Transmission genetic Flashcards

1
Q

Mendel’s First Law

A

Alleles will always segregate away from each other into

gametes.

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2
Q

Chromosomal Theory of Inheritance

A

•Chromosomes are not visible during interphase
•Genetically males contribute traits equally with females, yet males only donate a nucleus during fertilization
- chromosomes are found in the nucleus
•The segregation of Mendel’s genetic traits mirror that of chromosome separation at meiosis
-Sex determination had a chromosomal basis

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3
Q

Chromosome of The Fruit Fly - Drosophila

A
  • 3 autosomes

- XX or XY

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4
Q

Crossing the white-eyed variant

A
Female red eye x male white eye
F1: all red eye
F2: Female all red eye, male 1/2 1/2
Female white eye x male red eye
F1: Female red eye, male white eye
F2: both gender 1/2, 1/2
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5
Q

Reciprocal cross

A

A Reciprocal cross is an important genetic test when assessing whether a trait is sex-linked . When performing this cross the phenotypes of the female and male parents are reversed . So if we consider body colour in flies where we have a yellow body mutant, then in one cross we would use body colour females and yellow body colour male . In a second cross we would use wildtype body colour male and yellow body colour female. If the outcome of the two crosses was the same then we would conclude that the trait is autosomal. If the outcome was different then we would conclude that the trait is sex-linked. The flies do not need to be pure breeding in these experiments.

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6
Q

Bridges noticed that

A

●Common F1 flies showing mother-son and father- daughter transmission had standard chromosome configurations
●Exceptional F1 flies showing mother-daughter and father-son transmission had abnormal sex chromosome configurations

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7
Q

Sex-linked traits were consistent

A

Sex-linked traits were consistent with the behaviour of the X chromosome
•Mother to son inheritance
•Father to daughter inheritance

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8
Q

Non-disjunction

A

◆Aberrant segregation of chromosomes during division
●Results in cells with more or fewer chromosomes than the standard set
◆Aneuploids
❖Monosomic (2n-1), trisomic (2n+1)
❖Polyploids
◆Occurs spontaneously at low frequency
◆Can be induced by drugs that affect spindle formation

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9
Q

Non disjuction in action

A
  • can occur in the first or second division of meiosis
  • non-disjuction at first division: AA,aa,Aa
  • Non disjuction at second: AA,aa,Aa,a or A
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10
Q

Non-disjunction in fruit fly

A

White eye female x Red eye male

  • female produce XX and 0 gamte
  • during cross, XXX and Y0 dies
  • X red eye and XXY white eye f survuve
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11
Q

Meiosis and Mendelian Genetics

A

◆Mendel showed traits are inherited in specific ways that are predicable
◆Bridges showed that these traits are on chromosomes
◆The behavior of Mendel’s genes is driven by the process of MEIOSIS which produces gametes.

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12
Q

Mendel’s Second Law

A

Alleles of separate genes will always segregate independently into gametes.
There will be nine possible genotype with 4 pheno type
9:3:3:1

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13
Q

Assumption of independent assortment

A
  • S and Y are dominant
  • No genetic interaction between genes
  • No linkage
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14
Q

Linkage

A

●Genes on the same chromosome are physically linked
●Genes on the same chromosome may be genetically linked
-the frequency of recombinant can use to find out how far two gene are from each other

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15
Q

Why is male fly used in linkage

A

Scoring only males because they are hemizygous
- Examining event that have happened on a single chromosome from the female parent
- Phenotype equals genotype
A test cross can be used to achieve the same thing for autosomal traits

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16
Q

Linkage – Test Cross

A

two pure type is cross with each other to create a heterozygous wild type
the wild type then combine with a homozyous recessive
if the ratio is 1/4 for each off spring, the gene is unlinked. not 1/4 result in link

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17
Q

The relationship between distance and linkage

A
  • the futher the two gene is from each other, the more likely they are to recombine
  • gene that alway recombine with each other is called genetically linked
  • maxium 50%
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18
Q

What is cis and trans link

A

AB|ab is cis Ab|Ba is trans

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19
Q

Map distance calculation form test cross

A

Map distance = 100 x Recombinants / total progeny

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20
Q

Linkage - Drosophila important fact

A

There is NO recombination in males regardless of distance

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21
Q

Effect of multipult cross over

A

Underestimate of the recombination frequency, corss over occur more than once, returning the two gene to their OG position

