<p>I would say cold-reading campbells is fine. That’s what I did my freshman year, and I managed to grasp most of the concepts. Yeah go for it :)</p>
<p>Don’t worry that’s exactly the thing i did (except i didn’t take ap bio) </p>
<p>would the Campbell 8th edition still be valid for the 2015 open exam, even though the10th edition of Campbell is out</p>
<p>Yes. Actually, I would think 8th edition will be even better in some ways because Campbell’s cutting out some information according to what I heard. </p>
<p>Hey, can anyone help me out with #77-91 on the 2009 semifinal exam? I realize that seems like a lot, but I don’t need detailed explanations for every question, just a general idea of how to do them (there are a bunch of similar problems). </p>
<p>2010 Semi - Can some one help? @Blueroses67 - Can you tell us where we can find more practice questions like the following?</p>
<ol>
<li>The Black/Brown locus in rabbits is linked to the Solid/Himalayan color pattern. If a homozygous Solid Black male is bred to a Brown Himalayan female. A mating of an F1 progeny with an individual that is homozygous recessive for both traits results in the following phenotypic distribution {Black-Solid 345; Black-Himalayan 123; Brown-Solid 254; Brown-Himalayan 452}. What is the genetic distance between these two loci?
A. 32 centimorgans.
B. 40 centimorgans.
C. 49 centimorgans.
D. 51 centimorgans.
E. 68 centimorgans.</li>
</ol>
<p>Answer for 2010 Semi 47 (above question) is A</p>
<p>The distance between the loci is equivalent to the recombination frequency. The recombinants are black-himalayan and brown-solid, so the answer is (123+254)/(345+123+254+452)=32. </p>
<p>@central -
Test Result 1 - The two columns in this gel have a difference of 0.4, the length of one of the introns in the provided gene. This suggests that we are looking at RNA (because the intron is removed) and thus a NORTHERN BLOT is being used here. Most likely, this is Allele X, since it’s the only one adjacent to the 0.4 intron (the spliceosome presumably is unable to effectively remove the intron because of the mutation). This implies that the LEFT band is the patient. Because this is a blot, we know that a Probe must be in use (oligonucleotides are here used as primers for PCR amplification). Thus, Probe K is the logical choice, since it encompasses much of the target region. Probe J is possible-ish, but it mostly covers the intron before Exon 1, so Probe K is more likely
Test Result 2 - Because there are two bands in the second column, a restriction endonuclease is cutting the (presumably DNA => SOUTHERN BLOT) gene, and this suggests Allele W (which is the only allele with a restriction endonuclease site). We can surmise that in the mutation, the endonuclease (which normally is able to cut the DNA as per the diagram) is no longer able to cut the DNA. Thus, the patient is the LEFT column. Again, Probe K is the logical choice
Test Result 3 - The small size of the molecules in question immediately suggests PCR, since with PCR we’re able to amplify targeted regions of DNA. Additionally, the 4 bp difference suggests Allele Y, in which the mutant is 4 bp smaller. Thus, the patient is on the LEFT (once again). Based on the size of the products, Oligos B and C are the logical choices (if you assume not to scale, I guess B and E is possible, but that isn’t a choice). I’m not quite sure why it says DNA and RNA though :P</p>
<p>Hope that helps!</p>
<p>Thanks so much! That was exactly the kind of explanation I needed… I just got frazzled by the amount of information, but it makes a lot of sense now. </p>
<p>@abacus1: get a genetics textbook. It doesn’t have to be a new one; this stuff was worked out decades ago. There are free ones available online (e.g. <a href=“http://gene.bio.jhu.edu/bm2whole.pdf”>http://gene.bio.jhu.edu/bm2whole.pdf</a> - large PDF warning!), or your local library should have a dead-tree version if you prefer.</p>
<p>Hey all! What dates have been assigned as semifinal test days for you?</p>
<p>28th!</p>
<p>28th for me as well!</p>
<p>27th here. What’s in it for the teacher? </p>
<p>some schools got extensions to next week…no discussion please
I think we should all get extensions >:D </p>
<p>Which area schools got the extension? </p>
<p>Can somebody explain how to do #42 - 44 on the 2012 Semis? </p>
<p>^same question as above</p>
<p>Does anyone happen to know what the cutoff scores for making camp have been for the past few semis?</p>
<p>@chrism0713: I did that one a few pages back; here you go.</p>
<p>“In questions 42 to 44, you have two true-breeding strains of pea plants with white flowers. Crossing these two strains yields an F1 generation that only has purple flowers. Self-crossing these purple flowered plants yields the following ratio of progeny with purple and white flowers: 9 purple: 7 white.”</p>
<p>This is where pattern recognition helps with genetics questions. This is a classic 9:3:3:1 ratio - what you get when you cross two double heterozygotes.</p>
<p>P1: AAbb and aaBB (true breeding, apparently you need A<em>B</em> to be purple)
F1: AaBb (all of them)
F2: 9 A<em>B</em> : 3 A<em>bb : 3 aaB</em> : 1 aabb (9 purple : 7 white)</p>
<p>From there, you can figure out pretty easily what fraction of the purples are AABB and AaBb.</p>