Tingnan ang lahat ng mga set ng pag-aaral

Genetics Final

Pataasin ang iyong marka sa homework at exams ngayon gamit ang Quizwiz!

NAPS typically associate with viral genomes to help with DNA compaction.

False, bacterial genomes

Translation

mRNA to a protein (convey phenotypes, main structural and functional units of our bodies)

M = mitosis. Why not meiosis?

meiosis cannot cycle back after sperm/egg, it leads to terminally differentiated cells

If you have the normal range of repeats, then the first generation that presents the expanded mutation > spontaneous (Pre-mutant or mutant size expansion > inherited)

Practice with DNA primers (one stranded and double stranded)

The following diagram represents one of the Christmas-tree-like structures shown in Figure 10.3. On the diagram, identify parts a through i: DNA molecule 5' and 3' ends of the template strand of DNA At least one RNA molecule Direction of movement of the transcription apparatus on the DNA molecule Approximate location of the promoter Possible location of a terminator Upstream and downstream directions Molecules of RNA polymerase (use dot to represent these molecules)

https://en.wikipedia.org/wiki/Prokaryotic_release_factors

A linear piece of DNA has the following EcoR1 restriction sites: ___2kb___(Site 1)___4kb___(Site2)_________5kb___________ If 500 bp of DNA between the two restriction sites were deleted, how would the banding pattern on the gel differ from the one that you drew in part a?

ok one fragment of 2kb one fragment of 4.5kb/4,500 bp one fragment of 5kb

A linear piece of DNA has the following EcoR1 restriction sites: ___2kb___(Site 1)___4kb___(Site2)_________5kb___________ If mutations that alter EcoR1 sites 1 and 2 occur in this piece of DNA, how will the banding pattern on the gel differ from the one you drew in part a?

ok one fragment that is 11kb

Loss-of-function mutations,

which are more common, result in reduced or abolished protein function.

Transduction

(a) Phage genome is DNA. All other parts of the bateriophage are protein. (b) Phage attaches to E. coli and injects its chromosome. (c) Bacterial chromosome breaks down and the phage chromosome replicates. (d) Expression of phage genes produces phage structural components. (e) Progeny phage particles assemble. (f) Bacterial wall lyses, releasing progeny phages. the process by which foreign DNA is introduced into a cell by a virus

Expanding Nucleotide Repeats

...mutations in which the number of copies of a set of nucleotides increases

DNA 'units' 1 base = Base-pair (pb) = 1 Kb = 1 Mb = 1 Gb =

1 base = 1 nucleotide (nt) Base-pair (pb) = pair of complimentary nucleotides 1 Kb = 1,000 bp 1 Mb = 1,000,000 bp 1 Gb = 1,000,000,000 bp

The three main regions of a mRNA are?

1. 5' untranslated region 2. 3' untranslated region 3. protein coding region

What are the three principal elements in mRNA sequences in bacterial cells?

1. 5' untranslated region 2. 3' untranslated region 3. protein coding region

Define PCR

1. A polymerase chain reaction is a DNA amplification technique that takes a specific copy of a DNA sequence and amplifies it to generate many copies of that particular DNA segment 2. this method needs heat stable DNA polymerase, 2 oligonucleotide (many) DNA primers, and a DNA template. 3. the first step is denaturation, where the DNA is heated to around 95 C in order for the H bonds connecting DNA strands to break. 4. the second step is annealing, which cools down the DNA to around 55 C. This step lowers the kinetic energy so primers can bind to each of the single-stranded DNA templates, these primers must be complimentary to each strand. 5. the third step is elongation/extension, which heats up the DNA to around 75C which is a temperature at which Taq can add complimentary nucleotides. dNTPs (nucleotides) allow the extension of your primed DNA strand. 6. Process goes through about 25 cycles and makes about one billion copies

Explain how gel electrophoresis is used to separate DNA fragments of different lengths.

1. A porous gel is made from agarose (a polysaccharide isolated from seaweed), which is melted in a buffer solution and poured into a plastic mold. As it cools, the agorase gel solidifies making a gel that looks something like stiff gelatin. 2. Small wells are made at one end of the gel to hold solutions of DNA fragments and an electrical current is passed through the gel. DNA fragments of known length are placed in one of the wells (ladder). Because the phosphate group on each DNA nucleotide carries a negative charge, the DNA fragments migrate toward the positive end of the gel. 3. During this migration, the porous gel acts as a sieve, separating the DNA fragments by size. Small fragments migrate more rapidly than large ones do, so over time, the fragments separate on the basis of their size. After electrophoresis, fragments of different sizes have migrated different distances. 4. A dye specific for nucleic acids is added to the gel (ethidium bromide). EB wedges itself tightly (intercalates) between the bases of DNa and fluoresces orange when exposed to UV light, producing orange bands on the gel. 5. By comparing the migration distance of the unknown fragments to the ladder, one can determine the approximate size of the unknown fragments. DNA fragments appear as bands on the gel.

6 Different types of crosses

1. AA x AA 2. AA x Aa 3. AA x aa 4. Aa x Aa (3:1) 5. aa x aa 6. Aa x aa

#7 EOC: Give the genotypic ratios that may appear among the progeny of simple crosses and the genotype of the parents that may give rise to each ratio

1. AA x AA > 1 AA 2. AA x Aa > 1 AA: 1 Aa 3. AA x aa > 1 Aa 4. Aa x Aa > 3 A_ : 1 aa 5. Aa x aa > 1 Aa: 1 aa 6. aa x aa > 1 aa

Summarize the different types of processing that can take place in pre-mRNA. 10.5

1. Addition of the 5' cap 2. Addition of the Poly(A) tail (addition of 50-250 or more A nucleotides at the 3' end, not encoded in the DNA but are added after transcription in a process called polyadenylation, signal is reached and proteins find the site and remove the 3' end and then A nucleotides are added without a template, confers stability, increasing the time during which mRNA remains t=intact and available for translation before it is degraded by cellular enzymes, facilitates, export of the mRNA into the cytoplasm) 3. RNA Splicing

What Monohybrid Crosses revealed (Mendel's first three postulates)

1. Alleles exist in pairs: (for every gene, you are going to have two copies of it) Every trait, there had to be two copies in every individual. 2. Dominance / recessiveness: In an individual that is a heterozygote, the allele that determines the phenotype is considered the dominant allele, whereas the other is considered the recessive allele. 3. Law of segregation: The two alleles of one gene will be separated into separate gametes by the process of meiosis. (*During gamete formation, the alleles for each gene segregate from each other so that each gamete carries only one allele for each gene, Two different alleles end up in separate gametes)

Name six different levels at which gene expression might be controlled.

1. Alteration of DNA or chromatin structure (primarily in eukaryotes) - mod. in DNA or its packaging may help to determine which sequences are available for transcription or the rate at which sequences are transcribed (ex. DNA methylation and changes in chromatin) 2. Level of transcription (for cellular economy, limiting the production of a protein early in the process makes sense) 3. mRNA processing (eukaryotic mRNA > 5' cap, 3' polyA tail, introns removed, 3' cleaved//determine mRNA stability, movement of mRNA into cytoplasm, whether mRNA can be translated, rate of translation, aa sequence of the protein) 4. Regulation of mRNA stability (amount of protein produced depends not only on the amount of mRNA synthesized, but on the rate at which the mRNA is degraded) 5. Level of translation (enzymes, protein factors, and RNA molecules affect the rate at which proteins are produced and therefore provide points at which gene expression can be controlled, trans. can be affected by sequence in mRNA also) 6. Post-translational modifications

Give three important characteristics of cloning vectors.

1. An origin of replication ensures that the vector is replicated within the cell 2. Selectable markers enable any cells containing the vector to be selected or identified 3. One or more unique restriction sites into which a DNA fragment can be inserted. Vector must have only ONE (unique) restriction site for each restriction enzyme used (if a vector is cut at multiple restriction sites, several pieces of DNA are generated, and getting these pieces together again in the correct order is difficult).

Differences between Prokaryotic and Eukaryotic Translation.

1. Characteristics of the Code: 2. Location of Translation: 3. mRNA life: 4. Presence/Use of aminoacyl tRNA synthetases: 5. Characteristics of the ribosomal subunits (importance?): 6. Attachment of small subunit during Initiation: 7. Amino acid specified by the initiation codon: 8. Factors used during Initiation, Elongation, and termination: 9. Presence of polyribosomes:

List the different types of chromosome mutations and define each one.

1. Chromosome rearrangements alter the structure of chromosomes. For example, a piece of a chromosome may be duplicated, deleted, or inverted. 2. In aneuploidy, the number of chromosomes is alter: one or more individual chromosomes are added or deleted. 3. In polyploidy, one or more complete sets of chromosomes are added. A polyplod is any organism that has more than two sets of chromosomes (3n, 4n, 5n, or more).

FISH

1. Design a probe (5,000 to 2 mill bp) of DNA (ds or ss) complimentary to something we are looking for in the genome (gene or of exon). When making the probe, label it with a fluorescent marker, e.g. fluorescein, by incorporating nucleotides that have the marker attached to them. 2. Prepare either a metaphase or an interphase preparation 2. On our glass slide, we put the liquid that contains fluorescent probe and let it heat up so both the genomic DNA and probe melts. 3. Let the genomic DNA and probe mix so they can hybridize (probes can find where there exact pieces of complimentary DNA are) 4. Leave it for a while and wash the excess off 5. Look at it under a microscope that is fluorescent and you will see where the probes were bound to. 6. You get a signal? FISH is useful, for example, to help a researcher or clinician identify where a particular gene falls within an individual's chromosomes. FISH is a kind of cytogenetic technique that uses fluorescent probes binding parts of the chromosome and provides a novel way for researchers to visualize and map the genetic material in an individual cell, including specific genes or portions of genes.

Arrange the following components of translation in the approximate order in which they would appear or be used in prokaryotic protein synthesis: 30S Initiation complex fMet-tRNAfMet Initiation factor 3 and Initiation factor 1 70S initiation complex Elongation factor Tu Elongation factor G Release factor 1 FIX

1. IF-3 and IF-1 bind to small subunit of ribosome 2. fMet-tRNA^fMet attaches to the initiation codon on the P site (this attachment requires IF-2) 3. 30S Initiation complex 4. 70S Initiation complex 5. EF-Tu (catalyzing the binding of an aminoacyl-tRNA (aa-tRNA) to the A site on the ribosome) 6. EF-G (translocation to the next codon) 7. RF-1 Initiation of protein synthesis begins with binding of a special formyl-Methionine charged tRNA to the start codon of the mRNA bound to the 30S subunit.

What are the three basic stages of transcription? Describe what takes place at each stage.

1. Initiation: 2. Elongation 3. Termination

Name the first step of translation.

1. Initiation: Formation of 70s translation initiation complex. Small ribosomal subunit binds to the mRNA transcript to the 5' end of the mRNA (Shine-Delgarno sequence that helps with ID of start codon is here) IF one and three help the mRNA bind to the small ribosomal subunit and prevent binding of the large ribosomal subunit. IF two brings in charged tRNA for f-met (formal methionine). IF one and two allows binding of large subunit. First tRNA is in P site.

Draw types of classifications of chromosomes

1. Metacentric: the centromere is located approximately in the middle, so the chromosome has two arms of equal length 2. Submetacentric: the centromere is displaced toward one end, creating a long arm and a short arm. (On human chromosomes, the short arm is designated by the letter p and the long arm by the letter q.) 3. Acrocentric: the centromere is near one end, producing a long arm and a knob, or satellite, at the other end 4. Telocentric: the centromere is at or very near the end of the chromosome

Elongation tranlsation bacteria

1. Once 70S initiation complex is complete, there is a spot on A site. EF-TU brings the next charged tRNA that has the anticodon that matches the codon on messenger, goes on A site. 2. First aa is cleaved and bound to the second amino acid. Aa in P site forms a peptide bond with the aa in A site. Aa in P site on that tRNA is cleaved, and bound on aa at the tRNA at the A site. 3. EF-G causes translocation (has a G domain which binds G, form that is similar to ATP but with guanine instead of adenine). The messenger needs to move right while the ribosome needs to move left. 4. EF-TS: purpose is to recycle EF-TU. EF-TU is used once and has to be recharged. Makes it ready so it can get another tRNA. Charged tRNA: aa covalently bonded to a tRNA

Study how to draw the summer squash biochemical/whatever pathway just in case. List the steps here.

1. Plants with genotype ww produce enzyme I, which conerts compound A (colorless) into compound B (green) 2. Dominant allele W inhibits the conversion of A into B. 3. Plants with genotype Y_ produce enzyme II, which converts compound B into compound C (yellow). 4. Plants with genotype yy do not encode a functional form of enzyme III. Conclusion: genotypes W_Y_ and W_yy do not produce enzyme I: wwyy produces enzyme I but not enzyme II: wwY_ produces both enzyme I and enzyme II.

Steps of PCR and Gel Electrophoresis

1. Prepare samples. 2. Create an agarose TAE solution. Heat. Place the solution in a mold tray, and fill with TAE buffer. Place the comb in the gel to form the wells. Cool. Once gel has cooled, add ethidum bromide to "stain" the DNA.. 3. DNA is placed into the wells using a micropipette. (- electrode on top. + on bottom). 4. Electrical current is turned on. Electrons run through the current and carry DNA molecules with them. Smaller fragments move farther through the gel.

What Dihybrid Crosses Reveal (Mendel's Fourth Postulate)

1. Principle of independent assortment: separation of alleles at one locus is independent of the separation of alleles in a second locus (for mendelian, very far apart on the same chromosome or on different chromosomes)

Differences between Prokaryotic and Eukaryotic Transcription.

1. Promoters: Prokaryotes have a -10 and -35 consensus sequences where RNA polymerase and the specific sigma factor for that promoter binds to begin transcription. Eukaryotes have transcription factors that bind to their promoter first and aid the RNA polymerase in binding to the promoter (many eukaryotes have a sequence called the TATA box which is recognized by one of these TF). 2. RNA Polymerases: Prokaryotes have one type of RNA polymerase, Eukaryotes have multiple types (RNA polymerase I transcribes rRNA genes, RNA polymerase II transcribes mRNA, and RNA polymerase III transcribes tRNA MAINLY!!) 3. RNAs produced from eukaryotic genes have to undergo posttranslational modifications (there is a pre-mRNA, 5' cap, Poly(A) tail, and splicing of introns). These events do not occur in prokaryotes. 4. RNAs produced from eukaryotic genes are monostronic, meaning they contain a single gene on an mRNA. RNAs produced from prokaryotic genes are polycistronic, meaning they contain multiple different genes on a single mRNA. 5. Transcription Initiation Complex: Prokaryotes have different proteins called σ factors which bind to promoter sequences and start transcription (they are part of the RNA polymerase, which is often considered as its "core" polymerase + a σ factor). Different σ factors bind different promoters and are responsible for a different set of genes, often functionally related. (GENE EXPRESSION!) In eukaryotes, a variety of other proteins and transcription factors are required to help recruit the RNA Pol to the promoter (this is partly because eukaryotic DNA is coiled around histones so needs to be "opened up" for a polymerase to bind it). 6. Transcription Termination: Termination in prokaryotes is done by either rho-dependent or rho-independent mechanisms. In eukaryotes transcription is terminated by two elements: a poly(A) signal and a downstream terminator sequence. Eukaryotes: termination can happen differently depending on the RNA polymerase. 7. Location: transcription in eukaryotes takes place in the nucleus, where transcription in prokaryotes takes place in the cytoplasm (nucleoid region)

What role do restriction enzymes play in bacteria? How do bacteria protect their own DNA from the action of restriction enzymes?

1. Restriction enzymes (also called restriction endonucleases) recognize and make double-stranded cuts in DNA at specific nucleotide sequences (called restriction sites, usually 4 to 8bp long, most are palindromic). 2. RE produced naturally in bacteria and protect against viruses. 3. A bacterium protects its own DNA from a RE by modifying the recognition sequence, usually by adding methyl groups to its DNA.

Why are mutations important?

1. Source of genetic variation = evolution (we don't always have disease states, we have variant states. Variation in the population = microsatellites: mutations, bigger or smaller number of repeats helps us study identity 2. Evolution: sometimes gives you an advantage to survive in a specific environment, or the disadvantage 3. Source of disease: how can we work towards how to counteract negative effects of the mutant variant? Once we understand it we can do this! How is it inherited? Inheritance of mutant alleles: Will my children have this disease if I have it? 4. Tools to probe biological processes: I may suspect that a specific mutated gene is involved in a condition, knock out that gene and make it useless and see if i can reconfigurate? the phenotype (mutate certain proteins/genes in model organisms and see if you get the diseased phenotype that you thought what you would get to confirm)

DNA Intiation

1. Starts at the OriC (origin of replication for E. Coli Cell), it is 245 bp that signals for DnaA 2. DnaA (initiator protein) binds to DnaA sites (R1, R2, R3, R4) (4 x 9-mers), DnaA interacts at these sites and wraps around DNA creating loops to relieve tension to melt 13-mers, opens DNA so that helicase can attach/load onto DNA (creates replication bubble) 3. DnaB (helicase) works with DnaC (loading protein) forming a DnaB/DnaC complex. Complex loads helicase onto the DNA strands 4. Helicase unwinds/separates DNA strands 5. SSBP attach to sstrand, stabilize, protein binds to ssDNA in ss form to keep DNA from reannealing due to high chemical reactions between strands 5. Once helicase is loaded, DnaA leaves!

How can you tell whether your gene of interest is in the forward or reverse orientation?

1. Take the white colonies seen from blue-white screening and put them in test tubes. 2. Decide which RE would be most useful to determine the orientation using your knowledge of the RS. 3. Treat each tube with RE chosen. 4. Run each tube on a gel and compare the size of DNA fragments to determine the orientation.

Define blue-white screening.

1. Use a plasmid with the lacZ gene as a vector. 2. The lacZ gene contains a series of unique restriction sites at which a piece of DNA can be inserted. 3. In the absense of an inserted fragment, the lacZ gene is active and produces beta-galactosidase. When foreign DNA is inserted into the restriction site, it disrupts the lacZ gene and beta-galactosidase is not produced. 4. The plaasmid also usually contains a second selectable marker, which may be a gene that confers resistance to an antibiotic, such as ampicillin. 5. Bacteria that are lacZ- are exposed to the plasmids and plated on medium that contains ampicillin and X-gal. Only cells that have been successfully transformed and thus contain a plasmid with the amp-resistance gene will survive and grow. Some of the cells will contain an intact plasmid and others will contain a recombinant plasmid. 6. When cleaved, X-gal produces a blue substance. Bacteria cells with an intact original plasmid (without an inserted DNA fragment) have a functional lacZ (front end provided by the plasmid and back end by bacterium). These bacteria can synthesize beta-galactosidase, which cleaves X-gal and turns the bacteria blue. 7. Bacteria cells with a recombinant plasmid have the front end fo the beta-galactosidase gene disrupted by the inserted DNA, they do not synthesize beta-galactosidase and remain white. 8. Once cells with the recombinant plasmid have been identified, they can be grown in large numbers to replicate the inserted fragment of DNA.

List the process of molecular cloning. FIX

1. You must have a cloning vector.. An effective cloning vector has (1) an origin of replication, (2) selectable markers (antibiotic resistance gene and lacZ gene) and (3) one or more unique restriction sites. 2. The vector must have only one unique restriction site for each restriction enzyme used (if a vector is cut at multiple RS, several pieces of DNA are generated and getting them back together in the right order is very difficult). 3. Plasmids (that exist naturally in bacteria) are used as vectors for cloning DNA fragments in bacteria. 4. First, you must cut the foreign DNA (containing a DNA fragment of interest) and the plasmid with the same restriction enzyme. 5. If the RE makes staggered cuts in the DNA, complementary sticky ends are produced on the foreign DNA and the plasmid. 6. The DNA and plasmid are then mixed together (some of the foreign DNA fragments will pair with the cut ends of the plasmid). 7. DNA ligase is used to seal the nicks in the sugar-phosphate backbone, creating a recombinant plasmid that contains the foreign DNA fragment. 8. Once a DNA fragment of interest has been placed inside a plasmid, the plasmid must be introduced into bacterial cells. This is accomplished by transformation (mechanism by which bacterial cells take up DNA from the external environment). 9. Inside the cells, the plasmids replicate and multiple, and the cells themselves multiple, producing many copies of the introduced gene. 10. Once cells with the recombinant plasmid have been identified through blue-white screening, they can be grown in large numbers to replicate the inserted fragment of DNA.

The three main parts of a gene are?

1. a promoter 2. an RNA coding region 3. a transcription terminator sequence

What parts of DNA make up a transcription unit? Draw a typical bacterial transcription unit and identify its parts. Suppose that a mutation occurs in an intron of a gene encoding a protein. What will the most likely effect of the mutation be on the amino acid sequence of that protein? Explain your answer.

1. a promoter 2. an RNA coding region 3. a transcription terminator sequence (transcription start site - +1)

Post-transcriptional modifications, also called pre-mRNA processing, includes the following steps?

1. addition of the 5' cap 2. cleavage of the 3' end (addition of 3' poly-a tail) 3. splicing

What are the seven characteristics examined by Mendel?

1. flower color is purple or white (purple is dominant) 2. flower position is axil or terminal (axial is dominant) 3. stem length is long or short (long is dominant) 4. seed shape is round or wrinkled (round is dominant) 5. seed color is yellow or green (yellow is dominant) 6. pod shape is inflated or constricted (inflated is dominant) 7. pod color is yellow or green (green is dominant)

Eukaryotic mRNA

1. many diff types of promoter related sequeces, can have one type multiple or none, placed close to or FAR away upstream from transcription start site, regulatrory seuqnces downstream from of the transc start site 2. EXONS AND INTRONS!

What events bring about the termination of translation?

1. the ribosome translocates to a termination codon 2. (e. coli) release factor 1, 2, and 3 come in. RF-1 binds to the termination codons UAA and UAG, and RF-2 binds to the termination codons UGA and UAA 3. Binding of RF-1 or RF-2 to the A site of the ribosome promotes the cleavage of the tRNA in the P site from the polypeptide chain and the release of the polypeptide chain 4. RF-3 binds to the ribosome and forms a complex with GTP, brings about a conformational change in the ribosome, releasing RF-1 or RF-2 from the A site and causing the tRNA in the P site to move to the E site (final translocation) 5. In the process, GTP is hydrolyze to GRP. The tRNA is released from the E site, mRNA is released from the ribosome, and the ribosomal subunit disassociates STOP CODON IS FOLLOWED BY A NUMBER OF NUCLEOTIDES THAT CONSTITUTE THE 3'UTR OF THE mRNA STOP CODON IS NOT NOT NOT NOT NOT LOCATED AT THE 3' END OF THE mRNA

How many different mRNA sequences can encode a polypeptide chain with the amino acid sequence Met-Leu-Arg? (Be sure to include the stop codon.)

108 Met- 1 codon Leu - 6 codons Arg - 6 codons STOP - 3 codons 6 x 6 x 3 x 1 = 108

Name the second step of translation.

2. Elongation: For various cycles, EF-Tu (brings next charged tRNA that has an anticodon that matches the codon on the mRNA), EF Ts (helps recycle TU, TU can only be used one and needs to be recycled), peptide bond formed between aa in P (peptidyl binding site) site and aa in A site. EF-G helps translocate the ribosome. This leaves an open spot for the A site (aminoacyl binding site) so the next charged tRNA can be brought in and received by the A site.

All of the following cells, shown in various stages of mitosis and meiosis,come from the same rare species of plant. What is the diploid number of chromosomes in this plant? Give the names of each stage of mitosis or meiosis shown. Give the number of chromosomes and number of DNA molecules per cell present at each stage.

2n= 6 Picture 1= anaphase 1 of meiosis Picture 2= anaphase of mitosis Picture 3- anaphase 2 of meiosis Picture 1 Chromosomes- 6 DNA molecules- 12 Picture 2 Chromosomes- 12 DNA molecules- 12 Picture 3 Chromosomes- 6 DNA molecules- 6

A chromosome with 3 genes will have a ploidy number (2n) of ? During G1 Number of Chromosomes? Number of Sister Chromatids? Number of DNA Molecules? In Metaphase I Number of Chromosomes? Number of Sister Chromatids? Number of DNA Molecules? In Metaphase II Number of Chromosomes? Number of Sister Chromatids? Number of DNA Molecules?

3 genes, A/a, B/b, and C/c 2n = 6 if each of these genes are on separate chromosomes, n = 3 (3 different alleles, 3 different genes on each gamete) 6 chromosomes 6 sister chromatids 6 DNA molecules ------------------------------- 6 chromosomes 12 sister chromatids 12 DNA molecules ------------------------------- 3 chromosomes 6 sister chromatids 6 DNA molecules

The following sequence of nucleotides is found in a single-stranded DNA template: Assume that RNA polymerase proceeds along this template from left to right. Attgccagatcatcccaatagat Which end of the dna template is 5' and which end is 3'? Give the sequence and identify the 5' and 3' ends of the RNA copied from this template.

