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12

By frame size, do you mean sliding window? I know that if you want to predict a secondary structure of a transmembrane protein, then your window size should be 20 amino acids (this is the average length of 1 transmembrane alpha helix spanning through the membrane). I found this paper by Chen, Kurgan, and Ruan[1]. It basically says that the window size ...


10

This is a very subtle question and I encourage you to read the Wikipedia articles on these different subjects (t-test, chi-squared test, p-value, etc) because the authors worked hard to combat common misconceptions about these commonly used statistical tests. Here is a rather oversimplified rule-of-thumb for these different tests: t-test: Used when you are ...


8

Every nucleotide sequence has six possible reading frames, because each codon (determining one amino acid) consists of a base triplet (3 frames), and there is a complementary strand which could be coding (3 reverse frames). To find the possible open reading frames contained in your sequence, you have to look for start codons. That is: ATG. But, as said the ...


7

Assumptions: Blonde hair is Homozygous Recessive and that the traits are strictly Mendelian. The parental generation must be both heterozygotes as at least one child is Blonde (bb). So your cross is Bb x Bb. Your square is going to look like this: _B_ _b_ _B_ BB Bb _b_ bB bb So of the ...


7

There are actually very few situations where organs can be harvested from donors. For all deceased donor transplants, the donor must be confirmed as being brain dead (both brain stem and higher cortical functions). However, in order for the organs to remain viable they must not become ischaemic - which is obviously a huge problem when the patients heart has ...


6

DAPI (4',6-diamidino-2-phenylindole) preferentially binds AT-rich DNA (although it binds CG-rich DNA, too), which can give chromosomes distinctive banding patterns if they are polytene or in metaphase. In interphase condensed chromosomes, such as the inactive X chromosomes of female mammals (Barr Body), the relatively high concentration of tightly-packed ...


6

Primary oocytes are formed prenatally and reain suspended in prophase of meiosis I for years until the onset of puberty. An oocyte completes meiosis I as its follicle matures (during ovulation) resulting in a secondary oocyte and the FIRST polar body. After ovulation, each oocyte continues to metaphase of meiosis II. Meiosis II is completed only if ...


6

If the question is about the one and only most important difference between mitosis and meiosis, then the answer "meiosis reduces ploidy" is probably correct. But if the list of important differences is open, it would be critical to add that mitosis generates identical cells (identical to each other and any ancestral cells, barring rare new mutations), while ...


6

Starting with the left hand side of the diagram: III:2 is definitely a carrier (Tt) as one parent (II:2) is affected (tt). III:1 is also definitely a carrier (Tt) as when mating with III:2 they produce an affected (tt) offspring (IV:1) This means that we can work out the possibilities for IV:4 as we know the parent genotypes. It follows the standard ...


6

Let's take Question A: Both fathers have siblings with red ears, and red ears are an autosomal recessive trait. The grandparents did not have red ears, we know then that they were carriers of the recessive allele. Each grandparent was Nr (for Normal and red alleles respectively). The fathers have normal ears, so they could be NN (probability 0.33) or Nr ...


6

Still if you change your question as (If histidine is abundant, HisP's job is to stop the histidine pathway as a "repressor." If HisP binds less tightly to promotors, the pathway should not produce as much histidine.) Then it should be under another assumption that what is the effect of HisP binding promoter of enzyme's gene. Is it suppressing the ...


6

Your approach to translate the AA sequence, codon by codon, was correct. This was a bit of a trick question because to recognize it, you had to read the sequence backwards. UGG-CAA-GGT-CAC etc is read directly off of the 3'->5' strand of the circled answer, reading from right to left. The bottom left is a red herring because it starts with a start codon, ...


6

The mechanism is called "capillary action". It requires a tube of a small diameter and happens because of the adhesion of water to the walls and the cohesion within the water (=surface tension).


6

The ureters run from the kidneys to the bladder whilst the urethra runs from the bladder to exit the body: The renal papillae are the site where urine drains into one of the minor calyxes of the kidney. Multiple minor calyxes join together into a major calyx, and multiple major calyxes join together to form the renal pelvis which then drains into the ...


6

I got $26$% as the answer. To get a recombination between C and E, there are two possible mechanisms:- C and D produce a recombinant, but D and E remain linked, therefore the final genotype will be a recombinant considering C and E(Chiasmata between C and D). Here $P_1=P_{CD}\times P'_{DE}$ where $P$ is the probability of recombination and $P'=1-P$ ...


6

The question is a bit vague in some important parts, so I'll have to make a few assumptions about what the authors likely meant. RNAses are enzymes that degrade RNA. There are a few different ones that lead to different kinds of degradation. The type that you would use in an experiment like this is an RNAse that completely degrades RNA. The purpose of this ...


