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…)
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.
<ul><li>Exam I questions: </li></ul><ul><li>Here are the questions I am considering: </li></ul><ul><li>Structures of molecules and their significance. </li></ul><ul><li>know how to draw a G-C base pair, including hydrogen bonds of the base pair, and phosphodiester back bone (draw 2 nucleotide long structure). </li></ul><ul><li>( Whenever we draw recombination structures, we imagine in our minds these chemical structures!) </li></ul><ul><li>Know the structures of Alanine, glycine, lysine, and serine. Know which are hydrophobic, hydrophilic, and which has a plus charge. I may give you a DNA sequence with a basepair change that changes an alanine to a glycine (minimal likely effect, hydrophobic to hydrophobic) or alanine to lysine (larger effect, hydrophobic to charged…). </li></ul><ul><li>Know how to draw the peptide backbone, and where the R groups are! </li></ul><ul><li>[Whenever we think of proteins and function and mutational consequences, we think of a polypeptide with R groups that help it fold and function..and that mutational changes will have consequences depending on the nature of the change (frameshift and nonsense, or missense and kind of amino acid change - similar or very different- see question 2!)] </li></ul><ul><li>Gene expression, domains, mutation and consequences. </li></ul><ul><li>Recombination; mitotic and meiotic. Holliday structure and its lovely details as asked in the HW#2 question. </li></ul><ul><li>T/F short answer. Philly chr and fragile site may show up here, as well as protein interaction network, for example. </li></ul>
Page 1 <ul><li>Another gene expression, mutation and consequences. </li></ul><ul><li>The DNA sequence of the top strand is shown. Capital letters indicate consensus promoter sequence (TATA) or coding sequence (ATG…TAG). The expression of the protein requires p53. The function of the protein requires all the specific amino acids in the ORF shown. +1 of the mRNA is a C. There is no splicing. The ATG (at +5,6,7) is the start codon. Answer the following questions (with the help of the Genetic Code given below). </li></ul>p53 binding site gcgTATAccgctacgtagg c tttATGCGATTTAAACCCTAGcccgcaag +1 +5 Transcribed? Translated (#amino acids) Functional? a. p53+/+ YES YES (full length) YES b. TATAcc to TggAcc NO (less) NO (less, though 5) NO (less) c. Delete c at +1 YES YES (5) YES d. Insert T between 7G &8C YES YES (3) NO e. Delete ccc after TAG YES YES (5) YES f. AAA to AAG YES YES (5) YES Enter YES or NO. 1pt each. Enter # of amino acids if different from wildtype gene. You can explain ambiguities on the other side of this sheet (you shouldn’t need to!) g. Explain briefly your answer to c and d. There is no change in part c because deletion of +1 merely changes where RNA polymerase starts the mRNA; the ribosome still reads the first codon etc as in the unmutated gene. In d the insertion creates a frameshift mutation, the consequence of which is a stop/nonsense codon resulting in a 3 amino acid product, with shorter and different sequence; the 2nd and 3rd amino acids are different (MET Arg Phe to Met Ser Ill). Underlined not required but for your information)
Page 2 2. Recombination <ul><li>Refer to the chromosomes of a G2 cell shown below. -label two sisters with “S” </li></ul><ul><li>-label two homologs with “H” </li></ul><ul><li>b. Draw the recombination and segregation events to form an A+a- B+b- cell. </li></ul>b- a- B+ A+ c. Harder question : This AaBb mitotic cell (diagram) could generate either aaBb, Aabb, and aabb cells. Which of these three would be formed least frequently and why? An aabb cell would be formed least frequently, because it requires TWO CO, one between b and the centromere, and one between a and the centromere. Draw both CO between chromosome 2 and 3, then segregate centromere 2 and 4, and wala, you have an aabb cell. TWO crossovers occurs more rarely than ONE…”rare squared”… In a Mitotic cell-a-c . In a Meiotic cell d. d. Draw a double Holliday structure intermediate from strand invasion step before replication occurs. NEATLY! - Draw an arrow to the D loop . Indicate the polarity of all 4 strands in your diagram. - Circle each 3’ site where DNA polymerase synthesizes DNA, and place dots what it synthesizes. - Place a square around where ONE Holliday structure would be and a triangle around one site of potential heteroduplex (without further branch migration) -Place slashs showing one way this would be resolved to form a crossover. 1 2 4 3 This was tricky for some because it was so easy; no CO or CO with any segregation leads to a A-a-B+b- cell! In an exam I would ask for something more informative; like how would a a-a-B+b- cell form. Draw anything here! S H 5’ 5’ 3’ 3’ Note: I misworded the question a bit. To make sense of it…a Holliday structure only forms after replication and ligation/joining. You get the idea from my answer. And, only one rectangle was asked for, and resolution can occur by H-V as shown, or V-H (the reverse).