3' to 5' 5' - UAACGGUCUAGUAGGGUUAUCUA - 3'

What would be the effect on the DNA replication of mutations that destroyed each of the following activities of DNA polymerase I?

3'--5' exonuclease activity-- can not remove mismatched bases (no way of proofreading?) 5'--3' exonuclease activity- can not properly remove RNA primers 5'--3' polymerase activity- can not properly add DNA nucleotides

One nucleotide strand of a DNA molecule has the base sequence illustrated below. 5'--ATTGCTACGG--3' Give the base sequence and label the 5' and 3' ends of the complementary DNA nucleotide strand.

3'--TAACGATGCC--5'

Name the third step of translation. *fix

3. Once stop codon is reached (DO NOT code for an amino acid), has the capability to bind one of two RF. Total of 3 release factors, but usually two work to bind. One specifically binds stop codon and one works with first RF and the ribosome to cause the whole system to disassemble. First, there is one final translocation that allows the cleavage of tRNA off of the peptide. Final translocation puts the RF in the P site and that is when the complex disassembles.

Name the third step of translation.

3. Once stop codon is reached (DO NOT code for an amino acid), has the capability to bind one of two RF. Total of 3 release factors, but usually two work to bind. One specifically binds stop codon and one works with first RF and the ribosome to cause the whole system to disassemble. Then, there is one final translocation that allows the cleavage of tRNA off of the peptide. Final translocation puts the RF in the P site and that is when the complex disassembles.

#28 EOC Assume that long earlobes in humans are an autosomal dominant trait that exhibits 30% penetrance. A person who is heterozygous for long earlobes mates with a person who is homozygous for normal earlobes. What is the probability that their first child will have long earlobes?

3/20 will have long earlobes

Suppose that you are given a short fragment of DNA to sequence. You amplify the fragment with PCR and set up a series of four dideoxy reactions. You then separate the products of the reactions by gel electrophoresis and obtain the following banding pattern: Original sequence: 5' - ___________________ - 3'

5' - CGTAGTCAT - 3'

What does a prokaryotic mRNA contain? Draw it.

5' UTR - contains a shine-dalgarno sequence that is an identification for the ribosome to bind stop codon start codon protein coding region 3' UTR

The following diagram represents DNA that is part of the RNA-coding sequence of a transcription unit. The bottom strand is the template strand. Give the sequence found on the RNA molecule transcribed from this DNA and identify the 5' and 3' ends of the RNA. 5' - ATAGGCGATGCCA- 3' 3' -TATCCGCTACGGT-5'

5'-AUAGGCGAUGCCA-3'

What is an octomer?

8 units of histones and protein together

Write the consensus sequence for the following set of nucleotide sequences. AGGAGTT AGCTATT TGCAATA ACGAAAA TCCTAAT TGCAATT

?GCAATT

How will you create a version of your gene of interest without introns so that the mRNA will be translated correctly by E. coli? FIX

A eukaryotic cell transcribes the DNA (from genes) into RNA (pre-mRNA). The same cell processes the pre-mRNA strands through posttranslational modifications. Extracted from the cell. Reverse transcriptase is added, along with deoxynucleotide triphosphates (A, T, G, C). This synthesizes one complementary strand of DNA hybridized to the original mRNA strand. To synthesize an additional DNA strand, one would digest the RNA of the hybrid strand, using an enzyme like RNase H, or through alkali digestion method. After digestion of the RNA, a single stranded DNA (ssDNA) is left and because single stranded nucleic acids are hydrophobic, it tends to loop around itself. It is likely that the ssDNA forms a hairpin loop at the 3' end. From the hairpin loop, a DNA polymerase can then use it as a primer to transcribe a complementary sequence for the ss cDNA. Now, you should be left with a double stranded cDNA with identical sequence as the gene of interest.

Why do extra copies of genes sometimes cause drastic phenotypic effects?

A fly with the Bar mutation has a reduced number of facets in the eye, making the eye smaller and bar shaped instead of oval. The Bar mutation results from a small duplication on the X chromosome that is inherited as an incompletely dominant, X-linked trait: heterozygous female flies have somewhat smaller eyes, whereas in homozygous female and hemizygous male flies, the number of facets is greatly reduced. The expression of some genes is balanced with the expression of other genes; the ratios of their gene products, usually proteins, must be maintained within a narrow range for proper cell function. Extra copies of one of these genes cause that gene to be expressed at proportionately higher levels, thereby upsetting the balance of gene products

31. A horse has 64 chromosomes and a donkey has 62 chromosomes. A cross between a female horse and a male donkey produces a mule, which is usually sterile. How many chromosomes does a mule have? Can you think of any reasons for the fact that most mules are sterile?

A mule would have 63 chromosomes They have an odd amount of chromosomes so there would not be a complete even division of chromosomes during anaphase 1

Gain-of-function mutation:

A mutation that confers new or enhanced activity on a protein.

One way to do this is to microinject the DNA straight into the zygote or early embryo. Lets say you want your gene expressed in the pancreas. What would you do differently and what other factors should you consider? FIX

A retroviral vector must contain a RNA virus able to produce reverse transcriptase (All The genetic material in retroviruses is in the form of RNA molecules, while the genetic material of their hosts is in the form of DNA.) Then insert the gene of interest into a retroviral vector plasmid. In the vector, make sure you include a promoter sequence common to the pancreas. The DNA can then be taken up by the early embryo and into the nucleus. Other factors include the control of the new gene by the body. Another enzyme, integrase, can then enable the gene of interest to be integrated into the host cell's chromosome)

In corn plants, normal height, N, is dominant to short height, n. Draw Punnett squares showing all possible crosses that generate: A) Short offspring and B) Tall offspring

A) nn x nn B) NN x NN, Nn x NN, NN x nn

24. Which of the following relations or ratios would be true for a double-stranded DNA molecule? A + T = G + C A + T = T + C A + C = G + T A + T/C + G = 1 A + G/C + T = 1 A/C = G/T A/G = T/C A/T = G/C

A/G = T/C A + T = G + C?

If a double-stranded DNA molecule is 15% thymine, what are the percentages of all the other bases?

Adenine: 15% Guanine: 35% Cytosine: 35%

#4 EOC What characteristics are exhibited by an X-linked trait?

All males show the phenotype of their X-linked characteristic since they only have one copy This x is only inherited in the males' daughters, not sons since sons inherit the y chromosome from their father

What is alternative splicing? How does it lead to the production of multiple proteins from a single gene? 10.5

Alternative splicing is a type of altnerative processing in which the same pre-mRNA can be spliced in more than one way to yield multiple mRNAs that are translated into different aa sequences and thus diff proteins two introns can be removed to yield one mRNA OR two introns and exon 2 can be skipped to yield a different mRNA

How can inversions in which no genetic information is lost or gained cause phenotypic effects?

Although inversions do not result in loss or duplication of chromosomal material, inversions can have phenotypic consequences if the inversion disrupts a gene at one of its breakpoints or if a gene near a breakpoint is altered in its expression because of a change in its chromosomal environment, such as relocation to a heterochromatic region. Such effects on gene expression are called position effects.

Primase

An enzyme that synthesizes short stretches of RNA nucleotide or primers [5'->3'] provides a 3'-OH group to which DNa polymerases can attach nucleotides Forms a complex with helicase at the replication fork and moves along the template of the lagging strand

Transcription: Termination (Rho-Independent!)

As the RNA polymerase approaches the end of the gene being transcribed, it reaches the terminator. It contains a region rich in C and G nucleotides followed by a stretch of U nucleotides in the RNA. The RNA transcribed from this region folds back on itself, and the complementary C and G nucleotides bind together. The result is a stable hairpin that causes the polymerase to stall. The hairpin is followed by a series of U nucleotides in the RNA. The hairpin causes the polymerase to stall, and the weak base pairing between the A nucleotides of the DNA template and the U nucleotides of the RNA transcript allows the transcript to separate from the template, ending transcription. In a terminator, the hairpin is followed by a stretch of U nucleotides in the RNA, which match up with A nucleotides in the template DNA. The complementary U-A region of the RNA transcript forms only a weak interaction with the template DNA. This, coupled with the stalled polymerase, produces enough instability for the enzyme to fall off and liberate the new RNA transcript. 12. Folding of haprin structure disrupts the engagement of RNA polymerase with the DNA template strand 16. " A short complementary GC rich sequence (followed by several U residues, will form a hairpin structure, causing release of the poly U stetch"

#29 EOC In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The genes for these two characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The F1 are intercrossed to produce the F2. a. What are the phenotypic ratios in the F2? b. If an F1 plant is backcrossed with the bitter, yellow-spotted parent, what phenotypes and proportions are expected in the offspring? c. If an F1 plant is backcrossed with the sweet, non-spotted parent, what phenotypes and proportions are expected in the offspring?

B = bitter fruit b = sweet fruit S = yellow spots s = no spots BBSS x bbss F1: all BbSs a. F2: 1/16 BBSS, 1/16 BBss, 2/16 BBSs, 1/16 bbSS. 1/16 bbss, 2/16 bbSs, 2/16 BbSS, 2/16 Bbss, 4/16 BbSs 9/16 B_S_ : 1/16 bbss : 3/16 bbS_ : 3/16 _ss b. 1/4 BBSS, 1/4 BBSs, 1/4 BbSS, 1/4 BbSs All offspring are bitter and yellow spotted c. 1/4 BbSs, 1/4 Bbss, 1/4 bbSs, 1/4 bbss Offspring expected to be: 1/4 bitter fruit and yellow spots, 1/4 bitter fruit and no spots, 1/4 sweet fruit and yellow spots, 1/4 sweet fruit and no spots

#28 EOC In guinea pigs, the allele for black fur (B) is dominant over the allele for brown fur (b). A black guinea pig is crossed with a brown guinea pig, producing five F1 black guinea pigs and six F1 brown guinea pigs. a. How many copies of the black allele (B) will be present in each cell of an F1 black guinea pig at the following stages G1, G2, metaphase of mitosis, metaphase 1 of meiosis, metaphase 2 of meiosis, and after the second cytokinesis following meiosis? Assume that no crossing over takes place. b. How many copies of the brown allele (b) will be present in each cell of an F1 brown guinea pig at the same stages as those listed in part a? Assume that no crossing over takes place.

B = black fur b = brown fur B_ x bb 5 B_ and 6 bb This means parent one must have been Bb to produce a 1 B_:1 bb ratio, offspring would be half bb and half Bb a. G1: one G2: two Metaphase of Mitosis: two Metaphase 1 of Meiosis: two Metaphase 2 of Meiosis: two Second Cytokinesis following Meiosis: one b. G1: two G2: four Metaphase of Mitosis: four Metaphase 1 of Meiosis: four Metaphase 2 of Meiosis: two Second Cytokinesis following Meiosis: one

Both prok and euk have a _________

BOTH cases have a 5' and 3' non-coding region, also called UTR (untranslated region) Eukaryotes: called a leader (5'), 5' non-coding region, 5' UTR, 5' untranslated region and a trailer (3'), 3' UTR, 3' untranslated region, 3' non-coding region

Why is gene regulation important for bacterial cells?

Bacteria carry the genetic information for synthesizing many proteins, but only a subset of that genetic information is expressed at any time. When the environment changes, new genes are expressed, and proteins appropriate for the new environment are synthesized. To maintain biochemical flexibility while optimizing energy efficiency.

Bacterial Genome Organizationq

Bacterial DNA is not complexed to histone proteins and is circular Circular dsDNA NAPs (nucleoid associated proteins) to compact DNA

Transcription

Bacterial cells also produce a small protein called the sigma factor, sev sf that can be produced by the cells, sf binds the RNA core polymerase and gives its specificty for the promoter (diff sf can allow RNA pol to recognize slightly diff promoters) > how much little/much transcription we need in the cell 3. Core RNA polymerase binds the sf and together the promoter is recognized, once promoter is recognized it melts (break H bonds), promoter allows homoenzyme (term that allows us recognize the core polymerase and the sf), he can now recognize what the +1 transc start site is 4. Goes very slow and adds a few compl RNA nucleotides, (do not mem first addition in detail in book!), remove sf which allows the RNA pol to add compl nucleo at a very FAST rate!, uses one of the strands as a template strand (3 to 5 direction) and RNA pol adds compl ribonucleotides to the template strand (bottom) 5. Coding strand (DNA, top, 5 to 3), wherever the +1 Transc start site is, we begin to add compl nuceltodies (nucleic acids always in 5' to 3' direction 6. Template: copy of the coding strand (U's instead of T's), top (coding) strand because it gives you the 5 to 3 direction of what the RNA is going to have

Transcription (Bacteria) Initiation and Elongation

Suppose that a mutation occurs in the middle of a large intron of a gene encoding a protein. What will the most likely effect of the mutation be on the amino acid sequence of that protein? Explain your answer.

Because introns are removed prior to translation, an intron mutation would have little effect on a protein's amino acid sequence unless the mutation occurred within the 5′ splice site, the 3′ splice site, or the branch point. If mutations within these sequences altered splicing, then the mature mRNA would be altered, thus altering the amino acid sequence of the protein. The result could be a protein with additional amino acid sequence. Or, possibly, the altered splicing could introduce a stop codon that stops translation prematurely. If a mutation in the intron induced a frameshift, the reading frame and the amino acid sequence would be altered Or some degree of effect because maybe the spliceosome can only recognize a certain sequence to remove the intron and if this sequence has been altered, then maybe the spliceosome can not recognize it as an intron and would keep it in the mature mRNA transcript.

#36 EOC A summer squash plant that produces disc-shaped fruit is crossed with a summer-squash plant that produces long fruit. All the F1 have disc-shaped fruit. When the F1 are intercrossed, F2 progeny are produced in the following ratio 9/16 disc-shaped, 6/16 spherical fruit, 1/16 long fruit. Give the genotypes of the F2 progeny.

Because of the 9:3:3:1 ratio, we see that: Disc: A_B_ Spherical: A_bb or aaB_ Long: aabb

Red-green color blindness is a human X-linked recessive disorder. A young man with a 47, XXX karyotype (Kline-felter syndrome) is color blind. His 46, XY brother also is color blind both parents have normal color vision. Where did the nondisjunction that gave rise to the young man with Klinefelter syndrome take place? Assume that no crossing over took place in prophase 1 of meiosis.

Because the father has normal color vision, the mother must be the carrier for color-blindness. The color-blind young man with Klinefelter syndrome must have inherited two copies of the color-blind X chromosome from his mother. The nondisjunction event therefore most likely took place in meiosis II of the egg (which is when identical sister chromatids are separated) .

Using breeding techniques, Andreal Dyban and V.S. Baranov created mice that were trisomic for each of the different mouse chromosomes. They found that only mice with trisomy 19 completed development. Mice that were trisomic for all other chromosomes died in the course of development for some of these trisomics, the researchers plotted the length of development (number of days after conception before the embryo died) as a function of the size of the mouse chromosome that was present in three copies (see the adjoining graph). Summarize their findings and provide a possible explanation for the results.

Being trisomic for larger chromosomes affects the dosage of more genes and is more detrimental, causing development to cease at an earlier age.

Suppose that a future scientist explores a distant planet and discovers a novel form of double-stranded nucleic acid. When this nucleic acid is exposed to dna polymerases from E. coli, replication takes place continuously on both strands. What conclusion can you make about the structure of this novel nucleic acid?

Both strands on the nucleic acid must be oriented in the same direction.

#36 EOC In the california poppy, an allele for yellow flowers (C) is dominant over an allele for white flowers (c) . at an independently assorting locus, an allele for entire petals (F) is dominant over an allele for fringed petals (f). A plant that is homozygous for yellow and entire petals is crossed with a plant that is white and fringed. A resulting F1 plant that is then crossed with a plant that is white and fringed, and the following progeny are produced: 54 yellow and entire, 58 yellow and fringed, 53 white and entire, 10 white and fringed. a. Use a chi square test to compare the observed numbers with those expected for the cross. b. What conclusion can you make from the results of the chi-square test? c. Suggest an explanation for the results.

C = yellow flowers c = white flowers F = entire petals f = fringed petals CCFF x ccff F1 must be CcFf F2 must be 9:3:3:1

The percentage of cytosine in a double stranded DNA molecule is 40%. What is the % of thymine?

C pairs with G C = 40%, G must equal 40% 80%-100% 20% left. A = T therefore A and T both 10%

Huntington's Disease

CAG trinucleotide repeats, leads to the protein have a huge glutamine track (pathological > 35, normal is < 27) that render the protein useless, they have a gain of function toxicity (they make huge protein aggregates that the cell does not know what to do with and lead to toxicity in the cell and the cell does not work, if hypothetically neurons, you can imagine the problem) Autosomal dominant Neurodegenerative disorder lies in HTT gene (exon 1 of huntingtin gene) Expansion in exon 1 produces the polyglutamine track and that leads to protein aggregation in the nucleus and the cytoplasm. It adds a gain of function toxicity. Involved in a various amount of things within the cell. HD pathogenesis (the manner of development of a disease.) is incredibly complex. The HTT interactome is comprised of proteins involved in transcription, DNA maintenance, cell cycle regulation, cellular organization, protein transport, energy metabolism, cell signalling, and protein homeostasis (proteostasis). cell signaling > how do cells tell what the body to produce, and protein homeostasis > this protein is fresh and needs to be turned over have too much or too little, control and protein quality inside the cell) Segregate: present together genetically in the same individual

Inducible

Can produce transcription by providing a substrate to the environment. (Can have positive or negative control but do not define this way)\ In inducible operons, the presence of the substrate induces the transcription of the structural genes of the operon.

A line of mouse cells is grown for many generations in a new medium with 15N. Cells in G1 are then switched to a new medium that contains 14N. Draw a pair of homologous chromosomes from these cells at the following stages, showing the two strands of DNA molecules found in the chromosomes. Use different colors to represent strands with 14N and 15N.

Cells in G1 before switching to medium with 14N Cells in G2 after switching to medium with 14N Cells in anaphase of mitosis, after switching to medium with 14N Cells in metaphase 1 of meiosis, after switching to medium with 14N Cells in anaphase 2 of meiosis, after switching to medium with 14N

Name three essential structural elements of a functional eukaryotic chromosome and describe their functions. Define microsatellites.

Centromere Pair of Telomeres Origins of Replication MS are short tandem repeats (STRs), that are very short DNA sequences repeated in tandem. Tey are found at many loci throughout the human genome. People vary in te numer of copies of repeat sequences they possess at each of these loci. They are detected using PCR, using primers that flank the repeats so that a DNA fragent contianing the repeated sequences is amplified.

Define centromere

Centromere: constricted region on a chromosome that stains less strongly than the rest of the chromosome; region where spindle MT attach to a chromosome

Imagine Picture of Chromosome 3. How many chromosomes? Pairs of homologous chromosomes? Unrep or Rep? SC? DNA molecules?

Chromosomes: no new content, chromosomes are just replicated, there is 2 Pairs: 1 pair, sequence identical besides alleles Replicatied or unreplicated? replicated SC: 4, 2 for each chromosome DNA molecules: 4 DNA molecules, each chromosome is = 1 DNA molecule

Cis vs. Trans-acting element

Cis and Trans apply to regulatory elements! Cis: region of non-coding DNA which regulates transcription of nearby genes on the same DNA molecule (functions as a binding site for transcriptional factors) Trans: code for a gene that codes for a protein to be used in the regulation of another gene (acting from a different molecule)

Second Step of mRNA processing

Cleavage of the 3' end: specific signal that identifies that there is a portion at the 3' end that needs to be removed (contains some of the information that is the terminator, non-coding so it can be removed)

For the RNA molecule shown, write out the sequence of bases on the template and nontemplate strands of DNA from which this RNA is transcribed. Label the 5' and 3' ends of each strand. mRNA = 5' - UGAC - 3'

Coding (non-template): 5' - TGAC - 3' Template: 3' - ACTG - 5'

A nontemplate strand of bacterial DNA has the following base sequence. What amino acid sequence will be translated from this? 5'--ATGATACTAAGGCCC--3'

Coding/Nontemplate: 5'--ATGATACTAAGGCCC--3' mRNA: 5'--AUG-AUA-CUA-AGG-CCC--3' Amino Acid Sequence: Met-Ile-Leu-Arg-Pro

The following amino acid sequence is found in a tripeptide Met-Trp-His. Give all possible nucleotide sequences on the mRNA, on the template strand of DNA and nontemplate strand.

Coding/non-template: 5' ATG-TGG-CAT 3' or 5' ATG-TGG-CAC 3' Template/non-coding: 3' TAC-ACC-GTA 5' or 3' TAC-ACC-GTG 5' mRNA: 5' AUG-UGG-CAU 3' or 5' AUG-UGG-CAC 3'

X-linked traits commonly found in humans Dominant or recessive?

Color blindness, deutan type Color blindness, protan type Fabry disease G-6-PO deficiency Hemophilia A - recessive Hemophilia B - recessive Hunter syndrome Ichthyosis Lesch-Nyhan syndrome Ducheene musclar dystrophy

List at least five properties that DNA polymerase and RNA polymerases have in common. List at least three differences.

Common: 1. The complementary strand is synthesized in a 5′ to 3′ direction that is antiparallel to the template 2. Both use DNA templates 3. Both use nucleoside triphosphates as substrates 4. Their actions are enhanced by accessory proteins. 5. DNA templates are read in the 3′ to 5′ direction Differences: 1. RNA polymerase does not require a 3'OH group to immediately synthesize a strand of RNA nucleotides. DNA polymerase does! 2. RNA polymerases use ribonucleoside triphosphates as substrates, whereas DNA polymerases use deoxyribonucleoside triphophates 3. RNA polymerases are used during transcription to synthesize the mRNA strand, whereas DNA polymerases are used during DNA replication to synthesize copies of the DNA strands

Under which of the following conditions would a lac operon produce the greatest amount of B-galactosidase? The least? Explain your reasoning. Condition 1: (Lactose present) Yes, (Glucose) No Condition 2: (Lactose present) No, (Glucose) Yes Condition 3: (Lactose present) Yes, (Glucose) Yes Condition 4: (Lactose present) No, (Glucose) No

Condition 1 would have the most. For maximum transcription, the presence of lactose and the absence of glucose are required. Lactose (or allolactose) binds to the lac repressor reducing the affinity of the lac repressor to the operator. This decreased affinity results in the promoter being accessible to RNA polymerase. The lack of glucose allows for increased synthesis of cAMP, which can complex with CAP. The formation of CAP-cAMP complexes improves the efficiency of RNA polymerase binding to the promoter, which results in higher levels of transcription from the lac operon. Condition 2 would have the least. With no lactose present, the lac repressor is active and binds to the operator, inhibiting transcription. The presence of glucose results in a decrease of cAMP levels. A CAP-cAMP complex does not form, and RNA polymerase will not be stimulated to transcribe the lac operon. Beta-galactosidase breaks down lactose into glucose and galactose for E. coli to use as an energy source.

How does autopolyploidy arise?

Consider meiosis in an autotriploid. In meiosis in a diploid cell, two homologous chromosomes pair and align, but in autotriploid, three homologs are present. One of the three homologs may fail to align iwht the other two and this unaligned chromosome will segregate randomly. Which gamete gets the extra chromosome will be determined by chance and will differ for each homologous group of chromosomes. The resulting gametes will have two copies of some chromosomes and one copy of others. Even if all three chromosomes do align, two chromosomes must segregate to one gamete and one chromosome to the other. Sometimes, the presence of a third chromosome interferes with normal alignment and all 3 chromosomes move to the same gamete. No matter how the 3 homologous chromosomes align their random segregation will create unbalanced gametes, with various numbers of chromosomes. A gamete produced by meiosis in such an autotriploid might receive, say two copies of chromosome I one copy of chromosome 2 three copies of chromosome 3 and no copies of chromosome 4. When the unbalanced gamete fuses with a normal gamete (or with another unbalanced gamete), the resulting zygote has different numbers of the four types of chromosomes. This difference in number creates unbalanced gene dosage in the zygote, which is often lethal. For this reason, triploids do not usually produce viable offspring. In even numbered autopolyploids such as 4n, the homologous chromosomes can theoretically form pairs and divide equally but this event rarely takes place so these types of autotetraploids also produce unbalanced gametes.

Coupling vs Repulsion

Coupling refers to the case where dominant alleles are on the same homologue chromosome and both recessive alleles are on the other homologue chromosome. ... Repulsion refers to the case where each homologous chromosome has one dominant and one recessive allele from the two genes.

Define Crossing Over

Crossing Over: The exchange of genetic material between nonsister chromatids that takes place in prophase I of meiosis

Information I might need l8er

Crossing over results in recombination; it breaks up the associations of genes that are close together on the same chromosome Linkage keeps particular genes together and crossing over mixes them up, producing new combinations of genes Complete Linkage: genes that exhibit complete linkage are located very close together on the same chromosome and do not exhibit crossing over Incompletely linked: genes that exhibit crossing over are said to be incompletely linked After a single crossover has taken place, the two chromatids that did not participate in crossing over are unchanged; gametes that receive these chromatids are nonrecombinants. The other two chromatids (which did participate in crossing over) now contain new combinations of alleles; gametes that receive these chromatids are recombinants for each meiosis in which a single crossover takes place, two nonrecombinant gametes and two recombinant gametes will be produced

What similarities and differences exist in the enzymatic activities of DNA polymerases I and III? What is the function of each DNA polymerase in bacterial cells?

