation to antigen presentation by macrophages and activation of helper T cells leading to activation of B-cells which divide to form clones of antibody-secreting plasma cells and memory cells. Macrophages consume foreign bacteria with antigen molecules in their membranes. They then present these foreign antigens on their own membranes with the help of special protein structures. When helper T-cells come into contact with macrophages, they also pick up these antigens and incorporate them into their own protein structures, which they use to present the antigens to B-cells. The B-cell then divides to form clones of antibody secreting plasma cells and memory cells. 11.1.5 Describe the production of monoclonal antibodies and their use in diagnosis and in treatment. Production should be limited to the fusion of tumor and B-cells, and their subsequent proliferation and production of antibodies. Detection of antibodies to HIV is one example in diagnosis. Others are detection of a specific cardiac iso-enzyme in suspected cases of heart attack and detection of human chorionic gonadotrophin (HCG) in pregnancy test kits. Examples of the use of these antibodies for treatment include targeting of cancer cells with drugs attached to monoclonal antibodies, emergency treatment of rabies, blood and tissue typing for transplant compatibility, and purification of industrially made interferon. Production of monoclonal antibodies is certain to be a growth area in biotechnology, with many potential applications and consequent economic opportunities. Some of the applications will be of most use in developing countries, raising the question of how they will be paid for, whether commercial companies should be expected to carry out pro bono research and development, or whether national governments should provide funds for it through aid budgets. Historically,

the development of treatments for tropical diseases and parasites has lagged far behind those for the diseases prevalent in wealthier countries. 11.1.6 Explain the principle of vaccination. Vaccination is when you inject a weakened or killed version of a pathogen into the body, causing the immune system artificially to mount a primary response. The resulting immune response results in the production of B memory cells, which can then rapidly produce antibodies in response to the pathogen. When the real disease strikes, a secondary response occurs immediately and prevents the appearance of illness caused by the pathogen. 11.1.7 Discuss the benefits and dangers of vaccination. Benefits include the total elimination of certain diseases, prevention of pandemics and epidemics, decreased health-care costs and prevention of harmful side-effects of diseases. The dangers should include the possible toxic effects of Mercury in vaccines, possible overload of the immune system and possible links with autism. For parents there are ethical decisions to be made, to minimize risk for one’’s own child, but also to help to prevent epidemics that could affect other children. This is an area where it is important to estimate accurately the size of risks, using good scientific data. The use of double-blind trials for vaccines or for drug treatments could be discussed. The placebo effect could also be considered, together with the complex interplay between mind and body in feelings of illness and health. Does the patient or the doctor decide whether the patient is well or not? There are also questions about the relationship between the scientific community and the general public. How can the general public be given clear information about the benefits and risks of vaccination? What went wrong in the recent case of misplaced fears about the measles, mumps and rubella (MMR) vaccine in the UK? There are ethical questions here about who should decide vaccination policy in a country, and whether it is ethically acceptable to have a compulsory vaccination program. Topic 11.2 - Muscles and Movement 11.2.1 State the roles of bones, ligaments, muscles, tendons and nerves in human movement. To produce movement, muscles often work in antagonistic pairs. This is due to fact that muscles can only do work when they contract. Relaxation is entirely passive. The biceps and triceps are obvious examples; but the circular and longitudinal muscles in the gut, blood vessels, anus and iris are also important. Muscles move the bone levers. Fibrous tendons made of cartilage attach muscle to bones. Hefty, rubbery ligaments, also made of cartilage, join bone to bone. Motor nerves initiate muscle contraction. Proprioreceptors monitor precisely how much the muscles are being stretched and how force they need to provide. 11.2.2 Label a diagram of the human elbow joint, including cartilage, synovial fluid, joint capsule, named bones and antagonistic muscles (biceps and triceps). 11.2.3 Outline the functions of the structures in the human elbow joint named in 11.2.2. Synovial fluid enclosed within ligaments provides a lubricating medium that prevents friction and wearing of bones. 11.2.4 Compare the movements of the hip joint and the knee joint. The hip joint is a ball and socket that allows rotation. The knee joint, like the elbow, works as a

