Questions about the course material for the final
Explanation: Since CAM plants use the Hatch Slack Cycle for primary fixation of carbon, they belong to the C4 plants. The only difference is that there is a temporal separation of the CO2 fixation reaction ( dark reaction , happens actually during the night in CAM plants) and the production of NADPH and ATP (light reaction) which is done during the day at closed pores to prevent moisture loss.
FADH2 is formed from FAD, but also 2 electrons and 2 protons are required. Thus :
FAD + 2 e- + 2 H+ = FADH2.
Where do the two electrons come from? Electrons are released in an oxidation reaction , these electrons can be used in the formation of FADH2 from FAD. Therefore FADH2 can only be formed in a reaction that releases electrons, but not in a reaction that consumes electrons. Reactions that consume electrons require an electron source, e.g. FADH2. Therefore, in reactions that consume electrons FAD is formed.
Explanation: The disintegration of the nuclear envelope is initiated by the phosphorylation of the envelope undercoat, called lamin. This event is always associated with the prophase. The disintegration of the nucleus happens during the end of the prophase or in the prometa phase. I mentioned this in class but it is not in the notes. During the metaphase the microtubules attach to the chromosomes, this would not be possible if the nuclear membrane still exists.
Oocytes are formed from their precursor cells (diploid sex cells) sometime during 3 rd and 8 th month of gestation. However the full development ( completion of meiosis I) does not happen until puberty and completion of meiosis II after fertilization. The fully developed oocyte is called ovum.
To answer this question you need to know the following:
The Na+ conc inside the cell ( in the cytoplasm) is very low, whereas the Na+ conc. outside the cell is high. The Na/K pump maintains this gradient since it pumps out Na and at the same time it moves K+ into the cytoplasm.
If an ion is moved from its low conc. side to the high conc. side, energy is required. If an ion moves from its high conc. side to the low conc. side, energy is released. Na+ is often a participant in the transport of metabolites in or out of the cell. For example glucose can be transported into the cytoplasm against its own gradient (i.e. form the low conc. side to the high conc.) if there is an energy source that can drive the transport. This energy is provided by a Na+ ion travelling together with glucose into the cytoplasm. Since Na+ moves from its high con. side to its low conc. side in this example , the energy released in this process can drive the glucose transport into the cell.
In picture 1 a hypothetical transporter is shown, in which Na+ drives the transport of a metabolite against its own gradient. The Na+ ion is shown to travel out of the cell . Such a process would require energy. Therefore a transporter which moves a metabolite from its low conc. to its high conc. side (a process that requires energy) could never be driven by a Na+ ion leaving the cell. If the Na+ arrow would point the other direction, then the transport would be possible.
Picture 4 shows a hypothetical transporter in which K+ drives the transport of a metabolite against its own gradient. Again this is impossible since the inward movement of K+ would require energy. If the K+ arrow would point the other direction, then the transport would be possible.
There is also an oscillatory relationship between the parasite and the prey. At high parasite concentration , the prey recovery from feeding episodes is more difficult. The prey is weakenes and is more prone to diseases, thus the death rate increases.
The third answer referred to the size difference. In predation the prey is consumed by the predator. In parasitism, the prey is not consumed during the food intake episode. Size plays a role, a parasite can not consume a prey during one feeding episode that is larger (has more biomass) than the parasite. Therefore in many cases the size argument is decisive, answer c was the only answer which came closest to the correct definition
Dark cycle" does not mean that the cycle occurs during the night or during a dark period. It means that the cycle does not require any pigments (such as chlorophyll) . What is the dark cycle? The dark cycle is the reaction of
1,5- RbBP + CO2 +H2O = 2 3-PG .
3-PG ( Phosphoglycerate) reacts further to form glucose, and some of the 3PG molecules reconstitute 1,5-RbBP, completing the dark cycle (also called the Calvin Cycle) . What else is necessary for the dark cycle? NADPH (reducing power) and chemical energy (ATP). NADP is necessary, since oxidized carbon (CO2 )is converted to reduced carbon (glucose: -CHOH-) and ATP provides the energy for the biosynthesis of glucose.
However, ATP and NADPH are synthesized during the light cycle. Thus the statement " the dark reaction and can operate without light " means that no pigment is necessary for the completion of the cycle. For example, if there would be an independent source of ATP and NADPH, the Calvin cylce could occur.
