1) Comparison between prokaryotic mRNA (polycistronic) and eukaryotic mRNA (monocistronic) 

- note that 5' UTR and 3' UTR is present in both prokaryotes and eukaryotes. The 5' UTR in prokaryote is important as it contains the Shine-Dalgarno sequence, while the 5' UTR in eukaryotes is important as it can regulate translation as regulatory proteins can bind to it and prevent binding of ribosomes so as to carry out translation.

 

- note that unlike the prokaryotes, there is RNA processing in eukaryotes. Hence eukaryotic mRNA contain the 5' GTP cap (important for recognition/binding by eukaryotic ribosome and prevent mRNA degradation) and the 3' poly A tail (facilitate export of mRNA from nucleus and prevent mRNA degradation)  



2) Another picture showing that the prokaryotic mRNA is polycistronic.

- The Shine-Dalgarno sequences (impt for recognition and binding of prokaryotic ribosomes) are in between the multiple start and stop codons on the prokaryotic mRNA.

 3) Another diagram that shows the Shine-Dalgarno sequences on prokaryotic mRNA more clealy.

- Since the prokaryotic mRNA has multiple ribosome binding sites called Shine Dalgarno sequences, thus can lead to the synthesis of several different polypeptides. This is unlike the eukaryotic mRNA with only one ribosome binding site/ start site, so it can only lead to the synthesis of one polypeptide.



Bacteia - Typo in structured Q3a

(3a)       A strain of F+ E. coli cells that are chloramphenicol-resistant were mixed with F- E. coli cells with ampicillin-resistance gene. The mixture of bacteria was then plated onto an agar plate with ampicillin chloramphenicol antibiotics. The colonies of bacteria that survived were then replica plated (transfer of microorganism colonies in the same spatial arrangement from a master plate onto another agar plate) onto an agar plate with chloramphenicol ampicillin antibiotics.  It was found that there exist bacterial colonies that are resistant to both chloramphenicol and ampicillin antibiotics.


To help you with this question, refer to the following diagrams:

Replica plating technique

Replica plating tool

Virus tutorial - Essay question 2 (HIV structure/function)

1.    The viral genome of HIV consists of two linear (+) single-stranded RNA, which carries the genetic information of the virus. [1]

2.    It has an cone-shaped capsid made up of capsid proteins/capsomeres to protect the viral genome and enclose two reverse transcriptase molecules, HIV protease and integrase. [1]

      (***Note: some diagrams of HIV are wrong. All 3 TYPES of enzymes are enclosed within the capsid protein.)

       (***Note: contrary to pictures of HIV, there are actually more than 1 protease and 1 integrase in each capsid so don't say that there is 1 only. To be safe, just exclude the numbers in your answer when you are talking about protease and integrase.)

3.    The viral nucleocapsid is surrounded by an envelope derived from the surface membrane of the host cell. There are glycoprotein spikes, gp 120 and gp41, embedded in the envelope. [1]

4. gp120 binds specifically to a complementary cell surface receptor protein CD4 on the surface of a T lymphocyte. gp41 helps in the fusion between the viral envelope with the host cell membrane.    

 

5.    Entry of the virus into the host cell occurs through fusion of the viral envelope with the host cell membrane, or receptor-mediated endocytosis (RME). [1]

6.    The internalised viral capsids are readily digested and removed by host cellular enzymes to release the viral (-) single-stranded RNA and viral enzymes (eg. reverse transcriptase and integrase).

7.    In the cytoplasm, the released reverse transcriptase (with RNA-dependent DNA polymerase)  uses the (+) single-stranded RNA as a template to synthesize a complementary single-stranded DNA (cDNA) during reverse transcription. Reverse transcriptase (DNA-dependent DNA polymerase) also subsequently converts cDNA to a linear double-stranded DNA. [1]

     (***Note: it is ok not to mention the type of polymerase as REVERSE TRANSCRIPTASE is the keyword/marking point).

8.    The double-stranded viral DNA migrates to the nucleus of the host cell and and is integrated into the the host genome by integrase, forming a provirus. [1]

            (***Note: Provirus refers to the integrated phage genome only, and does not include the bacterial host genome.)

