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A-Level Biology Tuition Singapore/H2/H1 Biology Tuition/JC Biology Tutor
Please post your Biology questions here to Ms Diana Goh our dedicated and awesome teacher
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Miss Diana Goh is an inspiring teacher who is always willing to encourage her students for the extra lessons or questions and a friend who teaches us life skills, right mindset and perseverance for the future endeavours. A teacher who truly loves her students and always ready to take extra step to change her students for the betterment of their lives. She has definitely symbolized the way I define howan excellent teacher should be.
Lee CK – Jurong Junior College
Ms Diana is a dedicated tutor who is willing to spend extra hours after tuition just to make sure we understand our texts. She is very resourceful and often spice the lesson up with interesting facts and most importantly, these facts are relate to our syllabus! Her passion for teaching and frequent words of encouragement greatly influences us to strive excellence in Biology. She definitely an inspiration to all her students.
Jocelyn Chew – St Andrew’s Junior College
Miss Diana is a wonderfully dedicated teacher who not only taught essential and useful skills in answering Biology questions but also encouraged us along the way. I have enjoy her lessons and the way she make lessons interesting by cracking lame jokes. Above all, I have learnt much from her. Thanks You!
Adelyn Lim – Anglo-Chinese Juniors College
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A-Level Biology Tuition Singapore/H2/H1 Biology Tuition/JC Biology Tutor
3 Classes of Biology
SAT 9am to 11am
SAT 3pm to 5pm
SAT 6pm to 7pm
We can customise to suit your schedule
From A-Level Biology Tutor Singapore
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A-Level Biology Tuition Singapore/H2/H1 Biology Tuition/JC Biology Tutor
Please be inform that J1 H2 Biology has started on 3 March and the feedback from the studens that the tutor Diana is Awesome and it make study Biology so much easy.
We are starting another J1 H2 Biology on 31 March Saturday from 5.30pm to 7.30pm. Please call 98639633 to book a place.
J2 H2 Biology will start soon on Saturday from 3.30pm to 5.30pm
From A-Level Biology Tuition Singapore – Admin
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admin
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A-Level Biology Tuition Singapore/H2/H1 Biology Tuition/JC Biology Tutor
Please be inform that J2 H2 Biology will atart on 31 March Saturday from 3.30pm to 5.30pm. Please call 98639633 to book a place.
From A-Level Biology Tuition Singapore – Admin
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Diana Goh
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A-Level Biology Tuition Singapore/H2/H1 Biology Tuition/JC Biology Tutor
J1 Class 21/04/2012
Check out the structure of hemoglobin in 3D! Have fun learning 
Do note that you’re supposed to know about Sickle Cell Disease and how it is due to changes in the structure of Hemoglobin.
Hemoglobin and Sickle hemoglobin structure
http://www.umass.edu/molvis/tutorials/hemoglobin/index.htm
Sickle cell disease
http://www.youtube.com/watch?v=R4-c3hUhhyc&feature=related
Cheers,
Ms Diana Goh
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Diana Goh
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A-Level Biology Tuition Singapore/H2/H1 Biology Tuition/JC Biology Tutor
J2 Class 22/04/2012
Cell Signaling class test 1hour 15mins. Answers will be given next lesson.
Having understood the molecular mechanisms of cell signaling and communication (how cells talk to each other), we are now moving on to a more macro chapter, where we look at the concepts of cell-communication can be applied to bigger systems (the endocrine and the nervous control), and how our body regulates homeostasis (a constant internal environment).
Do read up on Endocrine control, chapter 45 of Campbell and Reece, and be prepared for a lecture next Saturday!
In Endocrine control we focus on the regulation of blood glucose levels. Here’s a snippet of what we will be learning next week!
http://outreach.mcb.harvard.edu/animations/homeostasis10.swf
Cheers,
Ms Diana Goh
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Diana Goh
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A-Level Biology Tuition Singapore/H2/H1 Biology Tuition/JC Biology Tutor
J1 Class 23/04/2012
Today, we wrapped up the first part of the chapter on Cellular Functions (The Plasma Membrane). Covered some of the answering techniques required to tackle questions in this chapter.