However, in photosynthetic plants ATP and NADPH, can only be generated in presence of chlorophyll and light. Therefore the statement "without light, the dark cycle is inhibited." is correct. The reason for the inhibition is the enzyme ribulose bisphosphocarboxylase is inhibited by high ADP conc, low NADPH conc and pH=7 . All of these are conditions that arise when the light is off.
Before meiosis the precursor cells of sperm and ovum undergo DNA replication. Only only chromosome is shown e.g. the sex chromosome of a male (XY). Blue: replicated Y chromsome, red; replicated X chromosome. Meiosis I results in two daughter cells, one cell contains the replicated Y chromosome, the other the replicated X chromosome. In meiosis 2 the replicated Y and X chromosomes are separated to form the haploid sperm. The separation of the Y chromosome was normal, however in this example the x chromosome did not sepatate. Among the two resulting sperms one has no X chromosome, the other one has two X chromosomes. If an egg is fertilized with the sperm without the X chromosome a OX condition is exists (Turner syndrome) . If an egg is fertilized with the sperm that has the two X chromosomes a condition XXX exists .
Questions about the course material for Exam IV
Consider a gene that has two alleles, a and A . The genotype of individual in this population is either AA or aA or aa. The frequency of the allele a (or A ) in this population can be calculated from the genotype distribution :
p = (2NAA+ NAa) / 2N
q = (2Naa + NAa) / 2N
p = frequency of allele A
q = frequency of allele a
p + q = 1 (The frequency of finding a and A in the population = 1)
NAA = # AA ( #of individuals with genotype AA ) NAa = # Aa ( # of individuals with genotype Aa )
Naa = # aa ( # of individuals with genotype aa )
N = NAA + NAa + Naa ( N= total # of individuals in the population )
When you say "genetic frequency" I am assuming you mean genotype frequency.
In the above example, genotype frequency is the actual distribution of the alleles AA, Aa and aa in the population
frequency of AA = NAA / N
frequency of Aa =NAa /N
frequency of aa = Naa / N
If you mean the names in e.g. in Fig. 23.11, no you don't need to memorize them. If we discussed a species for a particular reason it may be good if you recognize (associate ) the name , e.g. Tragopogon was an example for tetraploidy.
Questions about the course material for Exam III
Hepatitis D ( sstr. (-) RNA and some circular RNA) by itself is not virulent, it requires the presence of hepatitis B virus for replication. When virulent Hepatitis D is often fatal. Hepatitis C virus ( sstr. (+) RNA) is virulent by itself, it causes chronic liver infection.
The base sequence of the transcript (RNA) near the terminal section of the gene contains a self complementary section which is capable to form a hair pin (very much like introns in a nRNA). These hairpin loop dislodges the RNA polymerase from the DNA and transcription is terminated. The stop codon you mentioned terminates translation ( = protein synthesis using a mRNA as template)
An alternative mechanism is the transcription of the transposon into a mRNA. With the help of a reverse transcriptase ( whose gene is also part of the transposon's DNA) the mRNA is reverse-transcribed intoa complementary DNA (cDNA). This cDNA can then be inserted at a different location into a chromosome with the aid of an intergrase, an enzyme whose gene may also be part of the transposon's DNA
They are both "sense RNA" that means they both can be mRNAs.
I answered several questions about the lac operon. Check out the description in the lecture notes. If it does not make any sense to you, send me a note. (yes, there will be question about it on the exam)
Assume that you have a segment of DNA that is transcribed into an RNA ( the 3' to 5' strand is transcribed):
3' __________________________ 5' DNA
The RNA transcript it complementary to the above DNA and looks like this:
5'____________________________ 3' RNA
This particular RNA is a mRNA or a sense RNA or (+) RNA since it can act as a template in protein synthesis
Assume for a minute that you make an exact copy of the above DNA, but not as a DNA but as RNA . This RNA would look like this:
3' ____________________________ 5' RNA = (-)RNA
This RNA could not serve as a mRNA since it is not complementary to the DNA (only the complementary copy is a mRNA). Therefore this RNA is called antisense RNA or (-)RNA. Viruses can have (+)RNA as genetic material. In such a case the (+)RNA can also serve as a mRNA. Some viruses have (-)RNA as genetic material. This (-) RNA has to be transcribed first into a complementary (+)RNA which can act as a mRNA
3' ____________________________ 5' (-) RNA
transcription
5'____________________________ 3' mRNA
No need to memorize codons or anticodons. What you may want to know is that a codon consists of three bases and the anticodon of a particular aminoacyl t-RNA is complementary to the first two bases of its codon region on the mRNA. The third base of the anticodon (=wobble base) does not require an exact fit to the codon base. You also may want to know that there are three stop codons and that many amino acids have more than one codon (code is degenerate) and that there is one initiation codon (AUG)
Questions about the course material for Exam II
A chromosome (in eukaryotic organisms) is a very long piece of linear DNA. During replication (S-phase in the cell cycle) identical copies of each chromosome are synthesized. The identical copy of a chromosome is joined to the chromosome (from which it was copied) at the centromer region. As long as the chromosome and it's identical copy are joined, it is called a chromatid. As you know each of the human chromosomes (#1 to #22) occur as homologous pairs e.g. chromosome #1 (paternal copy) and it's almost identical copy (maternal copy) . After replication the paternal chromosome #1 is joined to it's identical copy and the maternal chromosome #1 is joined to it's identical copy. These two chromatids are referred to as sister chromatids.