hinge in a single plane only. Both the knee and the elbow have a limited range of movement they cannot bend backwards. 11.2.5 Describe the structure of striated muscle fibers, including the myofibrils with light and dark bands, mitochondria, the sarcoplasmic reticulum, nuclei and the sarcolemma Muscle fibers are strange. They are elongated multinucleate cells, more like fungal hyphae than regular animal cells. The fibers are arranged in bundles in muscle tissue. Electron microscopy has revealed that each individual fiber consists of multiple myofibrils surrounded by a special type of endoplasmic reticulum called the sarcoplasmic reticulum. Myofibrils are divided along their length into units called sarcomeres. They are made of alternating actin and myosin filaments. Actin is made of thin protein filaments and myosin of thicker, stubbled ones. 11.2.6 Draw and label a diagram to show the structure of a sarcomere, including Z lines, actin filaments, myosin filaments with heads, and the resultant light and dark bands. 11.2.7 Explain how skeletal muscle contracts, including the release of calcium ions from the sarcoplasmic reticulum, the formation of cross-bridges, the sliding of actin and myosin filaments, and the use of ATP to break cross-bridges and re-set myosin heads. Contraction of muscles results from the sliding of the actin and myosin fibers past one another. The appearance of the light and dark bands changes accordingly. Contraction requires the release of Ca2+ ions from the sarcoplasmic reticulum. Extending from the myosin to the actin filaments are tiny swivel heads that are best described as ““oars behaving as ratchets.”” When induced by ATP, the myosin heads swivel forward slightly. As they twist they drag the actin fibers past the myosin fibers, anchoring them temporarily by forming cross-bridges. ATP is used to break cross-bridges and re-set the myosin heads so the whole process can repeat itself. As a result the entire sarcomere shortens. The net result of millions of sarcomeres shortening simultaneously is large scale muscle contraction. 11.2.8 Analyze electron micrographs to find the state of contraction of muscle fibers. Muscle fibers can be fully relaxed, slightly contracted, moderately contracted and fully contracted. Topic 11.3 –– The kidney 11.3.1 Define excretion. Excretion is the removal from the body of the waste products of metabolic pathways. 11.3.2 Draw and label a diagram of the kidney. Include the cortex, medulla, pelvis, ureter and renal blood vessels. 11.3.3 Annotate a diagram of a glomerulus and associated nephron to show the function of each part. The microscopic nephron is the functional unit of the kidney. Ultrafiltration in the Bowman’’s capsule; selective reabsorption in the proximal tubule; maintenance of a hypertonic medulla region by the loop of Henle; and the final production of hypertonic urine in the collecting duct are the four essential annotations. 11.3.4 Explain the process of ultrafiltration, including blood pressure, fenestrated blood capillaries and basement membrane. The renal artery branches extensively inside the kidney. The tiny branch leading to each nephron is called the afferent renal arteriole. The afferent arteriole sub-divides into a dense network of capillaries called glomerulus inside each Bowman's capsule. The capillaries merge

again into a single blood vessel that leaves the Bowman’’s capsule called the efferent renal artery. The afferent arteriole is slightly more muscular and has a larger lumen than the efferent arteriole. This means that blood inside the glomerulus is under high pressure. The pressure forces plasma through the fenestrated walls of the capillary into the capsule. Podocytes are epithelial cells of the inner lining of the capsule which have tentacle-like processes that cling to the capillary cell basement membrane. They provide an intimate interface and a massive filtration surface. Blood cells and large molecules like blood proteins and are held back. However, the process is a crude one. Glucose, amino acids and other valuable end-products of digestion end up in the glomerular filtrate. 11.3.5 Define osmoregulation. Osmoregulation is the control of the water balance of the blood, tissue or cytoplasm of a living organism. 