Lecture 4

Question 1

Lethal alleles

Answer 1

• two copies of mutants alleles are inherited
• yellow x brown =1/2 yellow & 1/2 brown (always)

Question 2

Penetrance

Answer 2

• % of individual that shows mutation

Types
- incomplete or variables penetrance
- individual with mutant genotype that will not show the mutant phenotype eg Osteogenesis imperfecta dominant disorder in bone formation

Question 3

Expressivity

Answer 3

• differing levels that a phenotype is expressed.
  
  • variables expressivity : individual shows differing degree of phenotype eg Polydactyly in cats - dominant traits, but affect cats have differing numbers of extra toes
• one genotype could produce multiple phenotypes

Question 4

Why would individuals with the same mutation not show exactly the same phenotype?.

Answer 4

1. Environment: exposure affecting those tissues could increase the chance of having disease
   
   • not causing but increasing the chance
2. Other genes: genetical background
3. Subtlety of mutant phenotype

Question 5

Beadle and Tatum

Answer 5

They investigate on genetic control of cellular chemistry Neurospora( how neurospora make amino acids.
1. They mate two different wild-type (mutagenized Conidia) then it formed fruiting bodies which was dissected and transferred to tubes of complete medium. Transfer conidia (asexual spores) to minimal medial where no growth was observed (Cos autotrophs don’t grow on minimal media plate) . Then those that failed to grow were being transferred to other supplements medial where only arginine grow on minimal + amino acid media) arginine surged on minimal +amino acid media
- each arginine mutation behave as single gene and mapped the mutation relative to other genes and found that they map to three different “Loci”

Question 6

Given three genes from beadle and Tatum experiments Arg-1, Arg-2,Arg-3

Answer 6

If any/ either of them mutated, then arginine autotroph needs arginine to survive.

Question 7

Many genes work together

Answer 7

• one gen, one trait
• mutation in any gene can cause related phenotype

Question 8

In order to determine relationships between gene

Answer 8

1. Mutate gene to generate mutation ( obtain many mutants) lines.
2. Perform complementary test to determine gene mutated
3. Make doubles lines to determine genes interactions

Question 9

Complementation test

Answer 9

• mutant alleles in different gene
• failed to complement mutant alleles in same gene

Question 10

Double mutants interaction

Answer 10

• 1. No mutants interaction ( 9,3,3,1)
     
     2. Same pathway (9,7) working towards end product
        3 recessive epistasis (9,3,4) two products produce phenotype
        4..dominant epistasis (12,3,1) one mutant hide the other
     3. Suppressor mutation (13:3) two wrong make a right

Question 11

Lecture 5

Answer 11

Chromosomal mutation

Question 12

Two main theme underlying the observation on chromosomal changes

Answer 12

1. Karyotypes: tracking evolutionary history generally remain constant within species
   - most genetic imbalances result in selective disadvantage
2. Different karyotypes in related species
   - few differ in rearrangement in closely-related species
   - distantly-related species differ by many rearrangement

Question 13

Chromosomal rearrangement

Answer 13

• normal arrangement : ABCDE-FGH
  - deletion: removes BC region. ADE-FGH
  - duplication: copied of BC was added into the region ABCBCDE-FGH
  - Inversion: BC, it was broken out of gene but still balance ACBDE-FGH
• Translocation: part of chromosome exchange with other part. ABCDE-FGH, LMNOPQR—> ABCDPQR AND LMNOEFGH

Question 14

Effect of chromosomes rearrangement

Answer 14

• impact on phenotype or even viability by affecting gene balance
• Severity effects depend on wether the individual is homozygous or heterozygous
• all the changes can alter crossing- over affecting the fertility of individuals

Question 15

Deletion loop forms in the chromosome of deletion

Answer 15

• recombinants occurs at only homologous region
• no recombinant within the the deletion loop
• genetic map distance In deletion heterogeneous is inaccurate

Question 16

Duplication type

Answer 16

• tandem duplication and non- tandem
• Tandem : not disperse
• non- tandem is disperse

Question 17

Different aspect of the genome that can duplicate

Answer 17

• Exon duplicate/shuffle at lowest level ( exon duplication create new gene variants) and exon shuffling create new gene.
• at next level, entire gene duplicate and create multi- gene .
• gene family duplicate to produce gene superfamily