The dark reaction in photosynthesis only occurs in the night and dark and light reactions do not occur at the same time. And I would think if one of the answers is correct, then the other is, which wasn't a choice..
The dark reactions (Calvin cycle) do not require light directly. That means chlorophyll is not catalyzing any of the reactions in the dark cycle. However, the dark reactions require NADPH and ATP which are produced in the light reactions (require chlorophyll and light) Here is question 3
Q.3: Which statement is correct ?
a. the dark reaction in photosynthesis occurs mostly during the night
Even though it theoretically could occur during night (no light directly required for any of the reactions), it does not, since ATP and NADPH can only be produced during light exposure
b. RuBP - carboxylase is activated by light
RuBP - carboxylase is not activated by light, but by high pH, NADPH and low ADP conc
c. dark and light reactions do not occur at the same time
they do occur at the same time since the Calvin cycle can only run if ATP and NADPH is present
d. the fixation of CO2 to RbBP does not require light
correct, it requires only CO2, water , RbBP (ribulose bispohosphate) and the enzyme RbBP carboxylase
e. all are incorrect
false
Type O blood group requires that both recessive alleles are present : OO. If one parent has heterocygous alleles for blood group A such as AO and the other parent has e.g. the heterocygous alleles for blood group B : BO, then the offspring could have the following blood groups : AB, AO, BO and OO. OO is also possible if both parents have AO or BO. But OO can not be the blood group of an offspring if one parent is AB
18. The arrows indicate the direction in with the new DNA strands are synthesized during replication. Which diagram is correct ?
DNA synthesis can only proceed from 5' into 3' direction. This is explained in the notes and illustrated in the figure
Pairing of the newly synthesized strand and template strand has to be antiparallel. That means if the template strand is 3' to 5' direction, the newly synthesized strand is growing from its 5' to 3' end.
The 5' to 3' template strand has to be copied from its 3' side . The newly synthesized DNA strand grows from its 5' end. In fig. 18c. the newly synthesized strands grow form their 3' side toward their 5' side, which is not possible. DNA synthesis always proceeds from 5' to 3' direction
The only mechanism you should be familiar with is the reaction of fructose-P + ATP = P-fructose-P +ADP in glycolysis. The enzyme is called phosphofructokinase . You can find it in in the glycolysis lecture notes. Activity
Q.5: In the oxidation of glucose
the enthropy change is negative
the free energy change is negative
the enthalpy change is positive
two of the above are correct
the correct answer is: the free energy change is negative the answer : the entropy change is negative is false, since the disorder increases (crystalline glucose to random gas), thus entropy is positive
Which statement is false? A human X - linked recessive gene may be:
a) passed to daughters from their fathers
b) passed to sons from their mothers
c) passed to sons from their fathers
d) all of the above statements are correct
Aren't both c and d false?
I agree: Answer c is wrong since the father can only pass the Y chromosome to the male offspring. The last statement is also false, it should read none of the above are correct I guess I have not mastered the art of double negation.
Glycolysis
You need to know that glucose as a starting material and two phosphorylations yield P-fructose-P. P-fructose-P decomposes into 2 glyceradehyde-P . 2 glyceradehyde-P react to 2 pyruvate and 2 NADH and 4 ATP are formed. Pyruvate can be reduced to lactate with NADH (=fermentation)
TCA
pyruvate + CoA + NAD+ = acetyl CoA + NADH. The acetly CoA is fed into the TCA: acetylCoA + oxaloacetate = citrate + CoA Citrate reacts via many intermediates to oxaloacetate. In these reaction NADH, FADH2, 1ATP is formed and 2 CO2 are lost.
