CH 15, 16, 17 Learning objectives

Explain at least one line of evidence to support that DNA is the heritable information.

DNA is heritable information because it contains genes (segments of nucleotide bases) which code for proteins that will make up certain traits, and these are passed down generationally by reproduction.

• Explain why DNA replication is considered semiconservative.

DNA replication was proven to be semiconservative by an experiment which putting a heavier 15N istopes into DNA, and allowing it to replicate.

lighter 14N isotopes were then added to this resulting bacteria. after letting replication happen: the DNA produced has a strand of 14N and a strand of 15N. Suggesting that a DNA helix is made of a new strand and an old strand (from which it was replicated)

the result is thus the Semi conservative model

Generation 1: DNA with 15N (old helix).

Generation 2: Replicated DNA with one 15N strand and one 14N strand.

Generation 3: Two types of DNA: one with 15N strand + 14N strand and one with two 14N strands.

• Discuss how the structure of DNA facilitates its replication.

the DNA structure is of 2 strands which are untwisted making a replication fork. called a replication fork as the forks (unwinded DNA strands) are both used as templates for the replication of a new strand. making 2 DNAs molecules out of 1 DNA molecule.

• the DNA structure is of 2 strands which are antiparallel to eachother. goes from the 5’ 3’ end and the other goes from the 3’ to 5’ end.

• these strands are both used as templates to make new dna strands (ie. a new strand will be made along each old strand)

• the strand going 3’ to 5’ will have DNA replicated on it going from 5’3 uninterrupted. this replicated strand is called the leading strand

• the strand going in the 5’ 3’ direction, will have DNA replicated on it going from the 5’3 direction as well

  but, it cannot produce the DNA straight across/uninterrupted as it produces the DNA in the 5’3’ while its template is in the 5’3 direction. As a result it has to produce the DNA moving backwards in segments called Okazaki fragments. because of this it is called the lagging strand

• Describe the process of DNA replication; include the following terms: 

• antiparallel structure, 

  • DNA polymerases I and III, 

  • leading strand, 

  • lagging strand, 

  • Okazaki fragments, 

  • DNA ligase, 

  • primer,

  • primase,

  • DNA helicase, 

  • topoisomerase,

  • single-strand DNA-binding proteins,

  • dNTPs

DNA is made out of dNTPs: Deoxyribose, nucleoside triphosphate

in otherwords, 3 phosphate groups, a ribose sugar with 2 OHs, and a nitrogenous base (nucleotide either A, T, G, C).

DNA strands alternate between different dNTPs: purines and pyrimidines

and the opposite facing DNA strand has dNTPs with nucelotides that compliment the other side

the DNA is untwisted into 2 strands by the DNA helicase into a replication fork.

As helicase unwinds DNA, the strands have a tendency to supercoil, Topoisomerase relieves the tension that might make strands strands from supercoiling (as helicase unwinds it)

SSBPs or Single stranded DNA binding proteins attach to these single strands and hold them in place, keeping them from reforing a double helix aka reannealing

DNA strands run antiparalell to eachother one strand is facing the 3’ 5’ direction, and the other in the 5’ 3’ direction. they will also grow antiparalell to eachother

a leading strand will grow in the 5’3 direction from the 3’5’ template strand. a lagging strand will grow in the 5’3’ direction off of the 5’3’ strand in backwards segments.

new DNA is formed when Primase lays down an RNA primer on the strand. This gives DNA polymerase III a starting point for adding nucelotide bases (build new DNA strand)

Polymerase III can only construct strands in the 5’3’ direction. After primase puts down an RNA primer on 5’3’ strand, it will form the leading strand continously as its alreadly facing the 5’3’ direction

since Polymerase III can only build 5’3’ direciton, it will be building the lagging strand backwards in segments called okazaki fragments as newe strand naturally faces the 3’5’ direction. the primase will have to put a new RNA primer down every so often, so it can construct the okazaki fragments as the old strands continue being unwinded by helicase. After this, DNA Polymerase I removes the RNA primers from the new strands and and fills the gaps between okazaki fragments with DNA.

