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UA / Biological Sciences / BSC 315 / Why genes on the same chromosome violate independent assortment?

Why genes on the same chromosome violate independent assortment?

Why genes on the same chromosome violate independent assortment?


Study Guide Test 2

Why genes on the same chromosome violate independent assortment?

Topic 8

Why genes on the same chromosome violate Independent  Assortment.

∙ Two genes close together on the same chromosome tend to be  inherited together and are said to be linked

∙ Linked gene can be separated by recombination in which homologous  chromosomes exchange genetic information during meiosis; this  results in parental, or nonrecombinant genotypes, as well as a smaller  proportion of recombinant genotypes  

∙ Some combinations of alleles might show up more frequently because  they are the parental type combinations

∙ The F1 female is more likely to pass on these parental combinations of  alleles, rather than the recombinant combinations, to her own progeny  Parental and recombinant classes.  

What is the genetic nomenclature for linked genes?

∙ P: b c+ / b c+ x b+ c / b+ c

∙ F1: b c+ / b+ c x b c / b c

∙ F2: 2924 b c+ / b c

 2768 b+ c / b c

 871 b c / b c

 846 b+ c+ / b c

∙ The top two are parentals and the bottom two are recombinants  ∙ The parental type gametes are the most common F2 gametes ∙ The two less common F2 gamete classes are not produced by the P  generation and are called recombinant types  

Genetic nomenclature for linked genes.

∙ F2 parental classes have same phenotype as P generation ∙ Each parental class approximately equal in number

∙ Each recombinant class approximately equal in number  ∙ Recombinant classes are produced by crossing over between homologs during meiosis I

How to calculate map distances from recombination data?

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o Crossing over = non-sister chromatids break and rejoin,  producing new hybrid chromosomes

o Frequency of crossing over between chromosomes is a measure  of the distance between them

Calculating map distances from recombination data. ∙ The frequency of crossing over is a measure of gene distance  o Frequency of crossing over = number of recombinant F2s / total  number of F2s

∙ Genetic terminology:

o X% recombinants = X centiMorgans = X map units

∙ Linked genes: on the same chromosome

∙ Unlinked genes: on different chromosomes  

∙ Distinguishing linked and unlinked genes:

o If unlinked: heterozygotes produce equal numbers of parental  and recombinant gametes

o If linked: heterozygotes usually produce an excess of parental  gametes  

Exceptions when genes are far apart on the same chromosome. ∙ When genes are very far apart on the same chromosome, they obey  independent assortment  

∙ Independent assortment means that when genes are on different  chromosomes all gamete classes are equal  

∙ Recombinants never exceed 50%. If genes are far apart on the same  chromosome, there are multiple cross-overs between them, and they  obey independent assortment  We also discuss several other topics like What are symptoms of depression?

∙ Genes on different chromosomes, and genes very far apart on the  same chromosome both exhibit independent assortment  

∙ Key feature of independent assortment:

o Parental gametes = 50%

o Recombinant gametes = 50%

∙ “Very far apart” means greater than or equal to 50 map units ∙ How to measure genes that are very far apart:

o Rule: if gene A is linked to B, and B is linked to C, then A and C  are linked, even though they may seem to be unlinked when  tested against each other  

o Linkage = physical relationship (the genes are on the same  chromosome)

Mapping genes on the X chromosome. If you want to learn more check out What is the nebular theory of solar system formation?

∙ When genes are on the X chromosome, any cross can be a test cross if  you analyze only the F2 males

∙ If parental classes are greater than the recombinant classes these  genes are linked, that is, on the same chromosome Don't forget about the age old question of What is v in electrostatics?

Topic 9

The experimental evidence that the genetic material is composed of  DNA

∙ Griffith, 1928. 4 experiments: Don't forget about the age old question of What is sensory adaptation and how does it occur?

o Virulent Pneumococcus kill mice (smooth colonies)

o Mutant strains (rough colonies) are avirulent

o Heat-killed extract from virulent strain does not kill

o Mix heat-killed virulent extract + living avirulent bacteria  restores virulence. Smooth colony morphology

∙ Therefore, genes have been transferred from the non-living extract to  the living bacteria. Transformation changes the virulence and  morphological phenotypes  

∙ Avery, 1940s

o Purify the “transforming substance”

o Remove proteins, RNA, fats – still transforms

o Remove DNA – no transformation

∙ Therefore, genes are composed of DNA

∙ Bacterial viruses (bacteriophage) infect bacteria, make more of  themselves

o Grow phage in presence of radioactive phosphorus (32P) or sulfur (35S). purify phage

o Use radioactive phage for infection of non-radioactive bacteria.  Measure entry of radioactivity into bacteria

o 32P enters, 35S does not

∙ therefore, it’s DNA (DNA has P, no S. Proteins have S, little P)  Chemical structures of the sugars and phosphates and generic  structure of purine and pyrimidine bases.