DNA Polymerase I- synthesizes new DNA nucleotides from 3' OH DNA Polymerase III- replaces RNA primers with DNA nucleotides

DNA molecules of different sizes are often separated with the use of a technique called electrophoresis. With this technique, DNA nucleotides are placed in a gel, an electrical current is applied to the gel, and the DNA molecules migrate toward the positive pole of the current. What aspect of its structure causes a DNA molecule to migrate toward the positive pole?

DNA is negatively charged The electrons help push the DNA to the positive side Electrons are negatively charged as well

A conditional mutation is one that expresses its mutant phenotype only under certain conditions (the restrictive conditions) and expresses the normal phenotype under other conditions (the permissive conditions). One type of conditional mutation is a temperature sensitive mutation, which expresses the mutant phenotype only at certain temperatures. Strains of E. coli have been isolated that contain temperature-sensitive mutations in the genes encoding different components of the replication machinery. In each of these strains, the protein produced by the mutated gene is nonfunctional under the restrictive conditions. These strains are grown under permissive conditions and then abruptly switched to the restrictive condition. After one round of replication under the restrictive condition, the DNA from each strain is isolated and analyzed. What characteristics would you expect to see in the DNA isolated from each strain with a temperature-sensitive mutation in its gene that encodes in the following proteins?

DNA ligase- multiple breaks in the DNA, therefore not a perfect helix DNA polymerase I- hybrid of DNA and RNA still in the helix DNA polymerase III- no new DNA nucleotides synthesized, hybrid of RNA primers and parental strands Primase- unwound DNA with no new daughter strands Initiator proteins DNA is not unwound and is still the same as the parental strands

What is DNA methylation? DNA Polymerase III: name activities. DNA Polymerase I: name activities.

DNA methylation is the addition of methyl groups (--CH3) to certain positions on the nitrogenous bases on the nucleotide DNA polymerase I: 5' to 3' polymerase activity allow enzyme to synthesize DNA, 3' to 5' exonuclease activity to correct errors, 5' to 3' exonuclease activity to remove primers laid down by primase and replace them with DNA nucleotides by synthesizing in a 5' to 3' direction. DNA Polymerase III: '5 to 3' polymerase activity allows it to add new nucleotides in the 5' to 3' direction. 3' to 5' exonuclease activity allows it to remove nucleotides in the 3' to 5' direction, enabling it to correct errors. ***If a nucleotide with an incorrect base is inserted into the growing DNA strand, DNA Polymerase III uses its 3' to 5' exonuclease activity to back up and remove the incorrect nucleotide. Then resumes 5' to 3' polymerase activity.

Recombinant DNA

DNA molecules formed by laboratory methods of genetic recombination to bring together genetic material from multiple sources, creating sequences that would not otherwise be found in the genome NOT found in the parent genome!

DNA Replication: Termination in detail

DNA replication arrest is triggered by the encounter of a replisome with a Tus-Ter complex "Ter" = repetitive DNA sequences; sites of termination "Tus" = proteins that bind to Ter sites Has two different sides: 1. Nonpermissive face of Tus (indent) = DnaB binds to Tus and unwinding is arrested 2. Permissive face of Tus (square) = Tus is dislodged from Ter and replication continues

Define cDNA. Why use it?

DNA synthesized from a single-stranded RNA template in a reaction catalyzed by the enzyme reverse transcriptase When mRNA is made, RNA polymerase will transcribe the DNA sequences from the end of the promoter to the termination sequence. Furthermore, if the mRNA is isolated from the cytoplasm of eukaryotes, intron sequences will have been removed and will not be copied into a cDNA clone. The cDNA clone will only contain the sequences found in the mRNA. When scientists transfer a gene from one cell into another cell in order to express the new genetic material as a protein in the recipient cell, the cDNA will be added to the recipient (rather than the entire gene), because the DNA for an entire gene may include DNA that does not code for the protein or that interrupts the coding sequence of the protein (e.g., introns).

30 nm fiber

DNA wraps 2x around histone core, histones interact to create organized structure called 30nm fiber

Dominance Incomplete Dominance Codominance

Dominance: phenotype of the heterozygote is the same as the phenotype of one of the homozygotes Incomplete Dominance: phenotype of the heterozygote is intermediate (falls within the range) between the phenotypes of the two homozygotes Codominance: phenotype of the heterozygote includes of the phenotypes of both homozygotes

Recessive Epistasis vs. Dominant Recessive

Dominant epistasis is when only one allele of the gene that shows epistasis can mask alleles of the other gene. Recessive epistasis is where two alleles have to be inherited in order for the phenotype of the second gene to be masked

Most RNA molecules have three phosphate groups at the 5' end, but DNA molecules never do. Explain this difference.

During initiation of DNA replication, DNA nucleoside triphosphates must be attached to a 3′-OH of a RNA molecule by DNA polymerase. This process removes the terminal two phosphates of the nucleotides. If the RNA molecule is subsequently removed, then a single phosphate would remain at the 5′ end of the DNA molecule. RNA polymerase does not require the 3′-OH to initiate synthesis of RNA molecules. Therefore, the 5′ end of a RNA molecule will retain all three of the phosphates from the original nucleotide triphosphate substrate.

A mutant strain of E. coli produces B-galactosidase in the presence and absence of lactose. Where in the operon might the mutation in this strain be located?

Enzyme is produced whether or not lactose is present. This means that the operon is repressible. Presence or absence of lactose, transcription is on. Therefore, the operon is under negative control. Outside of the operon, the mutation might have occurred in the regulatory gene (lacI) that codes for the repressor protein (causing negative control to become inactive, leading to constitutive expression of the structural genes). The mutation might have occurred in the operator. Since the product binds to the repressor and together they bind to the operator, the operator might have had a mutation in which the product and repressor cannot bind the operator. If the lacI gene produces the inactive repressor, it cannot bind to operator and transcription continues. If the mutation occurs in operator region, the repressor protein can not bind to the operator, transcription continues.

What are epigenetic effects? How do they differ from other genetic traits?

Epigenetic effects: represents the inheritance of variation above and beyond differences in DNA sequence, refers to phenotypes and processes that are transmitted to other cells and sometimes to future generations, but are not the result of differences in the DNA base sequence Many effects are caused by (1) changes in gene expression that result from alterations to chromatin structure, (2) or other aspects of DNA structure, such as DNA methylation Epigenetic effects: refers to change in gene expression or phenotype that are potentially heritable without alteration of the underlying DNA base sequence, can be influenced by environmental factors (genes have memory, environmental factors acting on individuals can have effects that are transmitted to future generations)

#18 EOC White (w) coat color in guinea pigs is recessive to black (W). In 1909, W. E. Castle and J. C. Phillips transplanted an ovary from a black guinea pig into a white female whose ovaries had been removed. They then mated this white female with a white male. All the offspring from the mating were black. a. Explain the results of this cross b. Give the genotype of the offspring of this cross. Genotype of F1

Explain the results of this cross. The female had gametes that were black for coat color so it didn't matter what she was expressing because she passing down a trait that was dominant. We can tell that this trait was homozygous dominant because all the offspring expressed the trait even though they had a parent that was homozygous recessive. Ww

If you know the sequence of your gene of interest, but it is not surrounded by "useful" restriction sites, you can "create" restriction sites by including them in the 3' end of the primer, and then proceed with PCR to make multiple copies of the segment of interest. T or F

FALSE 5' end doesnt have to be complimentary but 3' does

Transcription and Translation occur simultaneously in both prokaryotes and eukaryotes. T or F

FALSE! Transcription and translation occurs simultaneously in prokaryotes and in eukaryotes the RNA is first transcribed in the nucleus and then translated in the cytoplasm.

In a single nucleotide, the nitrogenous base is attached to the 3' carbon.

False, 1' Carbon

Bacterial genomes are made of single stranded DNA.

False, dsDNA

When a structural gene is under positive inducible control, what would be the result of a mutation that eliminates the activator protein? The transcription of a structural gene will not be affected, as an activator is not required. T or F

False, it will be required!!

All genomes contain high amounts of repetitive sequences.

False, only eukaryotic genomes

#23 EOC What is the most likely sex and genotype of the cat shown in Figure 4.12?

Female, X+X0

Translation Initation in Prok

Formation of the 70S Translation Initiation Complex in Bacteria 1. IF one and three help the mRNA bind to the small ribosomal subunit bind to the 5' end of the mRNA (Shine-Delgarno sequence that helps with ID of start codon is here!). and prevent binding of the large ribosomal subunit. 2. IF 2 aids the engagement of the start charged tRNA (start aa for bacterial translation, formal methionine specific for bacteria, f-met > methionine in us!). F-met attached to first tRNA that has the anticodon for AUG codon. IF brings that and makes a specific arrangement happen. 3. On the small ribosomal subunit, there is the A, P, and E site. tRNA binds the start codon, complex placed on P site. Large ribosomal subunit (50s) engages the rest of complex, IF one and two and three are removed. Large subunit also has grooves for A, P, and E site. Initiator tRNA is exactly at the P site. 4. 70S initiation complex: small ribosomal subunit, large ribosomal subunit, mRNA, and the charged initiator tRNA. Initiation is complete upon formation of the 70S initiation complex.

Forward vs. Reverse Genetics

Forward: Begins with a phenotype (a mutant individual) and proceeds to a gene that encodes the phenotype Reverse: Begins with a genotype (a DNA sequence) and proceeds to the phenotype by altering the sequence or inhibiting its expression

Examples of Trinucleotide Expansions

Fragile-X Syndrome - X linked recessive disorder 1. CGG repeat (between 50-1500 times is disease range) Form of inherited mental retardation (more severe in males, X chromosome constricted on long arms of chromosome, males only have one X and females have two X, females can have one undamaged and one damaged) 2. X chromosome has the fragile-X phenotype (area is so highly methylated that it creates a constriction on the long arm of the X chromosome), methylation leads to DNA compaction! Additional copes of the nucleotide repeat cause the DNA to become methylated which blocks transcription 3. Mutation: A CGG trinucleotide repeat is located at the 5'-UTR of FMR1 gene. Gene cannot be transcribed! 4. FMRP (fragile-X mental retardation protein) is an RNA-binding protein that shuttles between the nucleus and cytoplasm. This protein has been implicated in protein translation as it is found associated with polyribosomes and the rough endoplasmic reticulum. 6. In most affected individuals, CGG repeats are massively expanded (> 200) over 230 repeats and becomes abnormally hypermethylated, which results in FMR1 silencing. Normal: under 44, pathologic > 200 7. CpG (p=phosphate group is, 5' to 3' direction, methylation signal for the human genome, take these bases and methylate them, DNA becomes compact and unavailable for transcription) 2 features: compact and unavailable 8. Unavailable: if compact, unavailable for most molecules that need to interact with DNA (Transcription factors, any regulator of transcription, or transcription apparatus)

The amount of DNA per cell of a particular species measured in cells found at various stages of meiosis, and the following amounts are obtained: Amount of DNA per cell in picograms (pg) ___3.7 pg ___7.3 pg ___14.6 pg Match the amounts of DNA above with the corresponding stages of meiosis (a through f, below). You may use more than one stage for each amount of DNA . G1 Prophase I G2 After Telophase 2 and Cytokinesis Anaphase I Metaphase II

G1: 7.3 Prophase I: 14.6 G2: 14.6 After Telophase II and Cytokinesis: 3.7 Anaphase I: 14.6 Metaphase II: 7.3

A cell in G1 of interphase has 12 chromosomes. How many chromosomes and DNA molecules will be found per cell when this original cell progresses to the following stages?

G2 of interphase: Chromosomes: 12 DNA molecules: 24 Metaphase 1 of Meiosis: Chromosomes: 12 DNA molecules: 24 Anaphase 1 of Meiosis: Chromosomes: 12 DNA molecules: 24 Anaphase 2 of Meiosis: Chromosomes: 12 DNA molecules: 12 Prophase 2 of Meiosis: Chromosomes: 6 DNA molecules: 12 Prophase of Mitosis: Chromosomes: 12 DNA molecules: 24 After cytokinesis following mitosis: Chromosomes: 12 DNA molecules: 12 After cytokinesis following meiosis 2: Chromosomes: 6 DNA molecules: 6

Flow of genetic information Define and explain this term.

Gene > mRNA > protein (term depicts these three processes in gene expression) 1. Gene is transcribed from DNA to mRNA 2. Translated into a specific protein that impacts the phenotype 4. Eukaryotes produce a pre mRNA (begins with transcription start site and ends with terminator) 5. stop (between protein coding region and 3' UTR) and start codon (between protein coding region and 5' UTR) 6. Mature mRNA: a. In order: 5' cap, 5' UTR, protein coding region, 3' UTR, and polyA tail

#9 EOC: What is the principle of independent assortment? How is it related to the principle of segregation?

Genes encoding different characteristics (genes at different loci) separate independently; applies only to genes located on different chromosomes or to genes far apart on the same chromosome. This can be seen in anaphase I of meiosis as each pair of homologous chromosomes separate independently of all other pairs. To further add, if the genes are sufficiently far apart from one another, crossing over can take place to produce recombinants.

What is the difference between genes in coupling configuration and genes in repulsion? How does the arrangement of linked genes (whether they are in coupling or repulsion) affect the results of a genetic cross?

Genes in coupling/cis configuration: wild type alleles are found on one chromosome and mutant alleles are found on the other chromosome (When the alleles are in the coupling configuration, the most numerous progeny types are those with a normal green thorax and normal brown puparium // those with a recessive purple thorax and a recessive black puparium Genes in repulsion/trans configuration: each chromosome contains one wild-type and one mutant allele (when the alleles of the heterozygous parent are in repulsion, the most numerous progeny types are those with the normal green thorax and recessive black puparium // those with a recessive purple throax and normal brown puparium)

List the different protein and enzymes taking part in bacterial replication. Give the function of each in the replication process

Gyrase- a type 2 topoisomerase which controls supercoiling of DNA creating double strand breaks which reduces the torsional strain (torque) that builds up ahead of the replication fork as a result of unwinding (DNA Initiation). Gyrase creates a double stranded break, passes another segment of the helix through the break, and then reseals the broken ends of the DNA.

The term haploid depicts a genome that is composed of a single stranded nucleic acid.

Haploid refers to the number of complete sets of chromosomes in a cell. (A single set of unpaired? chromosomes).

Define the secondary structure of a protein. **Two main types, but know there are more.

Hydrogen bonds between atoms in the polypeptide backbone interactions between amino acids cause the primary structure to fold into a secondary structure, such as an alpha helix or a beta pleated sheet.

#19 EOC In cats, blood type A results from an allele I^A, which is dominant over an allele i^B that produces blood type B. (There is no O blood type, as there is in humans). The blood types of male and female cats that were mated and the blood types of their kittens are presented in the following table. Give the most likely genotypes for the parents of each litter. Male Parent Female Parent Kittens a. A B 4 w/A, 3 w/B b. B B 6 w/B c. B A 8 w/A d. A A 7 w/A, 2 w/B e. A A 10 w./A f. A B 4 w/A, 1 w/B ??????????

IA = type A iB = type B a. Almost a 1:1 ratio, therefore Male is IAiB and Female is iBiB b. Duh iBiB for both c. Female is IAIA and Male is iBiB d. Both heterozygous, Male is IAiB and Female is IAiB e. Male can be either IAiB or IAIA (or female too) and female is IAIA f. About half have A and half have B, therefore parents are IAiB and iBiB

#3 EOC How is sex determined in humans?

If an individual has two copies of the X chromosome, they are female If an individual has one copy of the X chromosome and one copy of the Y chromosome, they are male Presence of SRY gene on the Y chromosome determines maleness

What would be the most likely effect of moving the AAUAAA consensus sequence shown in Figure 10.19 ten nucleotides upstream?

If it was moved upwards, then there is risk that part of RNA-coding region would be cut off and so all of the polypeptide chain can not be synthesized.

In mammals, sex-chromosome aneuploids are more common than autosomal aneuploids, but in fishes, sex-chromosome aneuploids and autosomal aneuploids are found with equal frequency. Offer a possible explanation for these differences between mammals and fishes. (Hint: Think about why sex-chromosome aneuploids are more common than autosomal aneuploids in mammals) Define dosage compensation. Why is it important?

In mammals the humans have onne X chromsome and females have two X chromosomes. They have a process called dosage compensation one of the X chromosome is inactive and will not express the phenotype associated with the chromsome. Due to this aneuploidy is well tolerated in sex chromosomes and will be able to produce better offsprings. Autosomal aneupliods are more lethal as they are associated with more health risks. In fish, they do not have doasge compensation mechanism due to which the aneuploids are equal in sex chromosomes and in autosomes. the process by which organisms equalize the expression of genes In order to neutralize the large difference in gene dosage produced by differing numbers of sex chromosomes among the sexes, dosage compensation is needed to equalize gene expression among the sexes. (X-inactivation, For example, in humans, females (XX) silence the transcription of one X chromosome of each pair, and transcribe all information from the other, expressed X chromosome. Thus, human females have the same number of expressed X-linked genes as do human males (XY), both sexes having essentially one X chromosome per cell, from which to transcribe and express genes)

Positive Control

In operons under positive control, the regulator protein acts as an activator and binds to the promoter to enable transcription by aiding the binding of the RNA polymerase to the promoter.

Repressible

In repressible operons, the presence of the product represses the transcription of the structural genes of the operon. (can be positive or negative, dont define this way)

The blob operon produces enzymes that convert compound A into compound B. the operon is controlled by a regulatory gene S. Normally, the enzymes are synthesized only in the absence of compound B. If gene S is mutated, the enzymes are synthesized in the presence and in the absence of compound B. Does gene S produce a repressor or an activator? Is this operon inducible or repressible?

In the absence of compound B, the enzymes are synthesized. This means that the operon must be repressible. If gene S is mutated, enzymes are always present regardless of presence/absence of compound B. This means that the operon is under negative control. Gene S produces a repressor and the operon is repressible. Negative Repressible

Eukaryotic Transcription TERMINATION (protein-coding genes)

In the case of protein-encoding genes, the cleavage site which determines the "end" of the emerging pre-mRNA occurs between an upstream AAUAAA sequence and a downstream GU-rich sequence separated by about 40-60 nucleotides in the emerging RNA. Once both of these sequences have been transcribed, a protein called CPSF in humans binds the AAUAAA sequence and a protein called CstF in humans binds the GU-rich sequence. These two proteins form the base of a complicated protein complex that forms in this region before CPSF cleaves the nascent pre-mRNA at a site 10-30 nucleotides downstream from the AAUAAA site. The Poly(A) Polymerase enzyme which catalyzes the addition of a 3′ poly-A tail on the pre-mRNA is part of the complex that forms with CPSF and CstF.

A geneticist discovers a new mutation in Drosophila melanogaster that causes the flies to shake and quiver. She calls this mutation spastic (sps) and determines that it is due to an autosomal recessive gene. She wants to determine whether the gene encoding spastic is linked to the recessive gene for vestigial wings (vg). She crosses a fly homozygous for spastic and vestigial traits with a fly homozygous for the wild-type traits and then uses the resulting F1 females in a testcross. She obtains the following flies from this testcross: Sps sps vg vg X sps+sps+ vg+ vg+ F1: sps+sps vg+vg Are the genes that cause vestigial wings and spastic linked? Do a chi square test of independence to determine whether the genes have assorted independently?

In the cross between a heterozygous and homozygous tester the precentage of recombinants is less than 50%, then the genes are said to be linked. In this case, both genes are linked. For independent assortment the phenotypes would be in the ratio of 1:1:1:1 X2 value is 102.5 No, the X2 value indicates that the observed progeny are significantly different from what would be expected with independent assortment of the two genes.

Phosphorus is required to synthesize the deoxyribonucleoside triphosphates used in DNA replication. A geneticist grows some E. coli in a medium containing non radioactive phosphorus for many generations. A sample of the bacteria is then transferred to a medium that contains a radioactive isotope of phosphorus (32P). Samples of the bacteria are removed immediately after the transfer and after one and two rounds of replication. Assume that newly synthesized DNA contains 32P and that the original DNA contains non radioactive phosphorus. What will be the distribution of radioactivity in the DNA of the bacteria in each sample? Will radioactivity be detected in neither strand, one strand, or both strands of DNA?

In the initial sample removed immediately after transfer, no 32P should be incorporated into the DNA because replication in the medium containing 32P has not yet occurred. After one round of replication in the 32P-containing medium, one strand of each newly synthesized DNA molecule will contain 32P, while the other strand will contain only nonradioactive phosphorus. After two rounds of replication in the 32P-containing medium, 50% of the DNA molecules will have 32P in both strands, while the remaining 50% will contain 32P in one strand and nonradioactive phosphorus in the other strand.

#7 EOC How do incomplete dominance and codominance differ? THIS IS ONLY IN ONE GENE!!

Incomplete dominance results from an intermediate expression of two dominant alleles (phenotypes) Codominance results when both phenotypes are equally expressing resulting in two phenotypic expressions

The mmm operon, which has sequences A, B, C, and D (which may be structural genes or regulatory sequences), encodes enzymes 1 and 2. Mutations in sequences A, B, C, and D have the following effects, where a plus sign (+) indicates that the enzyme is synthesized and a minus sign (-) indicates that the enzyme is not synthesized. Is the operon inducible or repressible? Indicate which sequence is part of the following components of the operon:

It is a positive/negative repressible operon. Regulator Gene: B Promoter: D Structural gene for Enzyme 1: A Structural gene for Enzyme 2: C

#17 EOC Red-green color blindness in humans is due to a X-linked recessive gene. Both John and Cathy have normal color vision. After 10 years of marriage to John, Cathy gave birth to a color-blind daughter. John filed for divorce, claiming that he is not the father of the child. Is John justified in his claim of nonpaternity. Explain why. If Cathy had given birth to a color-blind son, would John be justified in claiming nonpaternity?

John and Cathy = normal vision X+Y (John) and X+X^m Produced a color-blind daughter. Claim of nonpaternity? John would give his daughter the X+ allele and Cathy could give his daughter either an X+ or an X^m allele. Regardless, John's X+ allele given to his daughter would mask the X^m allele received from Cathy. They could not have a color blind daughter, therefore, his claim is justified. Produced a color-blind son. Claim of non paternity? NO, it could be his son. John would give his son the Y allele and son can receive either a X+ or a X^m allele. Therefore, if the son received the X^m allele, there is no possibility of masking the color blindness and the son would be color blind. John is not justified in this claim.

How do translocations in which no genetic information is lost or gained produce phenotypic effects?

Like inversions, translocations can produce phenotypic effects if the translocation breakpoint disrupts a gene or if a gene near the breakpoint is altered in its expression because of relocation to a different chromosomal environment (a position effect).

List some similarities and differences between meiosis and mitosis. Which differences do you think are most important and why?

Meiosis: 46 chromosomes (2n=46), two rounds of division, different genetic variability (crossing over, independent assortment), homologous pairs line up in metaphase 1, only for gametic cells Mitosis: 23 chromosomes, one round of division, same genetics passed on, homologous pairs never line up (only chromosomes), only for somatic cells

Duchenne muscular dystrophy is caused by a mutation in a gene that encompasses more than 2 million nucleotides and specifies a protein called dystrophin. However, less than 1% of the gene actually encodes the amino acids in the dystrophin protein. On the basis of what you now know about gene structure and RNA processing in eukaryotic cells, provide a possible explanation for the large size of the dystrophin gene.

Much of the gene could consist of mainly the promoter and introns that are regulatory. Suppose that a mutation occurs in the middle of a large intron of a gene encoding a protein. What will the most likely effect of the mutation be on the amino acid sequence of that protein? Explain your answer.

#6 EOC: What are the addition and multiplication rules of probability and when should they be used?

Multiplication rule should be used when two independent events occur simultaneously. The combined probability of the two outcomes is equal to the product of their individual probabilities of occurrence. Addition rule is used when the probability that any one or two or more mutually exclusive events will occur is calculated by adding their individual probabilities.

The fruit fly Drosophila melanogaster (below left) has four pairs of chromosomes, whereas the house fly Musca domestica (below right) has six pairs of chromosomes. In which species would you expect to see more genetic variation among the progeny of a cross? Explain your answer.

Musca domestica there are more genes, indicating that there is a larger chance for variability when these cross over more genes, more chances that the chromosomes will cross over

Briefly explain how the polymerase chain reaction is used to amplify a specific DNA sequence

NEED: a DNA template, a pair of primers with a 3'-OH group to which new nucleotides can be added Step 1: Begin with a solution containing target DNA, DNA polymerase (Taq), four deoxyribonucleoside triphosphates (dNTPs - the substrates fro DNA polymerase), primers, and magnesium ions and other salts that are necessary for the reaction to proceed. - DNA is heated to a high temperature (90-100 C) which breaks the hydrogen bonds between the strands and produces single-stranded templates Step 2: DNA solution is cooled quickly, usually to between (45-65 C) which allows the primers to attach to the template strands Step 3: solution is heated to 72 C, the temperature at which DNA polymerase can synthesize new DNA strands At the end, you have two new double-stranded DNA molecules that were produced from each original molecule of target DNA. With each cycle, the amount of target DNA DOUBLES!!