• Explain the functions of telomeres and telomerase.

Basically they are needed to keep the DNA from shortening too much over time from replicating

at the end of building, DNA polymerase III cannot replicate the ends of the new DNA strands. without ends to the DNA strands, the nucelotide sequence would then shorten with each replication as Polymerase III cannot replicate the templates ends. eventually this would make the DNA disappear

telomeres and telomerase however help to combat this.

Teleomeres are protective caps at the ends of chromosomes, made out of repetetitive sequences of DNA. when new replicates are made, the telomere cap gets shortened instead of the actual DNA strand.

However, this cannot go on forever, so telomerase enzyme, helps by adding new nucleotides to the Telemere’s repetitive sequence, lengthening it to prevent the DNA strand from shortening too quickly

cut → CH 17

• Describe the process of transcription. Include the following terms: 

• mRNA

  • RNA polymerase

  • the promoter 

  • initiation

  • elongation

  • termination

  • primary transcript

Transcription is the creation of mRNA off of a DNA template strand

the primary transcript is the final RNA sequence produced

RNA is transcribed in the 5’3’ direction

RNA is replicated from DNA’s genes one after the other.

Each gene has a promoter region (start of the gene)

and a terminator region (end the gene)

The promoter region tells the RNA polymerase where to start building RNA. RNA polymerase builds the RNA from the gene’s promoter region to its terminator region

Initiation: when RNA polymerase binds to promoter region, it opens the DNA helix,

then it adds complimentary nucleotide bases along DNA template strand along the 3’ prime end: creating RNA/Elongating RNA

once it reaced terminator end of gene, the mRNA transcript is done, and the RNA polymerase, DNA strand, and RNA strand detach from eachother.

then

• there a regulatory sequences at promoter region

whether a gene is expressed or not, depends upon whether the RNA polymerase binds to promoter region.

when RNA

• Explain how prokaryotes and eukaryotes differ with respect to the fate of the primary transcript.

Readiness of Primary Transcripts

Prokaryote: the primary transcript serves as final mRNA sequence and can be used readily. since they dont have introns to be spliced

Eukaryote: the primary transcript serves as a pre-mRNA transcript. it needs to gain a 5’ cap, Poly A tail, and get spliced before becoming mature mRNA that can be used

Location of Translation

Prokaryotes: Do not have nucleus, has nucleoid region. DNA is stored in nucleiod region in the cytoplasm. This transcription occurs in the cytoplasm and so does translation (into protein)

Eukaryotes: Has nucleus: DNA is stored in nucelus, so transcription occurs in nucelus, and mRNA leaves and goes to cytoplasm for translation

• Differentiate between introns and exons.

exons are segments of the Gene which contain code for proteins

introns are segments of the gene which lack the code for proteins.

• Describe three ways that the primary transcript is modified in eukaryotic cells.

A 5’ cap is added. a Guanine nucleotide (G).

this protects the mRNA from degreation, it tells the ribosome where to bind to mRNA, helps transport mRNA from nucleus to the cytoplasm

A Poly A tail is added on 3’ end. it is a series of A (Adenine) nucleotides

it also protects mRNA from degretion, tells ribosomes where to bind to mRNA, helps transport from nucleus to cytoplasm

An before Eukaryotes can become mature mRNA they go through splicing in which the introns are removed from RNA transcript and exons are joined together

Spliceosome

series of proteins which carry out splicing (removal of introns and joins exons

• Describe the process of translation. Include the following terms: 

•  tRNA 

  • wobble

  • ribosomes

    • A, P, and E sites 

  • initiation

  • elongation

  • termination

  • release factors

  • codon

  • anticodon

  • aminoacyl-tRNA synthetase

mRNA is made up of codons: 3 nucleotides that carry code for an amno acid

tRNA is made up of anticodons:3 nucelotides bases which compliment the mRNA codons, and bind to them. the tRNA decodes the mRNA codons to make sure the correct amino acids are produced (by ribosomes)

Translation happens inside ribosome:

ribosome has 2 sub units: large sub unit S60 and small sub unit S40

small ribosome subunits together bind to initiator tRNA with initiator MET amino acid

(small sub unit + tRNA) then binds to mRNA’s 5’cap

it goes down and finds the mRNA’s start codon. AUG

then large ribosomal sububit is added

initiation factors released

A site is where new tRNA enters (located in area after start codon.