Calvin
ribuloseBP +H2O +CO2 (RbBPase) = 2 Phosphoglycerate . Phosphoglycerate, in separate reactions, is transformed into glyceradehyde-P. These reactions require ATP and NADPH. 2 glyceradehyde-P react to form glucose. Other glyceradehyde-P molecules are used for the synthesis of ribuloseBP
Photrespiration
ribuloseBP +H2O +O2 (RbBPase) = 1 Phosphoglycerate + a 2 carbon compound. This 2 carbon compound undergoes a series of reaction to form ge into to form Phosphoglycerate (three organelles are invoved) These reaction consume ATP and the loss of one carbon in form of CO2
The enthalphy change in a reaction
is the heat evolved or consumed at constant volume
free available energy released or absorbed in a reaction
is the heat evolved or consumed at constant pressure
has to be negative for a reaction to occur
The term enthalpy is reserved for " the heat evolved or consumed at constant pressure . Therefore " the heat evolved or consumed at constant volume" is wrong. The answer "has to be negative for a reaction to occur" is also wrong since delta G has to negative. If the entropy change is very large (disorder increases), delta G can still be negative, even if the enthalpy change may be positive.
Questions about the course material till Exam I
Unfortunately I ran out of time, while trying to explain H-bonds. An H-bond is formed ONLY between a H-atom which is covalently bonded to an O (or N) and an O (or N) which belongs to another molecule:
residue-O-H ..... N-residue or residue-O-H ...... O-residue or
residue-N-H ..... O-residue or residue-N-H ....... N-residue
residue = consists of atoms which are part of the molecule, ..... = H-bond
The H-bond forms only if -O-H ..... N- has a 180 degr. geometry. The H and N are in very close proximity ( about twice the distance of a covalent bond ). This makes the interaction very strong, stronger than a mere polar interaction. Polar interactions are not directed. As long as the molecules have a dipole they will associate radomly, i.e. they can associate at any angle ( the pos. site of one molecule lines up with the negative site of another molecule).
Thus, H-bonds are very specific interactions, first the molecules have to have a dipole and in addition a 180 dgr. geometry of the particpating atoms (N and O) is required.
The book will be useful in two ways. It is recommended that the student reads the specified chapters (see syllabus) before a particular lecture. This will make it easier to follow the material presented in class. You also should use the book to read up on topics which are presented in class to get a better understanding, as well as to prepare for the quizzes and exams. Go into details, as much as discussed in class. The book will cover more material than we can cover in class and occasionally the topics are presented in more detail than in the lecture, alternatively I may give more detailed explanations than the book. In any case, you are responsible only for the material covered in class at the level it was discussed. That means the exams and quizzes will deal only with material presented in class. The web notes will address all topics mentioned in class.
When egg is boiled, it denatures its proteins. When denatured proteins enter living system, it enters into Chaperonin cage and the protein folds into its appropriate shape and is released. Now would this be considered reversed protein(back to normal fuction)? Would this reversed effect apply to living systems such as humans? By eating this boiled egg, could an individual still otain amino acids of protein from it?
Chaperonins are a family of special enzymes that mediate protein folding. Many newly synthesized proteins can not fold on their own into their 3 dimensional shape, but require the help of chaperonins. Chaperonin molecules associate with the newly synthesized protein. Through this association the protein is forced to fold into a particular shape, which is then finalized through H-bond , polar, hydrophobic , ionic interactions and S-S bond formation. When a globular (folded) protein is denatured ( e.g. pH change, exposure to high salt concentration or temperature increase ) it unfolds or assumes another configuration. When the original conditions are re-established, the protein may return spontaneously into its natural configuration, or with the aid of a chaperonin if required. In such case the denaturization is reversible. However, if the protein concentration is sufficiently high during denaturization, the proteins will form large aggregates. Protein aggregates can not be re-natured even in presence of chaperonins. For example this is the case when an egg is boiled. The proteins in egg white form insoluble aggregates and can not be re-natured. Structural changes may also have occurred, e.g, rupture of S-S bonds or even peptide bonds. Similar, if tissue is burned, it can not repair itself. When protein, in form of a boiled egg is ingested, the enzymes of the stomach will aid in the hydrolysis of the egg protein into individual amino acids. These amino acids are absorbed by the epithelial cells lining the intestine. The amino acids are then discharged into the blood from which they are absorbed by tissue.
