<p>“12. You are looking for a lactose analog to induce long-term expression of a gene
under the control of the lac promoter. Which of the following molecules is the
most suitable choice?”
2012 semi test, why is the answer what it is? </p>
<p>Answer: A</p>
<p>Also, anyone think the answer key for question 14 is wrong?</p>
<p>@Lucky139 IMO, that one was sort of weird. You can eliminate B, C, and D right off the bat because they don’t look anything like lactose. I think E actually IS lactose. At first I just answered E, but the hint is in the title: “long term expresssion.” A is a real lactose analog called IPTG: <a href=“Isopropyl β-D-1-thiogalactopyranoside - Wikipedia”>http://en.wikipedia.org/wiki/Isopropyl_β-D-1-thiogalactopyranoside</a>. β-galactosidase cannot break down IPTG. Hence, it’ll just hang around and keep triggering the production of β-galactosidase.</p>
<p>D actually is lactose, which has beta 1,4 bonds. E has alpha bonds, which would not work as an analog. I honestly have no clue why A is right, but I would assume that the sulfur attached to the galactose-like base somehow prevents degradation by beta-galactosidase. Lol, that’s all i got, sorry</p>
<p>Yeah that does make sense, but since it only accounts for the galactose part of the sugar, by that reasoning I thought then why would cleaving lactose in the first place turn on the repressor again if you only needed part of lactose for it to function (since free floating galactose, which has the active part, would still be there).
Obviously this was wrong so I was left in the dark for this question 
And for 14, I’m pretty sure that short day plants mainly flower in fall and early spring? </p>
<p>The carbon chain leading out from the sulfur might serve as the analog of the glucose part</p>
<p>@lucky - i c wut you did there :P. Uhh for 14 we can immediately rule out D and E (vernalization isn’t necessary for all and interrupting day with dark is useless). Intuitively, a short-day (long night) plant would flower as nights get longer, so summer through winder solstice. Spring doesn’t make sense since days are getting longer then. Thus, B</p>
<p>IPTG is used because (unlike allolactose) it can’t be broken down by beta-galactosidase. According to wikipedia and @botherme, this is because of the sulfur atom in place of the normal oxygen atom in the glycosydic linkage. Honestly I’m not sure why this is, since the thiolate is a way better leaving group than an alkoxide (pKa of thiol << pKa of alcohol, and thiolate is a big fuzzy cloud of electrons). I guess it has something to do with steric interactions. @blueroses or @darksigma could probably explain this better.</p>
<p>According to <a href=“THE LAC REPRESSOR”>http://biology.kenyon.edu/BMB/Chime/Mat/MASTER.HTM</a>, it appears that the binding of IPTG to the Lac repressor involves the formation of hydrogen bonds (similar to that in allolactose) and hydrophobic interactions between a tryptophan residue and the isopropyl group (not similar). Sterically speaking a sulfur atom is huge compared to carbon and oxygen, so maybe sterically it’s about the same size as the lactose (not sure). Honestly though I have no idea. Interesting question to pose to a professor. </p>
<p>@BeanDelphiki O.O I never learned this in AP chem… I assume you’re making chem camp too?</p>
<p>hahaha I wish. Chem campers are absurdly skilled at organic chemistry (it’s actually pretty frightening how good they are). I picked up a bit from ChemWOOT last year (nucleophilic substitution!) and biochem reading. </p>
<p>For those doing both chem and bio, which one are you focusing on right now and which do you think is easier?
(for olympiad)</p>
<p>Sorry, but how do I view my USABO scoreo no the CEE website? Thanks</p>
<ol>
<li>go to <a href=“http://www.usaboo-trc.org”>www.usaboo-trc.org</a></li>
<li>Login with your ID and password. If you don’t have your ID, your teacher should have it.</li>
<li>Hover over Student Resource Center and Click “View Exam Score”</li>
<li>Voila</li>
</ol>
<p>your welcome you guys because you can count me out of getting like, above a 6 on the semifinal exam. I forget what i got last year on it but it was like 14 or something I think, something absurdly low. I got a 31 on open tho which is cool</p>
<p>What does the USABO website mean when it says “March 17 (Tentative): Semi-finalists posted to website (top 10% or 500 students)”? Are they going to post a list, or is it done now that scores and statuses are posted in the exam score section?</p>
<p>Could anyone elaborate on this question?
51(2011 semi). In foxes, there are 9 coat colors: red, standard silver, Alaskan silver, double-black, smoky red, cross-red, blended-cross, substandard silver, and sub-Alaskan silver. A red fox was crossed with a double-black fox and their offspring were then crossed with each other. The F2 phenotypes were; 10 red : 18 smoky red : 20 cross- red : 39 blended-cross : 9 standard silver : 19 substandard silver : 12 Alaskan silver : 22 sub-Alaskan silver : 8 double-black. How many genes are involved in this cross?
A. 18 genes
B. 4 pairs of genes
C. 9 genes
D. 9 pairs of genes
E. 4 genes
In addition, does anyone have good sources for practicing these multigenic problems?</p>
<p>as @blueroses said, an undergrad genetics book might help</p>
<p>Ok, they tell us that there are 9 different phenotypes for this trait. We can modify a hardy-weinberg expanded formula a slight bit to suit our current purposes. The actual equation says: (P+Q)^n=1 where n is the ploidy number (number of chromosome sets). However, if we set n as the number of genes involved instead, the expansion of (P+Q)^n only gives us 9 different terms if n=8. Each of those 9 different terms (P^8, Q<em>P^7, Q^2</em>P^6, etc.) codes for a different phenotype, so there are 8 genes involved, or 4 pairs.</p>
<p>@zinwell - Idk how to properly solve it either. However,</p>
<p>F2 ratio can be approximated to 1:2:2:4:1:2:1:2:1. We notice that this is basically (1+2+2+4):(1+2):(1+2):(1) = 9:3:3:1, suggesting one of the two (pairs of genes) answers, since we’re getting an answer that breaks down into the characteristic pattern for a dihybrid cross. I’m not sure what exactly a pair of genes is, but I’m wondering if it’s where it can choose one gene per pair to express (pseudo-epistasis, kinda). Assuming that’s the case, we basically have (x1/y1)(x2/y2)…(xn/yn). We should then be able to break down the ratio into some sort of 1:4:6:4:1 ratio if it’s 4 genes, and (whatever the pascal’s triangle for 9 is) if it’s 9. We can do 1:4:6:4:1, so B. The second part of the solution is bleh though - any USAMO people out there may be able to shed some light on this. </p>
<p>Maybe they just put the “pairs” to confuse us lol</p>
<p>Are anyone else’s scores not showing up when they click “View Exam Scores”? and if this isnt normal wat should i do…</p>
<p>Check and see if your teacher has your scores. If she doesn’t, email CEE</p>