A site is charge with amino acid aminoacyl-tRNA synthase

P site has petidyl tRNA, carries poly peptide chain

T site is exit for tRNA

small sub unit binds to an mRNA. tRNA also binds to mRNA, first binding at its start codon

after tRNA binds to mRNA’s start codon, the large ribosomal subunit binds to small sub unit

this causes intiation: “translation initiaiton complex”

after start codon, next anticodon of tRNA enters the ribosome (A site) and binds to next codons of mRNA

the amino acid of first anti codon then binds to the amino acid of 2nd anticdon (P site, peptide bond)

then first anticodon and amino acid leaves (T site), and next tRNA amino acids moves over one, from A site to P site so new tRNA anticodon can enter/ be added. This process is repeated. as each tRNA leaves it grows the polypeptide chain (elongation). this is repeated unti A site reaches mRNA’s stop codon (termination)

release factor attaches

finished polypeptide will detatch, ribosome breaks apart into subunits

cut ch14

• Differentiate between the following sets of terms:

  • genotype and phenotype

  • homozygous and heterozygous

  • dominant and recessive

  • monohybrid and dihybrid

• genotype represents the pair of alleles/traits one holds for a genes: the traits can be dominant or recessive.

• phenotype is the outcome of one’s genotype, which trait they show based upon their genotype.

• homozygous is when one’s genotype is made of 2 dominant, or 2 recessive allels

• heterozygous is when one’s genotype is made of 1 dominant, and 1 recessive allele

• for a recessive trait to show up in one’s phenotype, one needs to have 2 recessive allels in their genotype

• for a dominant trait to show in one’s phenoytpe, one only needs to have 1 dominant allele of the trait in their genotype

• monohybrid: a crossing of 2 pairs of alleles of the same gene

• dihybrid: crosses 4 pairs of alleles. from 2 different genes. uses 16 punnet square for possible combonations. to do crosses, use the FOIL method (first, outside, inside, last)

• Describe Mendel’s two principles of inheritance.

Law of Segregation

states that a gamete will carry only 1 allele (for 1 gene) becuase during meiosis organisms chromosomes are halved and resulting gamete can only hold 1 half of genetic info (only 1 allele from organisms pair of allele)

Law of Independent Assortment

states that when you have the allele pairs for 2 genes, only 1 of each goes into the gamete. and they assort into the gamete independent of eachother (ie. the alleles from each gene are not connected to eachother ergo they do not affect which alleles of other genes edn up in the gamete)

What is the F2 generation?

The F2 generation is the result of crossing genotypes from F1 generation

• Use a Punnett square to predict the results of a cross and to state the phenotypic and genotypic ratios of the F₂ generation. 

how do you get results of the Genotypic ratio

(# Homozygous D, # heterozygous, # Homozygous r)

how do you get results of phenotypic ratio

# of Homo D and Het : Homo recessive

• Interpret the results of a testcross.

• Explain the relationship between Mendel’s two principles of inheritance and the events of meiosis.

• Differentiate between the following sets of terms:

  • Sex-lined genes vs. linked genes

  • X-linked genes vs. Y-linked genes

• Describe how sex is determined in humans.

• Characterize the patterns of inheritance for sex-linked genes.

• Discuss why Y chromosomes are useful for describing patterns of human ancestry and migrations.

• Describe the patterns of inheritance for linked genes.

• Explain how recombinants are formed.

• Describe the relationship between frequency of recombination and the relative locations of genes along a chromosome.

• Explain how the frequency of recombination between genes can be used to draw gene maps.

• Explain how phenotypic expression in the heterozygote differs with complete and incomplete dominance. 

• Define and give an example of multiple allelism and epistasis.

• Distinguish between the patterns of inheritance observed with dominant vs. recessive genetic disorders.