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22
Q

Three Factor Cross

A
The two most frequent genotype is called parental phenotype, the two least is called recombinant class
the recombinant class is used to infer the order position of two gene (the odd one in the middle)
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23
Q

Three Factor Cross calculation

A
for short strand: find  total of the recombinant class of the two strand/total offspring times 100
for whole strand: sum two short strand or total recombiniation+2(Double recombinant class)
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24
Q

Interference

A

◆Recombination events are not fully independent
●A crossover may inhibit the formation of a second nearby crossover or it may enhance it
●This will affect the frequency of the double crossover class in a three factor cross.
Interference is defined as I = 1 - CC
CC = Coefficient of coincidence
if the answer is positive, one event hinder another
if negative one event boost the other

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25
Q

Coefficient of coincidence calculate

A

Observed DR / Expected DR

expected DR = mu of the two strand : 100 x total off spring

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26
Q

Haploid and diploid in genetic analysis

A

●Mendel’s laws, and the process of recombination, apply equally to eukaryotic haploid organisms that undergo a sexual cycle as they do to diploid eukaryotic organisms because they undergo meiosis.
●In many instances, these haploid eukaryotes are small and have experimental advantages similar to bacteria and bacteriophage for addressing genetic questions.

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27
Q

Generic life cycle of Haploid Eukaryotes

A
  • Vegetative growth/ mitotic growth (n)
  • Mating via cytoplasmic and nuclealar fusion (2n)
  • Transient diploid (2n) undergoes meiosis
  • four haploid (n) spores (tetrad)
28
Q

Genetics of Haploid Eukaryotes

A

◆There are no requirements for a test cross as the progeny are haploid and so the phenotype is the genotype
◆In all other respects the data can be handled as for diploids

29
Q

process of Tetrads development

A
  • First meiotic division
  • Second Meiotic division
  • post meiotic mitotic division
  • development of spore
30
Q

Centromere Mapping of tetraids

A

-Determine the Parental order and the Possible recombinant orders
-total number of recombiniant/ total x 100 x 0.5
◆Distance is recombination frequency (number of recombinant chromatids) not number of recombinant meioses.
◆Although this is a recombinant meiosis only half of the spores contain a recombinant chromosome the other half are still parental. So we have to multiply by 0.5 to get the true recombination frequency.

31
Q

Ordered Tetrads ab lenght

A

sum (amount of a class w/ a-b+ or a+b-(recombination) x their total number)/ total x number of n per class

32
Q

Mapping Function, distance between genes in ordered tetrad

A

Poisson distribution best describes potentially frequent but actually
infrequent events
◆e = natural logarithm base
◆m = mean number of crossovers
RF = (1 - e^-m1 )
2
m = -In (1 - 2RF)
◆Maximum map distance (recombination frequency) which can be measured is 50.
◆Therefore, the corrected map distance is 50 x m

33
Q

Tetrad nomenclature

A

●Asci which contain two genotypes are called ditypes

●Asci which contain four genotypes are called tetratypes

34
Q

Unordered Tetrads

A
  • One cross over: Tetratype
  • Two stranded cross over Tetratype
  • Three stranded crossover Tetratype
  • Four stranded cross over Nonparental di type
35
Q

Map distance for unorders tetrads

A

◆Map distance could be calculated by simple RF using parental and recombinant ascospore genotypes.
◆However, for large distances this is not accurate.
◆The key is the NPD class which can only be produced by a double crossover involving all four chromatids.
Map distance = 50 (T + 6 NPD)
if over 100 the it is unlinked

36
Q

Somatic Cell Genetics

A
  • Fusion between two cell, a human cell(HGPRT-, TK+) and a mouse cell (TK-, HGPRT+) via Virus induced cell fusion
  • the mixture is then placed into a HAT selective medium to kill all non hybrid cell
  • result in an unstable hybrid cell TK+, HGPRT+
37
Q

Somatic Cell is unstable

A

◆Hybrid cells are chromosomally unstable
◆As the hybrid cell divides the human chromosomes are lost in a random fashion until a stable hybrid is produced
◆Stable hybrids contain only a small number of human chromosomes

38
Q

Somatic cell mapping

A

◆Somatic cell genetics can be used to map genes to chromosomes
◆The nature of the variation used in these experiments is usually biochemical and involves ways in which we can identify the protein products of genes.
◆However the variation we are interested in is between human and mouse!
◆Examples of types of variation:
●Electrophoretic variation of proteins
●Antigenic variation in protein

39
Q

Electrophoretic variation of somatic mapping

A

●Electrophoresis involves the separation of proteins through a gel matrix (agar/acrylamide gel) under the influence of a voltage potential. The proteins are separated on the basis of their size and charge. The gel is then stained for total protein AND for a specific enzyme/protein (activity or antigenicity).it will include both human and mouse enzyme product

40
Q

Radiation hybrids

A

-because hybrid gene are unstable at first, we wait until it loss human gene into a stable state.