11.3.6 Explain the reabsorption of glucose, water and salts in the proximal convoluted tubule, including the roles of microvilli, osmosis and active transport. After ultrafiltration the reabsorption of valuable nutrients and homeostatic restoration of osmotic balance is necessary. The inside of the proximal tubule is lined with microscopic microvilli that provide a large surface area for selective reabsorption. Amino acids and glucose are regained by active transport. The reabsorption of salt molecules pulls back H2O by passive osmosis. 11.3.7 Explain the roles of the loop of Henle, medulla, collecting duct and ADH (vasopressin) in maintaining the water balance of the blood. The descending loop of Henle reabsorbs water by osmosis. The hair-pin bend of the loop enters the medulla section. The medulla is hypertonic compared to the cortex. This concentration gradient is maintained by the loop of Henle which acts as a counter-current multiplier. This is contrived by pumping Na+ ions by active transport across from the ascending limb to the descending limb. This results in accumulation of Na+ in the bend of the loop. As fast as they might be carried away by the ascending limb they are pumped back over to the descending limb where they travel back down. The collecting duct runs towards the medulla parallel to the loop of Henle. H2O is removed from the urine in the collecting duct by passive osmosis as it travels through the hypertonic medulla region. Fine tuning of H2O balance is controlled by hormone ADH (vasopressin). Secretion of ADH increases the permeability of the collecting duct wall concentrating the urine. As the name suggests ADH has an anti-diuretic effect. 11.3.8 Explain the differences in the concentration of proteins, glucose and urea between blood plasma, glomerular filtrate and urine. The renal artery enters the kidney with unfiltered blood containing urea and other unwanted materials. It also carries much needed oxygen to the highly metabolically active kidney. The renal vein leaves the kidney with blood that contains adjusted levels of salt, and water and dissolved CO2 in the form of HCO3-. No urea remains since it is eliminated in the filtrate. 11.3.9 Explain the presence of glucose in the urine of untreated diabetic patients. In primitive conditions a doctor might diagnose diabetes by tasting the urine of a patient! If urine is present above a certain threshold selective reabsorption is insufficient and some excess sugar appears in the urine. This is by no means sufficient to prevent dangerous symptoms.



Topic 11.4 –– Human Reproduction 11.4.1 Annotate a light micrograph of testis tissue to show the location and function of interstitial cells (Leydig cells), germinal epithelium cells, developing spermatozoa and Sertoli cells. Each testis contains about 500 tightly packed seminiferous tubules, each between 30 and 70cm long. The germinal epithelium lining the tubules is the site of sperm production. Unraveled, the seminiferous tubules from both testes would reach 500m. This represents an enormous surface area for spermatogenesis. Sertoli cells surround form niches for of dividing and differentiating sperm cells. They provide support, nutrients and contain binding sites for testosterone. Leydig cells are the interstitial cells, visible in cross section, between the circular seminiferous tubules. They secrete testosterone. 11.4.2 Outline the processes involved in spermatogenesis within the testis, including mitosis, cell growth, two divisions of meiosis and cell differentiation. Spermatogenesis and oogenesis follow the same essential scheme. Both produce haploid gametes by meiosis; but the detailed strategies are different——and for good reason. In spermatogenesis the objective is to produce as many motile sperm as possible. In oogenesis the production of colossal eggs, in small quantities, is the goal. Human egg cells (ova) are 1mm in diameter and are visible to the naked eye! One reason for this huge size is that the ovum provides all of the cytoplasm and organelles to the zygote. The sperm only contributes its haploid share of the genetic message. The other reason is that the zygote can rapidly and repeatedly cleave itself to make a hollow ball of cells (the blastula) without growing between successive divisions. By the time the blastocyst is ready for implantation the normal cell ratio of nucleus to cytoplasm is restored In spermatogenesis diploid germ cells first multiply by mitosis (several times) to create large quantities of potential sperm. Next, each diploid primary spermatocyte undergoes two meiotic divisions take place to produce four haploid spermatids. Spermatids then differentiate into mature spermatozoa. Mature sperm break away from the interior wall of the seminiferous tubule and travel down the lumen for storage in the epididymis. 