9.    When the host cells are activated, the integrated provirus is transcribed into viral mRNA molecules within host cell nucleus. This viral mRNA serves as new viral genomes and for translation into viral polyproteins in the host cell cytoplasm. [1]

10.    HIV protease then cleaves the viral polyproteins to form functional proteins (capsid proteins, viral envelope glycoproteins, and viral enzymes - protease/reverse transcriptase/integrase). [1]

Tamiflu Mode Of Action (MOA)

Errors in Organization and Control of Eukaryotic Genome lecture notes

1) Section 2.1 The DNA End replication problem (pg 9)

In Fig 7, under step (ii), the shortening of the ends of DNA after the removal of the last RNA primer should happen on the LAGGING STRAND instead of the leading strand. Please make the relevant changes.


2) Section II(A)1 Introns (pg 13)

As mentioned during the tutorial, in Fig 11, the pre-mRNA should NOT have the 5' cap and the poly A tail. Please cancel them out.

Clue to previous blogpost "Enzymes...have you wondered"

Click to enlarge

Section 7.3 DNA and Genomics Lecture notes (ERROR)

7.3.1  Sickle Cell anaemia

·         On chromosome 11, the 17th nucleotide of the gene that codes for the beta chain of haemoglobin is changed from an 'A' to a 'T'.  (GAG to GTG) à WRONG!!!

·         This changes the mRNA codon from 'CUC' to 'CAC' à WRONG!!!, resulting in the 6th amino acid of the chain being changed from glutamic acid to valine. The R side chain of glutamic acid is a polar whereas that of valine is hydrophobic. The mutant haemoglobin is known as Haemoglobin S (HbS).

·         This substitution of one single base in the beta globin gene alters the quaternary structure of haemoglobin, which has a profound influence on the physiology and wellbeing of the individual.

To change the above to:

7.3.1  Sickle Cell anaemia

·         On chromosome 11, the 17th nucleotide of the gene that codes for the beta chain of haemoglobin is changed from an 'T' to a 'A'. (template 3’ CTT 5’ to 3’ CAT 5’) à CORRECT

·         This changes the mRNA codon from 5' GAA 3' to 5’ GUA 3' à CORRECT, resulting in the 6th amino acid of the chain being changed from glutamic acid to valine. The R side chain of glutamic acid is a polar whereas that of valine is hydrophobic. The mutant haemoglobin is known as Haemoglobin S (HbS).

Chromatin remodeling

 

Click to enlarge

Organisation of Pro and Eu genome tutorial Q5 essay [8 marks]

1. Are non-coding DNA sequences that are bound by transcription factors

2. involved in regulating gene expression

3. 3 types of control elements include: promoter, enhancers and silencers.   

4. Promoter serves as recognition site for transcription factors and RNA polymerase

5. [more details ..] -  TATA box within promoter (consisting of 5’-TATAAA-3’) is recognized and bound by general transcription factors to recruit RNA polymerase to the promoter region

6. [relate back to how transcription is influenced] -  binding only produces basal / low levels of transcription 

7. Enhancers are distal control elements bound by activators.

8. a looping mechanism brings the bound activator to interact with the RNA polymerase  and transcription factors at the TATA site

9. [relate back to how transcription is influenced] -  Interaction of activators with RNA polymerase and/or general transcription factors increases the transcription of a gene.                                           

10. Silencers are distal control elements bound by repressors, 

11. [relate back to how transcription is influenced] -  Interaction inhibit transcription / gene expression

11. [Extra: some examples by which repressors work] - 

·          binding of repressor to DNA control elements mask the binding site of activators; OR

·         Binding of repressor to silencer and a certain transcription factor prevents further assembly of general transcription factors

Click to enlarge

(Diagram --> Ignore the coactivators)

Lecture test 2 Q2d: How pH changes affect enzymatic reaction

I have finished marking the lecture tests and I wish to highlight one question that was very poorly attempted in the test here. Although most of you are already aware that a change in pH from the optimum (more than or less than optimum) decreases enzyme activity, alot of you were unable to explain the mechanism behind this effect. Actually, the mechanism is already spelt out very clearly in your lecture notes so please go over it again to make sure you are clear about what I am saying here.