∙ Nucleoside structure = sugar + base

o Sugar = deoxyribose

o Nitrogenous base. 4 different bases

∙ Nucleotide = sugar + base + phosphate (PO4)

∙ Two families of bases:

o Purines = A and G

o Pyrimidines = C and T

∙ Sugar: 2’-doxyribode

o Numbering 1’ to 5’

o Pentose sugar. Derivative of ribose. 2’-doexyribose lacks –OH  group on 2’ carbon

∙ Phosphate: PO4 (+ ~ 3H). usually ionized = negative charge = acidic  Polynucleotide structure

∙ Phosphodiester bond linkes 5’ and 3’ carbons

∙ Polynucleotide “backbone: repeating sugar-phosphate-sugar phosphate…  

Structures that determine the 5’ to 3’ polarity of polynucleotides. ∙ Adjacent nucleotides are linked as: (3’-sugar-5’) – O – PO2 – O – (3’- sugar-5’)

∙ Sugars have uniform polarity throughout polynucleotide ∙ Every polynucleotide has a 5’ and a 3’ end

Hydrogen bonding, complementary bases and the double-stranded  structure of DNA.

∙ DNA is usually double stranded

∙ Hydrogen bonds between bases join strands

∙ Complementary strands are always anti-parallel (opposite 5’ – 3’  polarity)

o Two strands of DNA double helix

o DNA – RNA during transcription

o mRNA – tRNA during translation

∙ Hydrogen bonds between complementary bases

o Hydrogen bonds = electronegative atoms (O and N) share a H  between them

∙ H bonds are easily broken: DNA strands are separated (melting or  denaturation) by heat or high pH

∙ Configuration of nitrogens and oxygens allows formation of hydrogen  bonds

o A pairs with T

o G pairs with C

∙ A and T are complementary bases to each other. G and C are  complementary bases to each other  

∙ DNA is helical

∙ In B-form DNA:

o 10 bases per helical turn

o 3.4 Angstroms between bases = 34 angstroms per turn ∙ Other configurations other than B (Z = left-handed)

∙ DNA molecules can be linear or circular. Also single-stranded DNA  ∙ RNA Structure

o RNA is a nucleotide polymer

o Ribose sugar

o Bases A, G, C, U

 Complementary base pairs: A:U; G:C

o Has polarity: 5’ and 3’ ends

o Usually single-stranded  

Watson and Crick modeled the structure of DNA based on  experimental data.

∙ 1940s: analytical chemistry shows [A] = [T] and [G] = [C] ∙ 1950s: X-ray diffraction shows a helical structure

∙ 1953: Watson and Crick build models that are consistent with helical  structure and hydrogen bonding pattern of nucleotides

Topic 10

Biochemistry of nucleotide addition during transcription. ∙ Transcription: RNA synthesis using a DNA template

∙ RNA product: transcript. In eukaryotes, where processing follows, often called the primary transcript or pre-mRNA

∙ Enzyme: RNA polymerase

∙ One of the two DNA strands is used as template to make a  complementary RNA copy

∙ Each transcribed region has a beginning and end. The entire DNA  molecule is not transcribed

∙ Short regions of DNA become single-stranded to allow RNA polymerase to make a copy, and revert to dsDNA afterward

∙ RNA polymerase “reads” the DNA template strand from 3’ to 5’ and  makes the transcript 5’ to 3’. The new complementary base is added  to the 3’ end

∙ Nucleotide added is a tri-phosphate, but cleaved to produce nucleoside monophosphate, plus free diphosphate

Relationship between the 5’ to 3’ polarity of DNA and of the RNA  product.

∙ RNA polymerase “reads” the DNA template strand from 3’ to 5’ and  makes the transcript 5’ to 3’. The new complementary base is added  to the 3’ end

∙ The transcript RNA sequence has the same sequence and polarity as  the non-template DNA strand  

What is a gene?

