Class

How to calculate % recombination generally

% recombination = # of recombinants/total progeny x 100

How to calculate % recombination for tetrads

- % recombination = NPD + ½ TT / total tetrads x 100% - % recombination = # of recombinant spores/ total spores x 100%

What does + symbol mean in karyotypes What does - symbol mean in karyotypes

- + symbol preceding a chromosome designation indicates that the chromosome or arm is extra - + symbol following a chromosome designation indicates that the chromosome or arm is larger than normal - - preceding a chromosome designation indicates that the chromosome or arm is missing - - following a chromosome designation indicates that the chromosome or arm is smaller than normal

What are the three types of segregation that can result from translocations and their frequencies

- 1 + 2 <--> 3 + 4 - 1 + 3 <--> 2 + 4 - 1 + 4 <--> 2 + 3 - in which 1 and 4 are the reciprocal translocation - Expected frequencies of adjacent 1:adjacent 2:alternate segregation are ¼:1/2:1/4 - Semisterility of genotypes that are heterozygous for reciprocal translocation results from lethality due to duplication and deficiency gametes produced from adjacent 1 and adjacent 2 segregation - Frequencies are strongly influenced by position of the translocation breakpoints, by number and distribution of chiasmata and interstial region between the centromere and each breakpoint and by wheather the quadrivalent tends to open out into a ring shaped structure on the metaphase plate

What is the maximum length of DNA in a general transduction for E.coli

- 1 min or 1% of the E.coli genome

Explain low copy plasmids

- 1 or 2 copies per host cell - replication is initiated only once per round of replication

How large are eukaryotic Okasaki fragments

- 100-200 bp

Explain the active genes in the inactive X chromosome

- 15% of X linked genes escape inactivation to some degree - inactive X is not completely silenced - transcribed genes occur in large blocks that tend to be located in the distal portion of the arms - escape from X inactivation may be related with distance from Xist - most gene that escape X inactivation have levels of transcription that range from 15-50 % of those observed for their homologs in the active X chromosome - number of genes that escape X inactivation and their levels of transcription could account for some of the differences in traits between males and females

How was DNA semiconservative replication proven

- 1958 - Meselson and Stahl - equilibrium density gradient centrifugation - allows separation of molecules and cellular components that differ only slightly in density - used isotopes of N to create heady and light DNA - started with E. coli growth in medium containing only heavy 15N - all of the DNA was heavy - E. coli transfered to 14N medium so new DNA synthesized will be light - DNA extracted and centrifuged to determine relative weights - found experiment results resembled the semiconservative model

How many different chromosomal arrangements can occur in metaphase

- 2^n - n is the haploid number

Explain pol I in prokaryotic organisms

- 3 enzymatic activities - polymerase for 5' to 3' direction - exonuclease for 3' to 5' direction proofreading - exonuclease for 5' to 3' RNA removal - RNA to DNA exchange is done base by base

Explain the human genome

- 3 x 10^9 bp - 6 billion base pairs per cell in human - 23 chromosomes - hundreds to thousands of genes per chromosome - 1.3% of DNA is protein coding - 98.7 % is noncoding sequences in genes such as repetitive DNA and pseudogenes - 99.9% of DNA sequences are the same in all people - 0.1% is different among individuals and important to study - each chromosome contains a single molecule of DNA - differences is genomes include mutations that are responsible for human diseases - only small portion of all differences in DNA is associated with disease - others are associated with differences in height, weight, hair color, eye color, facial features

Explain the important biological functions of histone modifications

- 40% recombination hotspots in the human genome are associated with DNA binding site for the histone methyl transferase enzyme which transfers 3 methyl groups to a lysine present at amino acid position 4 in the histone H3 tail creating H2K4me3 - presence of H3K4me3 renders the DNA in the region susceptible to double strand breaks which initiates the process of a double strand break repair associated with recombination

Explain Turners Syndome

- 45, X - female - short in stature - no sexual maturation - normal mental abilities - more than 99% of 45, X fetuses undergo spontaneous abortion - profound deleterious effects on development

Explain Patau's Syndrome

- 47 + 13 trisomy - incidence of approx. 1 in 10,000 live births - Severe facial anomalies such as cleft lip and cleft palate - Very small, absent or fused eyes - Polydactyly - Rocker bottom feet - Internal malformation of the heart , brain and nervous system - Death often occurs in the first month and mean survival is 6 months

Explain Edwards Syndrome

- 47 + 18 trisomy - incidence approx. 1 in 56,000 live births - many are late spontaneous abortions - Clenched fists - Overlapping second and fifth fingers - Rocker bottom feet with protruding heels - Heart malformations - Death usually results from heart failure or pneumonia by age 2-4 months - Most live births are female

Explain Trisomy-X

- 47, XXX - female - mentally and phenotypically normal - mild mental impairment may be somewhat more frequent than 46, XX females - incidence is 1/1000 female births - many are undetected - Most show some degree of developmental delays - Usually tall - May have some mental health issues - Premature ovarian failure - Some may be fertile

Explain Klinefelter Syndrome

- 47, XXY - male - incidence is about 1 in 750 male births - tend to be tall - do not undergo normal sexual maturation - sterile - some cases have enlarged breasts - mild mental impairment - epithelial cells of these individuals have 1 Barr body indicating the presence of two X chromosomes - Excess gene expression from extra X - Some have more then 2 X's - Severity of symptoms increase with number of X's - Number of X chromosomes = number of Barr bodies

Explain double-Y

- 47, XYY - Jacobs Syndrome - male - tend to be taller than average - some may have slightly impaired mental function - rate of criminality is slightly higher, these crimes are often petty theft and nonviolent - majority are mentally and phenotypically normal and no criminal tendencies - Y chromosome is small and carries few genes - Incidence 1/1000 male births - Effect on male phenotype is limited - Most are fertile - Chromosomally normal gametes produced

How common is red green color blindness and how is it inherited

- 5% of males have some form of red green color blindness - since mostly males are affected this suggests X linked inheritance - Affected males have normal sons and carrier daughters - Carrier daughter have 50% affected sons and 50% carrier daughters

In what direction does DNA synthesis occur

- 5' to 3' - nucleotides can only be added to the free 3' OH group

Explain high copy plasmids

- 50 copies per host cell - replication is initiated multiple times during replication

How much energy to copy DNA?

- 6 billion energy (ATP/GTP/etc) - cells do not replicate DNA until they are going to divide

What is first division segregation

- A spores will be together and a spores will be together - No crossing over occurs

What is a bacteriophage

- A virus that infects and kills bacteria - Has many protiens and nucleic acids in head - Only protien in tail - Infection proceeds through: - Attachment of phage particle by the tip of its tail to the bacterial cell wall - Entry of the phage material into the cell - Multiplication of this material to form hundreds of progeny phage - Then release of these progeny phage by bursting of the bacterial host cell

Explain PKU

- Absense or defect of enzyme Phenyalanine hydroxylase (PAH) results in phenylketonuria (PKU) - Nervous system development is impaired - Autosomal recessive - Increase in phenylalanine results in unusual metabolism - Phenylalanine becomes phenylpyruvic acid and damages the developing nervous tissue - Excess phenylalanine is broken down into harmful metabolites that cause defects in myelin formation that damage developing nervous tissue and cause severe mental retartation - dietary restrictions can prevent symptoms from developing - All newborns are tested for PKU

Explain epistais

- All cellular metabolism under genetic control - Each step in a biochemical pathway is catalyzed by an enzyme encoded by a separate gene - Many genes involved in creating a final observable/measureable phenotype - In a genetic cross in which two mutations that affect different steps in a single pathway are both segregating the typical F2 ratio of 9:3:3:1 is not observed - Examining 2 genes creating the same phenotype will disturb phenotypic Mendelian ration for assortment of 2 genes - Different interactions produce different rations but all are based on the same dihybrid genotypic ratio which follows Mendelian rules - All ratios derived from 1:2:1:2:4:2:1:2:1 genotypic ratio and most from 9:3:3:1 phenotypic ratio - Always fewer then 4 classes observed because 2 or more expected phenotypic groups have the same appearance

What are the concequences of translocations

- All genetic information present just reorganized - Activation and inactivation of genes due to position changes - Biggest effect on gametogenesis and fertility of carrier - During meiosis homologous regions of all chromosomes pair and cross over - Gene known to be located on non homologous chromosomes close to break point appear to be linked - Segregation of chromosomes cause problems - Unbalanced gametes produced - Centromere N1 always segregates from T1 and N2 always segreagates from T2 - The two segregation events are independent so balanced products are not always produced - Segregation 1 and 2 are equally probable - Reduction in gamete viability to exactly one half due to production of unbalanced gametes in meiosis

Explain inversions

- All genetic information present just reorganized - Formed by two break event in which middle segment is reversed in orientation before breaks are healed - Can be created by ectopic recombination between DNA sequences that are inverted repeats - No problems in mitosis but meiosis problems can arise due to homologous recombination in area that is inverted - Inversion heterozygotes produce looped chromosomal configurations during meiotic synapsis called inversion loops - crossing over within inversion loop allows the chromatids that are involved in crossing over to become physically joined and results in chromosomes with many deletations and duplications - Paracentric = beside the centromere and not involving the centromere - Pericentric: around the centromere and involves the centromere

Explain the Wobble Hypothesis

- Anticodon of a specific tRNa may recognize 2 codons - Pairing must be between a purine and a pryimidine - Wobble between the 3' base of the codon and the 5' base of the anitcodon

Explain Inborn Errors in Metabolism (Garrod)

- Any hereditary disease in which cellular metabolism is abnormal results from an inherited defect in an enzyme - Established link between hereditary, enzymes and disease - Mistake in gene (mutation) can result in a mistake (lack of function) in the corresponding protein - In biological pathways an enzyme is needed to catalyze the reaction for each step to occur - People with an inborn error of metabolism have a defect in a single step of a metabolic pathway because they lack a functional enzyme for that step - Frequent result of a blocked pathways is that the substrate of the defective enzyme accumulates Ex. Alkaptonuria (black urine disease)

Explain autism and schizophrenia

- Autism and schizophrenia have reciprocal symptoms - Certain stages in development of disorders in which the same gene has opposite effects - Enhanced expression in duplication predisposes to one disorder and reduced expression in deletion predisposes to the other disorder - CNVs and genes that show these opposite effects indicate the presence of shared casual links in the development of these conditions

Define Enzymes

- Biological catalysts that accelerate biological reactions in cells - Are proteins - All enzymes are proteins but not all proteins are enzymes

Explain the genetic consequences of inversions

- Can have inactivation of vital gene or activation of detrimental gene functions - problems arise when crossing over occurs within inversion - Crossing over results in the formation of a dicentric bridge and acentric fragment which is lost - Amongst the surviving chromosomal products are one normal chromosome, one inverted chromosome and two deleted products - Only non crossover chromosomes are viable - Reduced recombination frequencies for genes in the inversion since all recombinant products are non viable - Non viable gametes leads to reduced fertility levels - Recombinant products of pericentric inversion have duplications and deletations - Only non recombinant chromosomes produce viable gametes - Pericentric inversions also show reduced recombination frequency for genes within inversion

Explain Substitution Mutation

- Change in single base pair - Change in codon may code for the same amino acid and no mutation may occur

Explain mutations

- Changes in nucleotide sequence are responsible for mutations - Mutation yields a mutant gene which in turn produces a mutant mRNA, a mutant protein and a mutant organism which exhibits the effects of mutation

Why is there a polymorphism to taste PTC?

- Chemical structure of PTC resembles a large and heterogeneous class of molecules called glucosinolates found in some plants - Individuals carry the PAV form of the hTASR38 taste receptor do find broccoli to be significantly more bitter tasting then individuals homozygous for AVI form

Explain M1V Mutation

- Codon 1 is affected - Methionine changed to valine - Initiation cannot occur and no protein is produced

Explain R408W Mutation

- Codon 408 is affected - Arginine (R) replaced with tryptophan (W) - Protein produced but reduced enzyme activity of only 3%

What is the coefficient of coincidence

- Coefficient of coincidence compares expected and observed DCO's - Expected DOC will never be 0, observed DOC can be 0 - COC = observed DCO/expected DCO - can be greater than one due to hotspots on the chromosome

Explain Transcription

- DNA --> RNA - 5' to 3' end - RNA product is called transcript - DNA duplex opens up - Ribonucleotides pair with complementary rules - RNA strand has 5' phosphate end and 3' hydroxyl end - Only one strand of DNA transcribed for a specific gene - Both strands can be template for different genes - Promoter for transcription and termination site of gene outside coding part of gene - Eukaryotic genes much longer then mRNA transcript - Initial transcript is highly modified and intervening sequences are removed - Exons are saved and used in mRNA - Introns are spliced out - 1 fewer introns then exons

Explain Restriction enzymes

- DNA cannot be manipulated intact - physical sheering reduces size but random break points complicates analysis because fragments containing a particular gene will be of different sizes - reproducible fragments are required - restriction enzymes do this, they look for a specific enzyme to cut - restriction enzyme technique ensures that all DNA fragments that contain a particular sequence have the same size and each sequence is located at exactly the same position - uses classes of DNA cleaving enzymes isolated from bacteria - most restriction enzymes recognize only one short base sequence (4-6 nucleotide pairs) - enzyme binds with DNA at these sites and makes a break in each strand of the DNA molecule producing free 3'-OH and 5'P groups

Explain restriction fragment

- DNA fragment produced by a pair of adjacent cuts in the DNA

Explain blunt ends

- DNA fragments from symmetrical cleaves

How is linkage analysis done using transformation

- DNA is fragmented - transformation and co transformation frequency for specific markers is determined - if 2 markers can be co transformed they must be very close together

Explain gene expression

- DNA is the information storage molecule while protiens act to produce an effect - DNA --> RNA --> Protiens - Typical protien is made up of one or more polypeptide chains - Each polypeptide chain consists of a linear sequence of amino acids connected end to end - Only 1.5% of a gene is devoted to coding for amino acids - Noncoding part includes some sequence that control the activity of the gene

Explain the ligation of adjacent fragments for Okasaki fragments

- DNA ligase covalently joins 3' OH and 5'P

Explain DNA markers in pedigrees

- DNA markers are used extensively in human pedigree analysis - used to localize genetic defects when defective function is not understood - codominant behaviour of DNA markers make them very useful

Explain chromatin

- DNA of all eukaryotic chromosomes is associated with numerous protein molecules in a stable ordered aggregate called chromatin - determines chromosome structure - determines changes in structure that occur during the division cycle of the cell - play roles in regulating chromosome functions - simplest form of chromatin is in nondividing eukaryotic cells

Explain the S phase

- DNA of chromosomes is replicated - each strand of chromosome becomes double stranded and composed of two sister chromatids that are held together by the centromere - each chromatid is composed of a single piece of double helix DNA - two chromatids are genetically identical

Explain how its known that participation of RNA in protein synthesis is a relic of the earliest stages of evolution

- DNA replication requires an RNA molecule in order to get started - An RNA molecule is essential in the synthesis of the tips of the chromosomes - Some RNA molecules act to catalyze key reactions in protein synthesis

Explain transposable elements

- DNA sequences are capable of moving/transposition from one location to another within a chromosomes or between chromosomes - transposable elements are present in most genomes - spontaneous mutations occur from insertion of transposable elements into a gene

Explain how SNP's are analyzed

- DNA sequencing is required for analysis of SNPs - Microarrays set up to genotype individuals and are called SNP chips - identifying the particular nucleotide present at each of a million SNP's is made possible by use of DNA microarray containing millions of infinitestesimal spots of a glass slide about the size of a postage stamp - each tiny spot contains a unique DNA oligonucleotide sequence present in millions of copies synthesized by micro chemistry when the microarray is manufactured - each oligonucleotide sequence is designed to hybridize specifically with small fragments of genomic DNA that include one or the other of the nucleotide pairs present in a SNP - Consist of oligonucleotides that contain deliberate mismatches that are intended to guard against being misled by particular nucleotide sequences that are particularly sticky and hybridize too readily with genomic fragments and other sequences that hybridize poorly or not at all - short fragments of genomic DNA are labeled with fluorescent tag and then the single strands are hybridized with a SNP chip containing the complementary oligonucleotides as well as numerous controls -duplex containing the T-A will hybridize with complementary oligonucleotides, C-G will do the same - after hybridization takes place SNP chip is examined with fluorescent microscopy to detect spots that fluorescent due to the tag on genomic DNA - Genomic DNA from the individual whose chromosomes contain two copies if the T-A from the duplex will cause the two leftmost spots to florescent and the two rightmost spots will be unlabeled - Genomic DNA from the homozygous C-G individual will cause two rightmost spots to fluorescent but not the two left - genomic DNA from heterozygous individual will cause all 4 spots to fluorescent