Negative Inducible Operon vs. Positive Inducible Operon

NIO: negative control > repressor protein is an activate repressor that needs to be inactivated, inducible operon produces substrate that inactivates the repressor and transcription is ON PIO: positive control > regulator protein is an inactive activator that needs to be activated. Inducible operon produces substrate, which activates the activator and transcription is ON

Negative Repressible Operon vs. Positive Repressible Operon

NRO: negative control > repressor protein is an inactive repressor that needs to be activated, the repressible operon produces a product that activates the repressor and transcription is OFF PRO: positive control > activator protein that is an active activator that needs to be repressed, the repressible operon produces a product

What happens when an operon is under negative control? Positive control?

Negative - In operons under negative control, the regulator protein acts as a repressor and binds to the operator to prevent transcription by preventing the binding of the RNA polymerase to the promoter. Positive -In operons under positive control, the regulator protein acts as an activator and binds to the promoter to enable transcription by aiding the binding of the RNA polymerase to the promoter.

Briefly describe the lac operon and how it controls the metabolism of lactose.

Negative inducible operon Structural genes include: lacZ, lacY, and lacA that code for B-galactosidase, permease, and transacetylase respectively The presence of lactose (allolactose is made from lactose) turns transcription on by inactivating the repressor protein (binds to the repressor and causes the repressor to be released from the DNA) Lactose present? Repressor inactivated, binding of RNA polymerase no longer blocked, transcription of lacZ, lacY, and lacA takes place, and the lac operon proteins are produced! Therefore RNA polymerase can bind to the promoter and allow for transcription to occur Beta-galactosidase must break down lactose into glucose and galactose for E. coli to use lactose as an energy source. This enzyome also converts lactose into allolactose (regulates lactose metabolism). All genes have a common promoter. Upstream of lacP is lacI, which has its own promoter and is transcribed into a short mRNA that is translated into a repressor. In the absence of lactose, the repressor binds to the lac operator (lacO).

Huntington's Disease and Effects

Neurotransmitters allow synapsis. Mutant protein blocks transcription and production of all of these important neurotransmitters (damage the connection of neural circuits) Mitochondrial dysfunction: damaged protein blocking at least 4 different processes of the mitochondria (energetic function of the cell is damaged) Two main working hypotheses: Happens specifically in neurons (does not happen in other cells). What fails is the mind, they start to produce uncontrollable and violent muscle spasms, alters movement and speech and results in a severe form of dementia but it doesn't damage other tissues of the body. WHY is the question? Why do neurons have such a poor time dealing with this? Genetic anticipation: older and have been diagnosed with huntington's, partner would be homozygous for normal, already have children, half of their children have the chance of having the disease, no treatment only palliative care

If there are no replicated chromosomes, can you have SC?

No lol

Predict what would have happened if Griffith had mixed some heat-killed type IIIS bacteria and some heat killed type IIR bacteria and inject this mixture into a mouse. Would the mouse have contracted pneumonia and died? Explain why or why not.

No the mouse would have lived because there would be no living cell to replicate the pathogenic information to cause enough pathogenicity to kill the mouse There has to be a living cell able to use the genetic information of pathogenic strained bacteria and utilize in such a way that would cause harm to the mouse

Define the following terms as they apply to the genetic code: Nonsense codon Universal code Non-universal code

Nonsense codon: do not encode amino acids, another word for stop codon/termination codon, UAG UAA UGA Universal code: each codon specifies the same amino acid in all organisms Non-universal code: most exceptions are termination codons but there are few cases in which one sense codon substitutes for another, most exceptions are found in mitochondrial genes; a few nonuniversal codons have been detected in the nuclear genes of protozoans and in bacterial DNA

The picture below is a sequencing gel from the original study that first sequenced the cystic fibrosis gene. From the picture, determine the sequence of the normal copy of the gene and the sequence of mutated copy of the gene. Identify the location of the mutation that causes cystic fibrosis (CF). Hint: the CF mutation is a 3 bp deletion. FIX

Normal: 5' - ATCATAGGAAACACCAAAGATGATA - 3' Mutated: 5' - TTCATCATAGGAAACACCAATGATA - 3' CF mutation at: ??

Nucleases Endo? Exo

Nucleases: any enzymes that break down a nucleic acid (exo- eat from the edge, endo - can take a sequence, recognize it, and cut it out from anywhere)

Transcription Elongation SUMMARY

Occurs when a newly synthesized complimentary RNA strand is made as the template DNA is threaded through the RNA polymerase

Draw the Mouse Locus :)

On the maternal chromosome, the unmethylated DMD/ICR (DMR1 in this figure) binds the CTCF protein and forms an insulator, preventing the common enhancers from activating Igf2. Instead, the enhancers activate the nearby H19 promoter. On the paternal chromosome, the methylated DMD/ICR (DMR1 in this figure) cannot bind CTCF and an insulator does not form, and, therefore, the Igf2 gene is expressed.

Suppose that you have just graduated from college and have started working at a biotechnology firm. Your first assignment is to clone the pig gene for the hormone prolactin. Assume that the pig gene for prolactin has not yet been isolated, sequenced, or mapped; however, the mouse gene for prolactin has been cloned and the amino acid sequence of mouse prolactin is known. Briefly explain two different strategies that you might use to find and clone the pig gene for prolactin.

One strategy would be to use the mouse gene for prolactin as a probe to find the homologous pig gene from a pig genomic or cDNA library. A second strategy would be to use the amino acid sequence of mouse prolactin to design degenerate oligonucleotides as hybridization probes to screen a pig DNA library. Yet a third strategy would be to use the amino acid sequence of mouse prolactin to design a pair of degenerate oligonucleotide PCR primers to PCR amplify the pig prolactin gene.

In the snail Capaea nemoralis, an autosomal allele causing a banded shell (B^B) is recessive to the allele for an unbanded shell (B^O). Genes at a different locus determine the background color of the shell; here, yellow (C^Y) is recessive to brown (C^Bw). A banded, yellow snail is crossed with a homozygous brown, unbanded snail. The F1 are then crossed with banded, yellow snail (a testcross). a. What will the results of the testcross be if the loci that control banding and color are linked with no crossing over? b. What will the results of the testcross be if the loci assort independently? c. What will the results of the testcross be if the loci are linked and 20 mu apart?

P: B^BB^BC^YC^Y x B^OB^OC^BwC^Bw F1: B^BB^OC^YC^Bw x B^BB^BC^YC^Y a. F2: 1/2 BOBBCBwCY unbanded brown : 1/2 BBBBCYCY banded yellow (linked with no crossing over b. If the loci assort independently? F2 = 1/4 BBBBCYCY : 1/4 BBBBCBwCY : 1/4 BBBOCYCY : 1/4 BBBOCBwCY c. 20 mu apart = 20% recombinant frequency. Each recombinant must be 10%. Nonrecombinant progeny must each be 40% since the total nonrecombinants is 80%. 40% unbanded brown, 40% banded yellow, 10% unbanded yellow, 10% banded brown. With absolute linkage, there will be no recombinant progeny. The F1 inherited banded and yellow alleles (BBCY) together on one chromosome from the banded yellow parent and unbanded and brown alleles (BOCBw) together on the homologous chromosome from the unbanded brown parent.

Transcription terminator sequence (Intrinsic or self terminator, there are two types but we will only look at this one, euk and prok have different terminators) has a:

Palindromic sequence that can create a hairpin structure and disrupts the engagement of the RNA polymerase with the template strand, leads to termination of transcription.

What is the difference between a paracentric and pericentric inversion?

Paracentric: inversions that do not include the centromere, such as AB.C(FED)G Pericentric: inversions that incljude the centromere, such as A(DC).(B)EFG

What does it mean to have a 3:1 ratio, as in, what genotype would you expect the parents to have?

Parents must be heterozygous because we have a 3:1 ratio!

#37 EOC Some sweet-pea plants have purple flowers and others have white flowers. A homozygous variety of sweet pea that has purple flowers is crossed with a homozygous variety that has white flowers. All the F1 have purple flowers. When these F1 are self-fertilized, the F2 appear in a ratio of 9/16 purple to 7/16. a. Give genotypes for the purple and white flowers in these crosses. b. Draw a hypothetical biochemical pathway to explain the production of purple and white flowers in sweet peas.

Parents: AABB (purple) x aabb (white) Produces AaBb (Purple) Assuming that A_B_ produces purple! a. Purple: 1/16 AABB: 2/16 AABb: 2/16 AaBB: 4/16 AaBb // 9/16 A_B_ White: 1/16 aabb: 3:16 aaB_: 3/16 A_bb b. White precursor -> white intermediate 2 -> purple pigment

Define Partial Diploid/Merozygote:

Partial diploid/Merozygote: Directed mutagenesis, E. coli cell with chromosome (Insert a plasmid that I want to study with chromosomes, E. coli (TRANSFORMATION) now has diploid for genes f and g (in pic) NOW genome is a partial diploid/merozygote (two copies of f and g but one copy of e)) a cell carrying two copies of some, but not all, of its GENES. In BACTERIA where one copy of the genes of interest is carried on the chromosome, a second copy may be introduced on a PLASMID. The cells therefore carry one complete set of genes and a duplicated copy of part of the GENOME.

Point mutations vs. Genomic mutation

Point Mutations: change of one nucleotide/small number of nucleotides Genomic Mutations: rearrangement of nucleotides (Ex. head to head fusion of Ch 13 and 21, involves two full chromosomes that have been fused, down syndrome is a rearrangement > 3 copies of Ch 21)

What is the difference between positive and negative control? What is the difference between inducible and repressible operons?

Positive: In operons under positive control, the regulator protein acts as an activator and binds to the promoter to enable transcription by aiding the binding of the RNA polymerase to the promoter. Negative: In operons under negative control, the regulator protein acts as a repressor and binds to the operator to prevent transcription by preventing the binding of the RNA polymerase to the promoter. Inducible: In inducible operons, the presence of the substrate induces the transcription of the structural genes of the operon. Repressible: In repressible operons, the presence of the product represses the transcription of the structural genes of the operon.

After translation, additional modifications may be needed. What is the term for this?

Posttranslational modifications Chaperonins Phosphorylated Methylated Acetylated Glycosylation Phospholipids and fatty acids (lipo proteins) Small peptide that may need to be added Ubiquitinylation SUMOyltation Nitrosylation Sulfonation along with alternative splicing, these are the sources of variation for proteins!

Define the primary structure of a protein. **Understand how each structure is influenced by the primary structure

Primary: string of amino acids, starting in the N-terminus and ending in the carboxyl terminus (ALWAYS!) Sequence of amino acids! linear chain of amino acids

14. What is the end-replication problem? Why in the absence of telomerase, do the ends of linear chromosomes get progressively shorter each time the DNA is replicated?

Primer at the end of chromosome has been removed, it cannot be replaced by DNA nucleotides Produces a gap at the end of the chromosome which suggests that the chromosome should become progressively shorter with each round of replication, does not add telomeres and chromosomes eventually get degraded

#3 EOC: What is the principle of segregation? Why is it important?

Principle of segregation states that each individual organism possesses two alleles that can encode a characteristic These alleles segregate when gametes are formed, and one allele goes into each gamete (equal proportions)

Gene regulation is key for unicellular flexibility and multicellular specialization. Prokaryotes? Eukaryotes?

Prok: operons Euk: more complex

Difference between eukaryotic and prokaryotic mRNA

Prokaroytic mRNA is polycistronic (mRNA that code for multiple different polypeptides) Eukaryotic mRNA is monostronic, pre-mature mRNA has introns and exons, 5' end of eukaryotic mRNA is different from prokaryotic mRNA as it has a 5' cap, 3' end of eukaryotic mRNA is different from prokaryotic mRNA as it has a Poly-AAA tail

Diff betwen euk and prok promoters

Prokaryotes: -10 and -35 region on the promoter NO Introns and NO Exons 6. Eukaryotes: Promoters have different types of sequences (varied) Introns and Exons in RNA coding region

What are some genetic differences between prokaryotic and eukaryotic cells?

Prokaryotic Cells: no nucleus, cell diameter is small, genome usually has one circular DNA molecule, DNA is small, no membrane-bound organelles Eukaryotic Cells: nucleus present, large cell diameter, multiple linear DNA molecules in genome, large DNA, has membrane-bound organelles

Regarding the expression of a eukaryotic gene in a prokaryotic cell, are there any concerns? Will you get the product you expect and will it be functional? FIX

Prokaryotic cells lack membrane-bound organelles, which means that certain eukaryotic genes may not function in E. coli like they would in their normal environment. The husK gene might produce a protein in need of undergoing posttranslational modifications to become functional (Transcription factors, translation factors might not be available when needed)

In what ways is eukaryotic replication similar to bacterial replication, and in what ways is it different?

Prokaryotic: small size > one Ori, circular DNA, no packaging Eukaryotic: greater size > multiple Ori, linear DNA, histone proteins (assembly of nucleosome must proceed after replication)

A cell in prophase 2 of meiosis has 12 chromosomes. How many chromosomes would be present in a cell from the same organism if it were in prophase of mitosis? Prophase 1 of meiosis?

Prophase of mitosis: 24 Prophase I of Meiosis: 24

Proteins that bind 5' cap and poly A tail

Proteins that bind the 5; cap: ribosomal binding, removal of introns, prevents degradation Proteins that bind the 3' poly A tail aids in ribosomal binding, export of mRNA to cytoplasm, prevents degradation

Nitrogenous Bases: Purines? Pyrimidines?

Purines: 2 rings > A and G Pyrimidines: 1 ring, C, U, and T

Recombination frequencies between three loci in corn are shown here. What is the order of genes on the chromosome?

R-----------------W2------------------L2

What are blunt ends?

RE cuts in the middle of its recognition sequence, and the cuts on the two strands are directly opposite on another

Recessive epistasis vs. Dominant epistasis Ratios?

RE: epistatic gene is recessive (Ex. Labrador retrievers), 9:3:4 DE: epistatic gene is dominant (Ex. Summer squash), 12:3:1

Intrinsic termination of transcription occurs when, after transcribing the terminator, a hairpin structure is formed that leads to disruption of the interaction between ________ and _______, and removes the transcript from the ________. Template strand is also known as the _________. Coding strand is also known as the _______.

RNA core polymerase, DNA, DNA template antisense, sense

Prokaryotic mRNA

RNA have 3 segments 2. ss RNA, starts at 5' ends at 3' 3. 5' untraslated region, protein coding region, 3' untraslated region 4. untraslated: NEVERERERERE! translated 5. Untraslated regions purpose: to recruit the ribosome, we need an area to recruit ribosome and have the ribosome assemble, and ribsoome can identify start codon, ends in stop codon 6. " " ^ Regions: provide stability and fold to protect this mRNA (out of cytoplasm travel to ribosome > euk), prevents degradation, provides a place for the ribosome to assemble and identify the start codon (both euk and prok) 7. Shine-Dalgarno sequence in bacteria only in 5' untraslated region helps ribosome assemble in the right way around the start codon 8. AUG, met, ribosome stops with one of stop codons (UGA, UAA, UAG)

AFter addition of the 5' cap, 3' cleavage, and polyadenylation of 3' end, splicing, two products are formed (introns are byproducts):

RNA nucleotides (can be broken down by nucleases): can be reused for more synthesis of RNA MATURE RNA: 5' cap, poly A tail, 5' untrans region, 3' untrans region, protein coding region (what separates the protein coding region and untranslated regions is at the 5' end a start codon is there, 3' end has a stop codon) speicc proteisn bind to poly A tail and 5' cap

#23 EOC Hairlessness in American rat terriers is recessive to the presence of hair. Suppose that you have a rat terrier with hair. How can you determine whether this dog is homozygous or heterozygous for the hairy trait?

Rat terrier with hair= R_ You can cross it with a hairless rat terrier (recessive rr) = rr Then you can determine if the F1 generation is all Rr (homozygous dominant for presence of hair) or not If you see all F1 generation rat terriers with hair (Rr), then you would expect the parent to be homozygous dominant for the hairy trait If you see half of the F1 generation is hairless and half have hair, then you would expect the parent to be heterozygous for the hair trait

Define the following terms as they apply to the genetic code: Reading frame Overlapping code Non-overlapping code

Reading frame: each different way of reading the sequence, any sequence of nucleotides has three potential reading frames Overlapping code: would be one in which a single nucleotide might be included in more than one codon Non-overlapping code: each nucleotide is part of a single codon

Define the process of screening cells for recombinant plasmids.

Recombinant Plasmids: 1. Cells bearing RP can be detected by using the selectable markers on the plasmid. Genes that confer resistance to an antibiotic are commonly used as selectable markers; any cell that contains such a plasmid will be able to live in the presence of the antibiotic, which would normally kill bacteria cells.

No crossing over results in?

Reduction in genotypes in the offspring compared to crossing over! (ALL nonrecombinant) Decrease in genetic variability in offspring

The introduction to this chapter, which describes the sequencing of 400-year-old DNA, emphasizes DNA's extreme stability. What aspects of DNA's structure contribute to the stability of the molecule? Why is RNA less stable than DNA?

Relatively Strong phosphodiester linkages connect the nucletodies Helical nature of the dsDNA molecule results in the negatively charged phosphates of each strand being arranged to the outisde and away from each other (greater opportunity for DNA repair and fidelity during replication) Complimentary nature of the nitrogenous bases of the nucleotides helps hold the two strands of polynucleotides together (a mistake occurs on one strand, the complimentary strand can serve as a template for corrections) Stacking interactions of the bases (allow for any base to follow another in a given strand) help hold two strands together Ability of DNA to have local variations in second structure contributes to stability Position of H on 2' carbon (lack of oxygen here), less reactive than RNA molecules

Why is DNA gyrase necessary for replication?

Relieves torsional strain by creating double-stranded breaks in the helical structure

Describe the composition and structure of the nucleosome?

Repeating core of protein and DNA Core particle consisting of DNA wrapped about two times around an octamer of eight histone proteins

How does replication licensing ensure that DNA is replicated only once at each origin per eukaryotic cell cycle?

Replication licensing factor (protein) attaches to an origin, then the replication machinery initiates replication at each licensed origin As replication forks move away from the origin, the licensing factor is removed, leaving the origin in an unlicensed state, where replication cannot be initiated again until the license is renewed The ensure that replication takes place only once per cell cycle, the licensing factor is active only after the cell has completed mitosis and before the next replication is initiated

If beta-galactosidase production is repressed, how can lactose metabolism be induced?

Repression never completely shuts down transcription of the lac operon There is a low level of transcription even with the active repressor bound to the operator A few molecules of beta-galactosidase, permease, and transacetylase are synthesized. When lactose is present, permease transports a small amount of lactose into the cell. A few molecules of beta-galactosidase convert some of the lactose into allolactose, which then induces transcription!

How can you detect the orientation of your plasmid? What enzymes can you use? FIX

Restriction Digest. You must choose a restriction site(s) that give you a different length of DNA fragments so that you can tell the orientation!

Rho dependent vs. Rho independent (self-intrinsic) termination (Transcription)

Rho-dependent: large hexameric protein that physically interacts with the growing RNA transcript to remove it from the template DNA Rho-independent: Specific sequences that contain inverted sequences that fold the mRNA into a hairpin structure causing the RNA polymerase to disable from the DNA template and disassociate the mRNA from the DNA

Know rDNA

Ribosomal DNA a DNA sequence that codes for ribosomal RNA. Ribosomes are assemblies of proteins and rRNA molecules that translate mRNA molecules to produce proteins. As the DNA double helix unwinds, other RNA molecules read the template that is provided from this DNA sequence and an rRNA molecule is formed. Since these DNA segments do not provide the code for specific proteins, the rRNA products produced from these DNA genes are considered their end products. As a result, rRNA molecules produced by rDNA molecules are stable, long-lasting components of the ribosome.

Last Step of mRNA processing

SPLICING!!!!!!! (not going to study in detail yay): catalyzed by ribonucleoproteins (proteins with pieces of RNA) called SNRPs in charged of splicing

Define: Nonsense Mutation Sense Mutation

Sense: A mutation which changes the nucleotide sequence of a codon but does not change the amino acid encoded due to the degeneracy of the genetic code.

Short arm of chromosome is ? What letter is designated? Long arm of chromosome is? What letter is designated?

Short arm: left/top, p Long arm: right/bottom, q

How are transcription and replication similar and how are they different? FIX

Similar: Results in strands that are complementary to the original DNA molecule Requires DNA template, raw materials (ribonucleotide triphosphates) needed to build a new RNA molecule Transcription apparatus, consisting of the proteins necessary for catalyzing the synthesis of RNA Both processes can lead to errors if an incorrect nucleotide is incorporated. An error in either DNA replication or transcription can cause a change in the gene, by either changing the DNA sequence in one of the daughter cells leading to transcription of the incorrect mRNA sequence, or by causing the mRNA to incorporate an incorrect base pair resulting in the wrong protein sequence being translated. Differences: Transcription copies the DNA into RNA, while replication makes another copy of DNA In DNA replication, all of the chromosomes are replicated whereas in transcription, the sequences that are transcribed are highly selective Not all genes must be transcribed but in DNA replication, all of the sequences must be copied Time and length of segments in transcription vary among other transcripts and must therefore be regulated Some transcripts also have to be transcribed more frequently versus other transcripts that are not needed as frequently

Compare and contrast the process of protein synthesis in bacterial and eukaryotic cells, giving the similarities and differences in the process of translation in these two types of cells.

Similarities: 1. Almost the same genetic code (eukaryotic genes can be translated in bacterial systems and vice versa) 2. aminoacyl-tRNA synthetases attach amino acids to their appropriate tRNAs by the same chemical process 3. mRNAs are translated multiple times and are simultaneously attached to several ribosomes!(polyribosome) Differences: 1. bacterial cells AUG encodes a modified type of methionine, N-formylmethionine), while AUG in eukaryotic cells encodes unformylated methionine 2. Translation and transcription take place simultaneously in bacterial cells, nuclear envelope separates these processes in eukaryotic cells (allows for extensive modification of eukaryotic mRNAs, implications for the control of gene expression) 3. mRNA in bacterial cells is short-lived, typically lasting only a few minutes, but mRNA in eukaryotic cells can last for hours or days (5' cap and 3' polyA tail found on euk mRNAs add to their stability) 6. large subunit of eukaryotic ribosome contains three rRNAs where the bacterial ribosome contains only two (diff in size and compositions of ribosomal subunits) 7. in bacterial cells the small subunit of the ribosome attaches directly to the region surrounding the start codon through hydrogen bnding between the shine-dalgarno consensus seuqnece in the 5' UTR of the mRNA and a sequence at the 3' end of the 16s rRNA//in eukcells the small subunit of ribosome first binds to proteins attached tot he 5' cap on mRNA and then migrates down the mRNA scanning the sequence until it encounters the first AUG initiation codon, has a larger number of IF 8. Different EF and IF are used

How does allopolyploidy arise?

Species 1 (AABBCC, 2n=6) produces haploid gametes with chromosomes ABC and species 2 (GGHHII, 2n=6) produces haploid gametes with chromosomes GHI. If gametes from species 1 and 2 fuse, a hybrid with six chromosomes (ABCGHI) is created. The hybrid has the same chromosome number as that of both diploid species so the hybrid is considered diploid. However, because the hybrid chromosomes are not homologous, they will not pair and segregate properly in meiosis; this hybrid is functionally haploid and sterile. The sterile hybrid is unable to produce viable gametes through meiosis but it may be able to perpetuate itself through mitosis (asexual reprod.) Nondisjunction can take place in a mitotic division which leads to a doubling of chromosome number and an allotetraploid with chromosomes AABBCCGGHHII. This type of allopolyploid, consisting of two combined diploid genomes, is sometimes called an amphidiploid. The amphidiploid is functionally diploid: every chromosome has one and only one homologous partner. This amphidiploid can now undergo normal meiosis to produce balanced gametes with 6 chromosomes each.