I also want to clarify some terms here as some of you seem to be confused over these terms. The enzyme is a globular protein that has undergone FOLDING to achieve its respective secondary/tertiary/quaternary (if more than one polypeptide chain) structures. Hence it has a specific CONFORMATION/configuration/3D structure as a result of the folding. And it is due to this conformation that the enzyme has a specific SHAPE in its ACTIVE SITE, which is complementary to the shape of its substrate. To summarise, "conformation" is not the same as "shape" so please don't use them interchangeably. The enzyme has a specific "conformation" while the active site has a specific "shape". Hope that this is clear.

Now let's move on to the mechanism. As I have already mentioned in class,  a change in pH changes the concentration of H+ and OH- ions, which causes the neutralisation of polar/charged R groups in the enzyme.

But what you should also know is that this effect can occur at the follwoing 2 areas of the enzyme. 

(i) At the active site -

At the active site, the catalytic and contact residues have charged R groups.

Hence, NEUTRALISATION of these CHARGED R groups disrupts the formation of INTERmolecular IONIC AND HYDROGEN bonds between the active site on the ENZYME and the SUBSTRATE. 

(ii) At other regions of the enzyme -

The bulk of the globular structure of the enzyme contain polar R groups (that form hydrogen bonds) and charged R groups (that form ionic bonds), which stabilise/maintain the secondary, tertiary and/or quaternary structures of the enzyme.

Hence, NEUTRALISATION of these POLAR/CHARGED R groups disrupts the INTRAmolecular IONIC AND HYDROGEN bonds maintaining the SECONDARY, TERTIARY and/or QUATERNARY structure of the enzyme. This causes the enzyme to unfold and lose its specific conformation (ie. denaturation) of the enzyme, hence causing the active site to lose its specific shape.

With both disruptions, binding of the substrate to the active site of the enzyme to form enzyme-substrate complexes, and hence products are prevented. And that is why the enzymatic activity decreases!

Importance of time management

Thanks for the love people!

Hope to see all of you soon! :)

DNA and Genomics - Have you wondered...

Q1) DNA is a polymer of deoxyribonucleotides (that contain either A/C/G/T as its nitrogenous bases). For the genetic code contained in DNA to be able to code for all 20 amino acids, the genetic code has to be a triplet code.

Given the above information, calculate (means workings must be shown) how many codons there are altogether. How many of these codons code for amino acids? Explain your answer. 

Q2) Given that there are XXX number of codons coding for amino acids, there should be the same XXX number of tRNA molecules available, as the anticodon on tRNA complementary base pairs to the codon on the mRNA.


However, this is not true. Explain why. 

Enzymes tutorial Q4b - Although Km is not in your syllabus, you still have to draw the curves accurately!

Starch, amylase, and iodine test

1) Starch (amylose + amylopectin):

"Starch is generally insoluble in water at room temperature. Because of this, starch in nature is stored in cells as small granules which can be seen under a microscope. Starch granules are quite resistant to penetration by both water and hydrolytic enzymes due to the formation of hydrogen bonds within the same molecule and with other neighboring molecules.

However, these inter- and intra-HYDROGEN BONDS can become weak as the temperature of the suspension is raised. When an aqueous suspension of starch is heated, the hydrogen bonds break. "

(Source: http://www.eng.umd.edu/~nsw/ench485/lab5.htm)

Hence for boiled starch --> H bonds broken --> exposed amylose and amylopectin structure --> allow amylase to hydrolyse starch. 


In contrast, for unboiled starch --> amylase unable to hydrolyse starch as the starch granules are intact. 

2) Amylase:

Acts on α-1,4 glycosidic bonds (present in amylose and amylopectin). 

However, amylase is unable to break down amylopectin completely due to the α-1,6 glycosidic bonds present at the branch points. 

(Further reading: http://books.google.com.sg/books?id=xIj3YPKSpJkC&pg=PA453&lpg=PA453&dq=amylopectin+red+brown&source=bl&ots=uokPwYlqHY&sig=v_zrQ-kF7hCYj7So3lKARSbhBCo&hl=en&sa=X&ei=NM2dT6LCH8TJrAe8kcBm&ved=0CCcQ6AEwAQ#v=onepage&q=amylopectin%20red%20brown&f=false)

3) Iodine test 

The amylose (unbranched/linear portion of starch) forms helices, which allow iodine molecules to assemble, forming a dark blue/black color (refer to diagram below). 

