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Genetics > Ch. 5 Chromosome Mapping, Linkage, Crossing Over: The Chromosomal Basis of Mendelism > Flashcards

Ch. 5 Chromosome Mapping, Linkage, Crossing Over: The Chromosomal Basis of Mendelism Flashcards

1
Q

1833 Roberts Brown

A

describes the nucleus as a feature of all cells (Eukaryotes)

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2
Q

1866 Ernest Haeckel

A

argues that the nucleus is the bearer of the genetic material

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3
Q

1866 Mendel

A

publications do not receive much attention

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4
Q

1869 Freidrich Meischer

A

isolated DNA, and names it nuclein

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5
Q

1876 Oscar Hertwig

A

identifies meiosis in Sea Urchin eggs, describing the formation of 3 polar bodies and an egg.
(microscope pics clearer)

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6
Q

1879 Walther Flemming

A

describes mitosis in stained living salamander cells

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7
Q

1883 Edouard van Beneden

A

describes the behavior of chromosomes during meiosis, and notes a reduction in chromosomal number

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8
Q

1888 Heinrich Wilhelm Gottfried Waldryer

A

names the mitotic threads “chromosomes” (“colored beads”)

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9
Q

Theodor Boveri

A

First descriptions of meiosis; reduction of chromosome number in Ascaris

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10
Q

1903 Walter Sutton

A

Are chromosomes the “units” Mendel mentioned?

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11
Q

The chromosomal theory of inheritance

Walter Sutton, 1902-1903

A

Measured the 22 chromosomes of the lubber grasshopper, and found 11 specific sizes (homologous chromosome pairs).
Followed the chromosomes through cell division, and observed that they paired up in the first meiotic division.
Gametes only count 11 chromosomes.
Upon fertilization the chromosomes pair again.
Concluded that chromosomes exist as identical pairs.

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12
Q

Chromosomal Theory of inheritance

A

(was developed by several scientists over decades)

  1. All genes are located on chromosomes
  2. Mendel’s principles can be explained by the transmission of chromosomes during reproduction
  3. The meiotic behavior of chromosomes is the base for Mendel’s principles of Segregation and Independent Assortment
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13
Q

The relationship between alleles and homologous chromosomes

A

Homologous chromosomes carry alleles for the same genes at the same positions along their lengths.

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14
Q

The law of segretation

A

depends on the separation of homologous chromosomes in anaphase 1

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15
Q

The law of independent assortment

A

depends on alternative orientation of chromosomes in metaphase 1

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16
Q

Proofing that genes are located on chromosomes

A

By 1910 Sutton, Boveri, and many other biologists suspected that “genes” were situated on chromosomes, but definitive proof was missing
Requirements:
A certain gene needs to be unambiguously lined to a certain chromosome.
The gene must be recognized through a mutant allele and the chromosome must be distinguished morphologically.
Patterns of gene transmission have to reflect chromosome behavior during reproduction.

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17
Q

Thomas Hunt Morgan

A

Geneticist,

identification of genes

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18
Q

Edmund Beecher Wilson

A

Cytologist,

movement of chromosomes

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19
Q

Thomas H. Morgan introduced new model Drosophila melanogaster (fruit fly)
Reasons:

A

Flies are one of the most popular genetics model organisms.
Small size
Easy to maintain
Rotten banana and mild jars, fly formula
Reproduce very fast
Produce lots of offspring
Bred many flies in the hopes of finding a mutant fly which he could unambiguously link to a specific chromosome.

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20
Q

First mutant fly

A

1910 - Morgan’s wife, Lillian and his undergraduate student Calvin Bridges found the lab’s first mutant fly.
Its phenotype was white eyes instead of the wild type red eyes.

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21
Q

Morgan’s first experiments

A

Only males have white eyes - The gene for eye color must be located on the sex chromosomes.
W- w on the x chromosome and a y chromosome = white eyed male

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22
Q

Cross between a heterozygous female and a hemizygous mutant male

A 
w+w x w
w+w red eyed female
ww white eyed female
w+ red eyed male
w white eyed male
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23
Q

Cross between a homozygous mutant female and a hemizygous wild-type male

A 
ww x w+
F1 ww+ x w
F2 
ww
ww+
w
w+
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24
Q

Morgan showed that the gene for eye color was on?