Summary List of Molecular Cloning

Step 1: Choose a restriction site on your plasmid into which you want your gene of interest inserted. You want to have the same restriction site on your plasmid as the one surrounding your gene of interest. (You want to use a plasmid with lacZ gene as your cloning vector in order to determine the presence and orientation of your gene of interest.) Step 2: Insert your gene of interest into the plasmid. Do this by mixing the plasmid, gene of interest, and the chosen restriction enzyme correlating to the restriction site you want to cut at on both the plasmid and surrounding gene of interest. Step 3: Once your fragment of interest is placed inside the plasmid, you need to introduce the plasmid into bacterial cells. This is accomplished by transformation, (bacterial cells take up DNA from the external environment). Inside the cells, the plasmids replicate and multiple, and the cells themselves multiple, producing many copies of the introduced gene. Step 4: Screen the cells for recombinant plasmids. Using the lacZ gene and the antibiotic resistant gene as selectable markers, you can perform a blue-white screening test. (Cells with interrupted lacZ will appear blue because beta-g cleaves X-gal, Cells interrupted by lacZ will appear white because they do not produce beta-g < this is the one you want, you also have some bacteria that were not transformed). PLATE your bacteria on a medium! Step 5: From the white colonies observed, you need to determine which ones have your gene of interest in the correct (forward) orientation. (If its in the reverse orientation, you might have a completely different protein produced than the one you want). To do this, you must determine which restriction sites can be cut in order for you to determine a difference in the number of base pairs of the fragments. Step 6: Once chosen, you take each sample of bacteria and put it in a test tube. Introduce each bacteria to the restriction site(s) chosen and then perform gel electrophoresis with each sample. Step 7: By doing the gel, you can then confirm which bacteria contain your gene of interest in the forward orientation. Step 8: Next, you can grow the bacteria with your gene in large numbers to replicate the inserted fragment of DNA.

Structural vs. Regulatory Gene

Structural: DNA sequence that encodes a protein that functions in metabolism or biosynthesis role in the cell Regulatory: DNA sequence that encodes a protein or RNA molecule that interacts with DNA sequences and affects with transcription or translation or both

Transcription (Euk and Prok)

Structure that allows cell machinery to identify what a gene is (smallest unit of information that allows us to produce a specific RNA or specific protein, specific DNA sequences that helps us de-limit that information unit) + STRUCTURE (3 most important units that identify a gene): i. Promoter: sequences diff in prok and euk, not transcribed, for RNA polymerase to recognize where the information begins ii. RNA coding region: goes all the way until transcription term site (transc does not end until AFTER termination sequence is transcribed) iii. Transcription termination site

Tandem Duplications vs. Displaced Duplications.

Tandem: duplicated segment is immediately adjacent to the original segment Displaced: if the duplicated segment is located some distance from the original segment either on the same chromosome or on a different one

Sketch and identify four different types of chromosomes based on the position of the centromere.

Telocentric Acrocentric Submetacentric Metacentric

Eukaryotic Genome Organization

Telomeres Linear dsDNA Histones, Euchromatin, Heterochromatin, mtDNA, clDNA

300 nm fiber

The 30-nm fiber forms looped domains with the help from different proteins

What is the purpose of the dideoxyribonucleoside triphosphates in the dideoxy sequencing reaction?

The dideoxynucleotide triphosphate causes chain termination wherever it is incorporated into a growing DNA strand. The result is a collection of newly synthesized DNA molecules whose lengths correspond to the positions of the dideoxynucleotide base (T, for example) in the DNA strand complementary to the template DNA strand. ddNTPs are identical to dNTPs, except that ddNTPs lack a 3=-OH group. ddNTPs are incorporated into a growing DNa strand, but after a ddNTP has been added, no more nucleotides can be added because there is no 3'-OH group to form a phosphodiester bond with an incoming nucleotide, termination of synthesis!

A geneticist uses a plasmid for cloning that has the lacZ gene and a gene that grants/confers resistance to penicillin. The geneticist inserts a piece of foreign DNA into a restriction site that is located within the lacZ gene and uses the plasmid to transform bacteria. Explain how the geneticist can identify bacteria that contain a copy of a plasmid with foreign DNA.

The gene was successfully inserted if the lacZ gene cannot produce B galactosidase So even in the presence of X-gal, it is not forming a blue product Because lacZ has been interrupted and B galactosidase is not formed BLUE WHITE SCREENING!! All colonies will remain white! (if bacteria have a copy of the plasmid) if the bacteria do NOT have a copy of the plasmid, they will be blue, beta-galactosidase will cleave X-gal making the colonies appear blue

#8 EOC What is incomplete penetrance and what causes it?

The genotype does not always produce the phenotype Penetrance is defined as the percentage of individuals having a particular genotype that express the expressed phenotype

Explain why mutations in the lacI gene are trans in their effects, but mutations in the lacO gene are cis in their effects.

The lacI gene encodes the lac repressor protein, which can diffuse within the cell and attach to any operator. It can therefore affect the expression of genes on the same or different molecules of DNA. The lacO gene encodes the operator. The binding of the lac repressor to the operator affects the binding of RNA polymerase to the DNA, and therefore affects only the expression of genes on the same molecule of DNA.

Briefly explain how an antibiotic-resistance gene and the lacZ gene can be used as markers to determine which cells contain a particular plasmid

The lacZ gene contains a series of unique restriction sites at which a piece of DNA can be inserted. In the absence of the inserted fragment, the lacZ gene is active and produces beta-galactosidase. When foregin DNA is inserted into the restriction site, it disrupts the lacZ gene, and beta-galactosidase is not produced. The plasmid usually contains a second selectable marker, which may be a gene that confers resistnance to an antibiotic, such as ampicillin. Bacteria that are lacZ- are exposed to the plasmids and then plated on medium that contains ampicillin. Only cells that have been successfully transformed and thus contain a plasmid with the ampicillin-resitance gene will survive and grow. Some of the cells will contain and instact plasmid and others will hav ea recombinant plasmid. The medium contains X-gal, which produces a blue substance when cleaved. (beta-galactosidase cleaves it) Bacteria with an intact original plasmid (without an inserted DNA fragment) will have a functional lacZ gene (front end provided by the plasmid and back end by the bacterium) and can synthesize beta-galactosidase, which cleaves X-gal and turns bacteria blue. Bacteria with a recombinant plasmid, however, have the front end of the beta-galactosidase gene disrupted by the inserted DNA and do not synthesize the enzyme and remain white. When cells with the recombinant plasmid have been identified, they can be grown in large numbers to replicte the inserted fragment of DNA

Suppose that a mutation occurs in the middle of a large intron of a gene-encoding a protein. What will the MOST likely effect of the mutation be on the amino acid sequence of that protein?

The mutation may or may not affect the intron removal depending on whether nucleotides affecting splicing are altered.

Several experiments were conducted to obtain information about how the eukaryotic ribosome recognizes the AUG start codon. In one experiment, the gene that encodes methionine initiator tRNA (tRNA^met) was located and changed. The nucleotides that specify the anticodon on tRNA ^met were mutated so that the anticodon in the tRNA was 5'-CCA-3' instead of 5'-CAU-3'. When this mutated gene was placed in a eukaryotic cell, proteins synthesis took place, but the proteins produced were abnormal. Some of the proteins contained extra amino acids, and others contained fewer amino acids than normal. What do these results indicate about how the ribosome recognizes the starting point for translation in eukaryotic cells? Explain your answer. If the same experiment had been conducted on bacterial cells, what results would you expect? Explain why some proteins contained extra amino acids while others contained fewer amino acids than normal.

The ribosome recognises the start codon near the 5' end of the mRNA where there are specific translation sites present that incorporate the mRNA onto that site . These sites are known as translation initiation sites. The 5' terminal sequence has an untranslated region that consists of the non coding sequence. Following this is the start codon , which is AUG Coding for methionine. The factors eIF-1, eIF-1A, and eIF-3 bind to the 40S ribosomal subunit, and the factor eIF-2 gets associated with the initiator methionyl tRNA. Also, eIF-4G, binds to eIF-4E and to poly-A binding protein or PABP, which is associated to the poly-A tail at the 3' end of the mRNA. Thus, the initiation factors in eukaryotes recognise both the 5' and 3' ends of mRNAs. When , the initiation codon is not recognised correctly, there is a total distortion of the translation machinery, which also recognises the 3G' end, thus leading to incorrect polypeptide sizes, either smaller or larger than usual.

Haploinsufficiency

The situation that occurs when one copy of a gene is inactivated or deleted and the remaining functional copy of the gene is not adequate to produce the needed gene product to preserve normal function.

Alternative Splicung

This type of mRNA processing can skip over certain exons in a gene to produce different mRNA strands, and thus, different proteins. For example, the introns could be removed and two exons can be skipped over, or the introns can be removed and only one exon could be skipped over.

DNA Polymerase I

To remove & replace primers with DNA nucleotides Has 5'->3' polymerase activity (adds DNA nucleotides) & 3'->5' exonuclease activities (removes RNA primer) Unlike DNA polymerase 3, DNA polymerase 1 posses the 5'->3' exonuclease activity which is used to remove the primers laid down by primase + replace them with DNa nucleotides by synthesizing in the 5'->3' direction

In figure 9.7, which is the leading strand and which is the lagging strand?

Top strand is leading strand Bottom strand is lagging strand

Difference between transversion and transition.

Transversion: refers to a point mutation where a single purine is changed for a pyrimidine, or vice versa. A transversion can be spontaneous, or can be caused by ionizing radiation or alkylating agents. It can only be reversed by a spontaneous reversion Transition: efers to a point mutation that changes a purine nucleotide to another purine, or a pyrimidine nucleotide to another pyrimidine

Characteristics of the Genetic Code

Triplet Code: combination of three nucleotides codes for one amino acid, only one combination, 3 nucleotides to identify one amino acid The code is specific: every triplet will only code for one amino acid The code is degenerate: there are several different codons that can identify one single amino acid An initiation codon sets the reading frame (start signal) The code is non-overlapping: one nucleotide will only be included in one codon (3 nucleotides that comprise one codon are not reused in the second codon) Termination codons indicate the end of the coding sequence (stop signal) The code is universal/nearly universal: (nearly universal, mitochondria has one triplet that codes for a different aa)

A regulatory gene has its own promoter and is transcribed into an independent mRNA. T or F

True

A repressible gene is controlled by a regulatory protein that inhibits transcription. T or F ????????????????

True

A restriction enzyme is an enzyme that cleaves DNA into fragments at or near a specific recognition site, called restriction site, found on the DNA molecule. T or F

True

After transformation, different colonies should be tested for orientation of the insert using restriction enzymes. T or F

True

Between viruses, bacteria, and eukaryotes, viruses have the smallest genome. T or F

True

Correct base-pairing is achieved when a specific purine forms hydrogen bond with a specific pyrimidine.

True

Each colony on a plate grows from a single bacterium. T or F

True

For a gene under positive repressible control, the normal state is transcription of a gene, stimulated by a transcriptional activator. T or F

True

Presence of an operon where genes of related functions are clustered is common in bacteria, but not in eukaryotes. T or F Be careful with wording, does not say always it says "common"

True

Reverse transcriptase is an enzyme that uses an RNA template to produce cDNA. T or F

True

A group of 100 dogs have a specific mutation, but only 72 dogs seem to express the mutant phenotype. We can say that the percentage of this gene is 72%. T or F Suppose that a beetle population is composed of 1000 beetles and you determine that 845 of them have the mutation. You notice that not all individuals with mutations present spots, and that the spots can have different sizes and locations on the body. Can you say that the penetrance of this mutation is about 85%? Does it have variable expressivity?

True No, because not all beetles that have the mutation express the phenotype. This incomplete penetrance. In addition, it does have variable expressivity. This is because beetles with the phenotype expressed have different variations.

How can you use blue-white screening to help you determine which E. coli cells contain copies of pCM999 with husK? FIX

Upon inserting the new copy of DNA into the vector plasmid at the EcoRI restriction site, you will interrupt the lacZ gene. The beta-galactosidase enzyme will not be expressed. Then, place the bacteria on media containing ampicillin and Xgal. The colonies with copies of the husK gene will remain white and the copies without the husK gene will become blue (beta galactosidase interacts with Xgal to form a blue product)

How might 32P and 35S be used to demonstrate that the transforming principle is DNA? Briefly outline an experiment that would show that DNA, rather than protein,is the transforming particle.

Use the radioactive phosphorus to stain bacteriophage protein coats Use the radioactive sulfur to stain bacteriophage DNA Whichever one is able to replicate and express its radioactive illumination in the bacteria is the molecule that is able to transfer genetic information

#21 EOC Joe has a white cat named Sam. When Joe crosses Sam with a black cat, he obtains ½ white kittens and ½ black kittens. When the black kittens are interbed, all the kittens that they produce are black. On the basis of these results, which coat color (white or black) in cats would you conclude is a recessive trait? Explain your reasoning.

Using the knowledge of the first cross progeny, we can infer that the black cat is ww (recessive) while the white cat is Ww (dominant) Crossing ww x ww, you get all black BLACK IS RECESSIVE

Erwin Chargaff collected data on the proportions of nucleotide bases from the DNA of a variety of different organisms and tissues. Data from the DNA of several organisms analyzed by Chargaff are shown here: For each organism, compute the ratio of (A+G)/(T+C) and the ratio of (A+T)/(C+G). Are these ratio constant or do they vary among the organisms? Explain why. Is the (A+G)/(T+C) ratio different for the sperm samples? Would you expect it to be? Why or why not?

Vary among organisms Some organisms have more structural genes in comparison to regulatory genes while other organisms have the opposite Humans have a lot more regulatory genes than structural genes

Relative Size of Genome Viral: E. Coli: Eukaryotes:

Viral: (smallest) 4Kb - 200Kb E. Coli: (middle) 300Kb - 10Mb Eukaryotes: (largest) 8Mb - 900Gb

What are some of the important genetic implications of the DNA structure?

Watson and Crick model suggests that the genetic information or instructions are encoded in the nucleotide sequences Complimentary polynucleotide strands indicate how faithful replication of the genetic material is possible The arrangement of the nucleotides is such that they specify the primary structure or aa sequence of protein molecules

EOC #2 What effect does crossing over have on linkage?

When linked genes undergo some crossing over, the result is mostly nonrecombinant progeny and fewer recombinant progeny. For closely linked genes, crossing over does not take place in every meiosis. In meioses in which there is no crossing over, only nonrecombinant gametes are produced. In meiosies in which there is a single crossover, half the gametes are recombinants and half are nonrecombinants because a single crossover affects only two of the four chromatids. Because each crossover leads to half recombinant gametes and half nonrecombinant gametes, the total percentage of recombinant gametes is always half the percentage of meioses in which crossing over takes place Even if crossing over between two genes takes place in every meiosis, only 50% of the resulting gametes will be recombinants Freq. of recombinant gametes is always half the freq. of crossing over, and the maximum proportion of recombinant gametes is 50%

What functions does supercoiling serve for the cell?

When the strain of over rotating or underrotating cannot be compensated by the turning of the ends of the double helix Takes place when being overwound or underwound Helps with packing DNA

#16 EOC Coat color in cats is determined by genes at several different loci. At one locus on the X chromosome, one allele (X^+) encodes black fur; another allele (X^0) encodes orange fur. Females can be black (X^+X^+), orange (X^0X^0). Or a mixture of orange and black called tortoiseshell (X^+X^0). Males are either black (X^+Y) or orange (X^0Y), Bill has a female tortoiseshell cat named Patches. One night patches escapes from Bill's house, spends the night out, and mates with a stray male. Patches later gives birth to the following kittens: one orange male, one black male, two tortoiseshell females, and one orange female. Give the genotypes of Patches, her kittens, and the stray male with which Patches mated.

X+ = black fur X0 = orange fur Females can be X+X+ (black) or X0X0 (orange) or X+X0 (tortoiseshell) Males can be X+Y (black) or X0Y (orange) X+X0 (female tortoiseshell) x _____ (unknown stray male) This mating produces 1 X0Y, 1 X+Y, 2 X0X+, and 1 X0X0 Genotype of the male: X0Y

In placental mammals, dosage compensation is achieved by ________, which happens early in development. Define X inactivation Define Barr Body

X-inactivation Inactivated chromosomes is observable in cells = Barr body Some X-linked genes escape inactivation Females are mosaics for the expression of X-linked genes X-inactivation (also called lyonization) is a process by which one of the copies of the X chromosome is inactivated in some female mammals. The inactive X chromosome is silenced by it being packaged in such a way that it has a transcriptionally inactive structure called heterochromatin. a small, densely staining structure in the cell nuclei of female mammals, consisting of a condensed, inactive X chromosome. It is regarded as diagnostic of genetic femaleness.

#19 EOC Bob has XXY chromosomes (Klinefelter syndrome) and is color blind. His mother and father have normal color vision, but his maternal grandfather is color blind. Assume that Bob's chromosome abnormality arose from nondisjunction in meiosis. In which parent and in which meiotic division did nondisjunction take place? Assume that no crossing over has taken place.

XXY > color blind Mother and Father have normal vision Maternal grandfather is color blind

#5 EOC Explain why tortoiseshell cats are almost always female and why they have a patchy distribution of orange and black fur?

X^+ = non-orange, black X = orange Patchy distribution results from X-inactivation during the development of the embryo Different cells inactivate the X which results in black fur spots and other cells have X activated resulting in orange spots

#24 EOC Red-green color blindness is an X-linked recessive trait in humans. Polydactyly (extra fingers and toes) is an autosomal dominant trait. Martha has normal fingers and toes and normal color vision. Her mother is normal in all respects, but her father is color blind and polydactylous. Bill is color blind and polydactylous. His mother has normal color vision and normal fingers and toes. If Bill and Martha marry, what types and proportions of children can they produce? FIX

X^m = color blindness X^+ = normal vision P = extra fingers p = normal fingers Martha: X^+X^mpp Martha's mother: X^+X^+pp Martha's father: X^mYPp Bill: X^mYPp Bill's mother: X^+X^+pp Bill x Martha = X^mYPp x X^+X^mpp

Structural genes

a gene that codes for any RNA or protein product other than a regulatory factor (i.e. regulatory protein). It may code for a structural protein, an enzyme, or an RNA molecule not involved in regulation.

Pseudoallelism

a state in which two genes with similar functions are located so close to one another on a chromosome that they are genetically linked. This means that the two genes are nearly always inherited together. Since the two genes have related functions, they may appear to act as a single gene.

cis-trans test

a test in microbial genetics to determine whether two mutations that have the phenotypic effects, in a haploid cell or a cell with single phage infection, are located in the same gene or different genes the test depends on the independent behavior of two alleles of a gene in a diploid cell or in a cell infected with two phages carrying different alleles.

What is a repressor protein?

a type of regulatory protein that binds to a region of DNA related to the promoter called the operator and prevents transcription from taking place

#15 EOC: In cucumbers, orange fruit color (R) is dominant over cream fruit color (r). A cucumber plant homozygous for orange fruit is crossed with a plant homozygous for cream fruit. The F1 are intercrosses to produce the F2. a. Give the genotypes and phenotypes of the parents, the F1, and the F2. b. Give the genotypes and phenotypes of the offspring ofa backcross between the F1 and the orange-fruited parent. c. Give the genotypes and phenotypes of a backcross between the F1 and the cream-fruited parents.

a. Parent's genotype: RR x rr Parents' phenotype: orange fruit x cream fruit F1 genotype: Rr F1 phenotype: orange fruit F2 genotype: 1/4 RR: 2/4 Rr: 1/4 rr F2 phenotype: 3/4 orange fruit: 1/4 cream fruit b. Parent 1 genotype: RR F1 genotype: Rr Cross genotype: 2/4 RR: 2/4 Rr Cross phenotype: all orange fruit c. Parent 2 genotype: rr F1 genotype: Rr Cross genotype: 2/4 Rr: 2/4 rr Cross phenotype: 1 : 1 orange fruit: cream fruit

#33 EOC . In cucumbers, dull fruit (D) is dominant over glossy fruit (d), orange fruit (R) is dominant over cream fruit (r), and bitter cotyledons ((B) are dominant over non-bitter cotyledons (b). The three characteristics are encoded by genes located on different pairs of chromosomes. A plant homozygous for dull, orange fruit and bitter cotyledons is crossed with a plant that has glossy cream fruit and non-bitter cotyledons. The F1 are intercrossed to produce the F2. a. Give the phenotypes and their expected proportions in the F2. b. An F1 plant is crossed with a plant that has glossy, cream fruit and non-bitter cotyledons. Give the phenotypes and expected proportions among the progeny of this cross.

a. Dull, Orange, Bitter = ¾ X ¾ X ¾ = 27/64 Dull Orange, Non-bitter = ¾ X ¾ X ¼ = 9/64 Dull, Cream, Bitter = ¾ X ¼ X ¾ = 9/64 Dull, Cream, Non-bitter = ¾ X ¼ X ¼ = 3/64 Glossy, Orange, Bitter = ¼ X ¾ X ¾ = 9/64 Glossy, Orange, Non-bitter= ¼ X ¾ X ¼ = 3/64 Glossy, Cream, Bitter = ¼ X ¼ X ¾ = 3/64 Glossy, Cream, Non-bitter = ¼ X ¼ X ¼ = 1/64 b. 1/8 DdRrBb (dull, orange, bitter), 1/8 DdRrbb (dull, orange, non-bitter), 1/8 DdrrBb (dull, cream, bitter), 1/8 Ddrrbb (dull, cream, non-bitter), 1/8 ddRrBb (glossy, orange, bitter), 1/8 ddRrbb (glossy, orange, non-bitter), 1/8 ddrrBb (glossy, cream, bitter), 1/8 ddrrbb (glossy, cream, non-bitter)

#30 EOC Figure 3.8 shows the results of a dihybrid cross involving seed shape and seed color. a. What proportion of the round and yellow F2 progeny from this cross is homozygous at both loci? b. What proportion of the round and yellow F2 progeny from this cross is homozygous at least at one locus?

a. 1/16 b. 6/16

#30 EOC In the early 1900s, Lucien Cuenot, a French scientist working at the University of Nancy, studied the genetic basis of yellow coat color in mice (discussed on page 91). He carried out a number of crosses between two yellow mice and obtained what he thought was a 3:1 ratio of yellow to gray mice in the progeny. The following table gives Cuenot's actual results, along with the results of a much larger series of crosses carried out by William Castle and Clarence Little. a. Using a chi square test, determine whether Cuenot's results are significantly different from the 3:1 ratio that he thought he observed. Are they different from a 2:1 ratio? b. Determine whether Castle and Little's results are significantly different from a 3:1 ratio. Are they different from a 2:1 ratio? c. Combine the Castle and Little results with those of Cuenot and determine whether they are significantly different from a 3:1 ratio and a 2:1 ratio?

a. 3:1 Since the p-value is greater than 0.05, we fail to reject our null hypothesis. The high p value indicates that the obtained data fits the expected data very closely and that any fluctuation can be explained by chance. 2:1 Since the p-value is less than 0.05, our null hypothesis is false. There is a significant difference between the expected and observed data, rather than chance alone. We assume there is some other factor causing the difference in the data. b. i dont want to do it right now

#34 EOC Tatuo Aida investigated the genetic basis of color variation in the medaka, a small fish found in Japan. Aida found that genes at two loci (B, b and R, r) determine the color of the fish: fish with a dominant allele at both loci (B_R_) are brown, fish with a dominant allele at the B locus (B_rr) are blue, fish with a dominant allele at the R locus only (bbR_) are red, and fish with recessive alleles at both loci (bbrr) are white. Aida crossed a homozygous brown fish with a homozygous white fish. He then backcrossed the F1 with homozygous white parent and obtained 228 brown, 230 blue, 237 red, 222 white. a. Give the genotypes of the backcross progeny b. Use a chi-square test to compare the observed numbers of backcross progeny with the number expected. What conclusion can you make from your chi-square results? c. What results would you expect for a cross between a homozygous red fish and a white fish? d. What results would you expect if you crossed a homozygous red fish with a homozygous blue fish and then backcrossed the F1 with a homozygous red parental fish?

a. 4/16 BbRr: 4/16 Bbrr: 4/16 bbRr: 4/16 bbrr b. Since the p-value is > 0.05, we fail to reject our null hypothesis. The high p value indicates that the obtained data fits the expected data very closely and that any fluctuation can be explained by chance. c. idk d. idk its really late at night

#31 EOC In cats, curled ears result from an allele (Cu) that is dominant over an allele for normal ears (cu). Black color results from an independently assorting allele (G) that is dominant over allele for gray (g). A gray and homozygous for curled ears is mated with a homozygous black cat with normal ears. All the F1 cats are black and have curled ears. a. If two of the F1 cats mate, what phenotypes and proportions are expected in the F2? b. An F1 cat mates with a stray cat that is gray and possesses normal ears. What phenotypes and proportions of progeny are expected from this cross?

a. 9/16 G_Cu_ (black w/curled ears), 3/16 G_cucu (black w/normal ears), 3/16 ggCu_ (gray w/curled ears), 1/16 ggcucu (gray w/normal ears) b. 1/4 GgCucu (black w/curled ears), 1/4 Ggcucu (black w/normal ears), 1/4 ggCucu (gray w/curled ears), 1/4 ggcucu (gray w/normal ears)

#33 EOC In chickens, comb shape is determined by alleles at two loci (R, r, and P, p). A walnut comb is produced when at least one dominant allele R is present at one locus and at least one dominant allele P is present at a second locus (genotype R_P_). A rose comb is produced when at least one dominant allele is present at the first locus and two recessive alleles are present the second locus (genotype R_pp). A pea comb is produced when two recessive alleles are present at the first locus and at least one dominant allele is present at the second (genotype rrP_). If two recessive alleles are present at the first locus and at the second locus (rr pp), a single comb is produced. Progeny with what types of combs and in what proportions will result from the following crosses? a. RRPP X rrpp b. RrPp X rrPP c. RrPp X RrPp d. Rrpp X Rrpp e. Rrpp X rrPp f. Rrpp X rrpp

a. All RrPp = All walnut OR 1 RrPp, 1 Walnut comb b. 4/16 RrPP: 4/16 RrPp: 4/16 rrPP: 4/16 rrPp 8/16 R_P_ (Walnut): 8/16 rrP_ (Pea) c. 9/16 R_P_: 3/16 rrP_: 3/16 R_pp: 1/16 rrpp 9 Walnut: 3 Pea: 3 Rose: 1 Single d. do it later e. do it later f. do it later

#31 EOC In rabbits, an allelic series helps to determine coat color: C (full color), cch (chinchilla, gray color), ch (himalayan, white with black extremities), and c (albino, all white). The C allele is dominant over all others, cch is dominant over ch and c, ch is dominant over c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C > cch > ch > c. The rabbits in the list on the following page are crossed and produce the progeny shown. Give the genotypes of the parents for each cross. a. full color x albino = 1/2 full color, 1/2 albino b. Himalayan x albino = 1/2 Himalayan, 1/2 albino c. full color x albino = 1/2 full color, 1/2 chinchilla d. full color x Himalayan = 1/2 full color, 1/4 Himalayan, 1/4 albino e. full color x full color = 3/4 full color, 1/4 albino

a. Cc x cc = 1/2 Cc and 1/2 cc b. c^ch x cc = 1/2 c^hc and 1/2 cc c. Cc^ch x cc d. Cc x chc e. Cc x Cc

#17 EOC J. W. McKay crossed a stock melon plant that produced tan seeds with a plant that produced red seeds and obtained the following results: Cross F1 F2 Tan x red 13 tan seeds 93 tan, 24 red seeds

a. F1 generation is ALL tan and F2 generation is 93 tan : 24 red, which represents a 3:1 ratio, showing us that the F1 generation is heterozygous for seed color and that tan must be dominant to red b. T = tan, t = red P1 = TT x tt F1 = Tt F2 = 1/4 TT: 1/4 tt: 2/4 Tt If all of the children in the F1 generation are expressing only one phenotype, we know that this is the dominant allele. If we know that the father of these children do not express this phenotype, the father is homozygous recessive and therefore all the children when crossed with the mother are heterozygous.