The amylopectin (branched portion of starch) forms much shorter helices due to the branching present, and iodine molecules are unable to assemble, leading the color to be of red brown or red violet. 

As starch is broken down or hydrolyzed into smaller carbohydrate units, the blue-black color is not produced. Therefore, this test can determine completion of hydrolysis when a color change does not occur.

(Further reading: http://www.webexhibits.org/causesofcolor/6AC.html)

***Please refer to the SPA answer scheme (when it becomes available at IVLE) to see how the answers should be phrased for Q18,19 and 22.

Enzymes tutorial Q2ciii

Q) Suggest why the lysosomal membrane remains intact? [3]

Possible Answers:

[1st mark] 
- Lysosomal enzymes hydrolyse polysaccharides, lipids, proteins, and nucleic acids.

[2nd mark] 
- These differ from the constituents of the lysosomal membrane which comprise of phospholipids, membrane-bound proteins, glycolipids, and glycoproteins. [1]
OR 
- The inner surface of the lysosomal membrane is coated with an extensive glycocalyx (refer to pictures below).
OR
- The integral and inner peripheral membrane proteins and membrane phospholipids on the inner surface of the lysosomal membrane are highly glycosylated.
OR 
- There are large numbers of glycoproteins and glycolipids on the inner surface of the lysosomal membrane.

[3rd mark]
- As the constituents of the lysosomal membrane are not complementary to the active sites of the lysosomal enzymes, they cannot fit into the active sites and thus, remain unhydrolysed. [1]
OR
- The glycosylation on the inner lysosomal membrane prevents the lysosomal enzymes from accessing the membrane phospholipids and proteins.

Enzymes - Have you wondered...

1) How enzymes from thermophilic bacteria is able to remain stable/retain catalytic activity at high temperatures?

2) Why lysosomal enzymes can function at a pH of 5.5 but not at pH of 6.8, even though the numbers are quite close?

Any takers?

Announcement

1) SPA
ANSWERS FOR THE PAST PRACTICALS ARE ON IVLE. GUYS, YOU MAY NEED TO CHECK YOUR EYESIGHT............................................................................................................

2) Enzymes tutorial essay Q
I have received some queries regarding the collection of the Enzymes tutorial essay Q. Just to clarify, I will not be collecting this essay to mark (i.e. not compulsory to hand in) because this Q is rather factual (everything can be found in the lecture notes). But you are still strongly encouraged to attempt the Q so as to test your understanding of the topic. However, if you would like me to mark it, please hand it in to me in the earlier part of next week (Mon/Tues/Wed) as it is my last week in school and I need some time to mark.

3) Plan for next week
I plan to complete the entire ENZYMES TUTORIAL next week so please complete the tutorial and REMEMBER TO BRING IT! Also, a reminder that the practical next week will be a TIME TRIAL (75min) so be prepared for it!

Good luck for your lecture test on Monday! I am expecting you guys to do well for it. Study hard and do me proud! :)

Enzymes worksheet B (temperature) - Q1b

Please be specific in your answer/ note the context of the question (i.e. NAME the enzyme and substrate if the question stem mentions them)

- D: Initially at very low temperature, the enzyme activity (rate of enzymatic reaction) is very low.
- E: This is due to the inactivation of the enzymes, and the low kinetic energies of both the substrate and enzyme molecules, leading to a low frequency of effective collisions between them, and hence reduced rate of formation of enzyme-substrate complexes and products. 

- D: As temperature increases to optimum, the enzyme activity (rate) doubles for every 10C increase in temperature (Q10), and eventually reaches a peak/maximum rate at the optimum temperature (65C). 
- E: Increased temperature increases the kinetic energies of both enzyme and substrate molecules which in turn, increases the frequency of effective collisions between them. As a result, more enzyme-substrate complexes and hence products are formed.

- D: As temperature increases to beyond optimum, the enzyme activity (rate) decreases.
- E: This is due to the excess heat increases the vibrations of atoms within enzyme and disrupts the intramolecular hydrogen bonds, ionic bonds, and hydrophobic/hydrophilic interactions stabilising the secondary and tertiary structures of the enzymes, causing it to unfold and loses its specific 3D configuration/conformation (enzyme denatures). Hence its active site loses its specific shape, causing the substrate to be unable to bind to it. Hence, less enzyme-substrate complexes and products are formed.