A

The X chromosome in Drosophila.
With that he was bale to correlate the inheritance of that gene with the transmaission of the X chromosome during reproduction.
1933 - Morgan, his wife and his students: Nobel prize for their evidence for the chromosomal theory of inheritance

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25
Q

Morgan challenged Mendel’s law of independent assortment

A

Hypothesis: 2 genes only assort independently if they are NOT on the same chromosome
Proved the hypothesis to be true.
Byproduct, his student Sturtevant discovered how to map genes on a chromosome.

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26
Q

Bateson and Punnett’s experiment with sweet peas

A

Phenotypes in the F2 generation did not correspond to the expected 9:3:3:1 ratio (assuming Mendel’s principle of independent assortment) but a 24.3: 1.1: 1: 7.1
Parental types overrepresented. (look similar to P generation)
Nonparental types: underrepresented. (new combinations)

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27
Q

Genes on the same chromosomes are inherited together

A

They do NOT assort independently.
They are linked: linkage.
All genes on a chromosome are a “linkage group” - inherited together.
Alleles of chromosomally linked genes can be recombined by crossing over.
Results in recombination: non-parental types
The non-parental types are rarer, bc crossing over is a rare event.
The frequency of non-parental types depends on the frequency of recombination between the two genes.

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28
Q

Independent assortment

A

2 genes on 2 different homologous pairs of chromosomes
4 possible gametes with equal ratio (1:1:1:1)
Mendel’s law of independent assortment works here

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29
Q

No independent assortment

A

Linkage: 2 genes on a single pair of homologs; no exchange occurs (no crossover).
2 genes on one chromosome; no crossover.
2 possible gametes: Parental types only.
Linkage: 2 genes on a single pair of homologs; exchange occurs between 2 non sister chromatids.
Crossover.
4 possible gametes: Parental and recombinant types of unequal ratio (not 1:1:1:1).

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30
Q

Color gene and length gene are linked (located on the same chromosome)

A 
The heterozygote F1 plants form 4 diff kind of gametes: Nonrecombinant gametes (parental; more common)
Recombinant gametes (nonparental; through crossing over; less common)
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31
Q

Recombination frequency formula

A

Number of recombinant offspring/ total number of offspring

32
Q

Recombination frequency

A

For any 2 genes, recombination frequency can never exceed 50% (50% or more you would have 1:1:1:1 ratio).
50% would mean the 2 genes are on 2 different chromosomes and assort independently.
Frequency <50% implies that 2 genes are on the same chromosome.
The smaller the recombination frequency, the less common is a crossover event between the two genes.

33
Q

Crossing over happens when?

A

During prophase 1 of meiosis.

34
Q

Tetrad

A

homologous pair of chromosomes with a total of 4 chromatids. (usually cross over happens between tetrads)

35
Q

Frequency of cross over event is determined by?

A

How much space there is between the 2 genes.

Close together - cross over is less likely

36
Q

Multiple crossovers

A

Even less frequent.
Not all have genetic significance.
Double crossover, triple crossover, quadruple crossover - genetic significance.
Cross over between sister chromatids - no genetic significant (bc they are identical)

37
Q

Chromosome Mapping

A

One night in 1911, undergrad student Alfred “Hot dog” Sturtevant studied outcomes of Drosophila crosses.
By morning he had constructed a map showing the position of genes on the chromosomes.
Using Sturtevant’s method, Morgan and colleagues were able to determine which genes are located on which chromosomes.
They drew maps implying that a chromosome is a linear array of genes.

38
Q

Recombination Frequency and linkage maps

A

Genes that are closer to each other on the same chromosome have a lower recombination frequency (because crossover is less likely to happen)

39
Q

Recombination frequency between 2 genes is directly proportional to?

A

The distance between the 2 genes

40
Q

Map units: centiMorgan

A

Recombination frequency =0.18 or 18%
18 map units (mu), or 18 centiMorgans (cM)
100 centiMorgans = 1 Morgan (M)
1 cM = 1% recombination

41
Q

Puzzle work

A

Create a linkage map, showing the map unit distances between loci

42
Q

Collecting Recombination frequency data

A

Needed to calculate the genetic distance between gene loci on a chromosome.
2 point test crosses: for pairs of genes.
3 point testcrosses: for more than 2 genes.