The cytoskeleton is a network of tubular and filamentous structures, spanning the cytoplasm. The function of the cytoskeleton is to stabilized the shape of a cell, to support the plasma membrane from the cytoplasmic side and to guide vesicular traffic .
Correct
there are three aromatic amino acids Phe , Tyr, Trp. They have a benzene ring (C6H5) in their side chain.
Both, sugar and fats are substances that are metabolized in a process that yields chemical energy in form of ATP. The turn over rates of ATP are very high, that means ATP is not a long term storage form of chemical energy.
There is an OH group on five of the six carbon atoms of glucose and one CH2OH group. In the cyclic form of glucose these OH groups and CH2OH group assume a position (equatorial position) in which they all point away from the ring. These groups are large (much larger than the H atom that is also bonded to each C atom) and in this position there is no steric interference. Other sugars e.g. galactose is different by the orientation of one OH group. This OH group is now spacially closer to a neighboring OH group and the structure is not as stable as glucose. This is the reason why glucose and not galactose is the most abundant sugar.
Two Cys residues can form S-S- bridges, thus folding the peptide chain. This process places the two residues automatically toward the inside of the protein
This term is used for proteins that fold into a particular shape. Structural proteins e.g. keratin , collagen do not fold, their structure is mostly helical.
Naturally occurring fatty acids have cis double bonds. Cis double bonds introduce a kink into the carbon chain.Thus molecules fatty acids can not associate very closely, which results in a lower density and thus more fluidity
Amino acids are zwitter ion, they have a positive charge (amino group) and a negative charge ( carboxylic group) in the same molecule
Sugar portions on glycolipids play a role in cell -ell recognition, e.g. in tissue formation and are therefore needed on the cell surface.
General questions about Quizzes and exams
If your score on a quiz is less than 60% , then that score will be assigned. Should you retake the quiz and get 60% or more, then you will get full credit or 100% for this quiz. If you have taken the quiz more than once, the highest score will stand. You will not be able to view your grade immediately. The grade file will be updated at specific times , which are shown on the grade file.
If a quiz score is 60% or better, a student will get full credit (= 100%) for this assignment. If less than 60% are made on a particular quiz, e.g. 35% then this quiz will count only 35%. That means if all quizzes are 60% or better you will have earned 10 semester points (out of 100 pts).
As it is now, the grade file is only updated several times a day. I will be more explicite with the update times. You may have noticed that the "next update" time is now listed on the grade file, so you do not need to waste any time looking for your grade. There are several reasons why it is not immediately updated. I do all the programming for our web site and I just did not get around to write the progam for immediate update. The other reason is pedagogical. If there is an instant feedback, there is a temptation to continuously resubmit the quiz with some minor changes until a score is 60% or higher is reached. A student will not learn from such an approach. I rather see a student read up on the topics addressed in questions that were difficult to answer or questions whose answers were guessed. After that the quiz can be resubmitted. I have set the minumum score for full credit to 60% to take away any pressure to make maximum points for a good grade. The idea is study a little before taking the quiz. Your score will tell you how well you understand the course material and what grade you would have earned if this would have been a lecture exam. I will try to update more often and I also will post the quiz a little earlier.
We have usually between 50 and 55 questions. They will be multiple choice questions, similar (content and style) to the quiz questions.
will a periodic table be provided
should we assume the responsibility of memorizing all of the number introduced in lecture (weight of a proton, electron and neutron; the conversion between AMU
Calculators are allowed in exams, but I don't think we will have problems where you need a calculator. If there are any questions about the periodic table, I certainly would provide a copy of the table or other adequate information. There is no need to memorize any numbers or equations , such as the radioactive decay equation. However you should know that e.g. C-14 is a beta emitter with half life time of roughly 6000 years. You do not need to memorize any other half life time, just know that U decay products have very long half life times It may be useful to know that a proton or a neutron has an approximate mass of 1 amu, ( which of course is a result of the definition of amu, 1/12 of the mass of a C-atom). I don't expect you to know that e.g. sodium has an atomic weight of 22 amu, or oxygen 16 amu. If necessary, I always will provide such information. However given the atomic number and the atomic mass of an element you should be able to determine the # of protons, neutrons and electrons of the element. You should be able to determine the electron configuration of an element if the atomic number is given, e.g. F has an atomic # = 9 , therefore there are 9 electrons, 1s2, 2s2, 2p5 .