#27 EOC The LM and LN alleles at the MN blood group locus exhibit codominance. Give the expected genotypes and phenotypes and their ratios in progeny resulting from the following crosses. a. LMLM x LMLN b. LNLN x LNLN c. LMLN x LMLN d. LMLN x LNLN e. LMLM x LNLN

a. Genotypes: 1 LMLM : 1 LMLN Phenotypes: 1 LMLM : 1 LMLN b. Genotypes: 1 LNLN Phenotypes: 1 LNLN c. Genotypes: 1 LMLM : 2 LMLN : 1 LNLN Phenotypes: 1 LMLM : 2 LMLN : 1 LNLN d. Genotypes: 1 LMLN : 1 LNLN Phenotypes: 1 LMLN : 1 LNLN e. Genotypes: 1 LMLN Phenotypes: 1 LMLN

#18 EOC The following pedigree illustrates the inheritance of Nance-Horan syndrome, a rare genetic condition in which affected persons have cataracts and abnormally shaped teeth. a. On the basis of the pedigree, what do you think is the most likely mode of inheritance for Nance-Horan syndrome? b. If couple III-7 and III-8 have another child, what is the probability that the child will have Nance-Horan syndrome. c. If III-2 and III-7 were to mate, what is the probability that one of their children would have Nance-Horan syndrome? ????

a. I think the Nance--Horan syndrome is caused by an X-linked recessive trait b. 1/4 c. 1/4

Ellis Engelsberg and his coworkers examined the regulation of genes taking part in the metabolism of arabinose, a sugar. Four structural genes encode enzymes that help metabolize arabinose (genes A, B, D, and E). An additional gene C is linked to genes A, B, and D. these genes are in order D-A-B-C. Gene E is distant from the other genes. Engelsberg and his colleagues isolated mutations at the C gene that affected the expression of structural genes A, B, D, and E. In one set of experiments, they created various genotypes at the A and C loci and determined whether arabinose isomerase (the enzyme encoded by gene A) was produced in the presence or absence of arabinose (the substrate of arabinose isomerase). Results from this experiment are shown in the following table, where a plus sign (+) indicates that the arabinose isomerase was synthesized and a minus sign (-) indicates that the enzyme was not synthesized. On the basis of the results of these experiments, is the C gene an operator or a regulator gene? Explain your reasoning. Do these experiments suggest that the arabinose operon is negatively or positively controlled? Explain your reasoning What type of mutation is C? To do this, know key concepts of inducible and repressible operons. Repressible operons are originally ON and want to be OFF so produce a product. Inducible operons are originally OFF and want to be ON so produce a substrate.

a. If it was the operator (in the second genotype), then if a repressor came in and tried to bind the operator and it was not able to bind then all enzymes would be produced. And it shows that none of the enzymes would be produced. So it cannot be the operator. It is the regulator. b. c. constitutive mutation

For E. coli strains with the lac genotypes given below, use a plus sign (+) to indicate the synthesis of B-galactosidase and permease and a minus sign (-) to indicate no synthesis of the proteins.

a. If lactose is present, then transcription is on (+). If lactose is absent, then transcription is off (-). b. lacI^-, then repressor is inactive. If lactose is present, then transcription is on (+). If lactose is absent, then trasncription is still on (+). c. lacO^C, the operon is always on. If lactose is present, then transcription is on (+). If lactose is absent, then transcription is still on (+). (a mutation prevents the repressor from binding to the small molecule) d. lacI^-, the repressor is inactive. lacY^-, permease is not synthesized. If lactose is present, transcription of beta-galactosidase is on (+), but transcription of permease is not. If lactose is absent, transcription of beta-galactosidase is on (+), but transcription of permease is still not on. e. lacI^-, repressor is inactive. lacP^-, promoter is inactive. Transcription of EVERYTHING is off because of the promoter being inactive (-). f. This is a partial diploid. One good copy of lacI which means repressor is always active. Lactose is present: transcription is on. Lactose is absent: transcription is off. g. Lactose is present: transcription always on because repressor protein is always there and lactose (substrate) inactivates it. Lactose is absent: transcription is on h. Lactose is present: transcription of beta-galactosidase is on but with no permease. Lactose is absent: transcription of everything is off. i. Lactose is present: no permease, beta-galactosidase will be synthesized. Lactose is absent: no transcription of anything. j. Lactose is present: everything is transcribed. Lactose is absent: nothing is transcribed. k. lac IS : "super" repressor, repressor is always bound to the operator, always preventing transcription. There are two possible causes for lac IS mutations: repressor can't bind to the inducer or the repressor can bind the inducer but can't undergo the conformation change required to get the repressor to fall off of the operator Lactose is present: transcription is off Lactose is absent: transcription is off l. Lactose is present: transcription is off Lactose is absent: transcription is off

Which of the following amino acid changes could result from a mutation that changed a single base? For each change that could result from the alteration of a single base, determine which position of the codon (first, second, or third nucleotide) in the mRNA must be altered for the change to result?

a. Leu > Gln Possible codons for Leu: UUA, UUG, CUU, CUC, CUA, CUG Possible codons for Gln: CAG, CAA CUG > CAG or CUA to CAA b. Phe > Ser Possible codons for Phe: UUU, UUC Possible codons for Ser: AGU, AGC, UCU, UCC, UCA, UCG UUU > UCU or UUC to UCC c. Phe > Ile Possible codons for Phe: UUU, UUC Possible codons for Ile: AUU, AUC, AUA UUU > AUU or UUC > AUC d. Pro > Ala Possible codons for Pro: CCU, CCC, CCA, CCG Possible codons for Ala: GCU, GCC, GCA, GCG CCU > GCU, CCC > GCC, CCA > GCA, CCG > GCG e. Asn > Lys Possible codons for Asn: AAU, AAC Possible codons for Lys: AAA, AAG AAU > AAA, AAC > AAG, AAU > AAG, AAC > AAA f. Ile > Asn Possible codons for Ile: AUU, AUC, AUA Possible codons for Asn: AAU, AAC AUU > AAU, AUC > AAC

#29 EOC When a chinese hamster with white spots is crossed with another hamster that has no spots, approximately ½ of the offspring have white spots and ½ have no spots. When two hamsters with white spots are crossed, ⅔ of the offspring possess white spots and ⅓ have no spots. a. What is the genetic basis for white spotting in Chinese hamsters? b. How might you go about producing Chinese hamsters that breed true for white spotting?

a. Parents are Aa and aa F1 is 1/2 aa: 1/2 Aa A_ = white spots aa = no spots F1: 1/4 AA: 2/14 Aa: 1/4 aa You would normally assume out of 4 in progeny, and we can infer that when two dominant alleles come together, a lethal allele is formed, therefore killing the other hamster. Since we see a 2:1 ratio (out of 3) b. Not possible cause AA would kill hamster

A mutation at the operator prevents the regulator protein from binding. What effect will this mutation have in the following types of operons? a. Regulator protein is a repressor in a repressible operon b. Regulator protein is a repressor in an inducible operon

a. Repressible Operon: repressor protein binds to operator to prevent transcription but in a mutation where the repressor protein does not bind to the operator, transcription is ON b. Inducible Operon: presence of product induces transcription, binds to the repressor protein and inactivates it, but in a mutation transcription is ON (all the time, constitutive expression)

#38 EOC The direction of shell coating in the snail Lymnaea peregra results from a genetic maternal effect. An autosomal allele for a right-handed shell (s+), called dextral, is dominant over the allele for a left-handed shell (s), called sinistral. A pet snail called Martha is sinistral and reproduces only as a female (the snails are hermaphroditic). Indicate which of the following statements are true and which are false. Explain your reasoning in each case. a. Martha's genotype must be ss. b. Martha's genotype cannot be s+s+ c. All the offspring produced by Martha must be sinistral d. At least some of the offspring produced by Martha must be sinistral. e. Martha's mother must have been sinistral f. All of Martha's brothers must be sinistral

a. T b. T C. F D. T E. F F. F

The following anticodons are found in a series of tRNAs. Refer to the genetic code in Figure 11.5 and give the amino acid carried by each of these tRNAs. a. 5'--GUA--3' b. 5'--AUU--3' c. 5'--GGU--3' d. 5'--CCU--3'

a. UAC, Tyr b. Asn c. Thr d. Arg Anti-codons usually given 3' to 5, just remember anti-parallel and complementary

For each of the following types of transcriptional control, indicate whether the protein produced by the regulator gene will be synthesized initially as an active repressor, inactive repressor, active activator, or inactive activator. Negative control in a repressible operon Positive control in a repressible operon Negative control in an inducible operon Positive control in an inducible operon

a. active repressor b. inactive activator c. inactive repressor d. active activator

#13 EOC What will be the phenotypic sex of a human with the following genes or chromosomes or both? XY with the SRY gene deleted XY with the SRY gene located on an autosome XX with a copy of the SRY gene on an autosome XO with a copy of the SRY gene on an autosome XXY with the SRY gene deleted XXYY with one copy of the SRY gene deleted

a. female b. male c. male d. male e. female f. ? Male?

#25 EOC Miniature wings in Drosophila results form an X-linked gene (Xm) that is recessive to an allele for long wings (X+). Sepia eyes are produced by an autosomal gene (s) that is recessive to an allele for red eyes (s+). a. A female fly that has miniature wings and sepia eyes is crossed with a male that has normal wings and is homozygous for red eyes. The F1 are intercrossed to produce the F2. Give the phenotypes and their proportions expected in the F1 and F2 flies from this cross. b. A female that is homozygous for normal wings and has sepia eyes is crossed with a male that has miniature wings and is homozygous for red eyes. The F1 are intercrossed to produce the F2. give the phenotypes and their proportions expected in the F1 and F2 flies from this cross.

a. i dont feel like writing this out, copy and paste its also in the back of the book b.

Give all possible genotypes of a lac operon that produces B-galactosidase and permease under the following conditions. Do not give partial-diploid genotypes.

a. lacI+, lacP+, lacO+, lacZ+, lacY+ b. lacI+, lacP+, lacO+, lacZ-, lacY+ c. lacI+, lacP+, lacO+, lacZ+, lacY- d. lacI-, lacP+, lacO+, lacZ+, lacY+ OR lacI+, lacP+, lacOc, lacZ+, lacY+ e. lacI+, lacP-, lacO+, lacZ+, lacY+ OR lacI-, lacP-, lacO+, lacZ+, lacY+ OR lacIs f. lacI+, lacP+, lacOc, lacZ+, lacY- g. lacI+, lacP+, lacOc, lacZ-, lacY+

For each of the sequences in the following table, place a check mark in the appropriate space to indicate the process most immediately affected by deleting this sequence. Choose only one process for each sequence

a. ori site > replication b. 3' splice site consensus > RNA processing, splice site are important in differentiating between introns and exons and thus deletion of splice will lead to formation of a incorrect mRNA c. Poly(A) tail > RNA processing d. Terminator > Transcription, deletion of terminator sequence will result in improper in a peptide of undesired longer length e. Start Codon > Translation, start codon is important in initiation of translation process can lead to synthesis of protein of desired sequence. If start codon is deleted then an undesired protein will be formed. f. -10 consensus > Transcription, (-10) consensus sequence is vital for RNA polymerase binding and hence for transcription g. Shine-Dalgarno > translation, shine dalgarno(SD) sequence of mRNA is important for binding of mRNA with ribosome. So deletion of SD will not allow translation to happen

In E. coli, three structural genes (A, D, and E) encode enzymes A, D, and E, respectively. Gene O is an operator. The genes are in the order O-A-D-E on the chromosome. These enzymes catalyze the biosynthesis of valine. Mutations were isolated at the A, D, E, and O genes to study the production of enzymes A, D, and E. Levels of the enzymes produced by partial-diploid E. coli with various combinations are shown in the following table: Is the regulator protein that binds to the operator of this operon a repressor (negative control) or an activator (positive control)? Explain your reasoning. Are genes A, D, and E all under the control of operator O? Explain your reasoning. Propose an explanation for the low level of enzyme E produced in genotype 3.

a. reperssor (negative control), low concentrations of enzymes produced with wild type operator gene b. Yes all the genes are under the control of the operator. When operator and repressor are both functional, then the levels of expression are low. When the operator is nonfunctional the enzyme levels generally increase. c. The low level of enzyme E produced in genotype 3 is likely due to a polarity effect. Because they share the same transcriptional control regions, genes A, D and E are transcribed together producing a polycistronic mRNA molecule. Gene E is located downstream of gene D and is thus transcribed after gene D. in genotype 3, it is likely that the defect in gene D affects transcription elongation that occurs subsequent to the mutation.

The following nucleotide sequence is found on the template strand of DNA. first, determine the amino acids of the protein encoded by this sequence by using the genetic code provided in Figure 11.5. Then give the altered amino acid sequence of the protein that will be found in each of the following mutations: 3' - TAC TGG CCG TTA GTT GAT ATA ACT -5' 5' - AUG ACC GGC AAU CAA CUA UAU UGA -3' Mutant 1: a transition at nucleotide 11 Mutant 2: a transition on nucleotide 13 Mutant 3: a one nucleotide deletion at nucleotide 7 Mutant 4: a T -> A transversion at nucleotide 15 Mutant 5: an addition of TGG after nucleotide 6 Mutant 6: a transition at nucleotide 9

a. ser b. glu c. meth, thr, ala, ile, asn, tyr d. his e. met, thr, thr, gly, asn, gln, leu, tyr, stop f. gly

A polypeptide has the following amino acid sequence Met- Ser- Pro- Arg- Leu- Glu- Gly The amino acid sequence of this polypeptide was determined in a series of mutants listed in parts a through e. For each mutant, indicate the type of mutation that occurred in the DNA (single-base substitution, insertion, deletion) and the phenotype effect of the mutation (nonsense mutation, missense mutation, frameshift, etc.) Mutant 1: Met-Ser-Ser-Arg-Leu-Glu-Gly Mutant 2: Met- Ser Pro Mutant 3: Met-Ser-Pro-Asp-Trp-Arg-Asp-Lys Mutant 4: Met-Ser-Pro-Glu-Gly Mutant 5: Met-Ser-Pro-Arg-Leu-Leu-Glu-Gly

a. single base substitution -> missense mutation b. single base substitution from CGA to UGA which is stop -> nonsense mutation c. -> deletion of one nucleotide -> frameshift mutation d. Deletion of an entire codon e. -> insertion -> missense mutation

A chromosome has the following segments, where "." represents the centromere: AB.CDEFG What types of chromosome mutations are required to change this chromosome into each of the following chromosomes? (In some cases, more than one chromosome mutation may be required) ABAB.CDEFG AB.CDEABFG AB.CFEDG A.CDEFG AB.CDE AB.EDCFG C.BADEFG AB.CFEDFEDG AB.CDEFCDFEG

a. tandem duplication of AB b. displaced duplication of AB c. paracentric inversion of DEF d. deletion of B e. deletion of FG f. paracentric inversion of CDE g. pericentric inversion of ABC h. inversion of DEF, duplication of DEF i. duplication of CDEF, inversion of EF

#35 EOC J. A. Moore investigated the inheritance of spotting patterns in leopard frogs. The pipiens phenotype has the normal spots that give leopard frogs their name. In contrast, the burnsi phenotype lacks spots on its back. Moore carried out the following crosses, producing the progeny indicated. a. On the basis of these results, what is the most likely mode of inheritance of the burnsi phenotype? b. Give the most likely genotypes of the parent in each cross (use B for the burnsi allele and B^+ for the pipiens allele) c. Use a chi square test to evaluate the fit of the observed numbers of progeny to the number expected on the basis of your proposed genotypes.

a. third parent phenotypes produce about half burnsi and half pipiens, indicating that the genotypes of the parents must be BB+ and B+B+ In the first parent phenotypes, it is possible that it is BB+ and BB+ Burnsi is the dominant phenotype b. second parent phenotype must be BB+ and B+B+ c.

#14 EOC Joe has classic hemophilia, an X-linked recessive disease. Could joe have inherited the gene for this disease from the following persons? a. His mother's mother b. His mother's father c. His father's mother d. His father's father

a. yes b. yes c. no d. no

Bases that make up certain places of DNA can _______

act as signals

Bases that make up certain places of DNA can _______ Purpose of Crossing over? Effect of Crossing over? Draw Replication Fork (Helicase, DNA Polymerase III, DNA gyrase, Clamp Loader)

act as signals Purpose: genetic variation (natural selection) effect: recombinant (original chromosomes from mom and dad are no longer), genetical variation is an outcome not the underlying, in an offspring that has a genotype not found in either parent, due to crossing over in meiosis

Name the (most common) two secondary structures of a protein.

alpha helices: looping coils (a helices) develop from the hydrogen bonds that form between the oxygen of a carboxyl group and the hydrogen of the fourth amino group in the chain beta-pleated sheets: form when 2 or more strands of a polypeptide line up parallel to one another. Hydrogen bonds form between adjacent carboxyl and amino hydrogen groups

A cell has eight chromosomes in G1 of interphase. Draw a picture of this cell with its chromosomes at the following stages. Indicate how many DNA molecules are present at each stage.

alright

A geneticist discovers that two different proteins are encoded by the same gene. One protein has 56 amino acids, and the other has 82 amino acids. Provide a possible explanation for how the same gene can encode both of these proteins.

alternative splicing!!

The term ______ depicts a type of alternative processing of the pre-mRNA to produce different mRNA's that can be translated into different proteins.

alternative splicing?

What does true breeding mean?

an organism must be homozygous for every trait for which it is considered true breeding; that is, the pairs of alleles that express a given trait are the same

When a structural gene is under positive inducible control, what would be the result of a mutation that eliminates the activator protein?

as the transcription will require an activator protein, the transcription will be turned off

lacZ is ______. lac Y is _______. lacA is _____.

beta-galactosidase, permease, transcetylase

_______ have 5 untrasnlated and 3 untranslated regions\ - introns: _______, RNA processing: these are removed!!!! (euk RNA processing)

both prok and euk not coding

In German cockroaches, bulging eyes (bu) are recessive to normal eyes (bu+), and curved wings (cv) are recessive to straight wings (cv+). Both traits are encoded by autosomal genes that are linked. A cockroach has genotype bu+bu cv+cv and the genes are in repulsion. Which of the following sets of genes will be found in the most common gametes produced by this cockroach?

bu+ cv and bu cv+

In females, how does sexual differentiation occur?

by default different genes participate in ovary development

How is the reading frame of a nucleotide sequence set?

by the codons

Human genes are written how?

capitalized and italicized

Sex-linked characteristics

characteristics determined by genes located on the sex chromosomes X-linked or Y-linked

Synonimous

codons that specify the same aa

Karyotype

complete set of chromosomes presented as a picture of organized metaphase chromosomes

DnaC and DnaB

complex that loads helicase onto replication bubble once DnaB is attached to DNA strands, DnaC leaves

What is the function of the spliceosome? 10.5

consists of 5 mRNA molecules and almost three hundred proteins RNA components are small nuclear RNAs; these snRNAs associate with proteins to form small nuclear ribonucleoprotein particles (snRNPs) each snRNP contains a single snRNA molecule and multiple proteins spliceosome is composed of five snRNPs and some proteins not associated with an snRNA the spliceosome cuts the pre-mRNA at the 5' splice site, this cut frees the first exon from the intron and the 5' end of the intron attaches to the branch point (the intron folds back on itself forming a lariat) the guanine nucleotide in the consensus sequence of the 5' splice site bonds with the adenine nucleotide at the branch point through a transesterification reaction, and the 5' phosphate group of the G nucleotide becomes attached to the 2'-OH group of the A nucleotide at the branch point the spliceosome then cuts at the 3' splice site and the 3' end of the first exon becomes covalently attached to the 5' end of the second exon the intron is released as a lariat

lacOc

constitutive operator, repressor cannot bind

DNA Replication Process is _____.

continuous!

DNA Replication Process is _____.

continuous!!!!

What two processes unique to meiosis are responsible for genetic variation? At what point in meiosis do these processes take place?

crossing over in early prophase I independent assortment in metaphase/anaphase I

What is the function of IF two?

delivers fMet-tRNA^fMet to the initiation codon on the mRNA

Define DNA

deoxyribonucleic acid a self-replicating material which is present in nearly all living organisms as the main constituent of chromosomes It is the carrier of genetic information

Chromosome duplications often result in abnormal phenotypes because?

developmental processes depend on the relative amounts of proteins encoded by different genes

Isoaccepting

different tRNAs that accept the same amino acid but have different anticodons

What are isoaccepting tRNAs?

different tRNAs that accept the same amino acid but have different anticodons

Inducible/Repressible: ________.

effect on transcription

Epistasis

epistasis occurs when the alleles of one gene can mask the alleles of a second gene the epistatic gene masks the hypostatic gene

Dosage Compensation

equalizes the amount of protein produced by X-linked genes in the two sexes

Diff in prokaryotic mRNA and eukaryotic mRN

eukaryotic protein coding region only code for one protein But in the pre-messenger RNA stage there is the potential to create several different types of protein (once splicing, one mRNA, capable of only producing one single protein)

START and END with an _____! (Introns are exons - one) DON'T start with an intron or END with an intron (on a gene)

exon

Once the construct is made, you have to test it with different restriction enzymes to see if the segment of interest is in the proper orientation. T or F

false, blue-white screening first!

Define Quaternary structure.

final working protein (some), several peptides that are grouped together to create the final protein (hemoglobin has two peptides each of beta hemoglobin and alpha hemoglobin, when they aggregate they form a complex which is quaternary structure > functional!) the result of aggregating tertiary structures, many proteins contain multiple polypeptides (same or different) that assemble into functional macromolecule (hemoglobin for example), bonds and interactions occur between different polypeptide chains

Multiple Alleles Example

for many genes there are more than two alleles in the population Ex. ABO Blood groups (antigens): IA IB and i The ABO locus encodes specific glycosyltransferases that synthesize A and B antigens Gene FUT1 encodes a fucosyltransferase that produces H antigen

In figure 14.11, Bob and Joe are each homozygous for different restriction fragment patterns. How many bands would you expect to see on the gel if a person was heterozygous for the A and B patterns? Explain your reasoning.

four bands heterozygous person would display bands seen in both A and B Therefore, it would have three of Bob's and two of Joe's, but since one of Joe's bands is the same as one of Bob's bands, the person would display four different bands

What are the major results of meiosis?

four new daughter cells (haploid) each cell has half the amount of chromosomes as started out with genetic variability because of crossing over and independent assortment

#15 EOC In Drosophila, yellow body color is due to an X-linked gene that is recessive to the gene for gray body color. a. A homozygous gray female is crossed with a yellow male. The F1 are intercrossed to produce F2. Give the genotypes and phenotypes along with the expected proportions, of the F1 and F2 progeny. b. A yellow female is crossed with a gray male. The F1 are intercrossed to produce the F2. give the genotypes and phenotypes, along with the expected proportions, of the F1 and F2 progeny. FiniSH

gray body color = X^R yellow body color = X^r X^RX^R x X^rY+ Genotypes of F1: 1/2 X^RX^r : 1/2 X^RY+ Phenotypes of F1: Gray body color

Define Operon

group of structural genes, controlled by a common promoter and regulators, that are expressed simultaneously

Operon

group of structural genes, controlled by a common promoter and regulators, that are expressed simultaneously Proteins usually involved in the same molecular pathway Other regulatory sequences that produce regulatory proteins that help regulate gene expression (ONLY FUNCTION) Can be more than three or only two Promoter and Operator (regulatory elements)

Coat color in cats is a sex-linked trait. The gene for black or orange color is found on the X chromosome. The gene that codes for white hair color is on a different chromosome. Since females have two X chromosomes, they are able to display two colors (orange and black, or variations thereof) and white; creating the 3-color calico mix. Since males have only one X chromosome, they can only be orange OR black (__________).

hemizygous

Females are the XX, meaning they are the _______ sex. Males are the ________ sex. The presence of the _________ on the ________ determines maleness.

homogametic, heterogametic, SRY gene, Y chromosome

The HTT gene provides instructions for making a protein called _____.

huntingtin

SC are _____.

identical!!