43
Q

Testcross

A

Cross between two parents, one is heterozygote for all traits, the other one is homozygote recessive for all traits.
Ex: Long wings gray body (vg b/ ++)
Vestigial wings black body (vg b/ vg b)

44
Q

Recombination mapping with a 2 point cross

1) are wing gene and body color gene linked?
2) What is the frequency of recombination?
3) what is the distance between the 2 genes

A

1 - yes! how do you know? recombinants<50% (higher than 50% genes on 2 separate chromosomes and assort independently)
2 - 180/ 1000 = 0.18
3 - 18% or 18 cM

45
Q

Linkage phases

A

With 2 genes on the same chromosome 2 different kinds of heterozygotes are possible
Coupling heterozygote
Repulsion heterozygote

46
Q

Coupling heterozygote

A

Heterozygote with dominant alleles on one chromosomes (cis)
RL (vg b)
rl (++)

47
Q

Repulsion heterozygote

A

Heterozygote with mixed dominant/recessive alleles on one chromosome (trans)
Rl (vg +)
rL ( + b)

48
Q

Looking at the linkage phase of the parents helps to figure out?

A

Which phenotypic classes of F2 are the parental ones and which are the recombinant ones.

49
Q

How to determine the arrangement of the genes in the heterozygous parent - cis or trans?

A

See which one - cis or trans - makes the parental classes more frequent and the recombinant less frequent

50
Q

If there are more than 2 genes linked, how do we know the order of the genes on the chromosome?

A

Requires the analysis of multiple crossovers.
Double crossover - all genes must be heterozygous with 2 alleles each. The gene in the middle gets switched.
(If you known which gene gets switched, you known which gene was located in the middle)

51
Q

3 point mapping

3 criteria?

A

One parent must be heterozygous for all 3 genes under consideration, the other one homozygote recessive (testcross).
Phenotypic class must reflect genotype of gametes of parents.
Sufficient number of offspring must be produced for representative sample.

52
Q

3 point mapping
What are the cis/trans arrangements of the genes?
Regarding the arrangement of genes, which one is in the middle?

A

Compare the parental types with the double crossover types: the gene in the middle has “switched” places (which gene differs?)

53
Q

3 point mapping

What are the map distances between the 3 loci?

A

ABC
Region 1 is A-B, Region 2 is B-C
Pick out all of the recombinant offspring where a crossover event happened in region 1, then do the same for region 2

54
Q

Mapping estimates can be inaccurate

As the distance between 2 gene loci increases, inaccuracy?

A

Inaccuracy of the estimates increase.

This is bc double (four, six, and so on) crossovers are not always detected.

55
Q

Interference and Coefficient of Coincidence

Do crossovers in one region occur independently of a crossover in another region?

A

Cross over events in 1 region do not necessarily happen independently of a cross over even in another region

56
Q

To calculate frequency of double crossover?

A

multiply frequency of crossover for each region
Assuming independence, Expected frequency 0.091 x 0.105 = 0.0095.
The observed frequency of double crossovers was 2/3248= = 0.0006
Interference happened.

57
Q

Interference

A

A crossover in one region inhibited a crossover nearby

58
Q

Map distance

Why does counting not match recombination frequency?

A

Map distance sc-f calculated through adding up distances between genes: 66.8cM.
Recombination frequency sc-f: 50% (50cM)
Why does it not match? Undetected cross over events

59
Q

The true genetic distance depends on?

A

The average number of crossovers on a chromosome. (may not be able to count this bc you would have to take account for all cross overs.
Multiple crossovers can occur between 2 genes far apart.
Some of these crossovers do NOT produce recombinant chromosomes.
Ex: double crossover may not contribute to the recombination frequency, but it contributes to the average number of exchanges on a particular chromosome. (cross over would go undetected, but would have to be included in average)

60
Q

Recombination Frequency and Genetic Map Distance

Summary

A

Recombination frequencies can be used to infer genetic maps.
We can estimate where crossover happened.
With that we can estimate the distance between genes and map them. BUT this only works as long as the genes are close together.
When they are very far apart, recombination frequency may not reflect the true map distance. WHY?
Bc of undetected multiple crossovers between 2 genes far apart.

61
Q

Distance of 20 cM or less

A

recombination frequency is a good estimator of the true genetic distance

62
Q

Distances about 20 cM

A

discrepancies between recombination frequency and map distance are getting worse, bc multiple crossover events are more likely.