P.S. If you have any queries with regards to the 3 MCQs in this enzymes wksheet B, you can ask me under comments here, or during the next tutorial. Note that we will be working on the main enzymes tutorial next week so please complete it over the wkend (treat it as lecture test revision). I aim to finish Enzymes before I leave! Hope you have enjoyed and benefited from my lessons. Don't miss me.. HAHA! See you next week! :)

Nondisjunction in meiosis

http://www.sumanasinc.com/webcontent/animations/content/mistakesmeiosis/mistakesmeiosis.html

This is a very good video on nondisjunction in meiosis. Due to time constraints I don't think i'll be able to play the entire video in class, so please take some time to view it on your own.

P.S. I have created a chatbox on the right! Feel free to chat there. :)

Cell division - terminologies

TYPES OF CHROMOSOME

1) Autosome = a chromosome that is not a sex chromosome. For example, in humans there are 22 pairs of autosomes and 1 pair of sex chromosomes (XX in females or XY in males).

NUMBER OF CHROMOSOMES

2) Aneuploidy = a chromosome abnormality where there is an abnormal number (extra or missing chromosome(s) / +1, +2...etc or -1,-2...etc chromosomes) of chromosomes, due to nondisjunction in mitosis/meiosis I/meiosis II. Eg. Monosomy = presence of only one chromosome (instead of the typical two in humans) from a homologous pair.

3) Polysomy = when a diploid organism has at least one or more chromosome(s) than the normal (diploid) number, due to nondisjunction. Eg. Trisomy = 3 copies instead of 2 copies of a particular chromosome (eg. Trisomy 21 in down syndrome).

NUMBER OF SETS OF CHROMOSOMES

4) Polyploid = cells with more than two paired homologous sets of chromosomes (diploid, 2n). For example, triploid (3n/ cells with 3 times the haploid number of chromosomes/ 3 sets of chromosomes) and tetraploid (4n/ cells with 4 times the haploid number of chromosomes/ 4 sets of chromosomes).

Hope that this is less confusing for you guys now. Since I've already covered most of the terminologies of your SDL on nondisjunction here, i expect all of you to be able to answer most of my questions next week during the tutorial! ;)

Chromosome number and cancer (Cell division tutorial Q8c)

http://www.reuters.com/article/2009/05/20/us-cancer-downs-idUSTRE54J5IN20090520

Cell division tutorial Q15b

The diagram below is the more accurate representation of crossing over as it shows the BREAKING of non-sister chromatids, followed by the EXCHANGE and then the REJOINING of the corresponding sections of genetic material.






The diagram below is the less accurate representation of crossing over as it does not show the exchange taking place. So please don't draw this in your exam scripts.

To help you understand better, below is a pictorial representation of the mechanism of crossing over.

Note that before crossing over, the sister chromatids are genetically identical, for example the chromosome coloured red is homozygous dominant for the 'c' allele (CC) and homozygous dominant for the 'e' allele (EE).

However, after crossing over the sister chromatids are no longer identical, for example the chromosome coloured red is still homozygous dominant for the 'c' allele (CC) but is now heterozygous for the 'e' allele (Ee).

Another diagram to show the products formed after crossing over in meiosis. You should be a pro by now!


If you are clear now about crossing over, please re-attempt MCQ11. Hope you will get it right this time! 


Some tips for MCQ11: 


- Identify/label the pair of homologous chromosomes, and the sister chromatids that are held at the centromere in the diagram. You can use two differently coloured highlighters to help you trace the chromatids from the paternal vs the maternal chromosomes. 


- Has crossing over taken place? How do you know? (Hint: before crossing over, the sister chromatids of a chromosome must be genetically identical/have identical alleles.)


- What are the products formed for this Q? (Note that this question is asking you about the daughter cells formed after MEIOSIS I, and not meiosis II!!! There is a difference, because meiosis I separates the homologous chromosomes while Meiosis II separates the chromatids!) By the way, "segregation of alleles Q from q" means that the separated daughter cell carries only allele Q or allele q (and not both Q and q alleles). 

Let me know if you have any problems (you can raise the Q after tutorial/ leave a comment on the blog). 