41
Q

Mitotic Recombination example

A

Drosophila twin spots
●X-linked recessive traits
◆Yellow body (y) and singed bristles (sn)
y+sn x ysn+
y+sn y
F1 females mostly wild type
●Patches of cell with different phenotypes
◆Yellow spot (single spot) (crossing over of y gene)
◆Singed spot (single spot) (double cross over)
◆Yellow and singed spots (twin spots) (cross over of an arm)

42
Q

Result from Mitotic Recombination example

A

◆Twin spot > Single spot (yellow) > Single spot (singed)
◆Single spot (singed) requires two crossovers, so rarest
◆Twin spot > Single spot (yellow), then distance between snand
centromere must be greater than snand y.

43
Q

Bacterial and Bacteriophage Genetics

A

◆Bacteria and bacteriophage do not undergo meiosis
◆Crosses are performed by identifying recombination events occurring in parents to yield progeny
◆This requires a mechanism to bring together the parent genetic material

44
Q

Bacteria genetic method

A

Bacterial conjugation
◆Direct transfer of DNA from one parent (donor) to another (recipient)
Bacterial transduction
◆Transfer of DNA from one parent (donor) to another (recipient) by a bacteriophage
Bacterial transformation
◆Direct transfer of isolated DNA from one parent (donor) to another (recipient)
Co-infection of bacteriophage
◆Infection of a host cell by two bacteriophage

45
Q

Nutritional Mutants

A

◆Most bacterial species can grow on very simple defined medium (minimal medium) as they are capable of synthesizing all of the complex biomolecules that they require. This is known as prototrophy (prototrophic, prototrophs)

46
Q

Minimal medium consists of

A

●A simple source of carbon and energy such as glucose
●A simple source of nitrogen such as ammonium (NH4+) salt
●Some salts and trace elements
●From these basic components complex molecules such as nucleic acids (DNA and RNA), amino acids (proteins) and complex carbohydrates (cell wall) can be synthesized.

47
Q

Auxotrophic Mutants

A

◆Mutants can be isolated which can no longer grow on minimal medium but can grow on rich medium. These mutants are known as auxotrophs.
◆Auxotrophic mutants are isolated and identified by a process called screening.
●A rich medium normally has a mixture of yeast extract and other hydrolysed substrates, such as casein. It contains all of the basic amino acids, lipids, nucleosides and vitamins.

48
Q

Utilisation Mutants

A

◆Mutants can be isolated which are unable to grow on a particular compound in minimal medium. These mutants are known as utilisation mutants.
◆Utilisation mutants are isolated and identified by a process called screening.
◆Glucose and other sugars can be used as carbon sources by bacteria. The more complex sugars are degraded to produce glucose, which is the preferred carbon source.
●Utilisation mutants can be isolated which are unable to utilise some complex sugars but grow perfectly well on others and on glucose.

49
Q

Resistance Mutants

A

◆Mutants can be isolated which are resistant to the toxic effect of a particular compound. These mutants are known as resistance mutants.
◆Resistance mutants can be directly isolated by a process called selection.
◆Antibiotic inhibit the growth (bacteriostatic) or kill bacteria (bacteriocidal). Similarly, high levels of some heavy metals and bacteriophage can inhibit or kill bacteria.
●Resistant mutants can be isolated which are no longer susceptible to the effects of these compounds.

50
Q

Screening versus Selection

A

◆Screening
●To screen we look at large number of potential mutants to find one that has the characteristics we want.
◆Selection
●Selection only allows the survival of the mutants that we want.

