BIOL 264 Study Guide - Spring 2018, Comprehensive Midterm Notes - Dna, Gene, Rna
12 Oct 2018
School
Department
Course
Professor
BIOL 264
MIDTERM EXAM
STUDY GUIDE
Fall 2018
9/5/17
Medel’s Las
• People disoered Medel’s las ere ot alas true- one allele is not always
dominant over the other- ex: variable penetrance and expressivity- linked genes are on
the sae hroosoe ad a’t phsiall separate- two different genes can affect
the same trait (epistasis- genotype of one gene masks phenotype of other gene)
• Molecular basis of recessivity: mutant on one chromosome can make a non-functional
protein- if we see no difference then the wild type allele is dominant over the mutant
allele- if organism lacks enzyme completely and both chromosomes are mutated then
get a different phenotype
• Incomplete dominance: one allele is not completely dominant- heterozygotes have
phenotype that is intermediate between homozygotes- if only have one functional
enzyme, get half as much pigment
• Codominance: phenotypes of both alleles are expressed in the heterozygote
o Ex: I gene controls surface markers on blood cells- IA and IB will be the AB blood
type- IA and IB are codominant over i
• Dominance reflects interactions between alleles at a single locus
• Penetrance and expressivity:
o Variable penetrance: all individuals have same genotype but either phenotype is
hat e epet OR it’s ot there at all- describes proportion of individuals in
population who have genotype AND express phenotype
o Variable expressivity: not an all or nothing- a range of phenotypes (with
eggplants can be range from purple to white)- everyone has phenotype but at
different ranges- it’s there ut etet to hih it’s there can vary
o Variale peetrae AND epressivit: soe have purple, others do’t, some
have intermediate form
• Lethal alleles: alleles that in combination are actually lethal
o Ex: yellow mice- y gene affects coat color AND embryo development- Y is
recessive in regards to lethality component but dominant with regards to coat
color- have to have two Y to have lethality but only one gives yellow coat color
• Multiple alleles: more than two alleles
o Ex: ducks- MR > M > md
• Second law: law of independent assortment- ratios can be altered if multiple alleles
affect same trait- genotypic ratio will always stay same but phenotypic ratio can change
• Recessive epistasis: genotype at one locus masks genotype at other locus
o Ex: flowers- w+ has enzyme to convert from white to pink- m+ has enzyme to
convert pink to blue- w comes before m because enzyme 1 comes before
enzyme 2- w allele is recessively epistatic to alleles at m locus- get 9:3:4-
recessive w will block any expression of m gene (recessively epistatic)- the gene
that controls the expression of the other gene is the one that is epistatic to the
other one
find more resources at oneclass.com
find more resources at oneclass.com
o Ex: Labradors- actual coat color encoded at B gene- B is black and b is brown-
second gene at play here (E protein)- required to take pigment developed by B
gene and put it into the fur- homozygous ee (no functional E gene) will result in
a yellow lab because pigment does not get deposited into fur- E is epistatic to B
locus
o “eeig these ratios tells us e’re dealig ith a oleular patha- two genes
are biochemically linked somehow
• Dominant epistasis: epistatic allele is dominant rather than recessive
o Ex: plants with W_ will be white- end up with 12:3:1 phenotypic ratio
• Some genes in mitochondrial DNA (mtDNA) affect phenotypes- generally transmitted
only through mother lineage b/c orgaelles oe fro other’s gaetes- egg cells
have a lot of cytoplasm while sperm only carry nuclear genome- egg is carrying all of the
cytoplasm to the zygote- inherited randomly through mitosis
o Cytoplasmic inheritance: chloroplasts also have own genome- formerly free-
living bacteria that are now basically little cells inside bigger cell- transmission is
very different (randomly segregated during mitosis and maternally inherited)
o Multiple different genotypes of mitochondria will not be evenly distributed
throughout body
Questions on Slides
• Question 5: Dup is a ool used utat pheotpe i the eatode or C.
elegas. To Dup idiiduals are rossed to eah other and this cross produces 210
Dup ad 8 WT idiiduals. If oe of the Dup idiiduals used i this ross as
ated ith a WT, hat ratio of Dup:WT ould e osere i the offsprig?
o Dumpy is dominant- Dd x Dd
o Offspring would be 1:1 (using the parents so it would be Dd x dd)
• Question 10
o 9:3:4
• A random mutation ours…
o B (50%)
• I drosophila elaogaster…
o E- variable penetrance b/c not really a halfway point between curly and straight
wings
9/7/17
Molecular Genetics
• Physical basis of traits? What is gene made out of and how do they work?
• Central dogma: DNA → RNA → protein → molecular activity → trait
o DNA replicates → transcription → translation
• Genetic material must: store large amounts of information (develop/ grow/ support
cell), encode an orgais’s pheotpe, a e replaed faithfull, have the apait to
var a’t e stati iforatio storage
o Can be proteins, or carbohydrates, or nucleic acids (DNA or RNA)
find more resources at oneclass.com
find more resources at oneclass.com
Document Summary
Questions on slides: question 5: (cid:862)du(cid:373)p(cid:455)(cid:863) is a (cid:272)o(cid:373)(cid:373)o(cid:374)l(cid:455) used (cid:373)uta(cid:374)t phe(cid:374)ot(cid:455)pe i(cid:374) the (cid:374)e(cid:373)atode (cid:449)or(cid:373) c. elega(cid:374)s. t(cid:449)o (cid:862)du(cid:373)p(cid:455)(cid:863) i(cid:374)di(cid:448)iduals are (cid:272)rossed to ea(cid:272)h other and this cross produces 210 (cid:862)du(cid:373)p(cid:455)(cid:863) a(cid:374)d (cid:1010)8 wt i(cid:374)di(cid:448)iduals. Question 10: 9:3:4, a random mutation o(cid:272)(cid:272)urs , b (50%) I(cid:374) drosophila (cid:373)ela(cid:374)ogaster : e- variable penetrance b/c not really a halfway point between curly and straight wings. Information content in dna is just the sequence of nucleotides. Utr of transcript (between start codon and transcription start site: non-template: 3"-attgctgaggta-5". In prokaryotic systems usually see an abundance of negative regulators. Cap protein binding site is just upstream of rna polymerase site- facilitates binding of rna poly. 10/19/17: consider an e. coli partial diploid with the following genotype at the lac operon: i+ oc z+ y-/i- o+ z- y+ In addition, the cap binding site in the lac promoter has been deleted on each of the two chromosomes.