Explain the Hershey-Case Experiment

- Demonstrated that DNA is responsible for directing the production of progeny T2 phage - DNA has phosphorous and no sulfur - Most proteins contain sulfur and no phosphorus - DNA and proteins where labeled differently by radioactive isotopes of phosphorus and sulfur - E. coli cells where infected in a medium of 32P or 35S - Phage heads were removed in a blender and the infected bacteria where examined - Most of the radioactivity from 32P phage was associated with the bacteria while only a small amount of 35S phage was present in the infected cells - This implies that T2 phage transfers most of its DNA but very little protien into cells it infects, this proves that the genetic material in T2 phages is DNA

Explain nucleic acid hybridization/southern blot

- Denaturation: DNA becomes single stranded with heating or treatment with alkali to rase pH and disrupt H bonds - Hybridization/renaturation: incubation at mid range temperatures and high salt concentration which allow complementary sequences to anneal - researcher supplies complementary probe DNA which is radioactivly labeled to be followed - positions of molecules diffuse once taken out of electric field - DNA is transfered to nitrocellulose paper so diffusion does not occur - use specific probes to find specific sequences in DNA - in the three samples the size of the fragment is not the same as seen in the autoradiogram - cool down and increased salt allows things that are not exactly the same to stick - similar sequences are recognized by making hybridization condition less stringent - convienent and very sensitive

What does designation 1p34 mean What does 1p36.2 means

- Designation 1p34 indicates chromosome 1, short arm, division 34 - 1p36.2 means the second band in 1p36

How would you determine the ratio for epistatis problems

- Determine total progeny (80) - Divide by 16 = 80/16 = 5 - Divide each class number by this value: 44/5 = 8.8, 36/5 = 7.2 = 9:7

Explain Complementation test

- Determines allelic relationship between mutant forms showing the same phenotype - Requires that both mutant forms be together in the same cell/organism and usually requires sexual reproduction

Explain 12:3:1 ratio in epistatsis

- Dominant genotype at locus 1 masks the expression of all genotypes at locus 2 - Two true breeding Digitalis purpurea flowers crossed - A- genotype renders the B- and bb genotypes indistinguishable

Explain the DNA model

- Double stranded - Right handed helix - Anti parallel strands - Complementary nitrogen base pairs: Adenine with Thymine and Guanine with Cytosine - Stacked nitrogen bases - Sugar phosphate backbone - Single stranded DNA can exist but is not stable - One end has phosphate, other end has hydroxyl - Hydrophilic sugars and phosphate - Hydrophobic nitrogenous bases - Sequence of nucleotides stores information - Nature of nucleotide strands dictates replication - Change of nucleotides causes a mutation due to change in info

Explain asexual polyploidization

- Doubling of chromosome number takes place in mitosis - Chromosome doubling though abortive mitotic division is endoreduplication - self fertilization endoreduplication creates a new genetically stable species because if the chromosomes in the tetraploid can pair two by two in meiosis they can segregate regularly and yield gametes with a full complement of chromosomes, self fertilization will restore the chromosome number

What are some examples of multigenic deletations in Drosophila and humans

- Drosophila: notch wing, visible dominant effect and lethal when homozygous - Human: Cri-du-chat syndrome, terminal deletation of 5 p, 2 bands, 5p15.2 and 5p15.3 deleted

How is helicase produced in E.coli

- E.coli gene dnaB produces DnaB helicase

Explain Human ABO blood groups

- Example of multiple alleles and differing dominance relationships - 3 common alleles A, B O - 6 genotypes, 4 phenotypes - A and B are codominant and both dominant over O - AB is universal recipeient - O is universal donor - Gene codes for glycosyl transferase enzyme and attaches specific sugar to precursor carbohydrate of glycoprotein - A = sugar attached produces antigen A - B = sugar attached produced antigen B - O= inactive enzyme only precursor is present, neither antigen is produced - Transfusion of blood containing A or B red blood cell antigens into a person that makes antibodies against them results in an agglutination reaction in which donor red blood cells agglutinate many blood vesicles are blocked and recipient goes into shock and may die

Explain a PKU diet

- Excludes meat, poultry, fish, eggs, milk and milk products, legumes, nuts and bakery goods made with regular flour - Foods are replaced with expensive synthetic formula - Diet will be returned to during pregnacy

Define F1 and F2

- F1 is the first filial generation, only created by crossing true breeding parent generation - F2 is the second filial generation, only created by inbreeding F1

What progeny do you expect from a red eyed female and white eyed male parental cross What happens with a backcross of F1 females to original white male

- F1 will all be red eyed - F2 will show 3/4 red eyed and 1/4 white eyed - 1/2 of progeny will be red eyed female - 1/4 will be red eyed male - 1/4 will be white eyed male - Produced white eyed females though backcross of F1 females to original white male

What progeny do you expect from a white eyed female and red eyed male parental cross

- F1 will show half red and half white eyed, all red eyes will be female and all white eyes will be male - F2 cross shows half red and half white eyes with the sexes in equal numbers for both phenotypes

Explain the Frederick Griffith Experiment (1928)

- First demonstrated bacterial transformation - Identity of transforming principal was not known - Demonstrated that genetic material can be transfered from one bacterial cell to another - R strain does not cause pneumonia because with out a capsule the bacteria are inactivated by the hosts immune system - S strain results in pneumonia because slime coating protects the cells from the immune system - Heat killed S cells do not result in sickness - When mice injected with both living R cells and heat killed S cells they die - Bacteria that have been isolated from the dead mice show living S cultures even though the injected S cells have been killed - R cells are changed to S cells, this is the first documented change in genes

Explain the evolution of human chromosome 2

- G banding indicates that this chromosome was formed by fusion of the telomeres between the short arms of two acroscentric chromosomes - fusion created a dicentric chromosome with its two centromeres close together - new chromosome must have been sufficiently stable to be retained in human lineage - present day chromosome 2 has only one functional centromere

Explain Checkpoints within cellular division

- G1/S: DNA synthesis will not begin until sufficient cell size is attained and sufficient time elapsed since last cell division - G2/M : Mitosis does not begin until DNA synthesis and repair is complete - Mitosis metaphase: kinetochores must align on metaphase plate before cell progress to anaphase

Explain multiple alleles

- Gene of n nucleotides can theoretically mutate at any of the positions to any of the three other nucleotides - Number of possible single nucleotide differences in a gene of length n is therefore 3 x n - Most genes in most natural populations have multiple alleles which can be considered wildtype

Explain the segregation of two or more genes

- Genes can be on the same or different chromosomes - start with genes on different chromosomes - principles apply to genes that are on the same chromosome but so far away from eachother that they segregate independently - if genes are on different chromosomes they will behave as mendel describes - if genes are on the same chromosome they cannot behave as mendel describes

Explain the stability of chromosome complements

- Genes reside on chromosomes - Each individual within a species will have the same unchanging number of chromosomes - Chromosomes exist in pairs, this is the diploid number - Gametes posess a single copy of each chromosome, this is the haploid number - Unlinked genes undergo independent assortment - Nonhomologous chromosomes undergo independent assortment

Explain how we known crossing over occurs at the 4 strand stage with tetrad analysis

- Genotypes of ascus contents shows that crossing over must occur at the 4 strand stage because parental and nonparental products can be produced in an individual meiosis

Explain roles of Structural Proteins Give examples of structural protiens

- Give cell rigidy and mobility - Some form pores in the cell membrane to control the traffic of small molecules into and out of the cell ex. Karyotin, collogen, myosin, hemoglobin

Explain genotypes of recombinant phages

- H+ = can only infect strain 1 of E.coli - H- = can only infect strain 1 and 2 of E.coli - R+ = slow lysis and small plaques - R- = rapid lysis and large plaques - plaque morphology used to infer genotype of phage - phage with limited host range produced cloudy plaques indicating that only some of the bacterial cells were lysed

Explain a model organism

- Haploid organism has advantages, no question with genotype, phenotype directly expresses genotype - Can grow on minimal medium - Extreme and complete metabolic activities

Explain Alkaptonuria

- Homogentisic acid is excreeted in urine turning it black when the homogentisic acid is oxidized - Kidneys filter unwanted substances from blood into urine - Individuals produced homogentisic acid and it was filtered out by kidneys into urine - This is an inherited defect in a single gene - Patient has inability to break down phenyl ring of 6 carbons that is present in homogentisic acid - Ring is obtained as a breakdown product of amino acids phenylalaine and tyrosine which are are of normal protiens in the diet

Explain 9:6:1 ratio in epistatsis

- Homozygosity for either of 2 recessive alleles produces the same distinct phenotype - Double homozygous recessive and dominant genotypes have their own phenotypes

Explain the 9:7 ratio in epistatsis

- Homozygous recessive for either or both genes produces one mutant phenotype - At least one dominant allele for both genes required to produce wild type phenotype - Occurs due to complementation - must have enzyme C and P to be purple

Explain sexual polyploidization

- Increase in chromosome number takes place in meiosis though formation of unreduced gametes that have double the normal chromosome complement - Unreduced gametes are formed in many species at frequencies of 1-40% - Single recessive mutation that acts during pollen formation causes first and second division meiotic spindles to be oriented in the same direction - Pollen nucleus forms around each of the two adjacent groups of telophase 2 chromosomes yielding unreduced gametes - Mutation acts to eliminate second meiotic division during the formation of female gametes resulting in unreduced gametes - Two unreduced gametes will yield a tetraploid - Union of an unreduced gamete with a normal gamete will yield a triploid

Explain Deletation/Insertion mutations

- Insert or delete one base pair - Everything downstream from deletation/insertion is changed

What are the main features of all inversions

- Inversion loops during meiosis 1 - Reduction of recombination frequencies of linked genes often to 0% - Reduced fertility from unbalanced or deleteted meiotic products - Some inversion may be directly observed as an inverted arrangement of chromosomal landmarks and pericentric

Male or female? Heterogametic sex Homogametic sex

- Male is heterogametic sex - Female is homogametic sex

Chromosomes

- Means colored body - Presence of chromosomes in nucleus - Chromosomes carries genes - Chromosome splitting in cell replication in which each daughter cell formed receives an identical complement of chromosomes - Each organism of the same species has the same number of chromosomes - Chromosomes contain various types of proteins but amount and kind differ from one cell type to another - Amount of DNA in chromosomes of the same cell type is constant - Chromosome = 50% DNA, 50% protein - One double stranded molecule of DNA per chromosome

What traits did Mendel study?

- Mendel studied the inheritance of morphological characters - seed shape - seed color - flower color - flower position - pod position - stem length

Explain mRNA

- Messenger RNA carries coding information for gene being expressed - Most nucleotides in mRNA specify amino acids -Sequence of nucleotides in a gene specifies the sequence of nucleotides in a molecule of mRNA - Sequence of nucleotides in mRNA specifies the sequence of nucleotides in the polypeptide chain

Explain Neurospora

- Model organism - Grows on minimal medium - Filaments: consist of a mass of branched threads separated into interconnected, multinucleate compartments allowing free exchange of nuclei and cytoplasm - Each nucleus contains of a single set of 7 chromosomes - Can synthesize all metabolic components other then biotin - If the biosynthetic pathways needed for growth are controlled by genes then a mutation in a gene responsible so synthesizing an essential nutrient would be expected to render a strain unable to grow unless the strain was provided with the nutrient

Explain DNA replication

- Molecule is self replicating - Synthesis in only 5' to 3' direction - Strands of the original duplex separate and each individual strand serves as a template for the synthesis of a new strand - Replica strands are synthesized by the addition of successive complementary nucleotides - New nucleotides are added only to the 3' end of the growing chain - End result is a single double stranded molecule becomes replicated into two copies with identical sequences

Explain Robertsonian translocation

- Nonreciprical translocation - Centromeric fusion of acroscentric chromosomes with loss of small amount of material - 2 chromosomes fuse into one - in humans chromosome 14 and 21 most commonly involved in this event - leads to familial form of Down Syndrome which accounts for 1 in 5 Downs Syndrome births - Carrier is phenotypically normal - Chromosomal material lost is small and reduntant - Carrier has 45 chromosomes - Carrier will show decreased fertility - Increased risk of Down syndrome children due to aberrant segregation in meiosis - Down syndrome birth incidence is 1 in 3 - Actual incidence is about 15%

How does the cell nucleus contribute to inheritance

- Nuclei of male and female reproductive cells undergo fusion in the process of fertilization - Male gamete contributes only nucleus to next generation

How large is PARp How large is PARq

- PARp is 2.7 Mb - PARq is 0.2 Mb

Explain the ability to taste PTC

- PTC is an artificial chemical - Family studies showed ability to taste PTC is a trait inherited as Mendelian dominant - European populations 70% can taste, 30% cannot - African/Asian origin 90% tasters - Australian 50% tasters - Ability to taste PTC is quantitative , cut off is 0.2 mM - Most sensitive tasters can taste concentrations as low as 0.001 mM - Most insensitive tasters fail to detect concentrations as high as 10 mM - Most of the variation between tasters and non tasters is due to ajar taster polymorphism, also difference due to other genes, gender and environmental factors - About 5% of heterozygous tasters get classified as non tasters and at least 5% of homozygous non tasters get classified as tasters - Molecular basis of taster polymorphism is known to reside in a taste receptor protein known as hTAS2R38 - Several alleles of the gene - Most common forms of the protein differ by 3 amino acids at scattered positions along the protein - Allelic forms are known as PAV and AVI - Key amino Acids in PAV protein are proline, alanine and valine - 3 amino acids in AVI protein are alanine, valine and isoleucine - PAV form allows ability to taste PTC

What are some other autosomal recessive human conditions

- Phenylketonuria - Tay-sachs disease - Cystic fibrosis - Alkaptonuria - Sickle cell anemia - Xeroderma pigmentosum

Explain the Y chromosome

- Produces 26 distinct proteins involved in male development and fertility - critical region is SRY which is the sex determining region of the Y chromosome - codes for transcriptional factor and testis determining factor - many repeat sequences which can undergo gene conversion and serve as sites for homologous recombination - regions where there is no X-Y recombination theres is steady selection pressure for genes in the Y chromosome to undergo mutation al degradation into nonfunctional states - any Y linked gene that is nonessential will tend to degenerate gradually over time because of the accumulation of mutations

Explain sex lethal protein (SXL)

- Product of Sxl gene is sex lethal protein SXL - SXL is RNA binding protein that binds with RNA transcripts of several genes and causes female specific RNA processing - Absence of SXL transcripts undergo male specific processing - Small bursts of SEX activity in early female development leads to self sustaining production of SXL because of feedback in the processing of more Sxl transcripts

What are cohesins

- Proteins associated with the inside of the centromere - proteins that interact with the DNA of the centromere in a lasso like manner holding the two sister chromatids together - When cohesins dissociate nothing is holding the two sister chromatids together and results in separation during anaphase

Explain the difference between P, F1 and F2

- RFLP analysis shows directly the difference between P and F1 round seeds - RFLP and morphological analysis shows identical nature of F1 progeny of reciprocal crosses - mendel selfed F1 progeny to produce F2 progeny - hybrid nature of F1 progeny is revealed in F2 - F2 ratio = 3 round: 1 wrinkled - F1 hybrids express only dominant phenotype and all are heterozygous - 3:1 ratio produced in F2 progeny is central to transmission genetics

Explain Translation

- RNA --> amino acid/protein - Synthesis of polypeptide chain specified by the nucleotides of the mRNA - Translation occurs with the synthesis of charged tRNA molecules called amino acyl-tRNA synthetase enzymes - Matching of specific 3 base sequence with specific amino acid - If you don't have a particular synthetase enzyme you cant make any proteins - Protein synthesis occurs on ribosomes - Bases in mRNA read as non overlapping codons - Matched with anti codons - Amino acid carried by tRNA added to growing polypeptide chain - Each codon recognized by unique tRNA molecule and specific amino acid inserted - 61 codons specify amino acids - 3 termination codons - Genetic code is redundant in 3rd base codon - only 45 unique tRNA's

Differences between RNA and DNA

- RNA is single stranded - Sugar is ribose instead of deoxyribose - Uracil replaces thymine - 3 different types of RNA

Explain how a telomere is created

- RNA is used as a template to extend the 3' end of the telomere - after one tandem repeat is synthesized the enzyme is translocated so another can be synthesized - 5' strand is extended using the lagging strands synthesis protocal - final product still have a 3' overhand - the overhang is subject to degradation and must be protected - telomere forms a loop by opening a double stranded section to bind to its complement - numerous protiens associate to further protect the region from degradation

Explain the steps of southern blot

- Re digested DNA produces a smear on gel electrophoresis which is not visible at this stage - DNA is denatured in alkali and becomes single stranded by raising pH and disrupting H bonds - DNA is immobilized by transfer onto nitrocellulose paper by capillary action - single stranded radioactive probe is added to filter in sealed bag - hybridization at 68 degrees C for several hours with gentle mixing - excess unbound probe is removed by washing - x ray film exposed by filter reveals presence/location of desired fragment

Explain 9:4:3 ratio in epistatis

- Recessive genotype at locus 1 masks the expression of phenotype at locus 2 - Coat colors in mammals - Appearance of new phenotype in F2 is non Mendelian inheritance and not one gene inheritance

How are some mutations silent?