Microduplications

individuals can be homozygous or heterozygous for a duplication (one duplication from mom or from dad OR one duplication from mom and one from dad) Can arise by unequal crossing over Have arisen frequently throughout the evolution of many eukaryotic organisms and are a source of new genes

Arrange the following components of replication in the order in which they first act in the replication process: ligase, DNA polymerase I, helicase, gyrase, primase, single-strand binding protein, initiator proteins.

initiator proteins helicase SSBP Gyrase Primase DNA polymerase I Ligase

Oncogene

is a gene that has the potential to cause cancer. In tumor cells, they are often mutated or expressed at high levels. Most normal cells will undergo a programmed form of rapid cell death when critical functions are altered and malfunctioning.

Incomplete Penetrance

is observed when a particular genotype does not always produce the expected phenotype Do all mutant individuals present a mutant phenotype?

Giesma Staining

is the most common (stains AT rich regions)

Expressivity

is the phenotypic variability presented among individuals that have the mutation Do all mutant individuals present the same phenotype?

lacI+ vs. lacIS

lacI+: repressor is produced lacIS: superrrepressor, repressor cannot be inactivated by inducer

Low Copy repeats

lead to rearrangements during recombination pretty large (1 to 300kb) high sequence identity (> 95%) small in number and may be on different or same chromosomes Direct: sequences going same way (centromere to telomere) Inverted: sequence going towards or away from each other

Chaperonin

lid component (technically flat) and a "basket" component, has a hole that proteins can go into (has a specific environment condition that allows the protein to fold in a specific shape, ex. Hydrophilic protein, hydrophobic aa towards the outside) a protein that aids the assembly and folding of other protein molecules in living cells

Viral Genome Organization

linear or circular, dsDNA or ssDNA no genome associated proteins

Prokaryotic polycistronic RNA

mRNA can have one or several protein coding regions (when mRNA is produced, upon translation, is capable of synthesizing multiple proteins, in slides > 3), Type of mRNA that has multiple coding regions: polycistronic RNA!!!!!!! EXCLUSIVE OF BACTERIA, a cistron > coding units of DNA, open reading frames (another name for cistron)

Inversion

may disrupt specific genes, splice two genes together or alter gene expression by position effects also frequent in evolution

Positively supercoiled

molecules that are overrotated

Polyribosomes

multiple ribosomes attach to an mRNA molecule (cazac sequence helps ribosome, cap proteins, and polyA proteins help ribosomes assemble). FMRP protein is attached to proteins that help shuttle mRNA into cytoplasm and assemble of ribosomes for translation!

An operon is controlled by a repressor. When the repressor binds to a small molecule, it is released from binding to DNA near the operon. The operon is never expressed if a mutation prevents the repressor from binding to the small molecule. This type of control is: Positive or Negative? Inducible or Repressible?

negative inducible

Why are the two cells produced by the cell cycle genetically identical?

no crossing over or independent assortment takes place this means that all the genetic information is kept the same

Is DNA ever naked within the euk cell?

no, it is always bound to proteins which is why it is called chromatin

Free nucleotides:

nucleotide triphosphates, cleavage of those is where you get energy to add next nucleotide

Introns saty in the ?

nucleus

Nucleosome?

octomer + DNA (wrapped twice around it) 8 histones, protein, and 146 bp of DNA

#27 EOC In German cockroaches, a curved wing (cv) is recessive to a normal wing (cv^+). A homozygous cockroach that has normal wings is crossed with a homozygous cockroach that has curved wings. The F1 are intercrossed to produce the F2. Assume that the pair of chromosomes containing the locus for wing shape is metacentric. Draw this pair of chromosomes as it would appear in the parents, the F1, and each class of F2 progeny at metaphase 1 of meiosis. Assume that no crossing over takes place. At each stage, label a location for the alleles for wing shape (cv and cv^+) on the chromosomes.

#34 EOC Alleles A and a are located on a pair of metacentric chromosomes. Alleles B and b are located on a pair of acrocentric chromosomes. A cross is made between individuals having the following genotypes: AaBb X aabb. a. Draw the chromosomes as they would appear in each type of gamete produced by the individuals b. For each type of progeny resulting from this cross, draw the chromosomes as they would appear in a cell at G1, G2, and metaphase of mitosis.

22. The following diagram represents a DNA molecule that is undergoing replication. Draw in the strands of newly synthesized DNA and identify (a) through (d): Polarity of newly synthesized strands Leading and lagging strands Okazaki fragments RNA primers

A cell has two pairs of submetacentric chromosomes, which we will call chromosomes Ia, Ib, IIa, and IIb (chromosomes Ia and Ib are homologs and chromosomes IIa and IIb are homologs). Allele M is located on the long arm of chromosome Ia, and allele m is located at the same position on chromosome Ib. Allele P is located on the short arm of chromosome Ia, and allele p is located at the same position on chromosome Ib. Allele R is located on chromosome IIa and allele r is located at the same position on chromosome IIb. Draw these chromosomes, identifying genes M, m, P, p, R, and r as they might appear in metaphase 1 of meiosis. Assume that there is no crossing over. Taking into consideration the random separation of chromosomes in anaphase 1, draw the chromosomes (with genes identified) present in all possible types of gametes that might result from this cell undergoing meiosis. Assume that there is no crossing over.

A geneticist isolates a gene that contains eight exons. He then isolates the mature mRNA produced by this gene. After making the DNA single stranded, he mixes the single-stranded DNA and RNA. some of the single-stranded DNA hybridizes (pairs) with the complementary mRNA. Draw a picture of what the DNA-RNA hybrids will look like under the electron microscope. FIX

A linear piece of DNA has the following EcoR1 restriction sites: ___2kb___(Site 1)___4kb___(Site2)_________5kb___________ This piece of DNA is cut by EcoR1, the resulting fragments are separated by gel electrophoresis, and the gel is stained with ethidium bromide. Draw a picture of the bands that will appear on the gel. DRAW ON THE SAME ONE!!

Draw Open Replication Bubble and label polarity.

Draw a molecule of DNA undergoing eukaryotic linear replication. On your drawing, identify (1) origin, (2) polarity (5' and 3' ends) of all template and newly synthesized strands, (3) leading and lagging strands, (4) Okazaki fragments, and (5) location of primers. Would be like the one above but with multiple origins of replication

Draw a molecule of DNA undergoing replication. On your drawing, identify (1) origin, (2) polarity (5' and 3' ends) of all template strands and newly synthesized strands, (3) leading and lagging strands, (4) Okazaki fragments, and (5) location of primers.

Draw the cell cycle. Label where chromosomes condense.

Draw the different types of chromosomes Metacentric Telocentric Acrocentric Submetacentric

Draw the structure of a nucleotide.

First nucleotide is slightly different in the messenger, conserves ALL 3 phosphate groups, other nucleotides have one phosphate group that connects one to the next, difference in how RNA is synthesized, both euk and prok have this feature

A linear piece of DNA has the following EcoR1 restriction sites: ___2kb___(Site 1)___4kb___(Site2)_________5kb___________ If a mutation that alters EcoR1 site 1 occurs in this piece of DNA, how will the banding pattern on the gel differ from the one that you drew in part a?

ok one piece that is 6kb and one that is 5kb

A linear piece of DNA has the following EcoR1 restriction sites: ___2kb___(Site 1)___4kb___(Site2)_________5kb___________ If 1000 bp of DNA were inserted between the two restrictions sites, how would the banding pattern on the gel differ from the one you drew in part a?

ok two fragments of 5kb one fragment of 2kb

Draw a typical eukaryotic gene and the pre-mRNA and mRNA derived from it. Assume that the gene contains three exons. Identify the following items and, for each item, give a brief description of its functions: 5' UTR Promoter Aauaaa Transcription start site 3' UTR Introns Exons polyA tail 5' cap

ok :(

Each chromosome = ____ DNA molecules

one

When does Histone H1 attach?

outside of core after DNA is wound, Histone H1 attaches to keep everything in place

4 x 9-mers

place where DnaA binds

__________ mRNA is a term that depicts a prokaryotic mRNA that contains multiple coding regions.

polycistronic

Nucleic Acids

polymers consisting of repeating units called nucleotides; each nucleotide consists of a sugar, a phosphate group, and a nitrogenous base

G2 Phase

preparation for division, G2/M Checkpoint

Nondisjunction 2

problems with segregation, associated with advanced maternal age (>35 yoa) While the baby is still in the womb, the eggs start dividing for meiosis and arrest in a certain point. The girl hits puberty at 12, meiosis cycle happens and arrests again. Unless the egg is fertilized, it doesn't finish meiosis. Therefore, women who do not have children until 35 have had meiosis arrested for almost 35 years.

The -10 and -35 region can be found exclusively in?

prokaryotic promoters

The following diagram represents a transcription unit on a DNA molecule. Assume that this DNA molecule is from a bacterial cell. Label the approximate locations of the promoter and terminator for this transcription unit.

promoter is upstream of the transcription start site terminator is at the end of the RNA coding region of the gene

X and Y pair as homologues?

pseudoautosomal regions are short regions of homology between the mammalian X and Y chromosomes

Replication and Division go hand in hand. Why?

purpose of DNA replication is so cells can be divided. 2 daughter cells > need a set of instructions no division? no purpose for replication

Translocation

reciprocal non-reciprocal: one chromosome gives something to another chromosome and receives nothing in exchange. may lead to deletions or loss of chromosomal segments

Comparative Genomic Hybridization (Type of Microarray)

reference and test genome is compared On the slide you can print single stranded probes Each spot has several similar probes (important for technique) 1. Take test DNA and use reference DNA (agreed upon cell line, established) and break it up using restriction enzymes which cuts the genome into several pieces 2. Test DNA pieces are labeled with fluorescent red dye (Cy5) and Reference DNA pieces are labeled with fluorescent green dye (Cy3) 3. Take 500microL of one and 500microL of the other and mix them. (volume of equal proportion) 4. Same proportion of red and green = YELLOW! 5. Take mix of equal parts and pour it into the slide/array. 6. Let it incubate because you want the chopped up pieces to stick to the pieces that have been printed onto the slide (DNA complimentary). Wash off excess. Read it using a specific apparatus that has a laser. 7. The yellow spots (specific group of probes), equal amount of red and green probes stuck, individual and control have the same amount of DNA for that segment (NO CHANGE) 8. If you have a red spot, you have much more DNA of your patient than you do of your reference >> duplication of that specific segment of the genome (GAIN OR DUPLICATION on one chromosome, person has 3 copies fo that segment) 9. If you have a green spot, you have much more DNA of your reference than you do of your patient >> your patient has a deletion (or loss). Picture: center means no deviation Collection of probes that correspond to one specific spot of your genome 2 standard deviations to the left: 2 deletions: one deletion on each chromosome Microduplication: 1 standard deviation to the right: 1 duplication: for all of these probes the patient has one extra copy (duplicate on one chromosome and normal on the other)

A sigma factor is an example of a ________ protein./

regulator,only function is to regulate gene expression

Define RNA

ribonucleic acid a nucleic acid present in all living cells. Its principal role is to act as a messenger carrying instructions from DNA for controlling the synthesis of proteins (translation), although in some viruses RNA rather than DNA carries the genetic information

How would the deletion of the Shine-Dalgarno sequence affect a bacterial mRNA?

ribosome might not be able to recognize the sequence to bind to the mRNA, transcription would not begin

What are the genetically important results of the cell cycle and mitosis?

same amount of chromosomes after cell division each sister chromatid, when separated, becomes a chromosome all chromosomes are fully replicated 2 identical daughter cells

DNA ligase

seals nicks between the sugar-phosphate groups of the fragments

Ligase

seals the break in the sugar-phosphate backbone An enzyme that catalyzes the formation of the phosphodiester bond between the break of the 3'-OH group of the last nucleotide added by DNA pol 1 and the 5'-phosphate group of the first nucleotide added by DNA pol 3

DnaB/Helicase

separates/undwinds DNA strands DnaB codes for helicase

Purpose of DnaB

separates/undwinds DNA strands DnaB codes for helicase

What makes a pair of homologous chromosomes?

sequence identical besides alleles, location of centromere identical, banding pattern identical, gene content identical (diff versions/alleles but same genes)

What makes a pair of homologous chromosomes? How do they find their mates?

sequence identical besides alleles, location of centromere identical, banding pattern identical, gene content identical (diff versions/alleles but same genes) SEQUENCE IDENTITY (sequence lets them find each other!)

An RNA molecule has the following percentages of bases: A= 23%, U= 42%, C=21%, and G=14%. Is this single stranded or double stranded? How can you tell? What would be the percentages of bases in the template strand of the DNA that contains the gene for this RNA?

single stranded If it were double stranded (chargaff's rule), the complementary base pairs would be equal (A=U, G=C) 23% T 42% A 21% G 14% C

Origins of Replication

sites where DNA synthesis begins multiple origins of replication on each eukaryotic chromosomes on each eukaryotic chromosomes.

Aneuploidy and types

small number of chromosomes that are being changed (usually by a number of one), mistakes of nondisjunction Nullisomy: one of the homologous pairs is missing, 2n-2 Monosomy: one of the sex chromosomes is missing, 2n-1 Trisomy: full genetic compliment plus an extra chromosome 21, 2n+1 Tetrasomy: 2n+2

The 70S translational initiation complex includes the following proteins and molecules?

small ribosomal subunit, large ribosomal subunit, mRNA template, charged tRNA with formal methionine on the P site

The genetic code is ____ because every codon only codes for one amino acid.

specific

stable, nondividing period of variable length

How are tRNA's linked to their corresponding amino acids?

tRNAs serve as adapter molecules that bind particular amino acids and deliver them to the ribosome where the aa are then assembled into polypeptide chains Each type of tRNA attaches to a single type of amino acid aminoacyl-tRNA synthetases charge the tRNA, a covalent bond between aa and tRNA is formed

Thinking of a double stranded DNA molecule that may be used as a template for transcription, the 3' to 5' strand serves as the _______ strand, where as the 5' to 3' strand is called the _______ strand.

template, coding

Karyotype

the complete set of chromosomes possessed by an organism

Dideoxy method of DNA sequencing

the fragment to be sequenced is used as a template to make a series of new DNA molecules. In the process, replication is sometimes terminated when a specific base is encountered, producing DNA strands of different lengths, each of which ends in the same base RELIES ON THE USE OF A SPECIAL SUBSTRATE >> ddNTP is also used as a substrate with dNTP

Expression of SRY leads to...?

the production of testosterone and Mullerian-inhibiting substance

Polyploidy and types

the state of a cell or organism having more than two paired (homologous) sets of chromosomes. Autopolyploidy appears when an individual has more than two sets of chromosomes, both of which from the same parental species. Allopolyploidy, on the other hand, occurs when the individual has more than two copies but these copies, come from different species.

Chapter 1 considered the theory of the inheritance of acquired characteristics and noted that this theory is no longer accepted. Is the central dogma consistent with theory of the inheritance of acquired characteristics? Why or why not?

they are not consistent because inheritance of acquired characteristics demands reverse flow information which violates the central dogma. **The inheritance of acquired characteristics is a hypothesis that physiological changes acquired over the life of an organism (such as the enlargement of a muscle through repeated use) may be transmitted to offspring. The central dogma: **provides the basic framework for how genetic information flows from a DNA sequence to a protein product inside cells **This process of genetic information flowing from DNA to RNA to protein is called gene expression.

What is the function of IF one and three?

to prevent the binding of the large subunit of the ribosome to allow the mRNA to engage the small subunit of the ribosome

What is an enhancer? How does it affect the transcription of distant genes?

transcriptional regulator proteins bind here enhancers are regulatory elements that may be located some distance from the gene. The exact position and orientation of an enhancer in relation to the promoter it affects can vary. a short (50-1500 bp) region of DNA that can be bound by proteins (activators) to increase the likelihood that transcription of a particular gene will occur. transcriptional regulator proteins bind to the enhancer and cause the DNA between the enhancer and the promoter to loop out, bringing the promoter and enhancer close to each other so that the trans. regulator proteins are able to interact directly with the basal transcription apparatus at the core promoter

Non-allelic recombination

two loci that are not alleles are used for crossing over

Chromosome from 300nm fiber

ultimately continues to compact with help from scaffolding proteins to make a chromosome

Negatively Supercoiling

underrotated molecules

What are sticky/cohesive ends?

when a RE cuts the sugar-phosphate backbone of each strand at a point, it generates fragments with short, ss overhanging ends They are called sticky ends because they are complementary to each other and can spontaneously pair to connect the fragments these fragments can be 'glued' together, any two fragments cleaved by the SAME enzyme will have complementary ends and will pair

Define the tertiary structure of a protein

when all of these 2nd structure associate to give you the final 3D organization of that peptide, you are talking about tertiary structure, not all proteins are functional in this state (hemoglobin) Interactions between side chains create three-dimensional shapes a. main 3D shape of polypeptide chain b. several types of interactions that in combination form the tertiary structure (unique for each proteins) 1. non-polar side chains of AA cluster together to reduce contact with water molecules in the surrounding fluid (hydrophobic effect), clustering of hydrophobic groups away from water 2. H bonds can form between R groups 3. Covalent bonds can form between two sulfur atoms of the AA cysteine. These are usually called disulfide bridges 4. Ionic bonds between charged side groups 5. Van der waals: weak electrostatic forces add stability to the structure

Purpose of DnaC

works with DnaB to load helicase (DnaB) on replication bubble forms complex with DnaB that loads helicase onto replication bubble once DnaB is attached to DNA strands, DnaC leaves

For a gene under negative repressible control, a small molecule is required to prevent the gene's repressor from binding to DNA. T or F

False

5' cap helps with what? 3' Poly(A) tail?

5' cap: Aids: -Ribosomal binding -Removal of introns Prevents degradation 3' Poly(A) tail aids: -Ribosomal binding -Export of mRNA to cytoplasm Prevents degradation

Synthesis of the mRNA always occurs in the ____ direction.

5' to 3'

Give the amino acid sequence of the protein encoded by the mRNA in Figure 11.13.

5'--AUG-CCC-ACG-ACU-GCG-AGC-GUU-CCG-CUA-AGG-UAG--3' 5'--Met-Pro-Thr-Thr-Ala-Ser-Val-Pro-Leu-Arg-STOP--3'

Terminator (Bacteria)

8. Terminator sequence + Self/Intrinsic termination + Ro-dependent termination (NO!) + Palindromic complimentary sequence here, green and blue bases are G's and C's (two stretches, template strand) + Bunch of A's residues (template strand) + these sequences are palindrome, meaning that they are read the same way forward and reverse + once this RNA is removed off the template (only little bit of template RNA being transcribed is attached to Pol, rest of RNA is free) + Only piece of RNA that remains in the bubble produced by transcription machinery contains a segment called a heteroduplex meaning a piece of RNA is stuck to a piece of complimentary DNA, rest is free + HETERODUPLEX in BUBBLE!!

Chiasma

A chiasma is the point of contact, the physical link, between two chromatids belonging to homologous chromosomes. At a given chiasma, an exchange of genetic material can occur between both chromatids, what is called a chromosomal crossover.

Define a cloning vector.

A cloning vector is a stable, replicating DNA molecule into which a foreign DNA fragment can be inserted for introduction into a cell.

Explain why recombination is suppressed in individuals heterozygous for paracentric and pericentric inversions.

A crossover within a paracentric inversion produces a dicentric and an acentric recombinant chromatid. The acentric fragment is lost, and the dicentric fragment breaks, resulting in chromatids with large deletions that lead to nonviable gametes or embryonic lethality. A crossover within a pericentric inversion produces recombinant chromatids that have duplications or deletions. Again, gametes with these recombinant chromatids do not lead to viable progeny.

DNA Polymerase III

A large multiprotein complex that synthesizes nucleotide strands by adding new nucleotides to the 3' end of a growing DNA strand 5'->3': polymerase activity that allows it to add new nucleotides in the 5'->3' direction 3'->5': exonuclease activity that allows it to remove nucleotides in the 3'->5' direction thus enabling it to correct errors (proofreading capabilities)

#32 EOC The following two genotypes are crossed: Aa Bb Cc dd Ee X Aa bb Cc Dd Ee. What will the proportion of the following genotypes be among the progeny of this cross?

Aa Bb Cc Dd Ee = 1/32 Aa bb Cc dd ee = 1/64 aa bb cc dd ee = 1/256 AA BB CC DD EE = 0

In dogs, wire hair (S) is dominant to smooth (s). In a cross of a homozygous wire-haired dog with a smooth-haired dog, what will be the phenotype of the F1 generation?

All Wire haired!

Examine Figure 12.7. What would be the effect of a drug that altered the structure of allolactose so that it was unable to bind to the regulator protein?

Allolactose is the substrate here. When it is unable to bind to the regulator protein (repressor), the repressor protein binds to the operator and transcription is off. The lac operon is negative inducible. NO genes can be transcribed.

#10 EOC What is a complementation test and what is it used for?

An individual possessing two recessive mutation has a wild-type phenotype, indicating that the mutations are non-allelic genes Test this by crossing individuals that are homozygous recessive and dominant to see the probability of offspring with mutation

Transgenic

An organism that has been permanently altered by the addition of a DNA sequence to its genome is said to be transgenic, and the foreign DNA that it carries is called a transgene

What is the null hypothesis? p > 0.05%? p < 0.05%?

Any fluctuation between the observed and the expected data is due to chance. There is no difference between the observed and expected data. If the probability p is high, our hypothesis is true. We assume that chance alone produced the difference between the observed and expected ("true" null hypothesis, over 5%). We fail to reject our null hypothesis. "The high p value indicates that the obtained data fits the expected data very closely and that any fluctuation can be explained by chance." If the probability p is low, our hypothesis is false. We assume some other factor is causing the difference (false null hypothesis), under 5%.

Define Allopolyploidy

Arises from hybridization between two species; the resulting polyploid carries chromosome sets derived from two or more species

Hairpin stuff

As RNA pol moves across, segment before is released 10. bases are complimentary and have a strong affinity (G's and C's, palindromic sequence followed by A's leads to the folding (haripin structure) in DNA and hairpin structure followed by poly U tail (lot of U bases, weaker bonds than G's and C's), 10. hairpin binds to itself and allows for the formation of hairpin structure 11. RNA Poly attached to template strand, kink destabilizes the RNA polymerase (hairpin structure), hairpin disrupts engagement of RNA polymerase with template strand (AFTER RNA has passed) 12. Folding of haprin structure RIPS OFF RNA off the template strand 13. Hairpin disrupts heteroduplex 14. RNA goes over terminator, terminator binds, and snaps RNA off 16. " A short complementary GC rich sequence (followed by several U residues, will form a hairpin structure, causing release of the poly U stetch"

Define Autopolyploidy

Autopolyploidy: caused by accidents of mitosis or meiosis that produce extra sets of chromosomes, all derived from a single species Nondisjunction of all chromosoems in mitosis in an early 2n embryo, for example, doubles the chromosome number and produces an autotetraploid (4n) An autotriploid (3n) may arise when nondisjunction in meiosis produces a diploid gamete that then fuses with a normal haploid gamete to produce a triploid zygote All chromosome sets in AP are from the same species and therefore they are homologous and attempt to align in prophase I of meiosis (usually results in sterility)

Intrachromatid Recombination Pedigree Rules >:)

Autosomal Dominant: doesnt skip generations, affected parents can have unaffected children Autosomal Recessive: skips generation, unaffected parents can have affected children X-linked Dominant: disease never transfers from father to son, all daughters of an affected father will be affected X-linked Recessive: males are more affected, disease tends to transfer from mother to son and from father to daughter, disease never transfers from father to son

Huntington's 2

Autosomal dominant Diseases caused by trinucleotide expansions have genetic anticipation Large polyglutamine tract which leads to gain of function toxicity (on top) Transcriptional dysregulation TF must interact with DNA, nucleosomes. Huntington damages the transcription process by wrongly interacting with proteins and not allowing them to bind to the DNA Protein degradation: mutant proteins aggregate and make this huge mass and supposed to be degraded in lysosome but cannot be degraded, escaping protein degradation Alternate protein folding: can bind other proteins in other mass of proteins and starts to make a lot of different aggregates Altering neuron Over 35 repeats is considered pathological Housed on the short (p, petite) arm of Chromosome 4

Recombinant DNA

DNA molecules formed by laboratory methods of genetic recombination to bring together genetic material from multiple sources, creating sequences that would not otherwise be found in the genome

Why is primase required for replication?

