Central Dogma of genetics: DNA -> Transcription -> RNA -> Translation > Protein

Transcription

1 Genetics Midterm 1 Chapter 1: Purines: Adenine (double bond), Guanine (Triple Bond) Pyrimidines: Thymine (double bond), Cytosine (Triple Bond), Uracil Central Dogma of genetics: DNA -> Transcription -> RNA -> Translation > Protein The Genetic Code has 64 Codons (4^3), 61 of which code for amino acids and 3 for stop codons - Chargaff developed base pairing rules Chapter 2: Reciprocal Cross: to test cross alleles on both sexes of parents Test Cross: to cross an unknown genotype individual with a homozygous recessive phenotype Mendel s law of segregation: Alleles segregate independently into gametes Independent assortment: Genes assort independently of other genes into gametes Autosomal Dominant: Children must ve inherited the disease from a parent who had the phenotype of the disease Autosomal Recessive: Children appear to inherit the disease even though the parent s do not demonstrate the phenotype Chi Squared Test : X 2 = (O E)2 E Degrees of Freedom: (# of Phenotype/genotype classes) 1 If P<.05, reject the hypothesis; because this means only 5% of the time will we see similar Chi. Values If P.05 the hypothesis cannot be rejected, this means the Chi. Value obtained, will be seen more than 5% of the time in repeated tests.

2 When to use binomial probability? The female I-1 and her mate, male I-2, had four children, one of whom has albinism. What is the probability that they could have had a total of four children with any other outcome except one child with albinism and three with normal pigmentation? P=1/4 (albino chance) Q= ¾ (anything other than albino chance) P^4+4P^3Q+6P^2Q^2+4PQ^3+Q^4 Select the PQ^3 because this is the probability to get 1 albino and 3 normal Now, subtract from 1-(4*1/4*3/4^3) =.578 chance to get anything OTHER than this combination Chapter 3 Ploidy: how many copies of a chromosome an organism has - Humans have 2n=46, where n is the number of chromosomes o This means humans are diploid Cell division: - Somatic: replication of cells (Mitosis) - Germline: formation and joining of gametes (Meiosis) Tumor Suppressor Gene: Gene that checkpoints and regulates cell mitosis Proto-oncogene: Gene that stimulates cell division

3 X-linked Genes: - An X linked recessive gene may demonstrate itself mostly in men, because they only need 1 copy of the recessive X chromosome while females need both. - X linked Dominant: o men who are X linked dominant pass the phenotype to all daughters, but no sons; o women who are heterozygous pass it down to ½ of all offspring Chapter 4: Effects of Mutations - Loss of Function Mutations: decrease or lose the ability to produce a phenotype o Null/Amorphic Mutation No production of protein o Leaky/Hypromorphic Mutation Production of small amount of protein compared to wild type o Dominant Negative Mutation Mutation produces proteins that interact abnormally with other proteins compared to wild type - Gain of Function: express more activity per allele o Hypermorphic Mutation Alleles produce more than the wild type allele o Neomorphic Mutation Alleles produce a new phenotype Dominance: - Incomplete Dominance or Partial Dominance o When heterozygotes falls in between the homozygotes of both recessive and dominant alleles Ex. Pea Plant Height - Codominance o When heeterozygotes produce both products from both alleles Ex. Blood Types (AB, AO, BO) Lethal Mutations; - Yellow coat color in mice o AY produces Yellow o AA produces Wild Type o YY produces lethal (in this punnet square there is a 2/3 AY and 1/3 AA) Sex-Limited traits: genes that do not reside on the sex chromosomes, but do not exhibit themselves in both sexes equally - Beardedness on Females occurs only 1/4 th the chance of males despite similar genotypes Incomplete penetrance - When the genotype fails to produce the expected phenotype o Ex. Polydactyly - Variable Expressivity o When the genotype only produces less than all the possible phenotypes