- Redundancy in the genetic code - Mutation in non coding region of the gene - Change to chemically similar amino acid - If change is away from active site and change to similar chemically amino acid the enzyme will be fine

Explain roles of receptor protiens

- Regulate cellular activities in responce to molecular signals from the growth medium or other cells - Responsible for most metabolic activities - Essential for the synthesis of organic molecules and for generating the chemical energy needed for cellular activities

Explain the Avery-MacLeod-McCarty Experminet

- Repeated Griffith process and got the same results in a test tube - Gets a transformation even when solution had its proteins and RNA broken down by protease and RNase - When DNase is added to the solution to break down the DNA no transformation is observed - Therefore the transforming factor is believed to be DNA

Explain X-linked signal elements (XSE)

- SXL burst created by X-linked genes known as X-linked signal elements or XSE - If amount of XSE is high the early burst occurs and the organism becomes female - If amount of XSE is low early burst does not occur and organism becomes male

Explain how structure dictates function in the DNA molecule

- Sequence of bases could be copied by using each of the separate partner strands as a pattern for the creation of new strand with complementary bases - DNA could contain genetic info in coded form in the sequence of bases - Changes in genetic information (mutations) could result from errors in copying when the base sequence becomes altered

Explain incomplete dominance

- Single gene, 2 allele interaction - Distinct from codominance - Example of non Mendelian inheritance involving one gene and two alleles - The phenotype of heterozygous genotype is intermediate between the phenotypes of the homozygous genotypes - Amount of red pigment is determined by the amount of enzyme that the I allele produces - i allele encodes for an inactive enzyme and the ii flowers are ivory - amount of the critical enzyme is reduced in Ii heterozygotes and the amount of red pigment in flowers is reduce which makes them pink - Cross II with ii makes all pink flowers in F1 - Pink progeny by self pollination gives 1:2:1 ratio - Amount of pigment produced depends of the amount of enzyme present - Any phenotype resulting from incomplete dominance or codominance cannot be made into a true breeding line

Explain F'

- Sometimes Hfr strains revert to the F+ state, this is called F' - produced from incorrect F factor excision - resulting F factor carries some main chromosome genes - used to study bacterial genes in a particular diploid state - useful in determining dominance and recessive nature of bacterial alleles

What does suffix ter mean in karyotypes

- Suffix ter is used for terminal, therefore pter and qter refer to the terminal parts of the short and long arms

Explain tetraploids

- Tetraploids are represented as 4X - Possess four chromosome sets - Can be autopolyploid or allopolyploid

Explain triploids

- Three sets of chromosomes represented as 3X - Most triploids are autopolyploids - Arise spontaneously or by manipulation - Produced from fertilization of haploid gametes with diploid gamete - sterile - Exploited in commercial fruit production because sterilitiy results in no seed production - Sterility due to unbalanced gametes produced during meiosis - Prophase 1 results in two different pairing configuration of homolouges from - Trivalent or normal bivalent with univalent

Explain tRNA

- Transfer RNA - 45 different molecules responsible for ferrying amino acids to ribosomes during translation - 3 base sequence anticodon at one end of tRNA - Amino acid attached at opposite end of tRNA

Explain allotetraploid

- Type of tetraploid - Chromosome sets from more than one species by natural or artificial means - Plant breeders routinely try to mainiplulate plant genomes to imporive and combine favourable traits

Explain autotetraploids

- Type of tetraploid - Spontaneously doubling of chromosome number from 2X to 4X - Complete non disjunction during first cleavage of embryo development - Routinely induced using colchicine - usually fertile since even number of chromosomes produces mostly balanced gametes - Most of not all gametes have balanced 2X chromosome content

What does a highly concerved sequence mean

- When the variation between organisms is very small we say the sequence is highly conserved

Explain the evolution of the Y chromosome

- X and Y chromosomes began their existence as a pair of autosomes - stated divergance is DNA sequences and gene content 300-350 million years ago - once SRY had evolved the Y chromosome began divergence - region of possible X-Y recombination became progressively restricted to the telemetric regions - successive blocks of genes were removed from the region of X-Y recombination by chromosome rearrangements, mostly inversions - additional scrambling of gene order in the Y chromosome

Explain dosage compensation in mammals

- X chromosome inactivation - occurs at the 64-128 cell stage of embryogenesis - 1 X randomly remains active all others inactivated - descendent cells have same X inactivated - inactivation site marks a bend in the chromosome in mitotic metaphase which is a visible manifestation of X inactivation - structurally abnormal cells that lack inactivation would die - Females are a mosaic for X chromosome expression - X inactivation accomplishes dosage compensation

The X chromosome is in group blanks The Y chromosome is in group blank

- X chromosome is in group C - Y chromosome is in group G

Explain sex determination with exceptional flies

- Y chromosome is not male determining - XXY embryos developed into morhologically normal females - XO embryos develop into morphologically normal males but are sterile - Y chromosome is necessary for normal male fertility - Y chromosome contains 6 genes needed for normal sperm production - Genetic determination of sex depends on X-linked gene called Sexlethal because some mutant alleles are lethal when present in males - Sxl gene is active in females and inactive in males - Embryo with ratio of X:A of ½ develops as male - Embryo with X:A ratio of 1 develops as female - Embryo with X:A ratio between ½ and 1 is intersex - This was proven wrong - Sex is based on Sxl reacting to amount of XSE which is determined by number of X chromosomes - Autosomes do not play a role in sex determinination

What is interphase? What are the stages of interphase?

- a cell can only do its job in interphase, cannot do its job if it is dividing - G1, S and G2 phases

Explain large scale genome sequencing

- ability to determine the correct sequence of the base pairs that make up the DNA in an entire genome and to identify the sequences associated with genes

Explain a nondisjunction exception to normal sex linked eye color for flies

- about 1 in every 2000 flies were an exception to the expected - results from occasional failure of two X chromosomes in the mother to separate from each other during meiosis - some egg cells will have two X chromosomes and some will have none - exceptional white eyed females had two X chromosomes and a Y chromosome - Exceptional red eyed males had single X but no Y - Demonstrated validity of chromosomal theory of heredity

Explain the important gene activities in histone modifications

- acetylated histones tend to bind DNA more loosely and usually render chromatin more accessible to transcription - methylated histones can promote or impede transcription depending on the particular histone residue that is modified - histone modifications are important features of chromatin remodeling that takes place in the regulation of gene activity

Explain crossing over

- actual breakage and exchange of non sister chromatids of synapsed homologous chromosomes - occurs at the 4 strand stage after DNA replication has occured - when no crossing over occurs the alleles present in each homologous chromosome remain in the same combination resulting in 4 nonrecombinant chromatids - when crossing over occurs the outermost alleles in two of the chromatids are interchanged resulting in 2 recombinant and 2 nonrecombinant chromatids - sites of crossing over are random - probability of crossing over between two genes increases as physical distance between the genes increases

Explain antibiotic resistance mutants in bacteria

- add antibiotic to medium and only resistant bacteria will grow

Explain histone modifications

- addition of one or more acetyl, methyl or phosphate groups - histone acetylase, methylase or phosphorylase are reversible by histone deacetylase, demethylase, or dephosphorylase enzyme - have important biological functions and important gene activities

Explain 1 + 2 <--> 3 + 4 translocation

- adjacent 1 or disjunctional segregation - homologous centromeres go to opposite pole but each normal chromosome goes with part of reciprocal transloactaion - all gametes formed have a large duplication and defiencey for the distal parts of the translocated chromosomes - pair of gametes originating from 1 + 2 pole are duplicated for distal parts of blue chromosome and deficient for distal part of red chromosome - pair of gametes from the 3 + 4 pole have reciprocal defficency and duplication

Explain 1 + 3 <--> 2 + 4 translocation

- adjacent 2 or nondisjuctional segregation - homologous centromeres go to the same pole at anaphase 1 - gametes have large duplication and deficiency of the proximal part of the transloacated chromosome - pair of gametes from the 1 + 3 pole have duplication pf the proximal part of the red chromosome and a deficiency of the proximal part of the blue chromosome - pair of gametes from the 2 + 4 pole have reciprocal deficiency and duplication

What are some benifits of DNA markers

- advantages of multiple codominant alleles - often have large numbers of alleles - high levels of heterozygocity means most matings are informative - heterozygous genotypes distinguishable from homozygous - simple repeat polymorphism: each allele differs in size according to the number of copies it contains of a short DNA sequence repeated in tandem - differences in size are detected by electrophoresis after amplification of the region by PCR - usually have many codominant alleles and majority of individuals are heterozygous

Explain autosomal recessive inheritance for rare disease traits

- affected individual appears with no evidence of disease in earlier generations - affected child had normal parents but may have affected sibilings - males and females equally affected - affected child usually progeny of consanguineous mating - normal individuals marrying into family considered homozygous for the normal allele - affected individuals usually have normal progeny - pedigree illustrates characteristics of inheritance due to a rare recessive allele with complete penetrance - most homozygous recessive genotypes result from mating between heterozygous carriers in which each offspring has a 1/4 chance of being affected

Explain the semiconservative model expectations for the Meselson and Stahl experiment

- after transfer to 14N a band of lighter density will appear and gradually become more prominent as DNA is replicated and divided - after one round of replication each duplex consists of one light and one heavy strand so all molecules will be an exact intermediate between 14N and 15N - after two rounds of replication the duplexes containing an original parental strand are intermediate in density but there is also an equal amount of duplexes consisting of two light strands so a light band and an intermediate band will be seen - after a third round of replication DNA duplexes of light and intermediate DNA are expected in a 3:1 ratio - after 4 rounds of replication the ratio between light and heavy is 7:1

Explain a linkage group

- all genes on one chromosome form a linkage group but may not all be linked - number of linkage groups is the same as the haploid number of chromosomes of the species

Explain dosage compensation of X linked genes

- all organisms have a chromosomal basis of sex determination - males and females have a different number of sex chromosome genes - should result in anomalies but does not because dosage compensation is required - solutions include up regulate X chromosome genes in males or down regulate in females

What is autopolyploid

- all sets from one species

What is allopolyploid

- all sets from two or more species

Explain acrocentric centromere postion

- almost at the end - D, G groups and Y chromosome - Forms an I shape at anaphase

Explain 1 + 4 <--> 2 + 3 translocation

- alternate segregation - gametes are all balanced an euploid - none have duplication or deficiency - gametes from 1 + 4 pole both have reciprocal translocation - gametes from 2 + 3 pole both have normal chromosomes

Explain histone tails

- amino ends of the histone proteins - 25% of the total length - accessible to enzymes that modify particular amino acid residues

Explain polymerase chain reaction (PCR)

- amplifies a specific fragment/sequence of genomic DNA to a pure state - identifies experimentally manufactured replicas of fragments whose original templates were present in the genomic DNA - require extremely small amounts of DNA - some prior knowledge of DNA sequence is required -ex fingerprint analysis

How do you analyze traits to get ratios

- analyze traits separately because they are behaving independently - texture: (315 + 108) round : (101 + 32) wrinkled = 423:133 = 3.18:1 - Color = (315 + 101) yellow : (108 + 32) green = 416:140 = 2.97:1 - 315:108:101:32 = 9.06:3.11:2.91:0.92 - total number of progeny divided by 16 gives smallest number, multiply by 3 to get 3/16 number, etc - chi squared analysis would be needed to validate hypothesis - gametes carry one allele for each trait - segregation of alleles of each gene is independent of the other gene - heterozygous genotype is WwGg - 4 gametes produced in equal frequencies by F1 dihybrid heterozygote - use a punnett square to visualize events

What are the 3 classes of bacterial mutants that are often used

- antibiotic resistance - nutritional mutants - carbon source mutants

Explain the discontinuous replication of the lagging strand in prokaryotic organims

- antiparallel nature of DNA means template strands are different - template can only be read 3' to 5' - if template running 3' to 5' is direction of replication fork movement the new strand is synthesized continuously and called the leading strand - if template running 3' to 5' is opposite direction of replication fork movement the new strand is synthesized discontinuously and called the lagging strand - lagging strand synthesis is discontinuous moving from the replication fork backwards until previous fragment is reached - small pieces called Okasaki fragments of approx 1000-2000 bp long - each fragment requires primer synthesis

Explain genetic markers

- any difference in genetic content that can be traced though several generations - phenotypic difference - correlates with location in genome which may not be known - used to track disease states - need controled mating

Explain depolymerization of spindle

- as microtubules are formed new tubulin subunits are added to growing end of polymer - several proteins help regulate balance between polymerization and depolymerization - only chromosomal microtubules that make contact with a kinetochore will be stabilized

List the different patterns of inheritance in humans

- autosomal dominant - autosomal recessive - sex linked dominant - sex linked recessive

How are chromosomes arranged in a karyotype

- autosomes in metaphase are rearranged systematically in pairs from longest to shortest - numbered 1 to 22 with 1 being the longest and 22 being the shortest

Explain why peas were used in mendels experiments

- available in alternate forms - self fertilization - pure stocks - reproductive features - relatively small space needed to grow each plant and relatively large number of progeny that could be obtained gave opportunity to determine the number of different forms in which hybrid progeny appear and to ascertain their numerical interelationships

Explain rolling circle replication

- bacterial and eukaryotic viruses - does not include theta shape - replication starts with a single stranded cleavage at a specific sugar phosphate bond in a double stranded circle - cleavage produces a 3' end and a 5' end - DNA is synthesized by the addition of successive deoxyribonucleotides to the 3' end with simultaneous displacement of the 5' end from the circle - as replication proceeds around the circle the 5' end rolls out as a tail of increasing length - as single stranded tail is extended a complementary chain is synthesized with the single strand as the template which results in a double stranded tail - displaced strand is covalently linked to the newly synthesized DNA in the circle - replication does not terminate - forms a tail that may be many times longer than the circumference of the circle - long tail consists of tandem linear repeats of the DNA sequence within the circle - each complete replication around the circle adds another tandem repeat to the tail

Explain how semiconservation replication can be seen at the chromosomal level

- base analogue 5-bromo-deoxyuridine replaces thymidine and base pairs with adenine - DNA replication takes place in the presence of BUdR and is efficiently incorporated into the newly synthesized strands in the place of thymidine - after 2 rounds of DNA replication sister chromatids have different chemical content - chromosome that has undergone one round of replication have DNA molecules that are each half labeled - second round of replication allows the DNA duplex in one chromatid to be fully labeled and the other to be half labeled

Explain proof reading in DNA replication for prokaryotic organisms

- base pairing rules determine base acceptance - mistakes occur in every 10^5 nucleotides - 3' to 5' exonuclease activity of the polymerase is the proofreading function - must check every nucleotide before moving on because can only look back one base pair

Explain what Chargaff found

- base pairing rules recognized before Watson and Crick - A = T, G = C, A + T = G + C - number of purines = # of pyrimidines - this is true for all organisms

What are purines

- base that has double ring structure - adenine and guanine - heterocyclic aromatic compound

What are pyrimidines

- base that has single ring structure - heterocyclic aromatic compound - thymine and cytosine

Explain the nucleosome

- bead like units of chromatin - 2 of each histones H2A, H2B, H3 and H4 and 200 nucleotide pairs forms the bead - one molecule of H1 forms bridge between beads - nucleosome fibre supercoils next and is stabilized by the H1 proteins forming a 30 nm coiled fibre

Explain 9:4:3 ratio with Harebells

- blue is wild type and usually dominant - recessive mutant strains are white and magenta - new phenotype appearing in F1 suggests non mendelian genetics - due to enzymes

Explain proofreading in DNA replication of eukaryotic organisms

- both polymerases proof read new DNA synthesized - recognition of the appropriate incoming nucleoside triphosphate depends on base pairing with the opposite nucleotide in the template chain - DNA polymerase usually catalyzes the polymerization reaction that incorporates the new nucleotide at the primer terminus only when the correct base pair is present - major DNA polymerase also have exonucelase activity that breaks a phosphodiester bond in the sugar phosphate backbone and removed the terminal nucleotide from the 3' end which is a 3' to 5' exonuclease - presence of an unpaired base activates 3' to 5' exonuclease activity which cleaves unpaired nucleotide from the 3' end - proof reading can only look back one base

Explain burst times and burst sizes of the lytic phage

- burst time varies between 20-40 mins - burst sizes varies from 50-200 phage particles

How can histones be dissociated

- by placing chromatin in a high salt environment - eliminates the electrostatic attraction - causes histone s to dissociate from DNA

Explain how mosaicism of dosage compensation can be expressed phenotypically

- calico cat: one hair color gene is X linked - human X linked recessive absence of sweat glands