Page 3 <ul><li>3. True/False, short, relevant answer to all questions. </li></ul><ul><li>RNA polymerase stops transcription when it comes to a stop codon. False. Ribosome stops when it comes to a stop codon. Or RNA polymerase stops at some signal sequence that occurs after the stop codon sequence . </li></ul><ul><li>Bases at the 5’ end (upstream) of the start codon set the reading frame for the ribosome. </li></ul><ul><li>False. Bases at the 5’ end of the start codon are not read by the ribosome, that reads the first ATG to start translation. </li></ul><ul><li>The “Protein Interaction Network” is a bunch of worthless hooee. </li></ul><ul><li>FALSE! The “PIN” as some students refer to it, indicates how we think proteins interact; in clusters, and those clusters either share proteins or have proteins that interact between clusters. The implication is; when one protein is altered in one cluster, there can be an affect on other clusters. A mutation can be “pleitropic”--can have multiple effects .(Underlined not needed for this question as phrased, but the underlined is the significance!). </li></ul><ul><li>Spo11 and RecA cut DNA to make a doublestrand break to start mitotic recombination. </li></ul><ul><li>False. Spo11 makes DSBs in meiotic cells; or False, RecA pairs broken DNA with an intact homologous sequence . </li></ul><ul><li>In a hypothetical population of bees, the p53+/- diploids and p53+/-/- triploids get cancer at about the same frequency. </li></ul><ul><li>Either: True, each has one gene that needs to be mutated, so the frequency of mutation dictates equal frequencies of inactivation . OR </li></ul><ul><li>False: Mitotic recombination could lead to formation of cells with no functional p53, and since the triploid has more mutant strands than the diploid, the frequency of mitotic recombination might generate a higher frequency of mutant cells in the triploid versus diploid. Draw it out to convince yourself! </li></ul><ul><li>The Philadelphia chromosome is an example how a base pair change by DNA polymerase leads to a drastic genetic change. False. It is a translocation. </li></ul><ul><li>Drugs that target a specific domain are unlikely to specifically interact with the protein that are designed to interact with. True. Since domains are shared with many other proteins, those other proteins may also be affected by the drug. </li></ul>
4. Consider the cell 1 and cell 2 with the chromosomes shown. In Cell 1, Gene A is 50cM from the centromere, and in Cell 2 Gene C is 10cM from the centromere. In both cells the wildtype allele is on the top homolog, and the mutant allele on the bottom homolog. This problem involves mitotic recombination . - <ul><li>a. Will Cell 1 or Cell2 form homozyous mutants at the highest frequency? Explain. Cell 1 will, because Gene A is further from its centromere than Gene C is. Since errors are random, and crossover occurs due to random error, and the larger the region the more likely the error, Cell 1 will become homozygous a-a- more frequently than Cell 2 becomes c- c-. </li></ul><ul><li>A new Cell 3 is like Cell 1 except that it has a recessive lethal mutation. d- 10cM to the left of the GeneA+ allele. Where would the D+ allele be in Cell 3? (this should be obvious) </li></ul><ul><li>ah..ah..ah..it will be directly below d- on the other homolog! It’s the allele of d-. See diagram. </li></ul><ul><li>Will Cell 1 or Cell 3 give the most viable a-/a- mutants? Explain. This was tricky. See diagram below. It turns out, when you draw out the options, Cell1 and Cell3 give the same number of viable a-a- cells! In cell 1, there are 50cM of DNA, anywhere in that region to form an a-a- cell. In Cell 3, a CO between the centromere and the D locus gives a-a-D+D+, and between the D locus and A locus gives a-a-D+d-! Since d- is recessive lethal, the D+d- cell is alive and happy. [ Consider what the result would be if d- were linked to a- instead (say in Cell 4)! Then, cell 4 generates fewer a-a- viable cells, because a CO between the centromere and the D locus forms a-a- d-d-, and therefore dead cells.] </li></ul>Cell 1 Cell 2 + d- a- alleles + + 1 2 3 4 (each line is a double strand duplex!) Consider crossover here or here..then cosegregate centromeres 2 and 4! Gene C + - Gene A+ + d- D+ D+ D+ Gene A d-
