GED L13 Notes Flashcards

Question 1

Q

What does a gene map illustrate?

Answer

A

• Gene Map illustrates:

Relative order -> genes on chromosome
Distance between genes.

Question 2

Q

Name the types of gene map.

Answer

A

Physical
Cytogenic
Linkage

Question 3

Q

What do physical gene maps illustrate?

Answer

A

Physical maps:

Illustrate distances between genes / DNA markers -> direct measurement of DNA

Question 4

Q

What do cytogenic gene maps illustrate?

Answer

A

Cytogenetic maps:

Indicate positioning of genes in relation to cytogenetic markers (Banding patterns)

Question 5

Q

What do linkage gene maps illustrate?

Answer

A

Linkage maps:

Illustrate relative positioning of genes / markers on chromosomes -> meiotic recombination frequencies (centiMorgan, cM)

Question 6

Q

Describe physical gene maps.

Answer

A

Physical maps:
Illustrate distances between genes / DNA markers -> direct measurement of DNA
 1st maps made -> restriction enzyme map
Made using restriction enzymes -> cut DNA at specific sites.
 Human Genome Project -> collection of DNA seq.

Question 7

Q

Describe cytogenic gene maps.

Answer

A

Cytogenetic maps:
Indicate positioning of genes in relation to cytogenetic markers (Banding patterns)
 Human cytogenic maps:
G-Banding:
Mild proteolytic digestion w/ Gimesa
-»Characteristic patterns of light (G-Light) & dark (G-Dark) bands appear.
>Each band assigned specific number.
Human Chromosome 7:
Genes assigned -> Short p (petit) or long q arm
-> Region eg. q3
-> Band eg. q31
-> Sub-band eg. q.31.2
Eg. CF -> mutated gene -> cystic fibrosis -> 7q31.2

Question 8

Q

Describe g-banding on human cytogenic gene maps.

Answer

A

G-Banding:
Mild proteolytic digestion w/ Gimesa
-»Characteristic patterns of light (G-Light) & dark (G-Dark) bands appear.
>Each band assigned specific number.

Question 9

Q

Describe human chromosome 7 on human cytogenic gene maps.

Answer

A

Human Chromosome 7:
Genes assigned -> Short p (petit) or long q arm
-> Region eg. q3
-> Band eg. q31
-> Sub-band eg. q.31.2
Eg. CF -> mutated gene -> cystic fibrosis -> 7q31.2

Question 10

Q

What are the uses of gene maps?

Answer

A

• Uses of gene maps:

Identification -> genes responsible for diseases / traits -> positional cloning.
Aid in design & analysis of experiments studying gene function.
Effectively combine economically important traits -> plant & animal breeding.
Comparison -> genome organisation between organisms.

Question 11

Q

Describe genetic linkage

Answer

A

• Genetic Linkage:
- Genes -> same chromosome -> linked
- > May violate Mendel’s 2nd law -> Independent Assortment
- Alleles of genes -> same chromosome
> Segregate together in gametes during meiosis
&raquo_space;Unless crossing-over between them occurs.

Question 12

Q

Describe the linkage-mapping principles

Answer

A

• Linkage-mapping principles:
- Crossing-over -> random positions -> chromosomes
Eg. Humans -> 1-2 crossings each chromosome arm -> per meiosis
- Frequency -> crossing-over between 2 gene loci
> Proportional -> physical distance between them on chromosome.
&raquo_space; Measuring frequency -> crossing-over between 2 genes therefore indicates
measure of distance between them.

Question 13

Q

Describe characteristics of recombinant gametes

Answer

A

• Frequency -> recombinant gametes
 Proportional to frequency of crossing-over & distance apart on same chromosome.
 Independent assortment -> genes on diff. chromosomes
» produce 50% recombinant gametes

Question 14

Q

What recombination frequency does independent assortment cause?

Answer

A

 Independent assortment -> genes on diff. chromosomes

|&raquo_space; produce 50% recombinant gametes

Question 15

Q

What causes a 50% recombination frequency?

Answer

A

 Independent assortment -> genes on diff. chromosomes

|&raquo_space; produce 50% recombinant gametes

Question 16

Q

What does a 50% recombinant gamete indicate?

Answer

A

 50% recombinant gametes indicate
-» Genes -> diff. chromosomes
OR
-» Genes long distance apart -> chromosome

Question 17

Q

What does a gamete with <50% recombinant DNA indicate?

Answer

A

 <50% recombinant gametes indicate
-» Genes linked -> Same chromosome
OR
-» Smaller recombination frequency -> Closer genes.

Question 18

Q

What is a testcross?

Answer

A

 Testcross:
 Genetic cross -> genotypes of gametes determined from phenotypes of
offspring produced.

Question 19

Q

Why was testcross linkage mapping invented? / What was the problem with linkage mapping?

Answer

A

 Can’t easily & directly determine genotypes -> gametes.
 Invention of testcross

Question 20

Q

Describe how the distance between genes on the same chromosome is calculated.

Answer

A

 Determination -> distance between genes -> same chromosome:
1)
- Produce double heterozygote
AA BB x aa bb
-» F1 -> Aa Bb -> heterozygote
-» A & B -> one chromosome / a & b -> homologous chromosome.
- Gametes of F1:
-» No crossover -> AB & ab parental gametes
-» One crossover
-> Ab & aB recombinant gametes (50%)
&
-> AB & ab parental gametes (50%)
-> Relative frequencies enable calc. -> map distance.