How are lytic phages detected

- can not directly enumerate or characterize - presence of phage detected by mixing a phage sample with a sensitive bacteria in a soft overlay agar system - initial phage particles attach to cells and cause these bacterial cells to burst - released phage particles attach to cells and cause these bacterial cells to burst - released phage particles will infect adjacent cells repeating the process - phage only infects actively growing bacteria - after 15 hours bacteria stop growing because nutrients are exhausted - law of bacterial growth is spotted with clear plaques indicating reproducing of phage particles - phage can be characterized genetically

Explain autosomal trisomies

- can result in live births - size of chromosome is inversely related to frequencies and proportional to severity of anomalies - only trimsomy 13, trisomy 18 and trisomy 21 are compatible with live birth - trisomies for all autosomes except chromosome 1 are detected in spontaneous abortions - three chromosomes form a trivalent - distinct parts of one chromosome are paired with homologous parts of each of the others - at metaphase trivalent is oriented with two centromeres pointing toward one pole and other centromere pointing towards other pole - at end of meiotic divisions one pair of gametes have two copies and the other pair has a single copy - trisometric chromosome can form one normal bivalent and one univalent - in anaphase 1 bivalent segregate normally but univalent will proceed to one random pole - when mated with normal individual a trisomic is expected to produce trisomic or normal progeny in a 1:1 ratio

Explain how connections between nucleus and genes were established first and then chromosomes and genes

- cell nucleus first discovered in 1831 - nuclear division accompanies cell division known in late 1860s - gametic nuclear fusion in fertilization - 1880s discovery of chromosomes - chromosomes located in the nucleus - homologous chromosomes orderly segregate in cell division and gamete formation - chromosomal complement features

Explain G2 phase

- cell prepares for division - additional growth and accumulation of energy reserves

What is the G0 phase

- cells that never divide again or that will only divide at some time in the distant future with a specific stimulus remain in G0 phase

Explain meiosis 2

- centromere separation in anaphase 2 - 4 product cells that are haploid and genetically different - differences in genetic content arise from random segregation of homologous pairs and crossing over

Explain why cancer cells are not restricted by the Hayflick limit

- certain mutations that reactivate the telomerase gene - mutations that abrogate normal controls over cell division such as DNA double stranded break checkpoint

Explain CAF-1

- chromatin assembly factor 1 - ferries the histones into the replisome area and attaches to PCNA

What are chromatin loops

- chromatin fibres are folded into small chromatin loops with a DNA content of approx 100kb each

What are chromatin domains

- chromatin loops are further organized into chromatin domains with a DNA contend of approx 1Mb each

Explain eukaryotic chromosome problems

- chromatin structure must be replicated - chromosomes are linear - replicating the 5' ends after removal of the primer sequence cant be done

How do deletations occur

- chromosome breakage and reunion - low rate - can be induced by xrays and certain chemicals - homologous recombination between repeated DNA sequences known as ectopic recombination - if both copies of the DNA sequence have the same origin along the DNA then the configuration is a direct repeat - if the direct repeats undergo pairing and homologous recombination the result is deletation of the material between the direct repeats - inverted repeats have repeated sequences that are in reverse orientation and include genes that are important for male fertility

Explain how X inactivation occurs

- chromosome condensation initiated at a site called XIC near the centromere on the long arm - XIC incudes a transcribed region designated Xist - transcription of Xist is the earliest event observed in X inactivation - spliced transcript of Xist does not contain and open reading frame encoding a protein - functions as a structural RNA as transcription of Xist continues the spliced transcript progressively coats the inactive X chromosome spreading outward from the XIC - other molecular changes occur along the inactivated X chromosome that are associated with gene silencing - heavy cytosine methylation in 5' regulatory regions, deacelyation and methylation of histones and aggregation of heterochromatin specific DNA binding protiens

How can karyotypes be labeled

- chromosome paint: each chromosomes is a different color. labeled by different fluorescent dyes - Giemsa stain with G bands

Explain diakinesis phase

- chromosomes are maximally condensed - chiasmata are free of all synaptonemal complex - chiamasta undergo terminalization - terminalization moves the points of interwinding to the end of the chromosomes - nuclear membrane and nucleolous disappear - centrosomes migrate - spindle apparatus is assembled - kinetochore spindle fibres attach to the kinetochores - only one active kinetochore for each double stranded chromosome

Explain leptotene phase

- chromosomes become visible as slender threads as they condense - two sister chromatids are tightly aligned and cannot be identified individually - chromatin appears to be condensing in association with proteinaceous material known as the lateral element - ends of chromosomes (telomeres) associate with specific regions of nuclear membrane - association with nuclear membrane is necessary for the pairing of homologous chromosomes to occur - pairing of homologous chromosomes result in formation of synaptomemal complex

Explain the different chromosome sets within a karyotype

- chromosomes organized into groups by size and centromere position - seven sets denoted A-G - short arms = p and are upward - long arms = q and are downward - centromeres are aligned - within each arm the regions are numbered from the centromere to the telomere - some divisions can be subdivided further by the interbands in them which are numbered consecutively and indicated by a digit placed after a decimal point following the main division

Explain the E. coli chromosome How many base pairs? How many replicons? How many origins of replication?

- circular chromosome - 4.6 MB - 1 replicon - 1 origin of replication - origin of replication is 245 bp long

What are the two configurations of alleles that are possible in heterozygous individuals

- cis: both wild type alleles on the same chromosome resulting in pr+vg+/prvg genotype - trans: one wild type allele and one mutant allele on each chromosome resulting in pr+vg/prvg+ genotype -

Explain cotransduction

- co transduction of 2 markers indicates their close proximity - specific frequency calculations are used to determine distances between closely linked markers

Explain psudogenes

- collection of genetic sequences that have lost ability to produce proteins though evolution

Explain unequal crossing over in red green color blindness

- color vision is mediated by 3 light sensitive protein pigments present in the cone cells of the retnia - Light sensitivities of the cone pigments are related to blue, red and green - Genes for red and green pigments are on the X chromosome and separated by less than 5Mb - Red and green pigments arose from duplication of a single pigment gene and are still 96% identical they can pair and undergo unequal crossing over - This is the genetic basis of red green color blindness - Red and green pigment genes pair and crossover takes place in region of homology - Result is duplication of green pigment gene in one chromosome and deletation in another chromosome

Explain polyploidy

- common in plants - leads to increase in size - even number of chromosomes because odd number reduces fertility - rare in animals due to difficulty in regular segregation of sex chromosomes

Explain chromosomal scaffold

- composed of a number of nonhistone chromosomal proteins including a member of the SMC2 family - extends along the chromatid - 30 nm fibre becomes arranged into a series of loops radiating from the scaffold

Explain the synaptonemal complex

- composed of lateral elements of the two chromosomes which are separated by the central element of very dense protein material - synaptonemal complex is only visible under electron microsope

Explain prophase in mitosis

- condensation of chromatin into chromosomes by condensins - double stranded chromatids joined at the centromere - two chromatids held together by the centromere due to duplication of chromosomes in interphase - centromere is specific base pair sequence - when centromere base pair sequence becomes associated with proteins on the outside this forms the kinetochore - proteins associated with inside of centromere are cohesins - nucleolous disappears and nuclear membrane disintegrates - spindle assembly begins - chromosomal fibres attach to the kinetochore

Explain the circle method for complementation tests

- connect strains that are the same with a line - 2, 3, 4, 6 are all interconnected so they are the same - 1 and 5 are left over, they are also the same - so two different complementation groups in total - minimum is one complementation group - if you have 6 strains they can have 1-6 complementation groups

What is the spindle apparatus made of

- consists primarally of microtubules formed by polymerization of protein tubulin - at least one RNA-protein complex regulates tubulin polymerization and microtubule organization

Explain conjugation between Hfr and F- cells

- contact though pilus - conjugation bridge - DNA transfer begins within F factor - copy of main chromosome dragged along - F- cells retains F- because not all of the F factor is transfered since conjugation bridge will break before that can occur - recombination between transfered DNA and recipients DNA is possible creating a genetically recombinant F- cell

Explain recombination nodules

- control crossing over - lie across the synaptonemal complex - contain enzymes and proteins needed to accomplish crossing over

Explain G bands

- created by staining reagent called Giemsa - Bands that are specific for each pair of homologs - Bands allow smaller segments of each chromosome arm to be identified - replicate late in S phase and are gene poor - allows individual chromosomes to be identified - fine chromosomal detail visible

Explain a test cross

- cross between an organism with dominant phenotype but unknown genotype with homozygous recessive individual - expected progeny is 1/2 Ww and 1/2 ww - the phenotype of the progeny reveal the relative frequencies of the different gametes produced by the heterozygous parent because the recessive parent will only contribute recessive alleles

Explain the recombination of phages

- cross between different phages - both phage genomes must be in the same place at the same time for recombination to occur - suseptible bacterial cells injected at a high multiplicity of infection which is the ratio of phage to bacterial concentration with two strain of phage simultaneously - variants producing different plaque sizes and different host ranges

Explain Chiasma

- crossing over - physical breakage of DNA strands and an exchange of non sister chromatid fragments

What is second division segregation

- crossing over occurs - spores appear in pairs - distance from centromere to gene A is given by % recombination - % recombination = ½ SDS/total octads x 100 - ½ SDS because only ½ of spores are recombinant the other ½ parental centromere-allele combination

Explain pericentric inversion

- crossover products are all monocentric - could be included in a gamete but duplication and deficiency usually cause inviability - products of recombination are not recovered - among chromatids not participating in crossing over one carries pericentric inversion and other has normal sequence

Explain topoiomerase 2

- cuts both strands of DNA

What does del mean in karyotypes

- deletation

What are the different forms of chromosomal structural mutations

- deletations - duplications - inversions - translocations

Explain initiation in DNA replication

- deoxyribonucleotides can only be added to an existing partially double stranded polynucleotide sequence - short RNA primer - elongated by adding successive deoxyribonucleotides to the 3' end - at the end of synthesis primer RNA is removed and replaced with equivalent stretch of DNA

Explain different centromere positions

- determines the appearance of daughter chromosomes as they separate in anaphase - in the middle is metacentric - off center is submetacentric - almost at the end is acrocentric - at the end is telocentric but this is not seen in humans - chromosomes that do not form a centromere are acentric - chromosomes with two centromeres is dicentric

Explain the autosomes role in sex

- determining when in early development the commitment to sexual differentiation takes place - In normal XXAA or XAA embryos commitment takes place after 15 cycles of division - XA haploid embryos commitment takes place one cell cycle later when the amount of XSE is sufficient to induce Sxl - XXAAA embryos commit at cycle 13 and have too little XSE to induce Sxe - Others commit at cycle 14 and have enough XSE to induce Sxl - Result is intersex

Explain genetic map

- diagram of a chromosome showing the relative positions of the genes

Explain DNA markers

- difference detected directly in DNA - DNA sequence difference - Requires DNA sequencing - Require manipulation of genomic DNA - Serve as landmarks in DNA molecules which allow genetic differences in individuals to be tracked - using DNA markers as landmarks geneticitsts can identify positions of normal genes, mutant genes and breaks in chromosomes - detection of DNA markers requires that the genomic DNA be fragmented into peices of a managable size - Direct study of DNA eliminates the need for prior identification of genetic differences between individuals - DNA markers can be in noncoding DNA - Genetic markers need controlled mating, DNA markers dont - DNA markers behave in codon manner, not dominant or recessive manner

Explain the Dihybrid test cross

- direct test of independent assortment - use F1 double heterozygote and double recessive homozygote individual

Explain the lysogenic life cycle

- does not lead to immediate lysis of host cell - injected DNA is integrated into bacterial chromosome and becomes a prophage - few phage genes expressed - bacterial cell can not be infected by another particle of the same type - viral DNA replicated each time bacterial cells chromosome replicates - all progeny bacteria carry prophage - under stressful conditions prophage is induced the genome is excised and lytic life cycle beings

Explain what happens when a mutation is in a promoter

- does not recognize start codon and no protein will be produced

Explain seed color

- dominant yellow - recessive green - staygreen gene: gene product/enzyme involved in chlorophyll breakdown

Explain unwinding in DNA replication of prokaryotic organisms

- done by helicase - one helicase molecule for each replication fork

Explain the B for of DNA

- double stranded right handed helix - A pairs with T - G pairs with C - hydrogen bonds between bases of the two strands stabilize the molecule - one turn every 3.5 nm - 10 base pairs per helical turn - antiparallel - DNA molecule is amphipathic - Nitrogen bases are flat and hydrophobic - Stack one on top of another to exclude max amount of water - Hydrophobic core wrapped by sugar phosphate backbone which is hydrophillic - base stacking stabilizes the molecule - has major and minor groove - grooves are areas where sugar phosphate backbone does not protect nitrogenous bases - enzymes need to access nitrogen bases directly and do this at the major and minor grooves

What does dup, dir dup and inv dup mean in karyotypes

- dup means duplication - dir dup means direct duplication - inv dup indicates inverted duplication

How are pairs of centrosomes formed

- duplication of a single centrosome in interphase

Explain tandem duplications of chromosomes

- duplications are side by side, ex Bar - Can create even more copies of duplicated region by unequal crossing over which is actually type of ectopic recombination - When they undergo synapis these chromosomes can mispair with eachother - Crossover within mispaired part of the duplication will cause a triplication and a reciprocal product that has lost the duplication - Triplication will cause more severe abnormallies

Explain chromosome territories

- each chromosome arm occupies a chromosome territory - in cells cycling though mitosis the chromosome territories are disrupted when the chromosomes condense and the cell divides by reconstituted again in next interphase - chromatin is carefully folded in a way as to avoid knocks and tangles - stretches of DNA that are near each other are chemically crosslinked, isolated and sequenced to determine which parts of which chromosomes are in close proximity - chromosome territories are correlated with gene densities - territories of chromatin containing relatively few genes tend to be located near periphery of the nucleus or near nucleolous - territories of chromosome domains that are relatively gene rich tend to be located toward the inheritor of the nucleus

Explain sticky ends

- each end of the cleave site has a small single stranded overhang that is complementary to the other end

Explain the important points of the Watson and Crick model of replication

- each parental strand remains as an uninterrupted unit - each parental strand serves as a template for the synthesis of a complementary strand - end result of DNA replication is the formation of two daughter duplexes each identical in nucleotide sequence to the parental duplex

Explain experiment that showed linkage and recombination of genes in a chromosome

- early 1900's Bateson and Punnett were working with the sweet pea plant, different then the pea plants Mendel worked with - found certain dihybrid crosses failed to produce expected 9:3:3:1 ratio in F2 - some combinations showed up more frequently and others less frequently then expected - Morgan found similar deviations with Drosophila

What cells are not restricted by the hayflick limit

- embyonic stem cells - cancer cells

Explain the naming of restriction enzymes

- enzymes come from prokaryotes - italicize abbreviation of prokaryote - roman numeral for number in which it was found - normal letter means strain

Explain the error rate of DNA synthesis

- error rate of base pairing is 10^-5 resulting from tautomeric shifts in the bases - this is much too high so proofreading is required

Explain incomplete linkage

- evidenced by progeny who exhibited a combination of traits not exhibited in the parents - homologous chromosomes can undergo an exchange of segments when they are paired - allows parental genotypes and recombinant genotypes to be observed

Explain complete linkage

- expect genes on the same chromosome to always move together - all offspring inherit one of the two parents exact genotypes

Explain sex chromosome trisomy

- extra X chromosomes in human genotme is less deleterious due to X inactivation - all X chromosomes except one are inactivated by Barr bodies and epithelial cells

Explain sex chromosome abnormalties

- extra sex chromosome have phenotypic effects that are mild compared to autosome trisomies - single active X principal results in silencing of most X linked genes in all but on X chromosome - Y chromosome contain relatively few functional genes - Trisomy X (47,XXX) - Double-Y (47-XYY) - Klinefelter Syndrome (47, XXY) - Turner Syndrome (45,X)

Explain amount of X and Y chromosomes males and females have and how gametes fuse to become male or female

- females have two X chromosomes - male will have one X chromosome along with a morphologically unmatched chromosome called the Y chromosome - Y chromosome pairs with X chromosome in meiosis of males - Fertilization of X egg with X sperm becomes XX zygote and develops into a female - Fertilization of X egg with Y sperm becomes XY zygote and develops into male

Explain the two types of offspring an attached X female can produce

- females who have material attached X chromosome along with a paternal Y chromosome - males with material Y chromosome and paternal X chromosome

Explain the fractal globule

- folded interphase chromatin - chromatin thread follows a path that allows dense packing without knots or tangles - coils of interphase chromatin allow easy access and unwinding - folding enables regions of chromatin to undergo unfolding or reorganization with minimal disturbance to nearby regions

Explain the 95% confidence level

- for any individual progeny if the probability of success is 1/256 then the probability of failure is 255/256 - apply the product rule for sample size of n: probability of failure = (255/256)^n - then probability of sucess = 1 - (255/256)^n - for a confidence level of 0.95: 0.95 = 1 - (255/256)^n Solve 0.95 = 1 - (255/256)^n 0.05 = (0.9960937)^n log(0.05) = n x log(0.9960937) -1.30103 = n x (-0.0016998) 765 = n

Explain the spaces between chromosome domains

- form network of channels - channels are large enough to allow passage of molecular machinery for replication, transcription and RNA processing - molecules gain access to chromatin by passive diffusion

Explain chromosomal abnormalities in humans

- formation of unbalanced gametes results in abnormal chromosome complements in the zygote and embryo/fetus - 15% of recognized pregnacies result in spontaneous abortions and many are due to chromosomal abnormalities - probably underestimates since very sever abnormalities abort before implantation

Explain gene density and chromosome size

- gene density is not correlated with chromosome size - two of the smallest chromosomes 19 and 22 have the highest gene densities - two of the largest chromosomes 4 and 5 have the smallest gene densities

Explain symbolism of genetic crosses in genetics

- gene named for the mutant form - lower case letters indicates mutant is recessive - + indicates wild type allele - slash separates homolouges a+b+/ab - no punctuation between alleles on the same chromosome - alleles written in the same order on both homologues - semicolon separates genes on different chromosomes a+/a ; c+/c - dot separates genes with unknown linkage a+/a . d+/d

Explain R bands

- gene rich - replicate early

When doing a dihybrid test cross what does independent assortment indicate?

