Study Guides (248,551)
Canada (121,622)
Biology (1,112)
BIOL 308 (101)

Updated SQ 5-7.docx

5 Pages
Unlock Document

BIOL 308
Dragana Miskovic

STUDY QUESTIONS: LEC 5-7 1. How does complexity of bacterial genome differ from that of eukaryotic (calf) genome? Bacterial genome has more unique sequences while eukaryotic genomes have a lower proportion of unique sequences, and a large amount of highly repetitive sequences. However, calf genome is more complex as it has a greater amount of unique DNA overall. 2. Explain C value paradox. C value is the DNA content of a haploid cell. There is no correlation between the amount of DNA (size of genome) and the apparent complexity of organisms. 3. List and briefly explain factors that influence DNA renaturation kinetics.  Size of the DNA fragment (indirect)  the larger the size, the harder it is for complementary bases to find each other.  Complexity of genome (indirect)  a more complex genome will have more unique sequences that will renature very slowly as there are few similar copies, and so each strand must find its complement  DNA concentration (direct) at [higher], strands will be able to find each other faster.  Ionic strength (direct)  higher salt concentration will help neutralize –ve charge of DNA and therefore help annealing.  Time (direct) over time, more strands will renature  Temperature (indirect) 20-25 degrees below Tm.  pH  affects H-bonds 4. You have found a new species of insects. To evaluate the complexity of the genome of this species, you isolate genomic DNA from, fragment the DNA to uniform 500 base pair pieces, denature the DNA and measure the rate of reassociation. Your data is represented in the curve below (sorry for the bad drawing): (a) How many classes of DNA (in respect to sequence complexity) are found in this organism? 3; highly repetitive, moderately repetitive, and unique (b) What can you say about the relative complexity of each class? What fraction of the genome falls into each class? Highly repetitive Moderately repetitive Unique >1000 copies >10 copies 1-10 copies Less complex Moderately complex Very complex 25% 25% 50% 5. List three (3) differences between prokaryotic Topoisomerase I and Gyrase. Topoisomerases are enzymes that recognize and regulate supercoiling. They play an important role in replication and transcription. Topoisomerase I Gyrase Makes 1 cut Makes 2 cuts Removes –ve supercoils Introduces –ve supercoils; Relaxes +ve supercoils. L# = +1 L#= -2 Doesn’t use ATP Uses ATP 6. What are topological isomers of DNA? They are identical molecules of DNA that differ only in their degree of supercoiling (from relaxed to tightly supercoiled). 7. Explain the importance of DNA supercoiling for the cell survival? The only way to get the 2.5m long genome in a nucleus if the DNA is highly coiled. It is therefore important for packing of DNA and making it more condensed. It also reduces the induced stress by twisting the double helix. Coiling also prevents access of enzymes to DNA, which prevents unwanted processes from taking place. E.g., if DNA were to be unwinding at all times, DNAP would make multiple copies of the DNA at all times, costing the cell nutrients and energy and therefore, decreasing its viability. 8. What are the differences between primary (or secondary, or tertiary) structures of RNA and DNA? DNA RNA Primary Sugar-phosphate “chain” Similar to DNA except it can exhibit different with purine and pyrimidine bases as side conformations chain(s) SecondaryDouble helical structure (hydrogen Areas of regular helices and discontinuous bonding between A-T and G-C; stacking helices with stem-loops of hairpins. Frequently interactions; phosphate backbone fold back on themselves to form base-paired “outside”) segments between short stretches of complementary sequences. Tertiary Double stranded DNA (both circular AND Formed through interactions of secondary linear) makes complexes with proteins - structures: lack of constraint by long-range supercoil regular helices means RNA has high degree of Supercoiling = coiling of a coil. rotational freedom in backbone of its non-base- paired regions → capable of folding into complex tertiary structures Also, formation of unconventional U:A:U base triple is possible. Pseudo-knots can be formed due to base-pairing between sequences that aren’t adjacent. 9. You have a small 4800 bp long circular DNA. It has a linking number of 450 (L=450). What are the twist (T) and the writhe (W) of this DNA? What assumptions about the structure of the DNA have you made in your answer? (Hint: what is the definition of the twist #?) Ans: T = # of bp / # of bp/turn = 4800/10 = 480 turns L = T+ W W = -30 The DNA has 30 negative supercoils and protein can easily get in and transcribe it Also, the DNA has 480 complete turns assuming the DNA is in B form. 10. Thinking question 1: Many different mutations have been observed in almost all genes. However, only few have been isolated in histones. How would you explain this finding? Mutations that survive to be passed on to the next generations are in general favoured by natural selection, however, histone proteins are very essential for the cell, and their function is very well carried out by their structure as is. Therefore, natural selection is unlikely to favour mutations in the genes coding for histones and organisms with mutated genes are not likely to survive to pass on their genes. This could have led to the highly conserved nature of the gene sequence of histones. 11. Thinking question 2: would it be an easy task to produce polyclonal antibodies to h
More Less

Related notes for BIOL 308

Log In


Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Sign up

Join to view


By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.

Add your courses

Get notes from the top students in your class.