51
Q

Bacterial Transduction

A

◆Transduction can be divided into two types
●Generalised transduction can be used to map genes anywhere on the bacterial chromosome
●Specialised transduction is limited to certain regions of the bacterial chromosome
◆Transduction uses phage to act as a vector carrying genes from the donor bacterium to the recipient
◆The small genome size of phage means that they can only carry a small portion of the whole bacterial chromosome.
◆Closely linked genes can be separated or mapped by transduction

52
Q

Generalised Transduction pf P1

A

◆Generalised transducing phage P1
●Most commonly used E.coliphage
●Genome size of ~100kb (~2% of E.coli)
●At low frequency host chromosomal DNA instead of phage DNA is packaged.
●Resultant particles are infectious but can not make progeny phage
●A variant of P1 known as P1kc has lost the ability to recognise its own DNA and packages host DNA at a higher efficiency than the wild type.

53
Q

phage P1 mutated effect

A

●Infection by a phage particle which has packaged host DNA will produce a partial diploid E. colicell.
●This cell will not make more phage particles as there is no phage DNA, so the infection is not lytic.
●Inserted DNA can not circularise and is therefore unstable
●Homologous recombination between this DNA and the bacterial chromosome can occur to produce recombinants (transductants)
●A double recombination event is required (because bacterial dna is not linear)

54
Q

Generalised Transduction trend

A

◆Generalised transducing phage P1
●DNA fragments are generated at random
●Packaging of this DNA into P1 phage heads has a maximum size
◆Using transduction, genes which are closely linked on the chromosome can be mapped by determining cotransduction frequencies
◆The closer that genes are together then the more likely they are to be
within the same phage head
◆The closer that genes are together then the less likely that a cross over
occurs between them
◆The frequency of cotransduction measures both of these parameters

55
Q

Recombination frequencies are used as a proxy for physical distances.

A
Meiotic recombination
- recombination frequency
Bacteriophage recombination
- recombination frequency
these two would have lower recombination frequencies for closer and higher for futher
Transformation
- co-transformation frequency
Transduction
- co-transduction frequency
-closer would have higher recombiniation
56
Q

Non-Mendelian Inheritance

A

Inheritance of genetic traits that don’t follow the patterns of chromosomal assortment and segregation

57
Q

Variegation in the Four o’clock plant Non-Mendelian Inheritance

A

●Leaf and stem colour affected
●Progeny phenotype always the same as that of its female parent.
●Progeny phenotypes did not appear in Mendelian ratios.
◆Leaf colour shows maternal inheritance
●Reciprocal crosses yield different results
●All progeny have the same phenotype as the mother
●Variegation results from a mixture of chloroplasts (heteroplasmy)

58
Q

Neurospora crassa poky mutant Non-Mendelian Inheritance

A
●Slow growth
●Lacks cytochromes a and b
●Excess cytochrome c
●poky was only inherited when the female parent was mutant
◆Maternal inheritance
●Unequal contributions from parents
◆Nucleus & cytoplasm (female)
◆Nucleus only (male)
59
Q

Mutation of somatic cell leading to nonmendelian inheritance

A
  • Normally, a nucleus are nomopasmy
  • mutation cause it to become heterophasmon
  • cytoplamic segregation can revert it back
60
Q

Cause of Variable in Expressivity & Penetrance

A

◆Not all individuals display the disease phenotype or severity
●Environmental influence
●Genetic influence
◆Nuclear genes
◆Somatic variation due to cytoplasmic segregation

61
Q

Extranuclear Genomes inheritant

A

◆Inheritance of genetic determinants that are not associated with nuclear chromosomes
●Mitochondria and chloroplasts have their own genomes

62
Q

Nuclear and Cytoplasmic Genes of organelle

A

◆Nuclear and cytoplasmic genes contribute to the functioning of cytoplasmic organelles
●Evolutionary movement of genes, the movement of organelle gen into chromosomal gene. cannot survive on its own as some compound can stop tranlation and transcription of organlle dna but not mitocondrial dna

63
Q

Non-Mendelian Inheritance mechanism

A
◆Mammals
●Maternal
◆Flowering plants (Angiosperms) 
●Mostly maternal but some biparental and paternal
◆Conifers (Gymnosperms)
●Paternal chloroplasts
◆Yeast
●Biparental inheritance of the cytoplasm
64
Q

Extrachromosomal DNA in bacteria

A

Bacteria carry extrachromosomal DNA know as Plasmids
◆Circular DNA
◆Self-replicating, autonomous
◆Variable copy number (1 to >100)
❖High and low copy number control
◆Small (2 to >200 kb) relative to chromosomal DNA
◆Selectively advantageous

65
Q

Fertility factor (F factor)

A

●Single copy in each cell
●Encodes genes for conjugation
●Encodes genes for replication and transfer
●Integration into chromosome