∙ A gene is a segment of DNA that is transcribed into an RNA molecules  ∙ A chromosome-sized piece of DNA contains hundreds to thousands of  genes, each transcribed independently

Promoters and terminators: structure and function.

∙ Promoter = a DNA sequence that RNA polymerase and other proteins  recognize as the beginning of the gene

∙ Promoter: special sequence “upstream” from the transcribed region  marks the beginning of the gene. Orientation of sequence determines  direction of transcription and thus the template strand  

o Promoters differ between prokaryotes and eukaryotes  

o Prokaryotes: simple short sequences just upstream from the  gene

o Eukaryotes: more complex and variable

 Basal promoter: simple sequence. Adjacent to gene

 Enhancer: often not adjacent to gene. Can be multiple

∙ Terminator = a DNA sequence that RNA polymerase and other proteins  recognize as the end of the gene  

Mechanics of transcription: initiation, elongation, termination. ∙ Initiation: RNA polymerase, assisted by various regulatory proteins,  recognizes and binds to promoter

∙ Elongation: nucleotides are added one at a time to 3’ end ∙ Termination: sequences in the RNA (and DNA) molecule cause RNA  polymerase to terminate and release transcript  

∙ Transcription is underway

o DNA strands separate locally

o Nucleoside triphosphates are added to 3’ end only

o Phosphodiester bond formation

o Transcript made 5’ to 3’

o Typically only one strand is transcribed  

∙ Transcription termination

o Termination mechanisms are variable, but all involve a special  DNA sequence in the gene

o E coli: self-complementary (hairpin) structure in RNA is  recognized by a protein, causes RNA polymerase to stop &  release RNA

RNA processing: mechanisms and alterations of capping,  polyadenylation, splicing.

∙ Product of transcription = primary transcript

∙ Prokaryotes: primary transcript is used a mRNA with little additional  modification

∙ Eukaryotes: primary transcript is modified before becoming mRNA  (RNA processing)

o 5’ capping

o 3’ polyadenylation

o Splicing  

∙ Capping

o Unusual nucleotide is added to the 5’ end

o Atypical nucleotide (7-methylguanosine) and bond (a 5’ to 5’  phosphate linkage)

o Happens before transcription is finished  

∙ Polyadenylation  

o Part of the eukaryotic termination mechanism

o AAUAAA sequence. RNase cleaves 11-30 nucleotides  downstream

o Poly(A) polymerase adds 100-200 A’s. non-templated o Only after transcription is complete  

∙ Splicing

o Internal segments removed and bordering regions joined back  together

 Removed = introns

 Retained = exons

o Sites of cleavage/ rejoining determined by the sequence of the  RNA molecule at the exon-intron junctions

∙ Spliceosome: carries out splicing

o Components:

 Small nuclear RNAs (snRNAs) and proteins

 Complex of snRNAs & proteins = snRNAs

o Splicing mechanism

 Cut at 5’ end of intron

 5’ end of intron to A, via 2’ hydroxyl

 3’ end cut

 Exons joined together  

Generic amino acid structure. Peptide bonds.

∙ Proteins: polymers of amino acids

∙ Amino acid structure

o Amino: -NH2

o Carboxyl: -COOH

o Central (alpha) carbon: bonds to –H and –R. 20 different R  groups = 20 different amino acids

∙ Peptide bond: covalent bond between amino and carboxyl groups in  adjacent amino acids  

∙ Amino acids differ only in R group

∙ Properties of proteins are determined by properties of R groups  ∙ Proteins primary structure

o The sequence of amino acids

o Each protein (aka polypeptide) composed of 10s to 1000s of  amino acids

o Proteins have polarity: amino and carboxyl ends (N and C  termini)

∙ Proteins Secondary and Tertiary Structures

o Proteins fold into 3D shapes, determined by chemical  

interactions among R groups and peptide bond atoms

o Secondary structures = local 3D structures

o Tertiary structures = larger scale 3D structures

 Globular proteins

 Fibrous proteins  

∙ Quaternary structure

o Proteins contain >1 polypeptide. Chemical interactions among  groups maintain structures

Fundamentals of the genetic code: no overlap, no punctuation,  correct initiation is critical.

