GED L13 Notes Flashcards by Jade Benita McLaughlin

What does a gene map illustrate?

• Gene Map illustrates:

Relative order -> genes on chromosome
Distance between genes.

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Name the types of gene map.

Physical
Cytogenic
Linkage

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What do physical gene maps illustrate?

Physical maps:

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

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What do cytogenic gene maps illustrate?

Cytogenetic maps:

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

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What do linkage gene maps illustrate?

Linkage maps:

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

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Describe physical gene maps.

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.

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Describe cytogenic gene maps.

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

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Describe g-banding on human cytogenic gene maps.

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

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Describe human chromosome 7 on human cytogenic gene maps.

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

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What are the uses of gene maps?

• 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.

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Describe genetic linkage

• 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.

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Describe the linkage-mapping principles

• 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.

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Describe characteristics of recombinant gametes

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

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What recombination frequency does independent assortment cause?

 Independent assortment -> genes on diff. chromosomes

|&raquo_space; produce 50% recombinant gametes

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What causes a 50% recombination frequency?

 Independent assortment -> genes on diff. chromosomes

|&raquo_space; produce 50% recombinant gametes

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What does a 50% recombinant gamete indicate?

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

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

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

What is a testcross?

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

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

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

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

 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.

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

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

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

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%)

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

> > 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.

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

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

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

• Genetic Distance = Recombinant Frequency (RF)

What is the eqn for calculation of recombinant frequency?

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

• “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. 

 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+

 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

 RF= [(151 + 154) x 100] / [2839] = 10.7% | >>Map distance = 10.7cM

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

 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+

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+

 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

Why are genetic distances not exactly additive?

1. Multiple crossover events underestimate true distance over large distances 2. Crossover events not always independent Positive / negative inference (Crossover in one region -> influence crossover in adjacent region) 3. Frequency -> crossing over varies -> diff. regions of genome

Describe the reliability of map distance additivity in relation to distance

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