We will be working on Cell division SDL worksheet B and Enzymes tutorial/SDL (will probably start with the SDL) next week. I will be collecting the enzyme essay Q to mark as well so remember to do it on foolscap.

Study hard for your test!!! :)

Answers to Cell division tutorial - factual Q

(14di) How does A lead to genetic stability? [2m]

- A (mitosis) sepArates (note the spelling) sister chromatids of a duplicated chromosome to yield two daughter cells with exactly the same number of chromosomes and is genetically identical to the parent cell by having the same alleles/nucleotide sequence.

- No chiasma is formed and no crossing over occurs during A (mitosis), retaining genetic fidelity.


(14dii) How does B lead to genetic variation? [3m]

- During prophase I of B (meiosis), crossing over occurs between homologous chromosomes, resulting in new combinations of alleles on the chromosomes of the gametes.

[independent assortment]
- The arrangement and subsequent sepAration of the homologous chromosomes of each tetrad/bivalent in metaphase I and anaphase I of B (meiosis) respectively is completely independent of the other tetrad/bivalent, producing new combinations of chromosomes in gametes.

[independent assortment]
- The arrangement and subsequent sepAration of the chromatids of each chromosome in metaphase II and anaphase II of B (meiosis) respectively is completely independent of the orientation of other chromosomes, producing new combinations of chromosomes in gametes.


(Q15c) How does meiosis lead to the formation of four haploid nuclei? [2m]

- During anaphase I, homologous chromosomes separate, resulting in two daughter cells, each with a haploid nucleus, at the end of telophase I and cytokinesis.

- During anaphase II, the sister chromatids separate, resulting in four daughter cells altogether, each with a haploid nucleus, at the end of telophase II and cytokinesis.

What is ploidy?

Some of you seem to be confused about what ploidy means. Hope the following helps:

Ploidy = no. of single sets of chromosomes in a cell
- Haploid cells (gametes, for example either the sperm or the egg cell) --> one set of chromosomes (either from father/mother)
- Diploid cells (somatic cells) --> two sets of chromosomes (one set from father/paternal side, one set from mother/maternal side)

KARYOTYPE (number of chromosomes, and what they look like under a light microscope) OF A DIPLOID CELL (below)

P.S.

Please don't forget to refer to the learning outcomes for this topic when you study. Note that when you're describing the behaviour of the chromosomes during mitosis/meiosis, you have to mention the KEY WORDS that have been BOLDED (e.g. "centromere divides") in your lecture notes. :)

Cell division tutorial Q7aiii and Q7b

Q7aiii) How non-kinetochore microtubules help in mitosis (their function had been asked in the A levels bef0re):

The non-kinetochore microtubules from opposite poles slide past and push apart from each other, causing the distance between the poles to increase and hence, elongating the cell.

Animation: http://faculty.plattsburgh.edu/donald.slish/Polar.html

Q7b) Behaviour of chromosomes during mitosis vs meiosis.

Meiosis animation

http://highered.mcgraw-hill.com/sites/0072495855/student_view0/chapter28/animation__how_meiosis_works.html

Mitosis and Meiosis Lecture Notes Page 1 (Terminologies)

How DNA is packaged
http://www.youtube.com/watch?v=9kQpYdCnU14

Cell structure and function - SDL Wksheet B (Memb transport)

Fluid mosaic model animation:
http://www.stolaf.edu/people/giannini/flashanimat/lipids/membrane%20fluidity.swf

Gramicidin A animation:
http://en.wikipedia.org/wiki/File:Gramicidin_A.gif

Cell structure and function tutorial - Answer to Essay Q

1) The amino acid is taken into the cell by facilitated diffusion or active transport.

2) During protein synthesis by ribosome at the rough ER, the amino acid is added to the growing polypeptide chain, which eventually becomes anchored in the ER membrane.

3) In the ER, the protein may undergo chemical modification such as glycosylation, to form a glycoprotein.

4) ER vesicle carrying the protein buds off from the ER, travels towards the Golgi apparatus and fuses with the cis face of the Golgi apparatus.

5) As the protein travels through the Golgi apparatus through repeated budding and fusion of vesicle, it may undergoes further modification.

6) At the trans face of the Golgi apparatus, the modified protein is sorted and packaged, and eventually enters a Golgi vesicle which buds off from the Golgi apparatus.