∙ RNA intermediate (messenger RNA) has the same sequence (T~U) and  5’-3’ polarity as the non-template DNA strand

∙ A three nucleotide sequence in the mRNA specifies 1 amino acid ∙ Genetic code: which 3 nucleotide sequence encodes which amino acid ∙ Three nucleotide sequence in mRNA: codon  

∙ Most amino acids have >1 codon (genetic code is degenerate) o Most degeneracy is in the 3rd nucleotide of the codon

∙ Stop codons code for no amino acid = AUG, UAA, UGA

∙ Codons don’t overlap: i.e., no nucleotide is part of more than one  codon

∙ No punctuation, i.e., no skipped nucleotides between codons

∙ Starting right is critical – three possible reading frames produce three  different proteins depending on the start position

o Proteins begin with a methionine and protein synthesis with a  Met codon (AUG)

 The Met codon is never at the 5’ end of the mRNA

 In eukaryotes that start codon is usually the AUG codon  closest to the 5’ end of the mRNA

Ribosome structure.

∙ Ribosome = “protein factory”

∙ Composition: large and small subunits, composed of RNA molecules  (ribosomal RNAs) and dozens of proteins

Transfer RNA structure.

∙ Transfer RNAs: short structural RNAs, 70-90 nucleotides. Internal base pairing

Mechanisms of translation initiation in prokaryotes and eukaryotes. ∙ Initiation at AUG

o Prokaryotes: AUG adjacent to the Shine-Dalgarno sequence (=  AGGAAGG). Ribosome binds to internal SD sequence; does not  scan from 5’ end

o Eukaryotes: closest AUG to the 5’ end of mRNA> ribosome binds to cap, scans 5’ to 3’, finds the first AUG

∙ Ribosome small unit finds initiation codon as previous

∙ First amino acid is:

o Prokaryotes: formyl-methionine (fMet). fMet is used only as the  first amino acid in prokaryotes proteins

o Eukaryotes: ordinary Met

∙ Initiator fMet-tRNA or Met-tRNA binds

∙ Large subunit binds

Anticodons and tRNA charging.

∙ Anticodon: 3 nucleotides that pair with mRNA codon

o Non-standard bases, modified by RNA processing steps after  transcription

∙ A charged tRNA has a specific amino acid attached to the 3’ end Mechanics of translation elongation.

∙ Three functional / binding sites on ribosome

o Aminoacyl site = A site

o Peptidyl site = P site

o Exit site = E site

∙ Amino acid 1 – tRNA 1 in the P site

∙ Amino acid 2 – tRNA 2 enters the A site

∙ Peptide bond formation

o Amino acid 1 – tRNA 1 bond is broken, and amino acid 1 forms  peptide bond to amino acid 2 – tRNA 2

∙ Translocation: entire ribosome moves 3 nucleotides along mRNA, 5’ to  3’

o tRNA 1 now in E site

o Amino acid 1 – amino acid 2 – tRNA 2 now in P site

∙ tRNA 1 exits from E site

∙ Amino acid 1 – amino acid 2 – tRNA 2 in P site

∙ Amino acid 3 – tRNA3 enters A site

∙ Amino acid 1 – amino acid 2 makes peptide bond to amino acid 3 –  tRNA3

∙ Cyclical process: occurs over and over as long as coding codons are  encountered  

Effects of wobble and its significance.

∙ Base pairing between codon and anticodon determines which amino  acid is added to the protein

∙ A specific amino acid must be attached to specific tRNA  ∙ Enzyme: aminoacyl-tRNA synthetase. A family of proteins that attach  an amino acid to tRNA

∙ Specificity: each amino acid – tRNA synthetase works only with one  amino acid and one tRNA

∙ 61 “coding” codons, but <61 tRNAs. Some tRNAs recognize multiple  codons

∙ Mechanism: atypical base pairing (wobble) between 3’ nucleotide of  codon and 5’ nucleotide of anticodon

Topic 11

What is a gene?  

∙ A gene is a segment of a DNA molecule that

o Is transcribed

o Into an RNA,

o And the sequences that control its transcription

Parts of a gene: promoter and transcription unit, coding and non coding, etc.

∙ Promoter: just upstream from transcribed region. Generally, not  transcribed

∙ Enhancers: regulate timing/cell type/circumstances of transcription.  Not transcribed

∙ Transcription terminator: part of primary transcript

Alternative splicing generates multiple proteins from one gene. ∙ One gene can produce several different mRNAs by alternative splicing ∙ Alternative splicing: mRNAs contain different combinations of exons

Eukaryotic and prokaryotic gene structure, mRNA structure &  differences in translation initiation.