11.4.3 State the role of LH, testosterone and FSH in spermatogenesis. FSH stimulates the Sertoli nurse cells, which provide binding sites for testosterone. LH stimulates testosterone production by the Leydig interstitial cells. Testosterone is the hormone controlling the rate spermatogenesis. FSH and LH both come from the pituitary, the master endocrine gland linked to the hypothalamus of the brain. 11.4.4 Annotate a diagram of the ovary to show the location and function of germinal epithelium, primary follicles, mature follicle and secondary oocyte. It is conventional to show on one diagram the successive stages of egg development in time from primary oocytes, through ovulation from the mature Graffian follicle, up to and including the production of a corpus luteum. This diagram should be cross-referenced with the menstrual cycle outlined in 6.6.2 and 6.6.3. A baby girl is born with over a million immature egg cells already in her ovaries. At puberty about 300,000 of these diploid primary oocytes suspended in prophase of meiosis remain. Only a small percentage of primary oocytes will mature into eggs. Only about 400 eggs are released during a woman's reproductive life, usually one during each menstrual cycle. Until released an egg——at this stage a haploid secondary oocyte——remains dormant in its follicle suspended in prophase II of meiosis.

Eggs are the largest and amongst the most long-lived of all cells. 11.4.5 Outline the processes involved in oogenesis within the ovary, including mitosis, cell growth, two divisions of meiosis, the unequal division of cytoplasm and the degeneration of polar body. Typically each month, a primary oocyte completes meiosis I to form a haploid secondary oocyte. The meiotic divisions in oogenesis involve unequal cytokinesis. The first division produces a large cell and a much smaller polar body. The second meiotic division, which produces the ovum and a smaller polar body, occurs only if a sperm cell penetrates the secondary oocyte. At this time the second polar body also divides, yielding three polar bodies in total). After meiosis is completed and the second polar body separates from the ovum, the haploid nuclei of the sperm and the mature ovum, haploid at last from the final division, fuse together in the actual process of fertilization. After fertilization the three polar bodies degenerate and rapidly die. 11.4.6 Draw and label a diagram of a mature sperm and egg. 11.4.7 Outline the role of the epididymis, seminal vesicle and prostate gland the production of semen. The epididymis is where the sperms are stored before being added to the seminal fluid and ejaculated. The seminal vesicles and the prostate contribute to the seminal fluid which nourishes sperm and counters the hostile environment of the vagina. The vagina is quite acidic due to the production of lactic acid by native microflora, viscous with mucus secretions and patrolled by immune cells. Seminal vesicles produce a yellowish viscous fluid rich in fructose and amino acids. It is slightly alkaline and makes up about 70% of human semen. Prostrate fluid is whitish and thin. It contains proteolytic enzymes. The tiny bulbourethral glands secrete a pre-ejaculate during sexual arousal that lubricates the urethra and neutralizes traces of acidic remaining in the urethra. It is possible for this fluid to pick up sperm from previous ejaculations. This is one reason for the failure of the withdrawal method of contraception. 11.4.8 Compare the processes of spermatogenesis and oogenesis, including the number of gametes and the timing of the formation and release of gametes. See: 11.4.2 above. Spermatogenesis begins at puberty. Some men father children in extreme old age. 100 million sperm per day can be produced. A man can produce a trillion (1012) viable sperm cells in a lifetime! By contrast, a woman usually produces a single egg each month from puberty to menopause. No more than 500 in a lifetime. 11.4.9 Describe the process of fertilization, including the acrosome reaction, penetration of the egg membrane by a sperm and the cortical reaction. The acrosome reaction occurs when the sperm comes into contact with the protective coats surrounding the egg. Arrival at the zona pellucida region, which is chock full of jelly-like glycoproteins, initiates the process. The acrosome is vacuole filled with proteolytic enzyme located at the tip of the sperm head. The acrosome breaks through the zona pellucida and the plasma membranes of the sperm head and the egg fuse, allowing the sperm nucleus to enter the egg cytoplasm. The cortical reaction results as the sperm comes in contact with the egg cell membrane, triggering osmotic changes to the zona pellucida envelope that make it an impassable barrier for other sperms attempting to enter. The second meiotic division, which produces the ovum and a smaller polar body, is initiated by fertilization. See: 11.4.5.