- genes are in different chromosomes - possible metaphase I alignments of homologous chromosomes are equally likely - double heterozygote will produce all four possible gametes

Explain syntenic

- genes on the same chromosome - may not be genetically linked depending on relative positions

Explain the interpretation of mendels results

- genetic elements do not mix or blend - transmitted unchanged though multiple generations - can recombine to produce original true breeding parental like plants in F2 and beyond

Explain how structure is related to function in DNA

- genetic material must be able to be replicated accurately so that the information will be replicated and inherited by daughter cells - genetic material must have capacity to carry all of the information needed to direct the organization and metabolic activities of the cell - sequence of amino acids in protein determine its chemical and physical properties - a gene is expressed when a protein is synthesized - requirement of genetic material is that it direct the sequence in which amino acid units are added to the end of a growing protein molecule - genetic material must be capable of undergoing occasional mutations in which the information it carries is altered - mutant molecules must be capable of being replicated so mutations will be heritable - accounts for the evolution of diverse organisms though the slow accumulations of favourable mutations

X linked recessive inheritance shows

- grandfather to grandson transmission though daughters - more males then females affected - never see male to male transmission

Explain telophase 1 in meiosis

- haploid set of chromosomes consisting of one homolog from each bivalent is located near each pole - nuclear membranes reform and chromosome condensation may occur or cells may proceed directly into meiosis 2 - Cytokinesis may or may not occur

How are Drosophila different in the ends of their chromosomes

- have specialized non-LTP retrotransposable elements at the chromosome tips instead of tandem repeats

What are some examples of X linked human diseases

- hemophelia - red green color blindness - testicular feminization

Explain metaphase 1 in meiosis

- homologous pairs are moved to the center of the cell by attachment to kinetochore spindle fibres - two chromosomes align on opposite sides of equatorial plane

Explain segregation of a single gene

- homozygous genotypes are indicated experimentally by the observation that hereditary traits in each variety are true breeding which means that plants produce only progeny like themselves when allowed to self pollinate - outcome of genetic cross does not depend on which trait is present in the male or female, reciprocal crosses yield the same result

Explain gel electrophoresis

- hundreds of millions of fragments need to be separated before analysis - electrophoresis moves charged molecules in an electric field - for DNA agarose gel electrophoresis is the usual procedure used - DNA is charged negative - DNA moves to positive electrode/anode - movement depends on charge/mass ratio - for DNA this means size determines rate of migration - smaller fragments move faster - larger molecules move slower - small fragments are less pigmented then large fragments - DNA fragments are visualized by staining with ethidium bromide or other DNA specific stain, the potent mutagen interclases between bases - Ethifium bromide is a flat molecule and absorbs UV light and florecents orange - A red filter is used when taking a picture so white areas appear - has limited sensitivity so billions of copies of fragment are needed to see - nontoxic stains are SYBR but it is less sensitive - must run a lane of molecular weight markers as a control - multitude of factors that affect how far a fragment moves so molecular weight markers lane must be run everytime - plot log of size vs distance to get a linear line

What are some human autosomal dominant conditions

- huntington's disease - Marfan syndrome - Familial Hypercholesterolemia - Brachydactyly - most show reduced pentrance and variable exprestivity

What does helicase do

- hydrolyzes ATP to drive the unwinding reaction

Explain interference in three point test crosses

- i = 1 - COC - positive when fewer than expected double crossovers occur - negative if more double crossovers than expected occur - changes with distance between markers, the shorter the distance the greater the interference - when i = 1 there are no double crossovers observed - when i = 0 the expected number of double crossovers equals the number of observed double crossovers - For Drosophila when distance is less than 10 mu i = 1 - for most organisms when distance is greater than 30 mu i = 0

Explain the difference between the frequency of crossing over and the frequency of recombination

- if 1 meiotic cell in 50 has a crossover then the frequency of crossing over is 1/50 = 2% but the frequency of recombination is only 1% - each crossover results in 2 recombinant chromatids and 2 nonrecombinant chromatids - 2% crossing over corresponds to 1% recombination because only half of the chromatids in each cell with an exchange are actually recombinant

How can WW and Ww genotypes be distinguished from eachother

- if you inbreed WW then you only get WW - if you inbreed Ww then you get WW, Ww and ww

Explain the G1 phase

- immediately following cell division - cell is actively growing - ATP is consumed as subcellular organelles are synthesized and the cell performs its designated work

Explain the features of single gene inheritance

- in diploid organisms genes always come in pairs of 2 alleles - alleles may be the same or different, homozygous or heterozygous - gametes only carry one allele - both alleles are equally likely to be in the gametes - diploid state is returned to in fertilization

Explain dosage compensation in Caenobhabitis elegans

- in females protein complex decreases transcription of each X chromosome to ½ - Therefore the two X chromosomes in the female will have the same amount of transcriptional activity as one X chromosome in the male

Explain metacentric centromere position

- in the middle - in A, some C and F groups - arms are of approx equal length - forms a V shape at anaphase

Explain why risk of downs syndrome increases as the mother ages

- incidence 1 in 700 live births - at age 40 incidence is 1 in 100 - Believed to correlate to length of time that human oocyte are in prophase 1 - Homologous chromosomes remain synapsed - Longer time in prophase 1 increase risk of tetrad instability - Non disjunction in older women usually occur in anaphase 1 - If abnormality in chromosome structure is the cause risk of reoccurrence in subsequent children is very high up to 20% of births - High risk is caused by chromosomal translocation in one of the parents in which chromosome 21 has been broken and become attached to another chromosome

Explain incorporation of error in DNA replication

- incorrect or mismatched nucleotide is incorporated at a rate of about 10^-5 per template per round of replication - vast majority of incorporation error are corrected immediately after they occur by proofreading

What is a T even Phage

- infects E.coli

How are processes manipulated so that the genetic makeup of bacteria can be established

- interrupted mating - Wollman and Jacob in late 1950's Basic protocol - mix Hfr and F- stains together - Remove samples at timed intervals - samples treated to break up mating pairs - samples plated on selective medium - allows detection of recombinant - only F- cells that have received the ability to synthesize a specific nutrient will survive on the medium

What does inv mean in karyotypes

- inversion

Explain condensis

- involved in addition condensation required for interphase to metaphase state transistion

What does i mean in karyotypes

- isochromosome which is two identical arms attached to a single centromere

Explain chimeric gene

- joins 5' end of green pigment gene with 3' end of red pigment gene, - if crossover point is nearer 5' end then resulting gene is mostly red so deuteranopia is caused - if crossover point is nearer 3' end then most of green pigment gene remains in tact and deuteranomaly is caused - opposite causes red color blindness

Explain histones

- largely responsible for the structure of chromatin - small proteins made of 100-200 amino acids - 20-30% of their amino acids contain lysine and arginine which are both positive - positive charge allows histone molecules to bind to DNA that is negative - bind tightly to eachother - DNA is wrapped around an octamer of histone proteins - H1, H2A, H2B, H3, and H4 types - H1 associated with linker DNA - Apart from H1 histone molecules are remarkably similar between different organisms - associations with histones is first level of organization of eukaryotic DNA - Histones must dissociate from the DNA before replication can occur - Histones associate immediately as DNA replication is proceeding - histone octamers are formed from random assortment of old and new histone polypeptides

Explain multigenic deletations

- larger deletations - some deletations large enough to be visible in basic karyotypes - smaller ones visible with advance karyotyping stains - sever effects of deletations due to gene expression imbalances - because genes for structural proteins are distributed thoughout the genome any deletation of significance will delete structural protein genes - recessive mutations carried by homologous chromosome in deleted region is expressed

What the the different stages of prophase 1 in meiosis

- leptotene - zygotene-synapsis - pachytene - diplotene - diakinesis

Explain bidirectional replication in eukaryotes

- linear - replicates bidirectionally - multiple replication bubbles along the DNA molecule - each replication bubble originates at an origin of replication - bubble grows larger as replication fork proceeds outward from the origin of replication - replication forks of adjacent bubbles fuse where they meet

Explain Intragenic deletation

- loss of a small amount of DNA within a gene - loss of single genes activity - behaves as other null alleles with loss of function

Explain diplotene phase

- loss of most of the synaptonemal complex causes repulsion - prevention of complete separation of chromaids is caused by... - cohesins associated with the centromeres hold sister chromatids together - chiasmata are holding homologues together at one ore more positions where synaptonemal complex is still present

Explain what spindle fibres are Explain the microtubule organizing center Explain centrioles

- made of microtubules which are made of alpha and beta tubulin which form diamers that are taken from the cytoskeleton - microtubule assembly is directed by the microtubule organizing center - microtubule organizing center and centrioles are found in the centrosome of animal cells - centriole role in animal cells is not primary responsible for spindle assembly - centrioles are made of microtubules, 9 triplets and 2 singles - centrioles self replicate

Explain elongation of DNA replication in eukaryotic organisms

- major DNA polymerase and other proteins work together as the DNA polymerase complex - major DNA polymerase is polymerase delta and elongates the DNA - polymerase delta is recruited by other proteins in the DNA polymerase complex - elongation reaction is catalyzed by DNA polymerase is the formation of a phosphodiester bond between the free 3' OH group of the chain being extended and the innermost phosphorous atom of the nucleotide triphosphate being incorporated into the 3' end

Explain initiation of DNA replication in prokaryotic organisms

- major DNA polymerases can not initiate DNA synthesis on a single stranded template - requires short double stranded segment with free 3' hydroxyl - primase is the enzyme that produces RNA primer in prokaryotic cells - in E.coli primase is DnaG from the dnaG gene - 2-5 nucleotides long

How can a deletation in a chromosome be detected

- making use of the fact that a chromosome with a deletation no longer carries the wildtype alleles of the gene they are missing - Once a deletation is identified its size can be assessed by determining which recessive mutations in the region are uncovered by the deletation

Explain the relation between double crossovers, recombination frequency and map units

- map units measure how much crossing over takes place between genes - recombination frequency is how much recombination is actually observed in the experiment - double crossovers that do not yield recombinant gametes do contribute to map distance but do not contribute to recombination frequency - if the region between genes is so short that no more than one crossover can be formed in the region in any one meiosis then map units and recombination frequency are equal - for series of linked genes if the chromosomal region between each adjacent pair is sufficiently short that multiple crossovers are not formed the recombination frequencies between genes are additive

Explain conjugation

- mating between F- cells and F+/Hfr cells - initiated by physical contact via pilus which creates a conjugation bridge - genetic transfer of DNA occurs though the conjugation bridge - conjugation between F+ and F- cells converts F- cells to F+ - DNA replication will occur by rolling circle replication

Explain multiple initations in DNA replication of eukaryotes

- means of reducing total replication time - movement of each replication fork proceeds at a rate of approx 10-100 nucleotides pairs per second - origins are spaces approx 40,000 nucleotide pairs apart - allows each chromosome to be replicated in 15-30 mins - chromosomes do not all replicate simultaneously so complete replication of all chromosomes takes about 5-10 hours

Explain linkage

- measure of recombination between the gene pair

Explain monosomy

- missing copy of an individual chromosome - more frequent then a chromosome gain - more harmful to loose a chromosome then to gain a chromosome - abortions will take place so early that pregnancy is often not detected

What is mitosis? What is cytokinesis?

- mitosis is the division of the nucleus - nucleus can survive without division of the cytoplasm, this will produce a binucleus cell - mitosis produces genetically identical cells - mitosis can be conducted with haploid or diplod cells - cytokinesis is the division of the cytoplasm

X linked dominant inheritance shows

- more females than males display trait - females transmit to both females and males - all daughters of affected males are affected - never see male to male transmission - apparent in every generation

Explain linkage of chromosomes

- more genes than chromosomes - many genes located on the same chromosome - genes on the same chromosome cannot segregate independently

What is aneuploidy

- more or less of one chromosome or part of chromosome - unbalanced translocations - monosomy - trisomy - duplications - deletations

What does mos mean in karyotypes What does a / mean in karyotypes

- mos means mosaic and an individual composed of two or more genetically distinct types of cells - / separates karyotypes of clones in mosaics

Explain chain elongation in DNA replication for prokaryotic organisms

- multi protein complex responsible for DNA synthesis - called pol III holoenzyme - contains 2 DNA polymerase III proteins which is the major polymerase enzyme one for each template strand - at least 7 other proteins present and enables the assembly of complex beta clamp which holds pol III onto the template

Explain how Okasaki fragments are joined

- multistep process - removal of RNA primer nucleotides - Replacement with DNA nucleotides - ligation of adjacent fragments

Explain complementation groups

- mutants tested pairwise - Each complementation group represents mutations on a single gene

Explain acentric centromere position

- no centromere - unstable because they cannot be maneuvered properly in cell division - lost at cell division

Explain anaphase 1 in meiosis

- no separation of centromeres - homologous chromosomes separate from one another and proceed to opposite poles - chromosomes are still double stranded - at each pole there will be 1/2 of the original chromosomes - physical separation of homologous chromosomes in anaphase is the physical basis of Mendels principal of segregation - centromeres of the sister chromatids are tightly stuck together and behave as a single unit

Explain how a white female x white male --> whole brown litter

- non mendelian genetics - whiteness can be due to mutation in gene - these mice will also have a normal (brown) gene that it contributes along with the defective white gene - this results in brown babies

How is downs syndrome caused

- nondisjunction, one gamete will contain two copies of chromosome 21 and another will contain no chromosome 21 - If gamete with two copies of chromosome 21 participates in fertilization then trisomy is produced - Chromosome 21 is a small chromosome and is less likely to undergo crossing over - Noncrossover bivalents have difficulty aligning on the metaphase plate because they lack a chismata to hold them together, this increases risk of nondisjunction, 40% of trisomies occur this way - Nondisjunction of chromosome 21 is more likely to happen in oogenesis then spermatogenesis - 3% are not by nondisjunction but by abnormality in chromosome structure

Explain balanced translocations

- nonhomologous chromosomes have an interchange of part but not all of the parts are present

Explain chromatid interference

- nonrandom choice of chromatids in successive crossovers - excess of four strand double crossovers is positive interference and results in a maximum frequency of recombination greater than 50% - excess of two strand double crossovers is negative interference and results in maximum frequency of recombination less than 50%

Explain the choice of which X chromosome to inactivate

- not always random - in marsupials X chromosome inactivated is always the one contributed by the father - female marsupial expresses the X linked genes she inherited from her mother

Explain crossing over frequencies in males and females

- not always the same - extreme difference in Drosophila with no crossing over in males - human males show only 60% recombination frequencies compared to values in females

What are the two techniques for the separation and identification of genomic DNA fragments

- nucleic acid hybridization - polymerase chain reaction (PCR)

Explain 30 nm chromatin fibres

- nucleosome fibre compacts into a shorter thicker fibre with average diameter ranging from 300-350 A - string of chromosomes forms a series of stacked right handed coils in which each nucleosome is attached to the neighbor by linker DNA that stretches nearly linerally across the opposite side of the coil - in each revolution around the fibre axis the path of the linker DNA closely resembles the shape of a seven point star - nucleus of a nondividing cell 30 nm chromtain fibre is organized into higher order structures

Explain elongation in DNA replication

- nucleotide monomers can be added to the 3' end of the growing daughter strand - strand elongation can take place continuously only in one direction in which the DNA duplex is unwinding - continuously replicated strand is the leading strand - strand that has its 5' end facing the direction of unwinding is the lagging strand

Explain restriction site

- nucleotide sequence recognized for cleavage by restriction enzyme - some restriction enzymes cleave sites assemetrically and others cleave symmetrically, can result in sticky ends or blunt ends - restriction site of a restriction enzyme reads the same on both strands

Explain monoploid incests

- occur naturally in some incests in which males are derived from unfertilized eggs - monoploid males are fertile because gametes are produced by modified meiosis in which chromosomes do not separate in meiosis 1

Explain submetacentric centromere position

- off center - B, some C, E groups and the X chromosome - forms a J shape at anaphase

Explain map distances and frequencies of recombination

- one map unit = 1% recombination = 0.01 frequency of recombination = 1 cM - 1 map unit can be defined as the length of the chromosome in which on average 1 crossover is formed every 50 cells undergoing meiosis

What is a transposon

- one or more genes unrelated to transposition that can be mobilized along with transposable element - often cause of antibiotic resistance - mobilizes and inserts into conjugative plasmid

Explain translocations

- one part of the chromosome moved to another non homologous chromosome - reciprocal or centric fusion of two chromosomes - centromere position and chromosome size of one chromosome of each homologous pair is changed - two new recombinant chromosomes are produced - start with 4 chomosomes in 2 pairs - end with 4 chromosomes but 0 pairs

Explain specialized transduction

- only a few genes can be transfered - genes transferred varies with virus - one or the other is transferred - subsequent infection leads to transductants showing new phenotype or genotype - using gal- strain of E.coli transductants will be gal+

What are the three types of tetrads that can be produced

- only parental genotype is a parental ditype - only recombinant genotypes is a non parental ditype - one of each genotype is a tetratype

Explain the events of pre-initation for DNA replication in eukaryotic organisms

- origin recognition - helicase unwinds double stranded DNA - single stranded binding proteins protect the template - topoisomerase releases tension

what events occur in pre-initiation of DNA replication for prokaryotic organisms

- origin recognition - unwinding DNA double helix - stabilization of single strands - tension relief - synthisization of primer oligonucleotides

What does p mean in karyotypes What does q mean in karyotypes

- p is the short arm - q is the long arm

What are some factors that affect how far a fragment moves in gel electrophoresis?