We also started on the second part of the Cellular Functions chapter (The Cellular Organelles) – looking at the largest and smallest organelles in the cell – Nucleus and Ribosomes. Next week we would finish up the rest of the organelles in the cell. Read up chapter 6 of Campbell as well as the notes given before lesson begins next week!
An interesting topic of discussion fueled today’s lesson, the Endosymbiotic Theory, otherwise known as the Theory of Endosymbiosis. Do read up a little on it online and in your textbook.. and I shall tell you more next week!
Ms Diana Goh
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Diana Goh
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A-Level Biology Tuition Singapore/H2/H1 Biology Tuition/JC Biology Tutor
J1 Class 9-11am 1/5/12
Today we finished up the chapter on Cellular Functions! We have embarked on the first assignment of the cellular functions chapter. Here’s a review to all that we have learned!
http://www.cellsalive.com/cells/cell_model.htm
J2 Class 11-1pm 1/5/12
For today’s class we went through last week’s class test on Cell Signaling. Take note of the answering techniques you’ve learned. Remember, the Theory of Dissection! You know and I know what it means. Apply this to all the questions and concepts you learn, and you will find that you are actually blessed with an amazing depth and breadth of molecular details for even the simplest mechanism. For instance, describe the structure of the beta adrenergic receptor [3].
The answer is based on the three domains of the receptor protein, recall them!
Here is a video for Epinephrine GPCR signaling pathway and the steroid hormone receptor pathway. The difference is that the former is a membrane receptor and the latter is an intracellular receptor. Why this distinction? Think about the difference between Epinephrine and Steroids.
http://highered.mcgraw-hill.com/olc/dl/120109/bio48.swf
http://highered.mcgraw-hill.com/olc/dl/120109/bio46.swf
J1 Class 1-3pm 1/5/12
The second part of the biology marathon, we had a lecture on Enzymes today. We discussed about Enzymes, the protein molecules which orchestrate the multitude of biochemical processes in our system! Here is an enzyme of how sucrase catalyses the breakdown of sucrose, an example of an enzyme mediated reaction to better your understanding.
http://highered.mcgraw-hill.com/sites/0072507470/student_view0/chapter25/animation__enzyme_action_and_the_hydrolysis_of_sucrose.html
Cheers,
Ms Diana Goh
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This reply was modified 19 days ago by
 Diana Goh.
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This reply was modified 19 days ago by Diana Goh.
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This reply was modified 19 days ago by Diana Goh.
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This reply was modified 19 days ago by Diana Goh.
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This reply was modified 19 days ago by Diana Goh.
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Diana Goh
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A-Level Biology Tuition Singapore/H2 Biology Tuition/JC Biology Tutor
Here are the requested explanations to the questions in the Enzyme Speed Test, for JC1 class.
4. Which one of the following enzymes has the greatest substrate specificity?
A lipase
B nuclease
C pepsin
D sucrase
E trypsin
The answer is E, trypsin. A lipase is a name given to any enzyme that digests fats and lipids. A nuclease is a name given to any enzyme that acts on nucleic acids. A sucrase is a name given to any enzyme which can digest sucrose to its monomers glucose and fructose. Pepsin is slightly more specific, it digests proteins but it’s quite specific in that it cleaves peptide bonds between hydrophobic and aromatic amino acids. Trypsin is most specific because it specifically digests proteins only at the carboxyl side of lysine and arginine, when either of which is followed by a proline residue.