5 . The mutational spectrum found in cancer cells of the p97 protein and its domains is shown. (There are 4 “hotspots” in the DNA binding domain.) p97 mutant cells get cancer at a high rate. Amino Acid# 1 150 250 350 390 450 545 Transcriptional Activation DNA binding Trimerization domain <ul><li>What does “hot spot” mean in this diagram? These are amino acids that when changed inactive the p97 protein. All bases are mutated at similar frequencies, but when these particular ones are changed, the resultant protein is inactive. </li></ul><ul><li>The few mutations in the rectangle and circle domains were usually nonsense or frameshift mutations. Why might these mutations lead to mutant p97 while missense mutations in these domains do not typically lead to mutant p97? It may not matter as much what the amino acids are in these regions, rather that they are translated. Frameshift or nonsense mutations lead to failure of the ribosome to translate the rest of the sequence 3’ of the mutation…leading to a drastic change in the protein. </li></ul><ul><li>C. Would would the molecular consequence and phenotype be of a nonsense mutation in codon 5? Failure to translate essentially the entire protein, with the phenotype that the cells will because cancerous. </li></ul>Frequency Of mutations
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
8. Below are two genes that are present in a diploid cell on different chromosomes. Rare translocations occur between Gene A site 1 and Gene B site 3, or between GeneA site 2 and Gene B site 3. Site 1, 2 and 3 are all in exons. The recombination occurs within runs of polyA at each site. The length of As is show. One example of a site 1-site3 translocation is shown. <ul><li>Gene A is normally induced by DNA damage. Will the GeneA-GeneB fusion gene show in the diagram now be induced by DNA damage? Yes, because the DNA-damage inducible enhances, represented by p53 site, is in the fusion. Many did not know this because we have not discussed p53. FYI..p53 is an “enhancer”--it induces transcription of genes after DNA damage. </li></ul><ul><li>Only 1 in 3 translocations between Site 1 and Site 3 leads to an active GeneB protein fusion. Explain. Very tricky! Recombination occurs between the two runs of As. Each recombination event may form a product,and each product may have different numbers of As in the product (either 40As, in the example given, or 30As, or 33As, or a large number of options!). Once you understand that…then consider that the reading frame is critical to have a protein fusion produced. There is only one correct reading frame for the ribosome…so 1 in 3 fusions will have the “right number” of As to keep the frame correct!. </li></ul><ul><li>c. The black exon in Gene B has Gene B’s main function. Would you expect recombination between introns I-A and I-B to necessarily generate an active B protein fusion? Discuss . </li></ul><ul><li>It depends on whether Exon 1 in Gene A is inframe with exon 2(black) in Gene B; after the fusion gene is transcribed and spliced, if the ORFs in the exons are in fact in frame, an active B protein will be made . </li></ul>I-A I-B TATA Box TATA Box A30 +1 p53 site A50 A5 +1 p53 site A TATA Box +1 40 Site 1 Site 2 Site 3 GeneA GeneB GeneA-B