ch 4 mapping eukaryote chromosomes by recombination

Question 1

linkage

Answer 1

the association of two genes on the same chromosome

Question 2

coupling configuration

Answer 2

arrangement of linked alleles in a dihybrid such that the two dominant alleles are on a chromosome, the two recessive alleles are on the other chromosome (AB/ab)

Question 3

repulsion configuration

Answer 3

arrangement of linked alleles in a dihybrid such that a dominant and a recessive allele are together on each chromosome (Ab/aB)

Question 4

cis dyhybrid

Answer 4

a dihybrid in which the alleles are coupled (AB/ab)

Question 5

trans dihybrid

Answer 5

a dihybrid in which the alleles are in repulsion (Ab/Ba)

Question 6

two process that generate recombinant products

Answer 6

1. independent assortment of genes on different chromosomes
2. crossing-over between genes on the same chromosome

Question 7

what regions have higher recombinant frequencies

Answer 7

longer regions

Question 8

map unit or cM

Answer 8

a recombinant frequency of 1%

Question 9

why map the relative position of genes

Answer 9

calculated map distance is correlated to physical distance on a DNA molecule
- knowing the position of a gene on a chromosome allows us to find its sequence
- leads to an understanding of its structure and function

Question 10

map distance formula

Answer 10

precent recombinant progeny
= # recombinant progeny/total progeny x 100%

Question 11

recombination frequency never exceeds

Answer 11

50%

Question 12

3-point test cross

Answer 12

mapping 3 genes relative to one another

Question 13

rules for a 3 point test cross

Answer 13

1. there should be 8 phenotypic classes
2. progeny classes are grouped in pairs = reciprocal cross over events
3. largest classes are the parentals
4. smallest classes are the double cross over classes

Question 14

interference

Answer 14

a measurement of the independence of two cross over events (does one cross over interfere with the possibility of another cross over)

Question 15

interference formula

Answer 15

interference = 1 - (observed DCO/(p(SCO I) x p(SCO II) x total number of progeny))

Question 16

negative interference

Answer 16

more double cross overs than expected (cross over in one region promotes cross overs in another region)

Question 17

positive interference

Answer 17

(usually)
fewer double cross overs than expected (cross over in one region inhibits cross overs in another region)

Question 18

spores: division I - separation of

Answer 18

homologues

Question 19

spores: division II - separation of

Answer 19

sister chromatids

Question 20

distance between 2 points =

Answer 20

% meiotic products showing recombination between the 2 points
the frequency of crossing over between any two points on a chromosome is proportional to the distance between those two points
1/2 MII octads / total x 100%

Question 21

octad

Answer 21

an ascus containing eight ascospores, produced in species in which the tetrad normally undergoes a postmeiotic mitotic division

Question 22

first division segregation

Answer 22

different alleles go into different nuclei at the first meiotic division producing an M1 division pattern of ascospores (4:4)

Question 23

second division segregation

Answer 23

different alleles go into different nuclei at the second meiotic division producing an MII division pattern of ascospores (2:2:2:2)

Question 24

molecular marker

Answer 24

a site of DNA heterozygosity (difference), not necessarily associated with phenotypic variation, used as a tag for a particular chromosomal locus

Question 25

single nucleotide polymorphism (SNPs)

Answer 25

a nucleotide pair difference at a given location in the genomes of two or more naturally occurring individuals
- changes to a single nucleotide
- can be located within a gene or intergenic regions

Question 26

simple sequence length polymorphisms (SSLPs)

Answer 26

the presence of different numbers of short or simple repetitive elements at one particular locus in different homologous chromosomes
- become subject to expansion or deletion

Question 27

three methods of SNPs detection

Answer 27

1. DNA sequencing - not always feasible as prior knowledge of the sequence is required
2. RFLP analysis - restriction fragment length polymorphism
3. CAPS analysis - cleaved amplified polymorphic sequences

Question 28

RFLP analysis

Answer 28

restriction fragment length polymorphism
- restriction enzymes cleave DNA at a specific sequence
- detected by a Southern blot
- EcoR1 recognizes and cuts a specific sequence; but if the sequence is mutated, it will not cut

Question 29

restriction enzyme

Answer 29

an enzyme that will recognize specific target nucleotide sequences in DNA and break the DNA chain at those points

Question 30

4 steps of Southern blotting

Answer 30

1. separate genomic DNA (cute with the restriction enzyme) via agarose gel electrophoresis
2. DNA is transferred from the gel to a membrane
3. expose the membrane to a labeled probe (P-32), complimentary to the region of interest; short DNA sequences of P-32 will stick to regions we are interested in following
4. bound, radioactively labeled probe will expose x-ray film, and identify fragments to which the probe was bound

Question 31

3 steps of CAPS analysis

Answer 31

1. primers flanking a polymorphic R.E site are used to amplify region by PRC
2. amplified region is then cut by R.E
3. fragments are separated by electrophoresis
   (less DNA needed and eliminates Southern blotting)

Question 32

haplotypes

Answer 32

a genetic class described by a sequence of DNA or of genes that are together on the same physical chromosome (a chromosome segment defined by the array of markers it carries)

Question 33

microsatellites

Answer 33

very short sequences of repetitive DNA <10nts
different alleles due to variable number of repeats

Question 34

detection of microsatellites

Answer 34

use PCR to amplify the region containing the microsatellite and visualize fragments using agarose gel electrophoresis

Question 35

minisatellites

Answer 35

short sequences of repetitive DNA, 15-100nts
different alleles due to variable number of repeats

Question 36

detection of minisatellites

Answer 36

cut with R.E outside the minisatellite and southern blot using specific probe

Question 37

minisatellites can be used to generate a

Answer 37

DNA fingerprint

Question 38

microsatellites can show a linkage to a

Answer 38

disease gene