7. Which one of the following bonds permanently to the active site of an enzyme?
A allosteric inhibitor
B coenzyme
C competitive inhibitor
D non-competitive inhibitor
E substrate
The answer is D, noncompetitive inhibitor. Remember I mentioned that noncompetitive inhibitors bind allosteric sites? Here is an exception. Sometimes, like in this case, the question wants the best answer from you. Look at the list of choices, they are all impossible except for D. A binds allosteric site for sure, that’s why the name. B does not bind permanently, read up coenzymes in the notes given. C binds the active site for sure but definitely not permanent, that’s why known as competitive. E definitely not permanent too, if not there would be no product! So this is a tricky question. Non competitive inhibitors may bind active sites permanently, it is a form of Irreversible inhibition. It clogs up the active site.
14. The reaction rate of salivary amylase on starch decreases as the concentration of chloride ions is reduced. Which of the following describes the role of the chloride ions?
A allosteric inhibitors
B coenzymes
C cofactors
D competitive inhibitors
E noncompetitive inhibitors
The answer is C, cofactors. Refer to your notes on page 22, last sentence at the bottom. The exact same example is there. There are three types of cofactors – inorganic ions, prosthetic groups and coenzymes. Chloride Ion is an Inorganic Ion Cofactor.
16.Fructose syrup is used as a sweetener in the food industry and the scheme below outlines the major steps in its industrial production from starch. The process makes use of bacterial or fungal enzymes at steps 1,2 and 3. The three arrows are in chronological order, first one involves enzyme 1, second arrow enzyme 2 and third enzyme 3.
Starch suspension –> Maltose –> Glucose –> Fructose syrup in water
In the table below, Y means the step could be carried out by heating the subtrate with acid as an alternative to using enzymes. N means that it could not. Which is the correct combination?
1 2 3
A Y N Y
B Y Y N
C N Y Y
D N N N
The answer is B, YYN. This is because the treatment of heat with acid is actually stimulating the process of hydrolysis. Hydrolysis is the breakdown of a molecule into two parts by the addition of a water molecule. Enzyme 3 cannot be replaced by acid because in that reaction you are not catalyzing the breakdown of a substrate. You are catalyzing the CONVERSION of glucose to fructose. Something interesting here, Fructose is C6H12O6, same as glucose! Same but different, and the difference lies in that Glucose has an aldehyde group and Fructose has a Ketone group. The two are actually isomers of each other. Therefore, the conversion of glucose to fructose is an isomerism process and the enzyme involved is called glucose isomerase.
26.The diagram shows the initial rate of reaction using constant amounts of substrate and enzyme at different temperatures. X labels the part which is decreasing. What is the reason for the decline in the level of activity in region X, when temperature is very high?
A breaking of sulfur bridges and ionic bonds in the enzyme
B competition between substrate and product for the active site
C hydrolysis of peptide bonds and breaking of hydrogen bonds in the enzyme
D insufficient substrate to occupy all the active sites
The answer is A, breaking of sulfur bridges and ionic bonds in the enzyme. B is impossible, it has nothing to do with substrate product competition. D is impossible because temperature rise does not reduce substrate availability. C is right in the second part, hydrogen bonds break at high temperatures, disrupting the secondary and tertiary structures. However, the first part is wrong because heat does not break peptide bonds. The primary structure remains INTACT with enzyme denaturation, only secondary tertiary and quaternary structures are affected. A is right because if the temperature is high enough even strong covalent bonds like sulfur bridges and ionic bonds may be broken.
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Diana Goh
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A-Level Biology Tuition Singapore/H2 Biology Tuition/JC Biology Tutor
…continued.
28. In an investigation to determine the effect of temperature on the activity of an enzyme, the time for all the substrate to disappear from a standard solution was recorded. Which graph shows the results of this investigation?
The answer is C, a curve the shape of a smiley, with time at the y-axis and temperature on the x-axis. Remember that the temperature affects the rate of an enzyme catalysed reaction by a curve which has two parts – an exponentially increasing front portion and a steeply declining back portion? Refer to your notes on page 16. When the temperature is very low the reaction rate is very low, hence reaction time/duration is very long, thats why higher on the y-axis. As the temperature increases there is an exponential increase in reaction rate (page 16), therefore there is an exponential decrease in reaction duration (graph C). And there is this optimum temperature where the reaction rate is highest (page 16), which is when the reaction duration is shortest (graph C). Beyond that temperature, increasing temperature further causes denaturation of the enzyme hence a steep decline in the reaction rate (page 16) that is why there is a steep increase in reaction time (graph C). See the correlation between rate and duration? The graph on page 16 of your notes is actually like the graph in option C, hope you saw the link in their gradients!