Question 18

Chromosomes breakage can produce inversion

Answer 18

• peeicentric: occurs in the middle of the chromosome
• paracentric: parallel to each other
• Inversion can disrupts a gene or loss of function

Question 19

Inversion loop form in inversion

Answer 19

• this allows for tightest possible allignment of homologous region
• crossing over within inversion loops produces aberrant recombinant chromatids

Question 20

Why pericentric inversion heterozygotes produce few recombinant progeny

Answer 20

1. Each recomb chromatids has centromere but genetically unbalance
2. Zygote formed from union of normal gametes + gamete carrying recomb is non viable

Question 21

Why paracentric inversion heterozygotes produce few recombinant progeny

Answer 21

1. One recomb chromatids lacks centromere and other recomb chromatids has two centromere
2. Zygotes formed form union of normal gamete with gamete carrying the broken discentric recomb chromatids will be non viable

Question 22

Summary

Answer 22

• inversion don’t add or remove dna but only distrupt
• they are useful genetic tools
• balancer chromosomes have inversion
• no cross over with them and gene of interest
• in inversion, heterozygotes , recombinant within inverted segment result in in genetically unbalance

Question 23

Translocation

Answer 23

• attaching one chromosome to another

Question 24

Types of translocation

Answer 24

1. Reciprocal Translocation : two different chromosomes each break
   
   • then they exchanged fragments that replace each other
2. robertsonian translocation:
   - they break at or near centromereof two acrocentric chromosomes
   - generate one large metacentric chromosome and one small chromosome which usually lost
   - can reshape genome

Question 25

Lecture 6

Answer 25

The genetics of bacteria and their viruses

Question 26

Introduction

Answer 26

• human genome is approximately 20000
• prokaryotic genome is approximately 3000, densely packed, don’t go tru mitosis and meiosis. They only go tru binary fission.

Question 27

Bacteria genetic traits

Answer 27

minimal media: contains water, salt, carbon sources and agar
- prototroph : growth on it minimum media
- autotroph: can’t grow
- When minimum media+ histidine, growth occur. Genotype of histidine that grow =his +
- when arginine was + minimal media +autotrophs no growth occurs genotype of no growth =his -

Question 28

Bacteria DNA exchange

Answer 28

• they use horizontal gene transfer to exchange information without sexual reproduction within the same generation
  1. They use conjugation: direct contact bacteria cells via pilus, plastids, and genome
  2. Transformation: picking up DNA from environment/ dead bacteria cells.
  3. Transduction: virus mediates transfer to DNA from donor to recipient

Question 29

Bacteria conjugation

Answer 29

-Fertility factor: ability to produce pili
- HFR strain : te use F factor to integrate into chromosome

Question 30

What happen in Hfr x F- cross

Answer 30

• none of F-recipient converted to F+ nor Hfr strain
• integrated factor F in Hfr drives some or all the bacteria chromosomes. DNA transfer and replication begin where F factor was integrated.
  - fertility factor, origin (first), terminus was (last).
• recombination occur her donor fragment and recipient chromosomes.

Question 31

F interrupted mating to determine the order of bacteria genes in chromosomes

Answer 31

• F plasmid integrate at diff sites and in different orientations
  - diff interrupted mating experiments experiments will give diff results
• gene order of entrance will indicate location of F factor

Question 32

Chromosomes mapping

Answer 32

It start from the origin
- it bases on recombination frequencies
- recombination of bacteria: to keep genome intact, must be even number of recombination events.
- exogenote: a+ /- endogenote: a- : a single cross over can not produce a recombinant
The closer two gene are , the less likely crossover. The farther they are, the more likely crossover.

Question 33

Two types of genetic analysis with phases

Answer 33

1. viral genome with recombinant bet two phages.
2. Bacteria genome study linkage and gene

Question 34

Bacteria exchange transduction and types

Answer 34

• transfer of genetic material from bacteria donor to recipient

There are to types
1. Generalized transduction : random incorporation of bacterial dna into phage head.
- the closer the bacteria genes are in the chromosome, the more they are ti be packed together into a phage head and be transducted together ( cotransduced)
- the larger the recombinat frequency, the farther they are apart. The larger the cotransduction, the closer they are apart.
2. specificalized transduction :
- virulent phage: lyes and kill the bacteria host
- temperate phage: integrate their dna into host chromosomes without killing it.