        2) -	Cross heterozygote x tester strain (homozygous -> both recessive alleles) o	Tester strain -> homozygous  >>One type gamete produced (ab) &amp; any crossing over -> no effect -> 
 genotype of gametes.  >>Double recessive chromosome -> F2 phenotype always corresponds -> 
 genotype of gamete -> double heterozygote.     -> Easy classification of offspring as parental or recombinant with respect to 
   double heterozygote gametes.   F2 Phenotype -> Genotype -> gamete -> double heterozygote.

Question 21

Q

What gametes are produced in the F1 if no cross over occurs:

AA BB x aa bb

Answer

A

AA BB x aa bb

        - >> F1 -> Aa Bb -> heterozygote  - >> A &amp; B -> one chromosome / a &amp; b -> homologous chromosome.  - Gametes of F1: - >> No crossover -> AB &amp; ab parental gametes

Question 22

Q

What gametes are produced in the F1 if one cross over occurs:
AA BB x aa bb

Answer

A

AA BB x aa bb
-» F1 -> Aa Bb -> heterozygote
-» A & B -> one chromosome / a & b -> homologous chromosome.
- Gametes of F1:
-» One crossover
-> Ab & aB recombinant gametes (50%)
&
-> AB & ab parental gametes (50%)

Question 23

Q

Why is a tester strain (homozygous -> both recessive alleles) used in determination of distance between genes on the same chromosome?

Answer

A

> > One type gamete produced (ab) & any crossing over -> no effect ->
genotype of gametes.
Double recessive chromosome -> F2 phenotype always corresponds ->
genotype of gamete -> double heterozygote.
-> Easy classification of offspring as parental or recombinant with respect to
double heterozygote gametes.

Question 24

Q

Describe the resulting F2 generation from an F1 and tester strain cross in calculation of distance between genes on a chromosome?

Answer

A

 F2 Phenotype -> Genotype -> gamete -> double heterozygote.

Question 25

Q

What is the genetic distance equal to / how is it calc?

Answer

A

• Genetic Distance = Recombinant Frequency (RF)

Question 26

Q

What is the eqn for calculation of recombinant frequency?

Answer

A

• RF = [(No. recombinant offspring) x 100] / [Total no. offspring}
 Units -> centiMorgans (cM)
» 10% RF = 10cM

Question 27

Q

• “Two-point” mapping experiment:

pr -> purple eye
Pr+ -> wild type (red eye)
vg -> vestigial wing
Vg+ -> Wild type (normal wing)

F1 genotype -> Pr+Pr+ Vg+Vg+ **
-> But occasional crossing over 
Possible genotypes:
>> Pr+Vg
>> Pr+vg      -> (If crossing occurs)
>> prVg+      -> (If crossing occurs)
>> prvg

Pr+Pr+ Vg+Vg+ ** x prpr vgvg

Describe the subs. offspring produced; theor genotype, phenotype & whether they express the recombinant or parental genotype.



Answer

A

 Pr+Pr+ Vg+Vg+ ** x prpr vgvg
 Offspring:
» Pr+pr Vg+vg -> 1339 -> Phenotype > Pr+Vg+ -> WT -> Parental genotype
» Pr+pr vgvg -> 154 -> Phenotype > Pr+vg -> Vestigial -> Recombinant
genotype
» prpr Vg+vg -> 151 -> Phenotype > prVg+ -> Purple -> Recombinant
genotype
» prpr vgvg -> 1195 -> Phenotype > prbg -> Purple vestigial -> Parental
genotype

 Parental genotypes:
Pr+Vg+ & prvg
 Recombinant genotypes:
Pr+vg & prVg+

Question 28

Q

 Offspring:
» Pr+pr Vg+vg -> 1339 -> Phenotype > Pr+Vg+ -> WT -> Parental genotype
» Pr+pr vgvg -> 154 -> Phenotype > Pr+vg -> Vestigial -> Recombinant
genotype
» prpr Vg+vg -> 151 -> Phenotype > prVg+ -> Purple -> Recombinant
genotype
» prpr vgvg -> 1195 -> Phenotype > prbg -> Purple vestigial -> Parental
genotype

Calculate the Map distance

Answer

A

 RF= [(151 + 154) x 100] / [2839] = 10.7%

|&raquo_space;Map distance = 10.7cM

Question 29

Q

Identify the offspring genotype, parental genotypes present in this offspring and the recombinant genotypes in this offspring.
 Parental cross version 1:
Pr+Pr+ Vg+Vg+ x prpr vgvg

Answer

A

 Parental cross version 1:
Pr+Pr+ Vg+Vg+ x prpr vgvg
-» Pr+pr Vg+vg double heterozygote

           Parental types: Pr+Vg+ and prvg
           Recombinant types: Pr+vg and prVg+

Question 30

Q

Identify the offspring genotype, parental genotypes present in this offspring and the recombinant genotypes in this offspring.

 Parental cross version 2:
Pr+Pr+ vgvg x prpr Vg+Vg+

Answer

A

 Parental cross version 2:
Pr+Pr+ vgvg x prpr Vg+Vg+
-» Pr+pr Vg+vg double heterozygote

           Parental types: Pr+vg and prVg+
           Recombinant types: Pr+Vg+ and prvg

Question 31

Q

Why are genetic distances not exactly additive?

Answer

A

Multiple crossover events underestimate true distance over large distances
Crossover events not always independent
Positive / negative inference
(Crossover in one region -> influence crossover in adjacent region)
Frequency -> crossing over varies -> diff. regions of genome

Question 32

Q

Describe the reliability of map distance additivity in relation to distance

Answer

A

• Map distances are additive over short distances but not long distances.