- pH in buffer - how thick the gel is, the thicker a gel the slower things move - thick gels dont work with big fragments, decreased thickness of gel makes it difficult to work with - temperature and humidity - how electric field is set up, voltage and current - time

Explain zygotene-synapsis

- pairing of homolgous chromosomes facilitated by the synaptonemal complex which is a protein structure that helps hold aligned homologous chromosomes together - synapsed homologues are called bivalent

Explain complementation test with neurospora

- part of sexual reproduction is formation of heterokaryon - easy to examine phenotype of heterokaryon - fusion of filaments means both nuclei types in each cell and both genomes are expressed - hererokaryon is grown on minimal medium and scored - growth or returning to wild type phenotype = complementation = mutant form are from different genes - no growth mean mutant phenotypes are still expressed = no complementation = mutant forms in the same gene

Explain the activities of the protein coat in transduction

- particle can attach to a bacterial cell and inject DNA into the cell, these activities reside in the protein coat

Define penetrance

- penetrance of a genetic disorder is the proportion of individuals with the at risk genotype who express the trait

Explain temperate phage

- perform specialized transduction - display both lytic and lysogenic life cycles - gene transfer by temperate phage occurs during improper excision - removes bacterial genes and leave viral genes behind - genes transfered are adjacent to insertion site - different temperate phages transfer different genes because they insert at different locations - E.coli lambda is the first completely characterized genome - phage inserts into E.coli chromosome between gal and bio genes

Explain the events in lytic phage infection

- phage attaches to bacterial cell wall - injects its DNA into host cell - phage DNA shuts down synthesis of bacterial proteins and DNA - directs cell synthetic machinery to make viral proteins and copies of viral DNA - Viral proteins and DNA accumulate - Viral particles self assemble - Bacterial cell is broken open and lysed

Explain phage genetics

- phage can be characterized genetically - genes present on phage DNA control phage characteristics including host range and plaque size - host range is the different strains of bacteria that can be infected - plaque size is a measure of how fast the virus can reproduce

Explain the sex pilus

- physical connection between two mating cells - approx 20 genes are needed for assembly - these genes are present on the surface of F+ and Hfr cells

Explain the vital heads role in transduction

- piece of bacterial chromosome packages in a vital head produces a defective phage particle

Explain ordered tetrads

- planes of divison in meiosis 1 and meiosis 2 usually perpendicular and spindles overlap - some fungi produce ordered tetrads in which planes of division are parallel and spindles do not overlap - linear arrangement of spores results and allows gene centromere mapping - chromosomes at each end of the ascus arise from chromatids held together by one original centromere - order of phenotypes in the ascus is different for crossing over then for no crossing over

Explain the F factor

- plasmid in E. coli that confers state of fertility - F+ = carry F factor as a plasmid - F- = do not posses F factor - Hfr = carry F factor integraded into main chromosome - F+ x F- = call cells become F+ - Hfr x F- = Hfr cells unchanged, F- cells remain F- but some are genetically different

Explain the replacement of removed RNA primer nucleotides with DNA nucleotides

- pol 1 uses exonuclease activity to remove nucleotides from 5' end in a 5' to 3' endonuclease activity - pol 1 takes over and moves along removing the one primer ribonucleotide and replacing it with a deoxyribonucleotide one at a time

Explain telophase in mitosis

- pole to pole spindle fibres continue to lengthen - when chromosomes reach the pole they decondense though the dissociation of condensins - nuclear membrane reforms from the remnants of the parent cell nuclear membrane as well as from other portions of the cell endomembrane system - nucleolus reappears - spindle apparatus and astral rays disappear - cytokinesis occurs - in plant cells cytokinesis is cell plate formation and in animal cells it is cleavage furrow formation

What are the different types of spindle fibres

- pole to pole: lengthen cell, dont connect to chromosomes - chromosomal: connect to chromosomes and move them - centrosome anchor: hold centrosomes in place

Explain the molecular structure of DNA

- polymer of nucleotides - sugar and nitrogenous base connected first to form nucleosides - each base is chemically linked to one molecule of the sugar deoxyribose and forms the nucleoside - when a phosphate group is also attached to the sugar it becomes a nucleotide - Carbon atom to which the base is attached is 1' carbon - Nitrogen base on C1 - Phosphate on C5 - each nucleotide has phosphate groups attached to 5'C of sugar and free hydroxyl on 3'C of sugar - phosphodiester bonds between phosphate group and hydroxyl group of adjacent nucleotides - 5' phosphate at one end and 3' hydroxyl at other end - always add to 3' hydroxyl - only triphosphates are added - 3 H bonds between G and C - 2 H bonds between A and T - H bonds stabilize the DNA molecule - Cold climate organisms have more A and T - Hot climate organisms have more G and C

Explain PCR

- polymerase chain reaction - automated cyclical amplification of specific DNA sequence - uses oligonucleotide primers (18-22 bp) and heat resistant enzyme - repetitions of denaturization at 95 degrees, primer annealing at 50-60 degrees and primer extension DNA synthesis at 72 degrees - polymerase keeps going until temperature is bumped up - every round of DNA gets smaller until you only get fragment that you want - repeats several times so the small fragments you want will be in highest quantity - exponential increase in specific desired amplified sequence - 25-35 cycles leads to 99.99% of DNA in sample being the amplified sequence - essentially pure sample - PCR is highly sensitive with only 10-100 template molecules required - 10^4-10^5 times more sensitive then southern blot - large molecules not easily amplified

Explain pachytene phase

- presence of sister chromatids is apparent - chromosomes continue to thicken and shorten - evidence of crossing over

Explain initation for DNA replication in eukaryotic organisms

- primer produced by polymerase alpha - inserts 10-15 ribonucleotides followed by 15-20 deoxyribonucleotides - polymerase alpha is part of a multi protein complex composed of 15-20 polypeptide chains called primosome - main polymerase takes over after 35 nucleotides have been inserted

Explain the removal of RNA primer nucleotides in prokaryotes

- primer removal and replacement occurs with major polymerase complex and runs up against the RNA primer of the next precursor fragment - 3' end of growing strand faces 5' end of the RNA primer - enzyme DNA ligase cannot join these nucleotides because 5' end of primer carries triphosphate - ligase can only join 3' OH and 5' monophosphate - to join fragments DNA polymerase pol 1 is used

Explain RNA primer removal in DNA replication of eukaryotic cells

- primer segment removed in one piece - polymerase alpha reaches previous primer sequence - replication protein A joins the complex - replication protein A stabilizes displaced RNA-DNA strand and recruits endonucleases - endonuclease cleaves displaced fragment - polymerase alpha inserts replacement complementary nucleotides in continuous manner - nucleotides scavenged - ligase seals single stranded nick - RNA and DNA components removed are broken down by enzymes and recycled

Explain genetic distance

- probability of crossing over between any two genes serves as a measure of genetic distance between the genes

What is the basis of genetic mapping

- probability of crossing over between two genes increases as physical distance between the two genes increases

Explain bacterial transformation

- process of bacterial cells taking up DNA from the environment - first mechanism of bacterial recombination recognized - ability to take up DNA from environment depends on physiological state or degree of competency of the cell - normally few cells are competent - transformation can sometimes be used to determine linkage relationships

Explain inheritance of attached X females that are heterozygous

- produce some female progeny that are homozygous for the recessive allele - frequency with which homozygosity is observed increases with increasing map distance of the gene to the centromere - homozygosity can only occur if the crossover between the gene and the centromere takes place after the chromosome has duplicated - this implies that crossing over takes place at the 4 strand stage

Explain large genomes and gel electrophoresis

- produces thousands of bands which smear - even if we know how long the DNA fragment is we cannot distinguish it from others

Explain paracentric inversion

- products of crossing over are dicentric and acentric chromosomes - neither can be included in a normal gamete

What are the parental progeny by definition

- progeny classes in greater numbers - genotypes match the parental genotypes

What are the recombinant progeny types

- progeny classes that are in lower numbers - genotypes that do not match the parental genotypes

Explain PCNA

- proliferating cell nuclear antigen - eukaryotic equivalent of beta clamp so is in the heart of the replication zone

Explain origin recognition in DNA replication of prokaryotic organisms

- protein DNA A - recognizes E.coli ori which is 245 bp in length - begins strand separation

Explain dosage compensation in Drosophila melanogaster

- protein complex that includes at least 1 male specific protein alters chromatin structure of male X and increases transcriptional activity to equal the amount in the two female X chromosomes

Explain antibodies on ABO blood groups

- protein made by the immune system in response to stimulation molecule called an antigen and is capable of binding to the antigen - antibody only recognizes one antigen - defends against invading viruses and bacteria - may be stimulated by antigens that are similar to polysaccharides A and B and that are present on the surfaces of many common bacteria

What can standard interrupted mating be used for

- protorophic, antibiotic sensitive Hfr strain - auxotrophic, antibiotic resistant F- strain

Explain nutritional mutants in bacteria

- prototrophic or auxotrophic Phototrophic: - capable of synthesizing all of the necessary biochemical nutrients from simple inorganic nutrients and organic energy source - grows on minimal medium Auxotrophic - require one or more specific organic nutrients in order to survive - genetic defect in the biosynthetic pathway producing the nutrient - mutant require addition of the specific nutrient they cannot synthesize - amino acid, nitrogen base, vitamin

How are each tetrad types formed

- random alignment of homologous chromosomes and presence or absence of crossing over in synapsed chromosomes results in parental ditype or non parental ditype - independent assortment results in parental ditype or nonparental ditype - crossing over between one of the genes and its centromere results in a tetratype - if two genes are linked: - no crossing over will produce a parental ditype - a single crossing over produces a tetratype - a four strand double crossover yeilds a non parental ditype - two strand doiuble crossovers produce a parental ditype - three strand double crossovers result in a tetratype

Explain some differences between prokaryotic and eukaryotic DNA replication

- rates of synthesis - E.coli 1500 nucleotide pairs/second - single origin of replication - 30 mins for 4.6 Mbp replication - eukaryotic 10-100 nucleotide pairs/second - origins located every 40,000 bp -number of DNA polymerases - prokaryotic have 5 different polymerases - I though IV - III and I in main chromosome - eukaryotic have at least 15 different polymerases

Explain restriction sites

- read the same on either strand - sequence is recognized in all DNA - number of fragments depends of number of recognition sites and form of molecule, the number will be different between linear and circular

Explain characteristics of X-linked inheritance

- reciprocal crosses resulting in different phenotypic rations in the sexes - heterozygous females transmit each X-linked allele to approx half of their daughters and half of their sons - males that inherit an X-linked recessive allele exhibit the recessive trait because the Y chromosome does not contain a wildtype counterpart of the gene - Affected males transmit the recessive allele to all of their daughters but none of their sons - Male transmits his X chromosomes only to his daughters - Female transmits X chromosome to offspring of both sexes

What does rcp mean in karyotypes

- reciprocal translocation

What are the important characteristics of a restriction enzyme

- recognize a single restriction site - restriction site is recognized without regard to source of DNA - most restriction enzymes recognize unique restriction site sequence therefore the number of cuts in the DNA from a particular organism is determined by number of restriction sites present

How do recombinant genotypes come to exist

- recombinant genotypes result from recombination of the F1 females X chromosome in crossing over in meiosis

How is recombination affected by cis and trans configurations

- recombination between linked genes takes place with the same frequency weather the alleles of the genes are in trans or cis configuration - recombination is the same no matter how the alleles are arranged - recombination is characteristic of a particular set of genes

When is crossing over detectable?

- recombination between the marker genes takes place only when crossing over occurs between the genes - crossing over outside of the region between the two genes is not detectable though recombination because genetic markers stay in the nonrecombinant configurations - if two crossovers take place between marker genes and both involve the same pair of chromatids then neither crossover is detected because all of the resulting chromosomes are nonrecombinant

Explain genetic mapping

- recombination frequencies used to create linkage maps - genes close together are treated as additive and used to extend the map - definitive maps are built by using closely linked genes in sequence

Explain cointegrate

- recomposite plasmid caused by recombination - if one of the participating plasmids is nonconjugative and one in conjugative then cointegrate is also conjugative - after conjugation the nonconjugative plasmid can become free of the cointegrate by recombination between the same sequences that created it

Explain why embyonic stem cells are not restricted by the hayflick limit

- relatively high temomerase activity - undergo many cycles of cell division - can differentiate into many different types of specialized cells

Explain VNTR: variable number tandem repeats

- repeat sequence is 10-60 nucleotides - finger print analysis because different alleles are easier to distinguish - highly polymorphic, many have 10 or more common alleles and produces many possible genotypes - DNA fingerprinting uses 6-8 highly polymorphic VNTRs simultaneously - even with multiple alleles a particular chromosome must carry only one of the alleles defined by the number of tandem repeats - any individual genotype can carry at most two different alleles - large number of alleles implies an even larger number of genotypes - with n alleles there are a total of n(n+1)/2 possible end types of which n are homozygous and n(n-1)/2 are heterozygous - if no one allele is exceptionally common then each of the many genotypes in a population will have low frequency - if the genotypes at 6-8 highly polymorphic loci are considered simultaneously then each possible multiple locus genotype is exceedingly rare - very low frequency at any multiple locus genotype is what gives tandem repeat polymorphisms their utility in DNA typing

Explain SSRP: simple sequence repeat polymorphism

- repeat sequence is 2-9 nucleotides in length - these are more common - abundant in the human genome - highly polymorphic in human populations

Explain tandem repeat polymorphisms

- repeats of short non coding nucleotide sequences - repeat size and numbering of repeats varies - recognized as different lengths of DNA with southern blot or PCR - PCR is used more - Use this tequnique for finger print analysis - tandem repeats that are short in length are no useful because it is difficult to distinguish between 5 and 10 base pairs on electrophoresis - when any of the DNA molecules is cleaved with a restriction enzyme that cleaves at sites flanking the tandem repeat the size of the resulting restriction fragment is determined by the number of repeats it contains - molecules of duplex DNA containing the repeats can also be amplified by means of polymerase chain reaction using primers flanking the tandem repeat - whether obtained by endonuclease digestion of polymerase chain reaction the resulting DNA fragments increase in size according to the number of repeats they contain - two DNA molecules that differ in number of copies of the tandem repeat can be distinguished because each will produce a different sized DNA fragment that can be separated by electrophoresis

Explain plasmids

- replicate independently of the bacterial genome - segregate to the progeny when the bacterial cell divides - presence can be detected physically by electron microscopy or by gel electrophoresis - rely on DNA replication enzymes of the bacterial genome - contain sequences that promote their segregation into both daughter cells produced by fission of the host cell - often carry antibiotic resistance genes - carry genes not necessary for cell to survive

Explain carbon source mutants in bacteria

- require a carbon source for energy - glucose is preferred but other sugars can be used - lac- mutants cannot use lactose as a carbon source