7) The Golgi vesicle moves towards the plasma/cell surface membrane and fuse with it.

8) The transmembrane glycoprotein, which contains the radioactive-labeled amino acid, becomes part of the cell surface membrane.

*EXTRA*

For more information about how transmembrane proteins are held in the membrane, you can view this animation: http://www.youtube.com/watch?v=7XTpe7TRQEk

FYI only: Lysosome's role in autolysis

http://www.jci.org/articles/view/11829
A lysosomal protease enters the death scene (2001)

http://www.ncbi.nlm.nih.gov/pubmed/21914490
Lysosomes in cell death (2004)

http://www.ncbi.nlm.nih.gov/pubmed/21914490
Lysosomes and lysosomal cathepsins in cell death (2012)

Interesting read about lysosomes for the layman

http://www.nytimes.com/2009/10/06/science/06cell.html?_r=1

Self-Destructive Behavior in Cells May Hold Key to a Longer Life

"In recent years, scientists have also found evidence of autophagy in preventing a much wider range of diseases. Many disorders, like Alzheimer’s disease, are the result of certain kinds of proteins forming clumps. Lysosomes can devour these clumps before they cause damage, slowing the onset of diseases..."

Cell structure and function

This is a good recap video for the above topic:

http://www.youtube.com/watch?v=LP7xAr2FDFU&feature=related

How a light microscope works - FYI

http://www.microscopemaster.com/brightfield-microscopy.html

"In brightfield microscopy a specimen is placed on the stage of the microscope and incandescent light from the microscope’s light source is aimed at a lens beneath the specimen. This lens is called a condenser.

The condenser usually contains an aperture diaphragm to control and focus light on the specimen; light passes through the specimen and then is collected by an objective lens situated in a turret above the stage.

The objective magnifies the light and transmits it to an oracular lens or eyepiece and into the user’s eyes. Some of the light is absorbed by stains, pigmentation, or dense areas of the sample and this contrast allows you to see the specimen.

For good results with this microscopic technique, the microscope should have a light source that can provide intense illumination necessary at high magnifications and lower light levels for lower magnifications. "

Biological molecules Wksht B - Structured Q2

When checking through your Biological Molecules Tutorial, I found that many of you do not know how to phrase your answer to the above-named question. Since this is an important Q (data analysis Q), I think that it is crucial for you to understand how to phrase your answer. You will be seeing many more of these data analysis Q in future. As such, when you get your worksheets back, please check against your own tutorial worksheet and see that you got the answer below:

Q) DESCRIBE [D] and EXPLAIN [E] the data obtained (you should have two MEANINGFUL observations).

A)
[D1] The ratio of A:T and C:G is 1:1 in BOTH species X and Y.
[E1] This is because according to Chargaff's complementary base pairing rule, Adenine (A) always pairs with Thymine (T), while Cytosine (C) always pairs with Guanine (G).

[D2] In species X, the (C+G) content is 4 times higher than the (A+T) content, whereas in Species Y, both such contents are about the same.
[E2] This is because A and T forms two hydrogen bonds while C and G forms three hydrogen bonds,
making a C-G base pair stronger than a A-T base pair.
Since high temperatures denatures DNA, species Y that lives in a hot spring under high temperatures possesses higher C-G content to ensure stability of its DNA.

Clarification: Events at the rER (synthesized protein will end up in the rER lumen and later bud off in an ER vesicle)

- protein synthesis
- folding of protein
- chemical modification of protein (FURTHER chemical modification will occur in the GA later)

Lysosome vs Peroxisome

http://sln.fi.edu/qa97/biology/cells/cell7.html

"Peroxisomes contain oxidative enzymes. They are similar to lysosomes. Their enzymes have two functions; to convert fats to carbohydrates and to detoxify potentially harmful molecules which form in the cell.

Peroxisomes, in contrast to lysosomes, are produced only on the smooth ER system. They are found in the cytoplasm of many eukaryotic cells as well as prokaryotic cells, microorganisms, and plant cells.