63
Q

Recombination under genetic control

Male fly as the homozygote recessive testcross partner

A

Male Drosophila do not undergo genetic recombination in any of their chromosomes.
This fact greatly facilitated mapping experiments bc recombination only needs to be considered in one partner.

64
Q

Recombination events are dependent on

A

chromosome length, position of centromere, species dependent, and many other factors
(larger chromosome - more cross over can happen)

65
Q

Linkage maps (genetic map)

A

A map of the locations of polymorphic markers where order and distance is determined by recombination frequency (measured in map units, cM)

66
Q

Physical maps

A

A map of the locations of identifiable landmarks in the genome; there are many different types of landmarks. (measured in base pairs of the actual DNA sequence)
(can show you more than just the relative positions of genes)

67
Q

Physical maps

Resolution high to low

A

Cytogenetic (chromosome map) - based on distinctive banding patterns observed in stained chromosomes.
DNA map - locations of expressed DNA along the genome.
Radiation Hybrid map - order of DNA markers (STS) that uniquely occur in the genome.
Contig Map - order of overlapping DNA fragments spanning the genome.
Restriction Map - Describes the order and distance between DNA restriction enzyme sites.
Sequence Map - The complete DNA sequence of a genome.

68
Q

Cytogenic maps

A

Recombination frequencies can be used to estimate gene sequences on chromosomes (these are relative positions)
BUT: does not localize genes in respect to cytological landmarks, such as the centrosomes, p & q arms, and the banding patterns of chromosomes
(characteristic banding pattern that are named)

69
Q

Mapping of Human chromosomes

A

Large scale mating experiments are neither feasible nor ethical.
Ways of localizing genes on human chromosomes?
1950s: studying linkage through pedigree analysis.
X linked genes are the easiest to identify.
But genes which do not show x-linage could be on any of the 22 autosomes (wouldn’t know where it was located)

70
Q

1950s: Linage and pedigrees

A

Renwick and Lawler 1955: evidence for linkage between the gene controlling the ABO blood groups and the nail-patella syndrome (dominant mutation, autosomal disorder: affected people have abnormal nail and kneecaps)

71
Q

Gene for blood type and dominant mutation for Nail patella syndrome (NPS1)

A

are on the same chromosome.

Combining data from many pedigrees, the genetic distance between the ABO and NPS1 loci is 10 cM.

72
Q

Male pattern baldness (androgenic alopecia)

A

Defined as beginning balding by age 30. More than 40 million men in the US are affected. The trait is sex-influenced and stimulated by androgens (male sex hormones). Common folklore says that if a young man wants to know whether he will become bald, he should look at his mother’s father. With this background information, would you be able to localize the gene? On which chromosomes? And where on the chromosome?

73
Q

Gene location for androgenic alopecia

A

on X chromosome, position p12-22

74
Q

1980s: Detecting linkage with molecular markers

A

Until the early 1980s progress in human gene mapping was slow.
Technical progress: recombinant DNA technology, molecular markers, sequencing DNA and genomes.
It became possible to identify genetic variants in the DNA.
Variants results from differences in the DNA sequence itself.
Using molecular markers (which pick up these differences) combined with pedigree analyses made it possible to map large number of genes involved in humans disease.

75
Q

DNA polymorphisms are genomic landmarks

A

“Mile markers” throughout the genome.
We don’t know where the gene for our trait of interest lies but, if we can show that our trait is linked to a DNA polymorphism (landmark).. we’d know roughly were the gene is located

76
Q

Genome Wide association mapping (GWAS)

A

DNA markers (SNPs or microsatellites) are found throughout the genome and used as landmarks.
If a particular SNP or microsatellite is close to a gene, they are likely to be inherited together.
Similar, if a marker and a gene are far apart, they are more likely to be broken up by recombination.
GWAS studies try to find associations between these markers and genes.
EX: cystic fibrosis CFTR locus on chromosome 7

77
Q

Sequence Maps: Human genome project

A

2000s: Due to improvement in sequencing techniques, the human genome has been sequenced.
Sequence maps are fine scale maps.
The specific location of genes on a chromosome and the proximity of genes and noncoding sequences to each other, among other details.
Physical map: distance between genes on a chromosome is measured in base pair distances.

Genetics (15 decks)

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