Explain centromeres

- required for chromosome stability and transmission in mitosis and meiosis - all eukaryotic chromosomes are linear and have a single centromere

Explain the hayflick limit

- restricted number of doublings of normal cells - due to shortening length of telomeres - telomeres become shorter with each cell cycle - protiens that recognize double stranded breaks are activated and stop cell cycles when telomeres reach a length of 5kb - signal for shortened telomeres to be recognized as a double strand break appears to be the addition of two methyl groups to a lysine residue at position 20 in histone H4 in the telomere associated nucleosomes - most cells in adult human have telomeres of a length that permits only a few division until they are arrested by DNA double stranded break checkpoints

Explain the consequences of no recombination in Drosophila males

- result in all genes located on the same chromosome showing complete linkage - this leads to reciprocal crosses only showing nonrecombinant genotypes - all the alleles present in the male must be transmitted as a group without being recombined with alleles present in the homologous chromosomes

Explain recombination

- result of crossing over between homologous chromosomes - yields daughter chromosomes that carry combinations of alleles not present in the parental chromosomes

Explain the calico cat

- result of random X inactivation in females - two alleles affecting coat color are present in the X chromosome - one allele resulting in orange and one resulting in black - since a male has only one X chromosome a male is either orange or black - female can be heterozygous for orange and black resulting in calico - orange and black patches result from X inactivation - if the X chromosome bearing the orange allele is inactivate the X chromosome with the black allele is active and fur is black - if the X chromosome with the black allele is inactivated and the orange allele is active the fur is orange - white patches are due to an autosomal gene S for white spotting which prevents pigmentation - homozygous SS cat will have more white then heterozygotes

Explain telomerase

- ribonucleoprotien enzyme that is responsible for maintaining telomere length - contains a molecule of RNA that is complementary to the tandem repeat - DNA duplex has a single stranded overhang at the 3' end that is elongated further by telomerase - incorporates essential RNA molecule called guide RNA that contains sequences complementary to the telomere repeat that serves as a template for telomere synthesis and elongation

Explain rRNA

- ribosomal RNA - 3 different types in prokaryotes - 4 different types in eukaryotes - Structural components of ribosomes - Small ribosomal subunit has one piece of RNA - Large ribosomal subunit has 2-3 pieces of RNA - Mutation in rRNA is lethal

What does r mean in karyotypes

- ring chromosome

What does rob mean in karyotypes

- robertsonian translocation

Explain the difference between round and wrinkled seeds

- round is dominant and wrinkled is recessive - wrinkled seeds lack amylopectin which is a branched chain starch and not synthesized in wrinkled seeds because of a defect in the starch branching enzyme 1 which creates the alpha-1,6 glycosidic linkages to make amylopectin -round seeds shrink uniformly - wrinkled seeds shrink irregularly - seeds that are heterozygous have half as must SBE1 enzyme as the wild type homozygous seeds - molecular basis of wrinkled mutation is that the SBE1 gene has become interrupted by the insertion into a gene of a DNA sequence called a transposable element

What is euploidy

- same amount of genome content for all chromosomes - less severe phenotypic effects than aneuploidy - greater survivorship - balanced translocation - triploidy - tetraploidy - inversions

Explain two strand double crossovers

- same chromatids participate in both exchanges - all four products are parental - composes 1/4 of all double crossovers

Explain four strand double crossovers

- second exchange involves chromatids that did not participate in the first exchange - all four products are recombinant - composes 1/4 of all double crossovers

Explain chromosomal deletations

- segment of chromosome is missing - deletation at the end of a chromosome is a terminal deletation - deletation within a chromosome is a intersitial deletation - usually arise spontaneously during gametogenesis - parent does not carry the deletation - effects vary with size of deletation - very large deletation is lethal even when homologous with a normal chromosome - small deletations are usually viable when heterozygous with a structurally normal homolog because the normal homolog supplies the gene products necessary for surivial - if both members of a homolog carry the deletation it is lethal

Explain the different types of replication that were thought to be utilized by DNA

- semiconservative: used one old strand and one new strand - conservative: created one double stranded DNA with all new DNA and one double stranded DNA with all old DNA - Dispersive: created double stranded chunks of old DNA fused with double stranded chunks of new DNA in an alternating patturn

Explain autonomously replicating sequences

- sequences at which DNA replication can initiate - specific proteins in the nucleus recognize certain DNA sequences as origins of replication and bind with these sequences - when replication beings these regions serve as starting points for DNA replication - numbers and activities are coordinated so that each chromosomal region is replicated once in the S phase of each cycle

Explain interkinesis

- short interlude between meiosis 1 and meiosis 2 - similar to interphase - no DNA replication - Chromosomes are double stranded

Explain what BUdR is used for

- shows presence of sister chromatid exchanges - occurs during mitosis - frequency increases with mutagenic treatment such as xrays - correlates with chromosomal and genetic mutational events

Explain what happens when a mutation is at some other point in a enzyme that is not an active site

- silent sometimes, if changed from positively charged to negatively charged amino acid most likely the enzyme will not work

Explain bacterial plasmids

- similar to bacterial chromosomes - circular - one replion

Explain single nucleotide polymorphisms (SNP)

- single nucleotide sequence differences at a specific positioning in the genome occurring commonly in a population - most common in DNA - must be common to be considered a SNP - A-T in some members, G-C, C-G or T-A in other members - any two randomly chosen DNA molecules are likely to differ - defines two alleles for which there could be 3 genotypes, homozygous with T-A at the site in both homolous chromosomes, homozyogus with C-G at both sites in homoguolous T-A in one chromosome and C-G in the other chromosome - occurs in 1 in 1000 bp for non coding sequences and 1 in 3000 for coding sequences - most do not result in any phenotypic change due to redundancy in the genetic code - less common in coding sequences due to natural selection - in humans there are 3 million SNPs, 1 million are studied for diseases - allows genetic risk factors for disease to be identified - typical study compares the genotype of patients with a particular disease with healthy individuals matched for the patients sex, age and ethnic group - comparing the SNP genotype among these groups often reveals which SNPs in the genome marks the location of the genetic risk factors

Explain monoploid organisms

- single set of chromosomes - meiosis cannot take place normally - usually sterile

Explain the requirements for DNA amplification

- single stranded DNA template - all 4 deoxyribonuclotides - enzyme DNA polymerase which is responsible for polymerization of nucleotides - primers complementary to 3' ends of both strands of DNA, polymerase can only extend from partially double stranded DNA in 5' to 3' direction - fragment cant be too long - nucleotide sequence at each end must be known - must have triphosphates

Explain template stabilization in DNA replication of prokaryotic organisms

- single stranded binding proteins - single stranded DNA is unstable and tends to re anneal spontaneously - single stranded binding proteins coat single stranded DNA to ensure availability of template and protect single stranded DNA from degradation - single stranded binding protein has an affinity for single stranded DNA at least 10 fold greater than double stranded DNA

Explain sister chromatid exchange

- sister chromatids undergo a breakage and reunion - breakage and rejoining of strands that have the same polarity in both chromatids - because sister chromatids are identical in DNA sequence sister chromatid exchange normally has no genetic consequence - X rays can break down chromatids and show more sister chromatid exchanges

Explain pratical applications of nucleic acid hybridization

- small part of a DNA fragment can be hybridized with a much larger DNA fragment which is used for identifying specific DNA fragments in a complex mixture, when applied allows the tracking of genetic markers in pedigrees and the isolation of fragments containing a particular mutant gene - DNA fragment from one gene can be hybridized with similar fragments from other genes in the same genome, used to identify different members of families of genes that are similar but not identical in sequence and have similar functions - DNA fragment from one species can be hybridized with similar fragments from other species, allows isolation of genes that have the same functions in multiple species, used to study aspects of molecular evolution such as how differences in sequence are correlated with differences in function and the patterns and rates of change in gene sequences as they evolve

Explain restriction fragment length polymorphisms

- some SNPs occur within restrction enzyme site - can be analyzed using souther blot because RFLPs can change the number and size of DNA fragments produced by digestion with a restriction enzyme - faster and less costly then DNA sequencing - Homozygous and heterozyous genotypes are recognized easily - fragments of different length are produced - Sizes of restriction enzyme fragments produced will be different for A-T vs G-C alleles - in an EcoRI site 5'-GAATTC-3' - DNA molecules with T-A at the SNP will be cleaved at both flanking sites it also is at the middle site yielding two EcoRI restriction fragments - DNA molecules with C-G at the SNP will be cleaved at both flanking sites but not the middle - labeled probe DNA hybridizes near the restriction site at the far left and identifies the position of this restriction fragment in the electrophoresis gel - duplex molecule labeled allele A1 has a restriction enzyme site in the middle and when cleaved and subjected to electrophoresis yields a small band the contains sequences homolous to the probe DNA - duplex molecule labelled allele A2 lacks the middle restriction site and yields larger band - there can be 3 genotypes (A1A1, A1A2, A2A2) depending on which alleles are present in the homologous chromosome - Homozygous A1A1 yields only the small fragment - homozygous A2A2 yields only the large fragment - Heterozygous A1A2 yields both small and large fragments

Explain Pseuoautosomal inheritance

- some genes remain active on the inactive X - genes at tips of both arms - same genes located on Y chromosome - regions of X and Y synapse and undergo crossing over during meiosis - regions called pseudoautosomal PARp and PARq - crossing over in PARp ensures segregation os X and Y during spermatogenesis - mutant allele in the pseudoautosomal region is neither completely X linked or completely Y linked - can move back and forth between X and Y chromosomes because of recombination in the pseudoautosomal region - gene that shows autosomal like pattern of inheritance but is known to reside in a pseudoautosomal region is aid to show psudoautosomal inheritance

Explain what happens when an attached X female is mated with a male carry any X linked mutation

- sons receive mutant X chromosome - daughters receive attached X chromosome - inheritance of X linked gene in male passes from father to son to grandson which is the opposite of normal X linked inheritance

Explain selective replication of genomic DNA fragments

- southern blotting does not allow for isolation of the fragment - amplification is easy but requires DNA sequence information - obtaining a fragment in purified form requires cloning

Explain metaphase in mitosis

- spindle apparatus is fully assembled - poles of each spindle attach to centromeres - chromosomes moved to center of cell - kinetochores must align on metaphase plate before cell progresses to anaphase

Explain anaphase in mitosis

- splitting of the centromere by disolvation of proteins that hold the chromatids together producing two single stranded chromosomes each with their own centromere - Two chromosomes become separate and independent with the dissociation of cohesins from the centromere region - tension of the kinetochore spindle fibres will move the two daughter chromosomes towards opposite poles - pole to pole fibres lengthen which leads to the elongation of the cell - chromosomes reach the opposite poles of the cell

Explain chromosome condensation

- structure of chromatin varies with salt concentration, condensation only occurs when salt concentration is quite low - ordered energy consuming process - protien complex called condesins work together with topoisomerase II to actively coil chromatin - condensin complex composed of at least 5 proteins including two types of proteins called structural maintenance chromosomes, SMC2 and SMC4 - Histone H3 phosphorlation also plays a role - mammals have at least two types of condensin one of which is specific for regions near the kinetochore - partly unfolded DNA has many loops that extend from the central core of scaffold - length of metaphase chromosome reduced by factor of 10^4

What are nuclosides

- sugar and nitogenous base - each base is chemically linked to one molecule of the sugar deoxyribose and forms the nucleoside

Explain heterochromatin

- suppresses recombination - large blocks located near centromeres of chromosomes, some at telomeres and some dispersed throughout chromosome

Explain outcrossing

- surgical manipulation of immature flowers to reveal receptive female structures - excise and disgard undeveloped male pollen and undeveloped male pollen producing structures - expose female structures to mature pollen from another plant - disseced flowers were enclosed in a fine mesh bag to prevent stray pollen from accessing female structure - cross pollenation - creates hybrid by outcrossing plants that differ in one or more traits

Define Reciprocal cross

- swap phenotypes and gender - used to see if gene is located on autosome or sex chromosome

Explain the differences between mitosis and meiosis

- synapsis and crossing over in prophase 1 - crossing over exchanges peices of DNA in homologous chromosomes - synapsis matches homologous chromosomes to prevent frame shift mutations - segregation of homologous chromosomes in anaphase 1 - two nuclear divisions with no DNA replication between them - four cellular products each genetically distinct with 1/2 the amount of DNA of mitotic products - meiosis is part of sexual reproduction

Explain duplications

- tandem or insertional - less detrimental - less survival with increasing size of duplication - limited effect on phenotype - important role in evolution of gene families - synapsis in heterozygotes may show loops for unpaired region

Explain chromosome ends

- telomeres - end of chromosome after removing RNA primer a 3' overhang exists which is about 8-12 bp long - single stranded DNA is subject to degradation which results in shortening of the chromosome with each replication cycle - telomeres have short tandem repeats and proteins - as repeating telomere sequence is being elongated DNA replication occurs to synthesize a partner strand

Explain progenitor cells

- telomeres about 10kb long - can undergo 70 cell divisions before arrest - cells from children can divide more than adults and adult cells can divide more than elderly cells - progressive loss of ability to divide explains why healing process takes longer in the elderly

Explain telomere silencing

- telometric regions associated with proteins that inhibit gene expession

Explain Morgans dihybrid testcross

- testcross between F1 double heterozygote and double recessive homozygote individual - performed dihybrid test cross to analyze linkage and recombination - found that the heterozygote female produced all 4 expected gametes but not in equal proportions - original allele combinations from the true breeding parent predominated, this result was seen for both reciprocal; crosses - concluded - two parts of a single chromosome cannot segregate independently - shifting occurs during meiosis 1 during crossing over - incomplete linkage and recombination from crossing over allowed all 4 classes to be observed

Why is the drosophilla X chromosome well studied?

- the X chromosome was used because they knew many gene that were definitely on the X chromosome because all sex linked traits must be on the X chromosome - if crossing over occurs in either meiosis then 50% parental and 50% recombinant - the percentage of recombinant depends of how often crossing over occurs - some X linked genes show recombination frequency of 0.5 which would indicate independent assortment

Explain a quick mathematical way you can determine the frequency of progeny

- the expected frequencies of any genotype with independent assortment can be obtained by picking the corresponding term in the expression (1/4 WW + 1/2 Ww + 1/4 ww) x ( 1/4 GG + 1/2 Gg + 1/4 gg) - expected frequency of Wwgg in F2 is 1/2 x 1.4 = 1/8

Explain the expectation for unlinked genes on separate chromosomes is

- the gametes of the heterozygous will be 1/4 WG, 1/4 Wg, 1/4 wG, 1/4 wg - all gametes by recessive parent will be wg - the offspring will be 1/4WwGg, 1/4 Wwgg, 1/4 wwGg, 1/4 wwgg - mendel confirmed independent assortment using the test cross - the expected frequencies of any genotype with independent assortment can be

What is the sum rule

- the probability of either of two mutally exclusive events occuring is the sum of their individual probabilities - ex. rolling dice: p(of rolling two 4's or two 5's) = 1/36 + 1/36 = 2/36

What is the product rule

- the probability of independent events occuring together is the product of the probabilities of the individual events - ex. tossing a single die: p(of rolling a 4) = 1/6 -tossing two dice at the same time: p(of rolling two 4's) = 1/6 x 1/6 = 1/36

Define complete penetrance

- the trait is expressed in 100% of persons with the genotype

Explain autosomal dominant inheritance for fully pentrant traits

- trait is observed in every generation - males and females equally affected - affected individuals pass trait to 1/2 of progeny - affected individuals must have an affected parent - normal individuals must be homozygous for normal recessive allele - affected persons have HDhd becuase HD allele is very rare and HDHD genotype would abort

What is a generalization we can make with sexed linked inheritence

- trait will appear more often in one sex and reciprocal crosses produce different results with transmission varying with which parent displays the trait

Explain transduction

- transfer of bacterial gene between bacterial cells mediated by bacterial viruses or bacteriophages - lytic life cycle leads to generalized transduction - any host bacterial genes can be transfered - injected DNA can recombine with new hosts chromosome - DNA transducted is only as big as the size of the viral genome

What are some common bacterial gene transfer mechanisms

- transformation - conjugation - transduction

How is monoploid sterility overcome in plants

- treatment of meristematic tissue with colchicine - colchicine is an inhibitor of the meiotic spindle - when treated cells in the monoploid meristem begin mitosis and replicate normally - colchicine blocks metaphase and anaphase so endoreduplication results - colchicine is removed to allow continued cell multiplication - Creates small sector of tissue that is placed on agar surface where it will develop into a complete plant

Explain downs syndrome

- trisomy 21 - Mental retardation and physical abnormalities such as heart defects - Small in stature because delayed maturation of skeletal system - Muscle tone is poor resulting in characteristic facial appearance - Shortened life span usually less then 50 years - ¾ of trisomy 21 fetuses undergo spontaneous abortion

