Genetics Test 2 Review

• Linked genes


  Genes located on the same chromosome that tend to be inherited together in genetic crosses.

  
  

  Do not undergo independent assortment.
• Cytology


  the branch of biology that studies the structure and function of cells
• Unit Factors


  A gene that controls inheritance of a unit character

  
• Crossing Over


  Process in which homologous chromosomes exchange portions of their chromatids during meiosis.

  
• Consequences of Meoisis

  Independent assortment

  Linkage without crossing over

  Linkage with cursing over
• Independent Assortment


  Two pairs of chromosomes with heterozygous gene pair that do not exhibit linkage.

  Produces 4 different gametes!

  
• Types of Linkage

  w/o Crossing over

  -SAME genes linked on the SAME chromosome.

  -2 different gametes produced

  -Complete linkage creates PARENTAL gametes in equal proportions

  -HOWEVER, there are always some recombinant gametes meaning that we can use this small percentage to determine distance between two linked genes.

  
  

  w/ Crossing over

  -on two NONSISTER chromatids crossing over b/w two linked genes occurs

  -2 new allele combinations (RECOMBINANT gametes)
• Recombination Possibilities

  When loci of two linked genes far apart

  - Number of recombinant gametes nears 50%.

  - If 50% recombination occurs (two parental types and two recombinant gametes).

  - Transmission indistinguishable from two unlinked, independently assorting genes.
• Linkage Ratio

  -Complete linkage between two genes due to close proximity

  -Unique F2 phenotypic ratio results from HETEROZYGOUS cross

  -#linkage groups=haploid number
• Morgan and Sturtevant experiments


  Using flies, they studied mutations on the X chromosome.

  Looking at one trait showed X-linkage, but crossing two X-linked genes generated strange results...
• Chiasmata


  Source of gene separation. This creates the recombinant gametes.

  
• Chiasma


  site of crossing over between two synapsed homologous chromosomes.

  
• Sturtevant Mapping Experiments

  Based on Morgan's work on linked genes, Sturtevant realized that RECOMBINATION FREQUENCIES are additive and can estimate the relative distance between two chromosomes.

  
• Map Units (mu)

  a unit of distance on a chromosome equivalent to a 1% recombination frequency.

  
  

  REMEMBER: the higher the recombination frequency the farther the linked genes are, and the more likely recombinant gametes are formed rather than parental.

  
• Types of Crossovers


  Single

  -Linked genes <50 mu apart

  -Crossing over should theoretically occur nearly 100% of the time creating 4 equal proportion gametes.

  -2 parental and 2 recombinant gametes

  -Used for distance between 2 linked genes

  
  

  Multiple

  -Can facilitate more extensive chromosome map

  -Double cross overs (DCOs) can be used to conduct Three-point mapping.
• Three-point Mapping

  From a double crossover we can use the product law from probability to determine the relative distances of genes.

  
  

  The following conditions must be met though:

  1. Genotype MUST be heterozygous at all loci.

  2. Cross must be able to determine genotype from observable phenotype.

  3. Sufficient number of offspring is required for representative sample.

  
• How to solve an autosomal mapping problem

  1. Identify the Parental and Recombinant Offspring: The parental offspring are those that have the same phenotype as the parent, while the recombinant offspring have a different phenotype.

  2. Calculate the Recombinant Frequency: This is done by dividing the number of recombinant offspring by the total number of offspring and multiplying by 100 to get a percentage

  
• Example Autosomal Mapping problem:

  Suppose we have a heterozygous fly with the genotype a+b+c+/abc (where + indicates the wild type allele and abc indicates the mutant alleles) and we cross this fly with a homozygous recessive fly with the genotype abc/abc.

  We collect and count the following offspring:

  388 with phenotype a+b+c+

  362 with phenotype abc

  115 with phenotype ab+c

  108 with phenotype a+bc

  12 with phenotype a+b+c

  15 with phenotype ab+c+


  To solve this problem, we first identify the parental and recombinant phenotypes. The most common phenotypes are parental (from the heterozygous parent), and the least common are double crossovers. In this case, a+b+c+ and abc are parental, and a+b+c and ab+c+ are double crossovers.