4 Ex. If a gene produces deaf, blind, and muteness; but exhibits only deafness this is variable expressivity Pleiotropic Genes - Alteration of many phenotypes by 1 gene o Ex. Sickle cell disease is a mutation in the blood, but affects all body parts Genetic Pathways - Single gene mutation o When 2 genes interact on 2 different pathways to produce a phenotype When 1 of the genes are null, then only the other gene will be expressed Ex. Fly Eye Color - Two-Gene mutation o If both genes on the separate pathways are mutant and null, the phenotype; the end product is not produced Epistasis: - Altered ratio of expressivity from predicted mendelian ratios, this is because other genes effect the expression of that phenotype o Expected Dihybrid cross would be 9:3:3:1 Epistasis is easily detectable when this is no longer the case - Complementary Gene Interaction: o Ratio 9:7 o Genes must act in tandem to produce a phenotype Precursor 1>Gene A>A product>gene B> final product Missing on Gene A or B will not result in a final product - Duplicate Gene Interaction o Ratio: 15:1 o The dominant allele of either gene allow for the phenotype to be present Precursor1>Gene A or B> final product As long as one of Gene A or B are dominant final product is produced - Dominant Gene Interaction o Ratio: 9:6:1 Where if both genes are dominant, the wild type phenotype is produced; if 1 of the genes are recessive then a second phenotype is produced; if both are recessive a 3 rd phenotype is produced PrecursorA> GeneA>Protein A PrecursorB> GeneB>Protein B + A = Wild (9) PrecursorA> Gene(a)> Protein A + none PrecursorB> Gene(b)> Protein B + none = Mutant (6) PrecursorA>Genea>none PrecursorB> Geneb>none = Mutant recessive (1) - Recessive Epistasis o Ratio: 9:3:4 o Where the recessive gene reduces, or inhibits the expression of the dominant gene PrecursorA> GeneA> Protein Synthesis Precursor B> GeneB> Protein Deposition + Protein = Wild (9) PrecursorA>GeneA> Protein

5 PrecursorB>Geneb> no Deposition + protein = Reduced phenotype (3) PrecursorA> Genea>no protein PrecursorB>Gene(b)> deposition/no + no protein = no phenotype (4) - - Dominant Epistasis: o Ratio: 12:3:1 o A dominant allele of one gene masks or reduces expression of a second gene - Dominant Supression: o Ratio: 13:3 o Dominant allele of one gene suppresses the expression of the dominant other gene Chapter 5: Linked Genes - Recombinant Frequency is o r = #recombinant individuals/#total individuals - calculating genetic CentiMorgans o #recombinant/#compared gene This will give you the ratio of recombinant gene to the gene of interest The ratio is equal to the distance in cm on the Sturtevant Map A / in a genotype expression is described as these genes being on the same homolog chromosome Chapter 6: Bacteria F Plasmid = Fertility Plasmid R Plasmid = Resistance Plasmid (to antibiotics) Conjugation: Transfer of genes from Donor F+ cell to F- cell through a pilius - At the orit the F factor relaxes, and gets sent through the Pilius this is called the T Strand o The donor then undergoes Rolling Circle Replication (1 cycle) and restores the double stranded F plasmid o The recipient undergoes replication of the new F factor, and circularizes Hfr Chromosome: - Genes not on the F chromosome are now able to be transferred - At the Insertion Sequence the F chromosome integrates into the bacterial Chromosome o This is now a Hfr chromosome - The relaxosome cuts the new Hfr chromosome at the OriT site just as in a normal F factor, but now the entire chromosome can pass through the pilius o In the recipient cell, the implanted hfr chromosome can crossover with the recipient chromosome Interupted Mating: - Blending the mating process to stop conjugation midway o We can use this to produce time of entry maps, and map the genome of bacteria - Through the use of different Hfr strains, that have different OriT locations and transfer gene s in different directions we predict the genome is circular F cell:

6 - The Hfr chromosome could excise it s F factor, taking with it part of the chromosome o This new F factor can be conjugated to a recipient The recipient then now has a F factor with a donor partial chromosome; making it a partial diploid Transformation: - Recipient bacteria takes up double stranded DNA from surrounding medium o The bacteria breaks up one of the strands and recombines the remaining strand with it s own chromosome Transduction: - Transfer of genetic material from donor to recipient via a bacteriophage - Lytic Cycle o The bacteriophage replicates its chromosome in the bacteria and lyses the bacteria producing new phages - Lysogenic Cycle o Bacteriophage integrates its chromosome into the bacteria s chromosome; making the bacteria a prophage Eventually the prophage will excise and begin the lytic cycle - Generalized Transduction: o Durring lysis and production of phage protein, the bacterial chromosome is integrated into one of the phages This phage goes on to infect another bacteria, inserting the donor bacteria s chromosome o if the donor chromosome crosses over this is called cotransduction a higher cotransduction rate means the genes are closer together - specialized transduction o when a prohpage excises the phage DNA and takes with it part of the bacterial chromosome. These phage DNA are incapeable of lytic cycle and thus does not kill the bacteria