Lectures 9 and 10

Question 1

Syntenic Genes

Answer 1

Genes located on the same chromosome

Question 2

Calculating Genetic DIstance by Recombination Frequency

Answer 2

Observe the number of recombinant phenotypes with respect to non recombinant phenotypes. The percentage of recombination is expressed in map units or centiMorgans. 1.4% recombinant = 1.4 map units or cM

Question 3

Definition of Linked Genes

Answer 3

Genest in with a recombination frequency less than 50%

Question 4

Unlinked genes

Answer 4

Genes in which the parental types = recombinant types. RF = 50%. Unlinked genes are either on different chromosomes or are so far apart that the recombination frequency reaches 50% (when a crossover event virtually always occurs between the two genes.

Question 5

Reasons for Imprecision in GEnetic Mapping by Recombination Frequency

Answer 5

Double Crossovers: recombinant chromosomes from a double crossover event appear as parental chromosomes for distant genes. Leads to an underestimate of recombination frequency. Difference is crossover probability.

Question 6

Hotspots

Answer 6

Regions of a chromosome in which crossovers are favored.

Question 7

Recombination deserts

Answer 7

Regions of a chromosomes in which there is little crossing over. For example heterochromatic found near the centromere.

Question 8

Three Point Cross

Answer 8

This accounts for double crossovers and provides a more accurate estimate of distance between genes. Recombinants will be observed in which the gene which had its alleles switched must be the gene in the middle of the other two genes. Allows a determination of which of the three genes in in the middle by identifying the rarest recombinants.

Question 9

Restriction Endonucleases

Answer 9

Enzymes that cleave DNA at sequence-specific sites. Involved in bacteria as an antiviral defense. Cleave phosphodiester bond between adjacent nucleotides. Generally cleave both strands. Cleavage may depend on whether nucleotides are methylated. Each restriction endonuclease has a characteristic recognition sequence and cut site

Question 10

Sites in DNA at which a restriction endonuclease cleaves is determined by:

Answer 10

The enzymes recognition sequence (commonly 4-12 bp) Specific nucleotides between which the cleavage occurs (commonly within the recognition sequence) Commonly have palindromic recognition sequences.

Question 11

Blunt ends

Answer 11

Cuts on two DNA strands are directly opposite; no single-strand overhangs left on digestion fragments.

Question 12

Sticky Ends

Answer 12

Cuts are not opposite; generates digestion fragments with single-stranded overhangs = sticky ends. Sticky ends have either 5’ or 3’ overhangs. Sticky ends available for base pairing with complementary sticky ends.

Question 13

Digestion of Complex DNA: Frequency of cutting, ave length.

Answer 13

Larger recognition site, larger fragment. Rsal enzyme has a recognition sequence of GATC. This sequence will occur every 4^4 base pairs or 256 bp. Avg restriction fragment size will be 256 bp.

Question 14

Gel Electrophoresis

Answer 14

Takes advantage of the fact that DNA molecules are negatively charged. Utilizes a porous slab or thin layer of gel which DNA migrates. This technique is utilized for: separating DNA fragments in a sample by size, estimating the size in bp of DNA fragments, visualizing DNA in a sample.

Question 15

Determining the size of DNA fragments by gel electrophoresis

Answer 15

Run a standard with known-sized DNA fragments and compare to sample.

Question 16

Agarose gel vs polyacrylamide (PAGE) gell.

Answer 16

Polyacrylamide offers much higher resolution than agarose gels. DNA fragments differing by only one base pair can be separated in PAGE. Also used for proteins. A
Agarose: less hazardous, easier and quicker to pour and manipulate, quicker to run, lower resolution.