6

If really cystic fibrose is expressed only in homozygotes, then here are some possible explanations: Mutation very improbable Autofecundation very improbable False Negative when testing the father Because of the test False Negative Rate. @kmm made a very interesting comment below. He said that according to this article 75% of the mutations coding for ...


6

The chemist, Antoine Lavoisier said: Rien ne se perd, rien ne se crée, tout se transforme , which is usually translated in english into... Nothing is lost, nothing is created, everything is transformed. I don't think it is still valid with modern physics but for the purpose of this question it will hold perfectly. You are totally correct that ...


6

Liver does all those functions as far as I know. The liver hepatocyctes are stimulated to create the enzyme glycogen synthase which promotes the conversion of glucose to glycogen in the presence of insulin. Glycogen is stored in the liver after its production for further use. The whole process is explained in detail in this page. Regarding the storage of ...


6

This is not as hard, as it first seems. Lets have a look at the single enzyme digests first: The digest with enzyme A and B only leads to products which are 5kB (5000 bp) away from each other. Since they are of the same size, both equally sized restriction fragments appear as one band. So each enzyme cuts the plasmid exactly in half. The double digest is a ...


6

I will answer the questions one by one- Why does this miniaturization necessitates insulation ? Volume increases with a power of three with radius, while surface increases with a power of two. An organism's volume (and actually every (in)animate object) determines the total amount of heat that can be stored. It's surface determines how much heat ...


5

In humans, the amnion (amniotic sac) persists from the primitive amniotic cavity1. One side of this is formed from the cytoblast (a prismatic epithelium) and the plasmodioblast. Together these two layers are the ectoplacenta or chorion. They are also referred to as Rauber's layer. These replace the lining epithelium of the uterus, whereupon internal ...


5

You're right in saying that yeast is single celled. However, moulds are described to be filamentous fungi that are multicellular. The filaments of the mould give colonies "a woolly, fluffy, or velvety appearance, sometimes punctuated with a granular or powdery aspect that is produced by the formation of asexual reproductive structures"(1). Aspergillus is ...


5

Read "supply" as "carry action potentials to." When the action potential reaches the junction with the muscle (i.e., the neuromuscular junction), neurotransmitters are released into synapse. A similar membrane depolarization occurs on the muscle cell, ultimately leading to contraction. Nerves visible to the naked eye are actually bundles of individual axons ...


5

A) Here is the correct map: You made a mistake on your map at the PvuII site (it is not on 6.5kB from the start of the plasmid, but on 6kB). Can the Kpn I not go on the 8.5 site, it still creates the 2 and 8.5, so isn't there more than 1 correct option for plasmid map? Yes. What you need to do in order to make the correct map is try all possible ...


5

In high school, we did an experiment that showed this. Basically, if you take a glass of water, and let it sit out, perhaps in front of an open window, it will eventually lose water due to evaporation. It may take a few days/weeks to really see a large difference, but the level will go down. But, if you take a few flexible straws, put them in so the bendy ...


5

To be honest I know very few about enzymes and absolutely nothing about Michaelis–Menten. However, when I "took the Michaelis-Menten equation, replaced v with 0.4Vmax, canceled the Vmaxes (one on each side), and solved for [S]", my result is positive: 0.4 * Vmax = S * Vmax / ( Km + S ) 0.4 = S / ( Km + S ) S = 0.4 * Km + 0.4 * S 0.6 * S = 0.4 * Km S = Km * ...


5

Your reasoning is incorrect, because once you have the number of individuals surviving after the 50% mortality you can't just give them all one identical offspring, you have to mate them with one another. However, I can't get the answer given, instead I get 47.9% Here is my reasoning: f(A)=0.3 f(a)=0.7 calculate genotype frequencies: AA = ...


5

This is basically the same solution as @AlanBoyd's answer, but since he asked me to, I will post my solution as well (also a reason to try out $\mathcal{MathJax}$ a bit more). Assuming: $f(A)=p=0.3, \\f(a)=q=0.7$ and the standard formulas: $f(AA)=p^2\\ f(Aa)=2pq\\ f(aa)=q^2$ gives these genotypes in the parental generation (G1), before and after ...


5

Remember than Pvu1 cuts in the middle of AmpR, and when you insert your gene you'll be disrupting its function, meaning that any transformed colonies will no longer be able to grow on Amp-containing media. This is why you want to cut with BamH1 - the disruption to the TetR gene is irrelevant, and it leaves Amp resistance in place. The point of growing on ...



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