11.4.10 Outline the role of HCG in early pregnancy. Human chorionic gonadotropin is secreted in early pregnancy by the embryo and in later by the placenta of the developing fetus. It acts like pituitary LH maintaining the corpus luteum throughout the first trimester. The corpus luteum secretes progesterone the hormone of pregnancy. In the absence of the fetal HCG override, the decline in maternal LH due to inhibition of the pituitary by progesterone would result in menstruation and spontaneous abortion of the embryo. Levels of HCG in the maternal blood are so high that some is excreted in the urine, where it can be detected in pregnancy tests. 11.4.11 Outline early embryo development up to the implantation of the blastocyst. The elephantine zygote undergoes several mitotic divisions resulting in a hollow ball of cells called the blastocyst. At this juncture the individual cells are normal-sized and the ratio of cytoplasm to nuclear material is restored. 11.4.12 Explain how the structure and functions of the placenta, including its hormonal role in secretion of estrogen and progesterone, maintain pregnancy. During first month of development the embryo obtains nutrients directly from the thickened endometrium. Later tissues grow from the developing embryo and mingle with the endometrium to form the placenta. The placenta is disk-shaped and grows to about the size of a dinner plate. The surface of the placenta eventually forms chorionic villi which present a large surface area and intimate contact for the exchange of material between the fetal and maternal circulations. Note that fetal and maternal bloods do not mix. Diffusion of material between the maternal and embryonic circulations provides nutrients, exchanges respiratory gases, and disposes of metabolic wastes like urea for the embryo. Blood from the embryo travels to the placenta through arteries of the umbilical cord and returns via the umbilical vein directly to the liver of the embryo. 11.4.13 State that the fetus is supported and protected by the amniotic sac and amniotic fluid. 11.4.14 State that materials are exchanged between the maternal and fetal blood in the placenta. 11.4.15 Outline the process of birth and its hormonal control, including the changes in progesterone and oxytocin levels and positive feedback. Oxytocin is often known as the "hormone of love" because it is involved with lovemaking, fertility, contractions during labor and birth, and the release of milk in breastfeeding. Receptor cells allowing a woman's body to respond to oxytocin increase gradually in pregnancy, and then sharply in labor. Oxytocin is a potent stimulator of contractions, which help to dilate the cervix, move the baby down and out of her body, give birth to her placenta, and limit bleeding at the site of the placenta. During labor and birth, the pressure of the baby against the cervix and then against tissues in the pelvic floor stimulates oxytocin and contractions. Secretion is escalated by the stimulus of the contractions themselves, escalating its secretion by positive feedback. As suckling newborn has a similar but much less dramatic positive feedback effect on oxytocin secretion. Endorphins levels rise steadily and steeply during un-medicated labors. Endorphins produce an altered state of consciousness that helps women flow with the process of a long and arduous labor. Despite the hard work of labor and birth, a woman with high endorphin levels can feel alert, attentive, and even euphoric as she begins to get to know and care for her baby after birth. Endorphins may play a role in strengthening the mother-infant relationship at this

time. A drop in endorphin levels in the days after birth may contribute to the "blues" that many women experience at this time. Women who feel threatened during labor (for example by fear or severe pain) may produce high levels of adrenaline. Adrenaline can slow labor or stop it altogether. Earlier in human evolution, this disruption may have helped birthing women move to a place of greater safety. Progesterone maintains the uterine environment of the developing baby. Progesterone levels can rise to 15x the normal levels by the third trimester. After the second month of pregnancy, the placenta takes over the secretion of progesterone from the corpus luteum. After the baby is born, when the placenta is expelled as the afterbirth, the progesterone level drops dramatically.