1. Class. As I said in class, this homework was difficult. In part it was because it tested what you knew in tricky ways, as I clarify in the answers, and in part because you probably did not understand some of the questions clearly enough (also clarified in the answers). Hopefully the questions stimulated vigorous thought, and useful discussions amongst those who study with others! The virtue of hard homework problems is that you should discuss with others --what the heck is going on is a good start! If you truly understand the answers, you are in good shape for an exam. One way to see if you “truly understand” the answers is to manipulate the questions slightly and see if you still get it. I elaborate below. This serves as a study guide/extra problems. I hope to post more problems on Monday. Change the bases in question 1 and predict consequences. Explain other answers in 1 - write it out …so you have practice in formulating answers! For 1, you should also know which is the template strand for RNA polymerase, and be able to label the polarity of the strands shown, that the ribosome and RNA polymerases are “dumb” to each others function! For question 2, be able to draw out a mitotic recombinant to generate aa bb cells, for example. You can use a simple “X” to indicate where a crossover is, label your centromeres, and show a segregant that yields the desired genotype. Do it neatly!!!! And for meiotic recombination and the Holliday structure…learn how to draw it, using the secret handshake of strand polarity “switcheroo” between MOM and DAD homologs. Know how the mechanism of homologous recombination renders it error-free! Know how to draw the “Big Picture” as well…like we did on the board last week..G1 diploid to G2 diploid, one sister from each homolog pairs, recombines, and segregates in MI and MII. And of course, our blessed RecA and Spo11, where and when they do and don’t act. And homologs…the anaology of identical twins “searching for the identical face in tucson”--in an hour-- gives you an idea of how miraculously RecA searches for sequence homology/complementarity to align allelic sequences. Hypothetical problem; if a 10kb sequence is duplicated and inserted at a non-allelic sequence, does this pose a problem for RecA and the cell if either of the sequences experiences a DSB? Discuss briefly. True-false short answer question 3 is one of my favorite formats (which Bruce detests). Note: you only get full credit if you supply a reasonable factoid to explain the true or false. And, any one comment will do..you don’t have to write out all relevant comments, just one to let me know you get the idea(s). It allows be to have a look and see if you understand a broad range of ideas. That mitosis is a conservative process where recombination is rare and occurs following an error, for example, while meiotic recombination is programmed…DSBs made by the Spo11 “pacman” nuclease..and meiosis generates diversity by mixing alleles…not by creating them (which occurs by error…)

2. Question 4 is an elaboration of question 2, asking you to manipulate mitotic recombination, the linkage of different alleles, and the like. I put in some extra questions to answer in the answers I provide. Its mostly about some basics of chromosomes (knowing sisters, homologs, segregation patterns, and simple chromosome geometry (where DSB, alleles of interest, and centromere). Question 5 goes along with a question on the first HW..the idea of domains and mutational consequences. What does a “hot spot” mean? Review the question from the first HW. Note that we must speak in generalities…in general, missense mutations in domains may disrupt their function, while nonsense or frameshift mutations will disrupt their functions. Missense mutations in “linker” regions are less likely to disrupt function. Question 6 . The notion that cells that are heterozygous for a mutation, and that somatic cells will suffer a 2nd inactivating mutation is important to understand. The consequence of that 2nd inactivating mutation depends on the dysfunction of the cell; for p53, a -/- cell has a selective growth advantage over other cells, and will grow, while a CFTR-/- cell does not have a growth advantage, is dysfunction for pumping Cl- but lives amongst millions of other functioning cells so the tissue (lung) performs well enough. This is likely a general trend…mutations that cause cancer can be rare but are a bummer, while mutations that inactivate cell-specific functions but do not lend a selective growth advantage are not a problem. Question 7. Pedigrees . We will deal with these a lot more when Bruce talks about segregation ratios. The idea of what pedigrees would look like for dominant and recessive mutations is something you should be able to understand by classes end. We have not yet discussed dominant and recessive mutations much..we will after the exam. You should know what a recessive mutation is conceptually…an inactivating mutation (which most base pair changes in ORFs are)…and that if an intact, wildtype copy of the gene is present then the cell can function ok, usually (CFTR protein, for example). Question 8: This is was question asking you to integrate several concepts. That recombination could occur between short runs of repeat sequences is something you hopefully understand now…though this kind of reaction is more infrequent than between long streches of identical sequences (i.e between alleles!), it does happen. Then, the issue of reading frame comes up in several parts to this.

9. 6. Explain why a person with a germline genotype of p53+/- is prone to cancer while a person with a germline genotype of CFTR-/+ is not prone to cystic fibrosis. ( typo of CFTR corrected). Cells that are p53+/-, when the wildtype allele mutates by error, will form p53-/- cells that have a selective advantage in growth..and thus proliferate and form a tumor. For CFTR+/- cells, on the other hand, the rare mutant cell will be surrounded by functioning heterozygous cells; the lung tissue, for example, may have ~10 12 cells, and when 1000 become -/- (due to mutation rate of 1 in 10 9 ), the one trillion cells minus 1000 have sufficient function for good lung function. 7. In the pedigree shown below, compare the relative extent of heterology (between any two homologous chromosome) in individual 1, individual 2 and individual 3. I was looking for a qualitative answer here. And to clarify--compare the relative extent of heterology between two homologs in individual 1 (which is 0.1%) , with the heterology between two homologs in individual 2, and then individual 3. The idea is that if individual 1 is a normal, outbred individual (though the argument still holds even if he is inbred..), then individual 2 will inherit some of the same alleles from parent 1, risking being homozyous at many alleles. The perils of inbredding! Individual 3 will be more outbred than 2 because 3 has 1 more generation of outbred parents (see **) . So, homologs in individual 1 are most heterologous (~0.1%), individual 3 the next most (less that 0.1%) , and individual 2 is the least heterologous (much less than 0.1%). ** ** 1 2 3