The enzymes of peroxisomes remove hydrogen atoms from small molecules and joins them to oxygen creating hydrogen peroxide (toxic), which is later neutralised by a peroxisomal enzyme, catalase. In the liver this method is used to break down molecules of alcohol into substances that can be eliminated from the body. "

Clarification - primary vs secondary lysosomes

1) Origin of Primary lysosomes (page 17 of your notes I believe)

In your cell structure and function lecture notes, under the section on Lysosomes, it states that primary lysosomes bud off from the Golgi Apparatus OR Endoplasmic Reticulum.

Note that primary lysosomes largely originate from the GA, that is why we have emphasized this during the lecture/tutorial. In some cases, primary lysosomes can bud off the ER but you do not need to know this in detail.

Even if you mention just GA alone, it is still correct. :)

2) Primary vs Secondary lysosomes (tutorial structured Q1d)

- Primary lysosome are formed from the GA while the secondary lysosome is formed by the fusion of the primary lysosome and an endocytotic/phagocytotic vesicle.

- The primary lysosome does not release its contents out of the cell but the secondary lysosome may release useful products (which will serve as building blocks of new materials) into the cytoplasm of the cell via facilitated diffusion, or release waste products/products that cannot be digested out of the cell via exocytosis.

You can view a good diagram of primary and secondary lysosomes here: http://biology.kenyon.edu/HHMI/Biol113/lysosomes.htm

Cell structure and function (cell membrane) - SDL Wksheet A

Video of facilitated diffusion:
http://www.youtube.com/watch?v=I4123hUU8xo&list=FL-cFTJ8NxBjonnjuwSQgSsw&index=26&feature=plpp_video

Video of active transport:
http://www.youtube.com/watch?v=yz7EHJFDEJs&feature=autoplay&list=FL-cFTJ8NxBjonnjuwSQgSsw&lf=plpp_video&playnext=1

Query about cell membranes: Can water diffuse through the lipid bilayer?

Water, though small, is a hydrophilic molecule. Hence it cannot diffuse through the hydrophobic interior of the lipid bilayer. Aquaporin channels are required for water to pass through the lipid bilayer.

FYI: You can read about the experimental discovery of aquaporin in the links below:

http://www.ncbi.nlm.nih.gov/pubmed/11773613

http://www.bing.com/images/search?q=aquaporin+oocyte&view=detail&id=2BE08C8AEE232A3B3F05FE97719A64215101C3BE&first=0&FORM=IDFRIR

Reminder

HOMEWORK this weekend (besides studying VERY HARD for your lecture test on Monday :p):

1) For the current tutorial worksheet on Cell Structure, please copy out (since you should have already done it) your essay question (there's only one essay question, on the last page) onto a piece of foolscap for me to mark. Hand it in to your Bio reps and Bio reps, please put it in my *new* pigeonhole (labelled with my name) by 5pm the coming Monday.

I believe this is crucial as alot of students do badly in essays as they not know how to phrase their answers. As such, all essays (in future) must be done on a piece of foolscap and handed in for me to mark.

2) After the lecture test on Monday, I will be collecting your tutorial worksheets on Biological Molecules (carbo, lipids, protein tutorial worksheet + SDL worksheet B on nucleic acids) during our tutorial class next week. So please remember to bring them to school.

This is because all of you are new to the rigours of JC life, so I want to check that you are keeping up well with the course (and understand what is expected of you). :)

Good luck for your test!

Cell structure and function tutorial - origin of mitochondria and chloroplasts

Q1) How does the presence of double-membranes in these two organelles serve as evidence that these two organelles actually existed as free -living bacteria centuries ago?

A1) - The ancestor eukoryotic cells engulfed bacteria by phagocytosis/endocytosis, and the bacteria eventually became the mitochondria or chloroplast.
- Hence, the outer membrane is derived from the eukaryotic cell and the inner membrane is derived from the bacteria.

FYI only: This theory is actually named the "endosymbiotic theory", but you are not required to know it. You can watch an animation of the theory here: http://www.youtube.com/watch?v=EgaGh9-mFnQ

Note: Please do not be confused, prokaryotes (i.e. bacteria) have only one cell membrane. You will learn more about its structure in the later topics.


Q2) How does linear DNA (in eukaryotes) and circular DNA (in prokaryotes, mitochondria, and chloroplasts) look like under the electron microscope (EM)?

A2) Linear DNA (above) and circular DNA (below), in a relaxed state.