30. Which graph shows the effect of increasing enzyme concentration on product formation when there is an excess of substrate?
The answer is B, increase and then plateaus off. When there is an excess of substrate, the question implies that substrate is NOT a limiting factor. The limiting factor is therefore the enzyme concentration. (Note the synergistic relationship between substrate and enzyme, when one is limiting the other is definitely not-limiting and in excess. Analogy: gender ratio in the world population. If there is limiting number of males then the number of females is definitely in excess of the males and thus non-limiting). Back to the question. So if enzyme concentration is a limiting factor, increasing enzyme concentration would have an effect on the concentration of the product. Therefore we see changes in all four curvess. Enzyme concentration increase would increase product concentration but take note that it must plateau off because when enzyme concentration is so high that it is no longer a limiting factor, then substrate concentration would be. If you have just this amount of substrates in your reaction, no matter how much enzymes you add it will ultimately only form this amount of products. Hence the plateau.
31. The diagram shows how the enzyme glutamine synthetase removes the ammonia produced during plant metabolism.
Ammonia + glutamate —glutamine synthetase—> glutamine
Some herbicides contain an active agent which resembles glutamate. What is the likely mode of action of this agent?
A it acts as an end product inhibitor
B it acts as a competitive inhibitor
C it decreases levels of ammonia
D it increases levels of glutamate
The answer is B, competitive inhibitor. This is because the herbicide bears structural resemblance to glutamate which is the substrate. Therefore the herbicide agent would compete with the substrate for the active site. Hence most likely it would act as a competitive inhibitor. It cannot be an end product inhibitor because glutamate is not an end product, glutamate is the substrate. Bearing resemblance to glutamate must involve some substrate-inhibitor competition.
34. A medical scientise investigates four species of insects. He knows that one feeds on human blood and that the others feed on plants. As the insects look similar, he investigates the digestive enzymes present in their guts. Y indicates presence, N indicates absence. Which insect feeds only on blood?
Insect A is amylase Y lipase N protease Y sucrase Y
Insect B is amylase Y lipase N protease N sucrase Y
Insect C is amylase N lipase Y protease Y sucrase Y
Insect D is amylase N lipase Y protease N sucrase N
The answer is Insect A. Hemoglobin in red blood cells are often digested by these insects, and hemoglobin is a protein. Definitely the blood sucking bugs need proteases, which digest proteins. Weak concentrations of sucrase and amylase have been found to be present in their guts too. But they lack lipases because lipid and fat digestion is not needed/relevant. In contrast, plant feeding insects would need a bit of everything.
37.The diagram shows the structure of a competitive inhibitor of the enzyme lysozyme. Which substance is most likely to be the normal substrate of lysozyme?
A lipid
B polynucleotide
C polypeptide
D polysaccharide
The answer is D, polysaccharide. The answer is not B, please take note of the error! Does this solve your doubts? The normal substrate should bear resemblance to the competitive inhibitor of lysozyme. Hence it should look like the structure given. And that structure actually looks like a beta-glucose polymer. Hence the answer is polysaccharide.
Hope this helps!
Ms Goh
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A-Level Biology Tuition Singapore/H2 Biology Tuition/JC Biology Tutor
Please be inform that Biology Intensive Revision during the June holiday start on 11 June.
Schedule as follow
11 June Monday to 14 June Thursday – 10 am to 1 pm
Diversity and Evolution
11 June Monday to 14 June Thursday – 2 pm to 5 pm
Organisation and Control of the Prokaryotic and Eukaryotic Genomes
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