Explain dicentric centromeres

- two centromeres - unstable because it is not transmitted in a predictable fashion - often lost from a cell division in which centromeres proceed to opposite poles - chromosome will be stretched and forms a bridge between daughter cells - bridge may no be included in either daughter nucleus or it may break with the result that each daughter nucleus receives a broken chromosome - if the two centromeres are close enough together they can behave as a single unit and be transmitted normally - important in the evolution of human chromosome 2

Explain attached-X

- two chromosomes in female joined to a common centromere to form aberrant chromosome - attachment of two of these chromosomes to a single centromere results in a chromosome with two equal arms each consisting of a complete X - Females with an attached X chromosome usually also contain a Y chromosome

What is meiosis

- two divisions of the nucleus but only one replication of chromosomes - haploid cells cannot undergo meiosis - produces 4 genetically different cells

Explain three strand double crossovers

- two exchanges have one chromatid in common - result is indistinguishable from that of a single exchange - two products with parental combinations and two with recombinant combinations - composes 1/2 of all double crossovers

Explain classes in three point test crosses

- two largest classes show no crossing over and are parental - 2 smallest classes are produced by two concurrent crossing over events or double crossing over of gametes - 2 groups of intermediate size are produced by single crossing overs in each region

What is a reciprocal translocation of chromosomes

- two non homologous chromosomes exchange peices

Explain what Gregor Mendel did

- understood hereditary factors abstractly - Existence of genes and the rules governing their transmission from generation to generation were first articulated in 1866 - Showed the rules by which hereditary elements are transmitted from parents to offspring

Define replicon

- unit of DNA that replicates as an individual unit

Explain the X and Y chromosome

- unpaired chromosome - present in all somatic cells of males but only half of sperm cells, other half carry a Y chromosome

Explain the evidence behind patterns of genetic inheritance involving sex chromosomes important in linking genes to chromosomes

- unpaired chromosomes in males of some insects, two of the same in female called the X chromosome - heteromorphic pair in males and 2 of one in females - male specific Y chromosome

Explain tension relief in DNA replication of prokaryotic organisms

- unwinding DNA creates tension - Gyrase cleaves sugar phosphate backbone of both strands and swiveles the ends of the broken strands to release tension then reseals nicks - Gryase topoisomerases are enzymes that catalyze cutting and rejoining of DNA molecules

Explain theta replication

- used on circular molecules - autoradiography of DNA replication confirms circular nature before and during replication of E. coli and plasmid DNA - usual form of replication in prokaryotic systems - area of strand separation by DNA synthesis is the replication fork - circular DNA replication in theta mode can have one or two replication forks

Explain human pedigree analysis

- usually focuses on a rare disease - a diagram of a family tree showing the phenotype of each individual - females are circules - males are squares - diamond is used if sex is unknown (miscarriage) - people with phenotype of interest are colored symbols - recessive alleles are indicated with half colored symbols - mating is indicated by connection with horizontal line - vertical lines are offspring - offspring with sibship are connected by horizontal line left to right in order of birth - successive generations are represented with roman numerals - within a generation all individuals are numbered left to right - traits with different inheritance patterns show different characteristic pedigrees - pedigrees are used to determine likely mode of inheritance

Explain bacterial chromosomes

- usually only have one replicon - circular

Explain Monosomics

- very deleterious because of disturbances to gene product balance and all recessive alleles are expressed - result from nondisjuction in meiosis 1 or 2 during gametogenesis - nondisjuction occurs more in meiosis 1 then 2 - crossing over maintains tetrad stability for correct disjunction - no reported autosomal monosomy human births

Explain viruses

- very simple - not considered alive - obligate internal cellular parasites - do not posses essential metabolic machinery for virus replication - take over host cells protein and nucleic acid syntetic capabilities

Explain the DNA sequence of the amylopectin gene (W)

- w allele contains transposable element which interrupts the gene sequence - W and w are of different sizes which can be seen using a southern blot analysis which distinguishes the alleles as RFLP - molecular analysis allows definite genotypic identification - for all genes alleles are codominant at the DNA level and it is not always possible to easily distinguish between alleles in DNA - EcoRI fragment from the W allele would be smaller then that of the w allele because of the inserted DNA in the w allele, therefore it would migrate faster then the corresponding fragment from the w allele and move to a position closer to the bottom of the gel - Ww shows two bands with the same electrophoresis mobilizes as those observed in the homozygous genotypes - band in the homozygous genotypes is thicker because the single band in the homozygous genotypes comes from the two copies of the same allele and therefore contains more DNA then the DNA in the heterozygous genotype in which only one copy of each allele is present

Explain hydrogen bonds in DNA

- weak bond in which two participating atoms share a hydrogen atom between them - 3 H bonds between G and C - 2 H bonds between A and T - Hydrogen bonding between G and C is stronger and requires more energy to break - amount of heat required to separate the paired strands in DNA increases with the amount of G+C

Explain restriction enzymes and analysis with gel electrophoresis

- when digested with restition enzyme EcoRI the circular molecule yields bands of 4 kb and 6 kb - digestion of the circle with enzyme BamHI results in bands of 3kb and 7 kb - when both EcoRI and BamHI are used together the resulting DNA fragments reveal where the EcoRI and BamHI sites are located relative to eachother - digestion with both enzymes yields bands of 1 kb, 2 kb, 3 kb, and 4 kb - A restriction map is a diagram showing the sites of cleavage of one or more restriction sites along a DNA molecule - restriction enzyme digest shows order of fragments - EcoRI has two bands so one restriction site - separation is based on size, not identity, if two bands are the same size only one will show up - since 10 in total and 6 and 4 band = 10 we are not missing any fragments - BamHI also has one recognition site - using map place different enzymes and use trial and error to match with gel band placements

Explain replication forks in theta replication

- when one replication fork it proceeds around the circle in a single direction from the origin - when there are two replication forks they proceed in opposite directions from the origin and come together on the outer side of the circle at a region known as the terminus of replication

When is southern blot used

- when you want to know if a specific fragment or sequence is present

Explain the conservative model expectations for the Meselson and Stahl experiment

- will always show distinct heavy band and distinct light band - first generation will show ratio of 1:1 - second generation will show 3:1 ratio since heavy molecule will produce one full heavy molecule and one new light molecule and the light molecule will produce two light molecules

Explain the dispersive model expectations for the Meselson and Stahl experiment

- will always show hybrid weighted molecules - will always only show one band

Explain what happens when a mutation is at the end of an active site

- will produce an enzyme with some effectiveness but most likely not enough

Explain what happens when a mutation is at an active site

- will produce no effective enzyme

What does a nucleotide consist of

-2'-dexoyribose - phosphate group - nitrogen base - each nucleotide has phosphate groups attached to 5'C of sugar and free hydroxyl on 3'C of sugar - phosphodiester bonds between phosphate group and hydroxyl group of adjacent nucleotides

Use PKU as an example to explain how genes are affected by the environment

-Genes affect multiple phenotypes -PKU characteristics include: mental retardation, blond hair, blue eyes, fair skin, reduced melanin synthesis - Melanin synthesized from tyrosine - Lack of melanin due to reduced levels of tyrosine because protein not present -Seemingly unrelated trains connected biochemically -Traits influenced by multiple genes -Many genes involved in normal neurological development -Individuals with same PAH allele may have different outcomes -Difference in blood brain barrier development particularly important -Blood brain barrier not complete in children -Differences in other metabolic pathways affect final outcome -Gene interact with environment -Environment can aggravate or ameliorate genetic expression -Limiting phenylalanine intake prevents mental retardation in PKU individuals

Define P generation

-parental generation - true breeding

Explain nucleic acid hybridization

-recognizes presence of a specific DNA fragment or sequence in a sample of genomic DNA - identifies fragments that are present in genomic DNA itself - require greater amount of genomic DNA - relatively large fragments can be identified - no prior knowledge of the DNA sequence is necessary

what is the % recombination for complete linkage?

0%

How can downs syndrome be detected

15-16 weeks by aminocetesis - cells of developing fetus are obtained by insertion of a fine needle though the wall of the uterus and into sac of fluid called the amnion 10-11 weeks by chorionic villus sampling - uses cells from a zygote derived embryonic membrane called the chorion associated with the placenta - 3 fold higher risk of inducing miscarrage

How many different gametes can a specific individual produce assuming independent assortment of the 6 genes: AaBBCcddEeFf

2 x 1 x 2 x 1 x 2 x 2 = 16

What is the human tandem repeat used for telomeres

5'-TTAGGG-3'

Chromatin is blank % DNA and blank % protein by mass

50% DNA and 50% protein by mass

How is a gene defined experimentally?

A set of mutations that make up a single complementation group

Define hydroxlase enzyme

Adds hydroxyl group

Define Catabolic

Breaks things down

Define amino acid

Building blocks of protiens

Define Anabolic

Builds things up

Define complete medium

Complex medium enriched with a variety of amino acids, vitamins and other substances expected to be essential metabolites whose synthesis could be blocked by a mutation

Define minimal medium

Contains only the nutrients that are essential for growth of the organism

Sort in order of smallest to largest: DNA, Genomes, Genes

DNA --> Genes --> Genomes

True or False Genes involved in sexual differentiation are only located on the X and Y chromosome

False Genes involved in sexual differentiation located thoughout the genome, most genes on the X chromosome are not related to sexual dimorphism

True or False Template strand is what you would find online

False you would find the nontemplate strand online. The nontemplate strand is the exact same sequence as mRNA only difference is T is replaced with U

True or False Proteins are the genetic material?

False, DNA is the genetic material, However initially proteins were believed to be the genetic material and DNA was believed to be purely structural Proteins were thought to be the genetic material because they were exceedingly diverse collection of moleucles

What is the slime capsule on S strain cells?

Gelatinous capsule composed of complex carbohydrates

What is one problem with diploid?

If it has dominant trait we don't know if it is homozygous dominant of hererozygous

Explain the difference between incomplete dominance and co-dominance

Incomplete dominance:quantitative variation between the 2 homozygotes and heterozygote Co- dominance: heterozygote phenotype is distinct/unique At the DNA/protein level all alleles behave the same and are codominant

How can independent assortment be distinguished from linkage

Independent assortment = % recombination of approx 50% Linkage = % recombination of less than 50%

Define Central Dogma

Information flow from DNA to visible effect

Define mutant screen

Isolation of a set of mutants fretting any biological process

What is the first genome that was sequenced entirely?

Lambda

What did George Beadle and Edward Tatum do (1940's)

Made connections between inheritance (genes) and enzymes - Produced mutant strain of Neurospora with UV and X- rays - Irradiated spores where used in crosses with an untreated strain - Cultured ascospores on complete medium - Test samples on minimal medium, mutants cant grow and are of interest - Samples from each mutant culture were transferred to series of media to determine whether the mutation results in a requirement for an amino acid, vitamin or some other substance - Found that cells could grow in minimal medium + amino acid - Next tested mutant cells with minimal medium + one of the 20 amino acids for all 20 amino acids - Found that mutant strain could grow on minimal medium + arginine - Arginine is a biochemical pathway so therefore cannot say that arginine is the specific thing the mutant cells need to grow - More likely that strains need a particular enzyme within the arginine pathway to produce intermediate compounds - Tested mutant strains with compounds they believed to be in the arginine pathway - Minimal medium + arginine was used as a control - Found that mutant strains growed on minimal medium + citrulline

Define DNA

Material of genes and genomes

Can humans grow on a minimal medium?

No

What are the primary role of proteins?

Perform cellular activties

Explain how to calculate expected double crossovers

SCOx frequency x SCOy frequency = DCO frequency x total progeny = expected double cross overs

Define Metabolites

Small molecules on which enzymes act

How is the kinetochore formed

Specific base pair sequence becomes associated with proteins on the outside of the centromere

Define Genomics

Study of all genes in an individual organism, molecular organization, function, interaction and evolution

Definie Molecular Genetics

Study of differences between species through the comparison and analysis of DNA

Define classical genetics

Study of genetics through the analysis of the offspring from mating

Define Genetics

Study of individual inherited traits

Explain mendels second law

The Principle of Independent Assortment: segregation of the members of any pair of alleles is independent of the segregation of other pairs in the formation of reproductive cells - genes that are independent are unlinked - test cross is standard method of analyzing two or more traits simultaneously, tests independent assortment WwGg x wwgg - plants with doubly heterozygous genotypes produce 4 types of gametes in equal frequencies - plants with wwgg genotypes produce only wg gametes - possible progeny are WwGg, Wwgg, wwGg and wwgg in equal frequencies - easier analysis especially with genes located on the same chromosome

Explain mendels first law

The principle of segregation: in the formation of gametes, the paired heredity determinants separate/segregate in such a way that each gamete is equally likely to contain either member of the pair - molecular analysis of F2 progeny shows directly the implied genotypic ratio of 1WW:2Ww:1ww - he selfed F2 progeny individually to confirm the genetic content - if you inbreed WW then you only get WW - if you inbreed Ww then you get WW, Ww and ww - this is how you can determine genotype between WW and Ww - homozygous round and wrinkled seeds produced in F2 and F3 generations have phenotypes exactly the same as those observed in the parent generation this makes sense because DNA of each allele remains unaltered unless a new mutation occurs

Why do we need mutants for any kind of genetic analysis?

To understand how something makes a product we need an organism that can't do it

Define Genomic DNA

Total DNA extracted from cells of an organism

True or False Mutants are needed to study genetics

True

True or False Replication protein A is a single stranded binding protein

True

True or False Sex is determined chromosomally at fertilization and phenotypically much later

True

True or false DNA replication is much faster in prokaryotes than eukaryotes Why

True New round of replication is started at the origin every 20 mins even though previous round of replication is not complete

True or False Males phenotype reveals the genotype of the X chromosome that he inherited

True since each male receives his X chromosome from his mother

What amino acids only have one combination?

Trypotophan and Methionine

Do humans have one or two copies of every gene

Two copies because humans are diploid

Define Pleiotrophic Effects

Various, sometimes seemingly unrelated effects of a mutant gene

Define virus

Viruses are obligate internal parasites and cannot reproduce unless in a host cell

Using W as the enzyme explain SBE1

W = active SBE1 produced w = inactive SBE1 produced WW = active SBE1 and round seeds ww = inactive SBE1 and wrinkled seeds Ww = less enzyme sythesiszed but seeds are still round

Who determined the molecular structure of DNA

Watson and Crick

Would % recombination = # of recombinants/total progeny x 100 be true if genes are not linked?

Yes in this case the % recombination will be 50%

What is an example of a primarily haploid organism

algae, yeast

What is an example of a primarily diploid organism

animals

Explain Marfan syndrome

change in cologne, often occurs in tall people

Why does % recombination of genes far apart underestimate the true distance

due to double crossovers

Independent assortment show blank numbers of parental and recombinant classes

equal numbers

Explain insertional duplications of chromosomes

extra copy elsewhere in the genome

Explain tetraplody

four sets of all chromosomes

Explain familial hypercholesterolemia

high levels of cholesterol in young children

What is deuteranomaly

impaired ability to perceive green

What is protanomaly

impaired ability to perceive red

Define variable expressivity

in the population displaying the mutant phenotype the variation in phenotypes observed

What is duteranpia

inability to perceive green

What is Protanopia

inability to perceive red

Are the genes linked or not linked if PD=NPD

independent assortment and no linkage

Are the genes linked or not linked if PD are greater than NPD

linkage

When using codon chart should you match with mRNA codon or anticodon?

mRNA codon

How to calculate map distance for any two genes

map distance = (1/2) x (average number of crossovers in region per meiotic cell) x 100

what is polyspermy

one egg and two sperm cells

What is triticale

only human significant accomplishment using alloteraploid techniques

Define Complementation group

organisms with defects in same single gene

Explain trisomy

otherwise diploid organism that has an extra copy of an individual chromosome

What is a karyotype

pictorial representation of all chromosomes in a cell

What is an example of an organism that utilizes alternation of generations

plants

Explain tolerance in ABO blood groups

prevents an organism from producing antibodies against its own antigens

What is a ring chromosome

rare, results from breakage and rejoining of telomeres of a linear chromosome

Explain brachydactyly

shortening of the fingers

Explain unbalanced translocations

some part of the genome is missing

How can we find out what allele is dominant?

the phenotype of the heterozygote organism tells which allele is dominant

Define penetrance

the proportion of individuals with a specific genotype that actually show to any extent the corresponding mutant phenotype

Explain triploidy

threes sets of all chromosomes

True or False rRNA and tRNA are never translated

true

True or false prokaryotes dont have introns

true

Two plants with the following genotypes are crossed: AabbCcDdEe x AaBbCcddEe What is the probability of seeing an offspring that is homozygous recessive for all five genes assuming independent assortment of all 5

¼ x ½ x ¼ x ¼ x ½ x ¼ =1/256 - this is an average probability - the realistic progeny size needs to be determined by the 95% confidence level