  Next, we calculate the recombination frequencies:

  For genes a and b: (115 + 108 + 12 + 15) / total * 100 = 31.5%

  For genes b and c: (115 + 108 + 12 + 15) / total * 100 = 31.5%

  For genes a and c: (12 + 15) / total * 100 = 3%

  The gene order is determined by the smallest recombination frequency, which is between genes a and c. So, the order is a - c - b.

  Finally, we calculate the map distances in centiMorgans (cM), which are approximately equal to the recombination frequencies2:

  Distance between a and b: 31.5 cM

  Distance between b and c: 31.5 cM

  Distance between a and c: 3 cM

  So, our final gene map looks like this:

  a---3---c---31.5---b

  Copy

  This is a simplified example, but it gives you an idea of how to approach these problems.
• Interference


  Complete, positive, or negative

  If interference is complete I=1. meaning no double crossover can occur.

  
  

  if I is positive, positive interference has occurred. (two genes are closer together)

  
  

  If I is negative, then negative interference has occurred. (more DCO is expected)
• Coefficient of Coincidence (C)

  C = observed DCO / expected DCO

  
  

  -Quantifies disparities resulting from interference.

  -Observed number of DCOs divided by expected number of DCOs

  I=1-C
• DNA Marker


  a mixture of DNA fragments of known lengths and locations.
• Restriction Fragment Length Polymorphisms (RFLPs)

  Example of DNA marker

  
  

  polymorphic sites generated when specific DNA sequences are recognized and cut by restriction enzymes.
• Microsatelites

  Example of DNA marker

  
  

  Short repetitive sequences found throughout genome. (vary in number of repeats)

  -Identified by DNA sequences flanking them.
• single nucleotide polymorphism (SNP)


  Variation in single nucleotide found throughout the genome. These millions of variations allow genetics to identify and locate related genes.

  
  

  In combination with DNA markers multiple diseases can be diagnosed: Type 1 and 2 Diabetes, hypertension, coronary artery, bipolar disorder, and rheumatoid arthritis.
• Cystic Fibrosis


  Example of gene located using DNA marker.

  Life-shortening autosomal recessive exocrine disorder. (Located on chromosome 7 on the long arm.
• Cytological Markers

  Mapping in maize using cytological markers established that crossing over involves physical exchange of chromosome regions.
• Harlequin Chromosome


  Sister chromatids involved in mitotic exchanges.

  Patchlike appearance when stained for old DNA.
• Sister Chromatid Exchange (SCEs)


  reciprocal exchange similar to crossing over.

  Occurs during mitosis.

  Do not produce allelic combinations!

  Damage increases frequency of SCEs (caused by Viruses, X-Rays, UV-light, and chemical mutagens)
• Bromodeoxyuridine (BrdU)


  SCE stain used to tag Thymosin.
• Bloom syndrome


  SCEs elevated due to a recessively inherited disease. caused by prenatal and post-natal delays in growth.

  Have a disposition for malignant and benign tumors.

  
• Bacteriophages


  Viruses that use bacteria as a host for reproduction.

  
• Why are bacteria and viruses useful for genetic studies?

  They have:

  -short reproductive life cycles

  -studies in pure cultures
• Bacteria Culture conditions

  Bacteria are grown in liquid culture medium on a semisolid agar surface in a petri dish.
• Minimal Media

  Contains basic organic carbon source (a sugar like glucose or lactose) and ions Na, K, Mg, Ca, NH4, etc.

  
  

  If a bacteria is able to grow in the media and create essential organic compounds it is known as a prototroph.
• Phototrophs


  organisms that get energy from light and can manufacture their essential compounds.
• Auxotrophs


  cannot synthesize a needed nutrient.

  Will not grow on a plate that lacks the growth factor, but will grow on a complete medium that contains the growth factor
• Inoculum


  microbes that are introduced into a culture medium to initiate growth

  
• Method for quantifying Bacterial Growth


  Serial dilutions can be used to study colonies on a petri dish by multiplying the number of colonies and the dilution factors to find the number of bacteria in each milliliter of the initial inoculum.

  
• genetic recombination in bacteria


  Replacement of one or more genes in the chromosome of one cell with those from a genetically distinct cell.

  
  

  In eukaryotes this is crossing over.

  
• Three genetic transfer processes

  Conjugation

  Transformation

  Transduction