∙ Eukaryotic genes have one protein coding region per mRNA ∙ Prokaryotic mRNAs usually have several protein coding regions per  mRNA = polycistronic mRNA

∙ Prokaryotic genes are organized in operons: clusters of genes with  similar functions, transcribed as s single mRNA

o Example is the Trp operon

∙ A polycistornic mRNA each have an AUG initiation codon paired with its own Shine-Dalgarno sequence, and each coding region has its own  stop codon

∙ In prokaryotic translation the initiation codon is AUG, but the modified  amino acid formyl-methionine is incorporated as the first amino acid o Regular methionine is incorporated at internal AUG codons  

Topic 12

Semi-conservative DNA replication.

∙ Each old strand serves as the template to synthesize a new strand ∙ Was noted and predicted by the Watson-Crick model of DNA structure.  Each strand serves as the template to make a new complementary  strand  

Biochemistry of nucleotide addition.

∙ Enzyme: DNA polymerase

∙ The nucleotide next to be added is a nucleoside triphosphate ∙ Bond between first and second phosphate is broken, nucleotide (with 1 phosphate) is added  

∙ DNA is synthesized 1 nucleotide at a time

∙ The new strand is synthesized from 5’ to 3’ only  

Mechanisms of DNA replication on the leading and lagging strands. ∙ Replication begins at one or more origins, each called an origin of  replication. A small region is pulled apart into single strands, a  replication bubble

∙ Replication proceeds in each direction, at two replication forks  ∙ DNA polymerase synthesizes DNA using a ssDNA template and  dNTPs…

o … in the 5’ to 3’ direction only (difference in leading and lagging  strands)

o … only if part of the new strand is already present (necessity for  a primer)

∙ At the replication fork the two template strands have opposite 5’ to 3’  polarity

∙ Since DNA polymerase synthesize 5’ to 3’ only, synthesis of the two  strands relative to the fork is opposite  

∙ The two strands employ different replication mechanisms

o Leading strand: DNA polymerase is moving toward the  replication fork

o Lagging strand: DNA polymerase is moving away from the  replication fork

∙ On the leading strand, DNA polymerase simply follows the replication  fork as it moves right to left, continuous synthesis

∙ On the lagging strand, DNA polymerase starts over near the replication fork to synthesize a short discontinuous fragment left to right,  discontinuous synthesis

∙ Okazaki fragments are short DNA fragments on the lagging strand and  are temporary  

Function of the replication enzymes: DNA polymerase III, DNA  polymerase I, topoisomerase, helicase, primase, ligase, single  strand binding proteins

∙ DNA polymerase III

o All synthesis so far is carried out by enzyme DNA polymerase III ∙ Helicase

o Separate dsDNA into ssDNA

∙ Topoisomerase

o Relieves winding strain on DNA by:

 Cutting

 Unwinding

 Rejoining

∙ Single strand binding proteins

o Bind to ssDNA to prevent re-annealing (not an enzyme) ∙ Primase

o DNA polymerase cannot begin synthesis on a purely single  stranded DNA template – a small part of the new strand must be  already present

o Synthesizes short RNA primers 5’ to 3’

∙ DNA Polymerase I

o Extends Okazaki fragment, removes RNA primer, replaces with  DNA

∙ DNA Ligase

o Makes phosphodiester bond joining adjacent Okazaki fragments  Replication of large linear and circular chromosomes. ∙ Linear DNA (e.g., a chromosome): multiple origins of replication.  Replication bubbles grow and fuse until replication is complete o Result: 2 chromatids

o Each chromatid = 1 double stranded DNA molecule  

∙ Circular DNA (e.g., bacterial chromosome, plasmid): usually 1 origin of  replication, bidirectional replication forks meet on opposite side Why the replication of telomeres is a problem, and how the enzyme  telomerase solves that problem.

∙ Telomere: the end of a linear chromosome

∙ The special problem of telomere replication: at the ends of linear DNA  molecules a short segment can be primed but not copied into DNA  (because there is no upstream primer)

∙ Result: linear chromosomes become shorter at each replication by an  amount that equals the length of the primer

∙ Cells cannot avoid having their chromosomes lose DNA at each  replication

∙ Solution to this problem is the enzyme telomerase

o Telomerase: a non-template requiring DNA polymerase o Adds many copies of a short repeating “filler” sequence  5’TTAGGG-3’  

o Shortening of telomeres removes this filler sequence but not  gene sequences

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