DNA polymerase cannot synthesize without 3'OH available It cant add dna nucleotides without a primer Primase therefore fixes this job by providing the 3' OH

Remember to make mutations where?

DNA, template strand!

Example A (Ch 6) Describe and Draw the Mouse Locus

Deletion On the maternal chromosome, the unmethylated DMD/ICR (DMR1 in this figure) binds the CTCF protein and forms an insulator, preventing the common enhancers from activating Igf2. Instead, the enhancers activate the nearby H19 promoter. On the paternal chromosome, the methylated DMD/ICR (DMR1 in this figure) cannot bind CTCF and an insulator does not form, and, therefore, the Igf2 gene is expressed. Due to differential methylation on the maternal and paternal chromosome, some genes are expressed differently depending on the parent of origin The imprint is erased in the germ cells and reestablished during gamete formation

There are no obvious restriction sites surrounding the husK gene, yet you still need to insert this gene into pCM999. How will you do this? FIX

Design primers.

#9 EOC What is gene interaction? What is the difference between an epistatic and hypostatic gene?

Effects of genes at one locus depend on the presence of genes at other loci Products of genes at different loci combine to produce new phenotypes tha are not predictable from the single-locus effects alone Epistatic gene is the gene that does the masking- may be recessive or dominant Hypostatic gene is the gene whose effect is masked

A diploid human cell contains approximately 6.4 billion base pairs of DNA. How many nucleosomes are present in such a cell? Assume that the linker dna encompasses 40 bp. How many histone proteins are complexed with this DNA?

Every nucleosome has 140 bp H1 takes another 20 bp Linker DNA is another 40 Add all these together we get 200 bp 6.4 billion divided by 200 = 32 million 32 million x 9 histone proteins = 2.9 X 108 molecules of histones

Cytosine, Thymine, and Uracil are all purines.

F, Pyrimidines

EOC #1 What does the term recombination mean? What are two causes of recombination?

Gametes with new combinations of alleles are called recombinant gametes Progengy with new combinations of traits formed from recombinant gametes are termed recombinant progeny 1st Cause: Linkage/Crossing Over 2nd Cause: Independent Assortment

What is the difference between a genetic map and a physical map? Equation for recombination frequency.

Genetic Map: chromosome maps calculated by using the genetic phenomenon of recombination Physical Map: chromosome maps calculated by using physical distances along the chromosome (often expressed as numbers of base pairs) RF = (number of recombinant progeny / total number of progeny) x 100%

Diseases caused by trinucleotide expansions have a phenotype called ______. What does this mean?

Genetic anticipation_. This means that in every generation the number of repeats is greater and the phenotype is more severe. Why? These repeats (when being replicated) they are highly prone to errors, meaning that because of the bases they have they create these really interesting loop structures (FORM BONDS TOGETHER IN DIFF PLACES) When these kinks form, replication machinery becomes confounded and uses that as a template to make a new copy (structures are leading to errors in DNA synthesis!, called hairpin structures, alternate tertiary structures for DNA) Typically they can also break, need to be repaired during DNA replication, lining the repeats up exactly can be very tricky because there are so many, very easy to fix things in the wrong place (incorrect lining up with DNA repair), greater the expansion, more unstable and that is why it shows genetic anticipation (EARLIER ONSET, number of repeats greater in offspring)

#35 EOC E. W. Lindstrom crossed two corn plants with green seedlings and obtained the following progeny: 3583 green seedlings, 853 virescent-white seedlings, and 260 yellow seedlings. a. Give the genotypes for the green, virescent-white,and yellow progeny b. Explain how color is determined in these seedlings. c. Does epistasis take place among the genes that determine color in the corn seedlings? If so, which gene is epistatic and which is hypostatic

Green = A_B_ and A_bb virescent -white = aaB_ Yellow = aabb Explain how color is determined in these seedlings. If you have the dominant allele, the effect of B does not matter. Dominant allele will suppress the expression of B which will always result of the phenotype of green. If A is recessive, then two options can occur. This is when we see the effect of having dominant or recessive for gene B. if B is dominant, we see virescent-white. If b is recessive, we see yellow. Does epistasis take place among the genes that determine color in the corn seedlings? If so, which gene is epistatic and which is hypostatic? Yes A is epistatic whereas B is hypostatic

5 Types of Histones

H1 2 * (H2A, H2B, H3, H4)

Negative Control

In operons under negative control, the regulator protein acts as a repressor and binds to the operator to prevent transcription by preventing the binding of the RNA polymerase to the promoter.

#10 EOC: In which phases of mitosis and meiosis are the principles of segregation and independent assortment at work? ******

Independent Assortment: Anaphase I of Meiosis (alleles at different loci separate independently of one another) Segregation: Anaphase I of Meiosis (pg 51)

Microdeletions

Individuals can be homozygous or heterozygous for a deletion can arise by unequal crossing over one nonfunctional allele is enough to convey a diseased phenotype (haploinsufficiency)

Define the following terms as they apply to the genetic code: Initiation codon Termination codon Sense codon

Initiation codon: start codon, the first codon of the mRNA to specify an amino acid AUG Termination codon: stop codon, signal the end of the protein in both bacterial and eukaryotic cells, no tRNAs have anticodons that pair with termination codons Sense codon: A set of three nucleotides in a protein coding sequence that specifies individual amino acids 64 possible codons, but only 61 of these are sense codons because the last codon is a termination signal

Name the first step of translation.

Initiation: Formation of 70s translation initiation complex. First, IF 1 and IF 3 bind ot the small subunit, preventing the large subunit from binding. This allows the small subunit of the ribosome to bind to the mRNA. Second, initiator tRNA (that has fMet) binds to the mRNA through base pairing between the codon and anticodon. (requires IF-2) Third, the large ribosomal subunit joins the initiation complex when IF 1 and IF 2 disassociate from the small subunit.

What is the function of the Shine-Dalgarno consensus sequence? 10.5

It serves as the ribosome-binding site during translation it is found approximately seven nucleotides upstream of the first codon translated into an amino acid (called the start codon) It is complementary to sequences found in one of the RNA molecules that make up the ribosome and pairs with those sequences during translation In eukaryotes, ribosomes bind to a modified 5' end of mRNA (5' cap)

An operon is controlled by a repressor. When the repressor binds to a small molecule, it binds to DNA near the operon. The operon is constitutively expressed if a mutation prevents the repressor from binding to the small molecule. This type of control is: Positive or Negative? Inducible or Repressible?

Negative repressible

A gene encodes a protein with the following amino acid sequence: Met-Trp-His-Arg-Ala-Ser-Phe A mutation occurs in the gene. The mutant protein has the following amino acid sequence. Met-Trp-His-Ser-Ala-Ser-Phe An intragenic suppressor restores the amino acid sequence to that of the original protein: Met-Trp-His-Arg-Ala-Ser-Phe Give at least one example of base changes that could produce the original mutation and the intragenic suppressor.

Original: AUG-UUG-CAU-CGA-GCU-UCU-UUU Mutation example: AUG- UUG- CAU- AGA- GCU- UCU- UUU

In cucumbers, heart-shaped (hl) are recessive to normal leaves (Hl) and numerous fruit spines (ns) are recessive to few fruit spines (Ns). the genes for leaf shape and for number of spines are located on the same chromosome; findings from mapping experiments indicate that they are 32.6 mu apart. A cucumber plant having heart-shaped leaves and numerous spines is crossed with a plant that is homozygous for normal leaves and few spines. The F1 are crossed with plants that have heart-shaped leaves and numerous spines. What phenotypes and phenotypic proportions are expected in the progeny of this cross?

P: hlhlnsns x HlHlNsNs F1: All HlhlNsns x hlhlnsns F2: 33.7% HlhlNsns, 16.3% Hlhlnsns, 16.3% hlhlNsns, 33.7% hlhlnsns total recombinants = 32.6% total nonrecombinants = 67.4% Same chromosome? 32.6 mu apart = 32.6% recombinant frequency

#2 EOC: What is the difference between genotype and phenotype?

Phenotype: outer appearance, you can see it! appearance or manifestation of a characteristic (Ex. Ss means short hair, short hair is phenotype) Genotype: set of genes possessed by an individual organism (Ex. Ss is the genotype)

What does a prokaryotic transcription unit/gene contain? Draw it.

Promoter - contains consensus sequences that allow the RNa polymerase to bind to the mRNA (consensus sequences are short stretches of common nucleotides that posses considerable similarity) RNA coding region - a sequence of DNA nucleotides that is copied into an RNA molecule Terminator - a sequence of nucleotides that signals where transcription is to end, part of the RNA coding region, transcription stops after the terminator has been copied into RNA

Indicate the location of the promoters and terminators for genes a, b, and c.

Promoters are upstream from the gene Terminators at the end of the RNA coding region of the gene

The following diagram illustrates a step in the process of translation. Identify the following elements on the diagram. 5' and 3' ends of the mRNA A, P, and E sites Start codon Stop codon Amino and carboxyl ends of the newly synthesized polypeptide chain Approximate location of the next peptide bond that will be formed Place on the ribosome where release factor 1 will bind (stop codon)

RF-1 attaches to a site!!

Heinz Shuster collected the following data on the base composition of the ribgrass virus. On the basis of this information, is the hereditary information of the ribgrass virus RNA or DNA? Is it likely to be single stranded or double stranded?

RNA Single stranded ^ It wont follows chargaff's rule A not equal to U

Proteins are made up of ________.

RNA and ribosomes

The function of the promoter is to bind?

RNA core polymerase

lacP-

RNA polymerase cannot bind, therefore no transcription

When do chromsomes duplicate? Why?

S phase, to begin mitosis or meiosis

Somatic vs. Germ-line mutations

Somatic: ex. Skin cancer (does not passed down), happens at the level of somatic cells (radiation, etc), mutation happens in somatic cells. Germ-line: Mutation happens in a cell that produces gametes. Gametes have to undergo a round of DNA replication (accidents can happen here, with crossing over, etc) Next generation (egg/gamete) might already have mutations, will be in every single cell of that offspring of that individual

A strain bacteria possesses a temperature sensitive mutation in the gene that encodes the sigma factor. The mutant bacteria produce a sigma factor that is unable to bind to RNA polymerase at elevated temperatures. What effect will this mutation have on the process of transcription when the bacteria are raised at elevated temperatures?

The rna polymerase will not be able to form a holoenzyme with the sigma factor so therefore the DNA double strand cannot be separated to form a transcription bubble and the first few RNA nucleotides cannot be attached by RNA polymerase. RNA polymerase basically can not do its job. Promoter does not melt :(

Define Transcription AND translation.

Transcription: A process in which a complementary mRNA strand is synthesized from a template DNA strand. Translation: A process in which a peptide chain is synthesized from a from the mRNA strand

Explain how to determine, using the numbers of progeny from a three-point cross, which of three linked loci is the middle locus.

Use the one the allele that when switched would give you the genotypes of the nonrecombinant crossovers

After 3' cleavage, what happens?

Where this enzyme cuts, there is a polyadenylation enzyme that adds a stretch of adenines at the 3' end of the mRNA

Draw a picture illustrating the general structure of an operon and identify its parts.

____Promoter_Operator___Gene A___Gene B___Gene C

In silkmoths (Bombyx mori), red eyes (re) and white-banded wings (wb) are encoded by two mutant alleles that are recessive to those that produce wild-type traits (re+ and wb+); these two genes are on the same chromosome. A moth homozygous for red eyes and white-banded wings is crossed with a moth homozygous for the wild-type traits. The F1 have normal eyes and normal wings. The F1 are crossed with moths that have red eyes and white-banded wings in a testcross. The progeny of this testcross are: a. What phenotypic proportions would be expected if the genes for red eyes and for white-banded wings were located on different chromosomes? b. What is the rate of recombination between genes for red eyes and those for white banded wings?

a. 1/4 wild-type eyes and wild-type wings, 1/4 wild-type eyes and white banded wings, 1/4 red eyes and wild-type wings, 1/4 red eyes and white banded wings b. Recombination frequency = (recombinants)/(total progeny) x 100% RF = (35)/(879) x 100% = 4.0% The distance between the genes is 4 map units.

Refer to the diagram in Problem 22 to answer the following questions. a. What will be the anticodon of the next tRNA added to the A site of the ribosome b. What will be the next amino acid added to the growing polypeptide chain?

a. UGC b. Thr

What is the 5' cap? 10.5

consists of an extra modified nucleotide at the 5' end of the mRNA as well as methyl groups (CH3) on the 2'-OH group of the sugar of one or more nucleotides at the 5' end the cap functions in the initiation of translation cap-binding proteins recognize the cap and attach to it; a ribosome then binds to these proteins and moves downstream along the mRNA until the start codon is reached and translation begins presence of this cap increases the stability of mRNA and influences the removal of introns

First Step mRNA processing

ddition of 5' cap (modified methylated guanine base, 3-methylguanosine): has methyl groups attached and attached backwards to the 5' of the messenger RNA (prevents degradation from exonucleases)

The genetic code is _____ as each amino acid can be encoded by more than one codon.

degenerate

Positive or Negative Control

effect of regulator

Transcription

getting information DNA to mRNA

Prokaryotic Promoter

have two regions: -10 region (10 nucleotides going back from end of promotoer, first nucleotide transcribed is called +1, first nucleotide that RNA pol adds a compl nucleotide to) count 10 nucleotides back from this, -10 consensus sequence (TATA box, ta rich), go back 35 nucleotides, -35 consensus region (Both sequences conserved) + Why are they conserved? structure of RNA polymerase matches (has two binding points) on two sites + RNA polymerase engages double helix in two specific regions (-10 region and -35 region) < two consensus sequences are specifically where the proteins from the RNA pol will interact with DNA (high chemical affinity) - can mutate specific nucleotides and see how much transcription you get - the closer you are to a consensus , the easier the engagement of RNA pol with DNA (more transcription). FARTHER? RNa pol is harder to recognize them and transcription rate drops

Chromatosome?

nucleosome + Histone H1

Linker DNA

nucleosomes are separated by linker DNA stretch of DNA separting two nucleosomes

Do the table with Prok and Euk translation

ok!!

No advancement of helicase?

polymerases cannot come in, no replicaiton!

Example B (Ch 6)

translocation

Let's assume that the trait of bringing home shiny objects (H) is controlled by a single locus gene and is dominant to the trait of carrying home only dull objects (h). Suppose the two heterozygous individuals are crossed. How many of each genotype would be expected if only 4 offspring were produced?

¼ HH: ½ Hh: ¼ hh (3 bring home shiny objects, 1 is NOT!)

Purpose of DnaA

(initiator protein) binds to DnaA sites (R1, R2, R3, R4) (4 x 9-mers) DnaA interacts at these sites and wraps around DNA creating loops to relieve tension to melt 13-mers opens DNA so that helicase can attach/load onto DNA (creates replication bubble) open small segment of DNa so helicase can enter and load onto replication bubble

One SC is how many DNA molecules?

1 dsDNA molecule

Describe the compaction of DNA

1. 2nm fiber: simplest, chromatin is a double-stranded helical structure of DNA, DNA is complex with histones to form nucleosomes 2. 11nm fiber: the nucleosome, four histone core and 145bp.? 2. 30nm fiber: DNA wraps 2x around histone core, histones interact to create organized structure called 30nm fiber (nucleosomes form the 30nm fiber by folding up) 3. 300nm fiber: The 30-nm fiber forms looped domains with the help from different proteins 4. Chromosome from 300nm fiber: ultimately continues to compact with help from scaffolding proteins to make a chromosome, tight coiling produces the chromtid of a chromsome

Eukaryotic Genomes

1. Each chromosome (one molecule of linear, dsDNA) 2. usually diploid (2n) 3. Condensation state is dynamic (variable throughout the cell cycle) 4. DNA + proteins = chromatin 5. Euchromatin: gene rich = transcriptionally active, 'relaxes' during interphase 6. Heterochromatin: mostly devoid of transcription, remains highly condensed during interphase, maintains structural integrity/allows movement (centromeres and telomeres)

Mitosis Cell Division, DRAW and DESCRIBE

1. Interphase (nuclear membrane is present and chromosomes are relaxed) 2. Prophase (Chromosomes condense. Mitotic spindle forms, within each centrosome is the centriole which is composed of MT, some plant cells do not have centrosomes or centrioles but have a mitotic spindle) 3. Prometaphase (nuclear membrane disintegrates, Spindle MT enter nuclear region and attach to chromatids, when the end of a MT encounters a kinetochore the MT becomes stabilized, each chromosome becomes attached to the MT from opposite poles of the spindle) 4. Metaphase: (Chromosomes line up on the metaphase plate, centrosomes are now at opposite ends of the cell, a spindle-assembly checkpoint ensures that each chromosome is aligned on the metaphase plate and attached to spindle MT from opp. poles) 5. Anaphase: (Sister chromatids separate and move toward opposite poles becoming individual chromosomes) 6. Telophase: (Chromosomes arrive at spindle poles. The nuclear membrane re-forms and the chromosomes relax producing two separate nuclei within the cell) 7. Cytokinesis: (separation of the cytoplasm, forming two identical daughter cells, in plant cells the cell forms)

What are the different types of sequence classes?

1. Interspersed Repeats 2. Blocks of Tandemly Repeated Sequences 3. Low Copy Repeats or Segmental Duplications 4. Simple Sequence Repeats 5. Unique Sequence Repeats

Describe in detail DNA Initiation

1. Starts at the OriC (origin of replication for E. Coli Cell), it is 245 bp that signals for DnaA so it can bind 2. DnaA (initiator protein) binds to DnaA sites (R1, R2, R3, R4) (4 x 9-mers), DnaA interacts at these sites and wraps around DNA creating loops to relieve tension to melt 13-mers, opens DNA so that helicase can attach/load onto DNA (creates replication bubble), 13-mers are rich in A and T bases (2 hydrogen bonds vs. C and G 3 H bonds) which allows for easy melting or separation (breaking of H bonds) of DNA 3. DnaB (helicase) works with DnaC (loading protein) forming a DnaB/DnaC complex. Complex loads helicase onto the DNA strands 4. Helicase unwinds/separates DNA strands 5. SSBP attach to sstrand, stabilize, protein binds to ssDNA in ss form to keep DNA from reannealing due to high chemical reactions between strands, with help of SSBP helicase can work towards replication fork (stays and advances) 5. Once helicase is loaded, DnaA leaves!

How does an RNA nucleotide differ from a DNA nucleotide?

1. Sugars are different. RNA nucleotide has ribose sugar, DNA nucleotide has deoxyribose sugar. 2. RNA's sugar has a hydroxyl group attached to the 2' carbon. DNA's sugar has a hydrogen atom at its 2' carbon (one O fewer overall) 3. Both RNA and DNA have adenine and guanine. Cytosine is present in both RNA and DNA but thymine is only in DNA and uracil only in RNA. 4. In a nucleotide, the NB always forms a covalent bond with the 1' carbon atom of the sugar. A deoxyribose or a ribose sugar and a base together are called a nucleoside. 5. Phosphate groups found on every nucleotide and carry a negative charge, making DNA acidic. Always bonded to the 5'carbon of the sugar in a nucleotide. 6. RNA single stranded!! (NO CHARGAFFs RULE)

Bacterial Genomes (E. coli)

1. contained in the nucleoid region 2. dsDNA 3. Circular molecule 4. Plasmids with independent replication 5. Supercoiling helps pack DNA (less energy is required to melt DNA during replication and transcription, molecule occupies less space) 6. Nucleoid-associated proteins (NAPs in E. Coli) help compact the DNA molecule 7. VARIABILITY

Describe the structure of a DNA molecule

1. double helix, 2 antiparallel strands combine 2. every nucleotide has one pentose sugar (5 carbons), a phosphate group, nitrogenous base, and an oxygen atom 3. Collection of nucleotides (phosphodiester bond forms between 5' carbon of one nucleotide and the 3' carbon on another nucleotide) 4. sugar phosphate backbone creates double helix `. nitrogenous bases form hydrogen bonds 5. Nucleotides lie flat stacked on each other 6. 3 bonds with C and G, 2 bonds with A and T 7. width between sugar phosphate backbone is consistent 8. DNA and RNA have a negative charge because of phosphate groups 9. ALWAYS 3 rings in center (A > T, G > C, A > U)

DNA Replication (Termination)

1. in some DNA molecules, replication is terminated when two replication forks meet 2. in others, specific termination sequences (Ter sites) block further replication 3. a termination protein, called Tus in E. Coli binds to these equences creating a Tus-Ter complex that blocks the movemnt of helicase stallling the replication fork and preenting further DNa replication 4. each tus-ter complex blocks a replication fork moving in one direction, but not the other 5. tus protein binds terminator seq (7 X 20mers), stops the replisome (not letting each onme pass, slodes down mechanisms) 1. Nonpermissive face of Tus (indent) = DnaB binds to Tus and unwinding is arrested 2. Permissive face of Tus (square) = Tus is dislodged from Ter and replication continues 6. two circular rings end up intertwined called Catenanes 7. topoisomerase IV seperates Catenanes so they can migrate into two daughter cells through binary fission 8. as the fork goes, TUS gets out of the way

1. Interspersed Repeats

1. not repeated in tandem, scattered throughout the genome 2. Transposable elements (DNA seq. capable of moving from one genomic site to another, sequences that can multiple and move) 3. For ex., (a) SINES: short interspersed elements) = Alu family, 300 bp (b) LINES: long interspersed elements = L1 family

Viral Genomes

1. usually have only a single DNA or RNA molecule 2. RNA/DNA is largely devoid (lacking) of associated proteins 3. Circular OR Linear 4. Single or double stranded nucleic acid 5. Can become circular upon infection or not

3 x 13-mers

13bp sequences, rich in A and T bases (2 hydrogen bonds), allows for easy melting/separation (breaking hydrogen bonds) of DNA

How many centromeres does a replicated chromosome have?

2, a centromere is the region where spindle fibers attach to the Sister chromatids via the kinetochores, DNA sequence (specialized) of a chromosome that links a pair of SC

What is the prime on #1 end and #2 end?

3' end has a OH group, 5' end has a P group

Draw a short segment of a single DNA polynucleotide strand, including at least three nucleotides. Indicate the polarity of the strand by identifying the 5' end and the 3' end.

5'end: a free phosphate group is attached to the 5'carbon atom of the sugar in the nucleotide 3'end: free OH group attached to the 3'carbon atom of the sugar. RNA also connected by phosphodiester linkages!!!!

Which of the following relations will be true for the percentage of bases in doublestranded DNA? a) C + T = A + G b) C/A = T/G

A is true

Semiconservative Replication

Called SCR because each of the original nucleotide strands remains intact (conserved) despite no longer being combined in the same molecule; the original DNA molecule is half (semi) conserved during replication One old parental strand is used to synthesize a new DNA strand These two strands combined make up a new DNA molecule Consists of one parental and new daughter strand

Imagine Picture of Chromosome 3. One copy from mom and one from dad, both replicated. How many chromosomes? Pairs of homologous chromosomes? Unrep or Rep? SC? DNA molecules?

What four general characteristics must the genetic material posses?

Contain complex information- instructions for the traits of an organism Replicate faithfully- instructions must be copied with fidelity for descendants Encode the phenotype- genotype must have capacity to be expressed as phenotype Have the capacity to vary- different species and even in species, there is some variability

Chromatin

DNA + proteins

Explain the process by which a fully uncoiled 2nm fiber is coiled into a metaphase chromosome.

DNA is complexed with histones to form nucleosomes. A chromatosome consists of a nucleosome plus the H1 histone. The nucleosome fold up to produce a 30nm fiber, and that forms loops averaging 300nm in length. The 300nm fibers are compressed and folded to produce a 250nm wide fiber, and tight coiling of the 250nm fiber produces the chromatid of a chromosome.

Draw a table listing the differences and similarities between the Viral, Bacterial, and Eukaryotic Genomes.

DNA or RNA? ss or ds? Viral: DNA or RNA, ss or ds Bacteria: dsDNA Eukaryotes: dsDNA Relative Genome Size? Viral: smallest (200kb) Bacteria: middle (10Mb) Eukaryote: largest (900Gb) Ploidy Level? Viral: 1n (haploid) Bacterial: 1n (haploid) Eukaryote: 2n (diploid) Genome-associated Proteins? Viral: none Bacterial: NAPs Eukaryote: Histones, scaffolding proteins, kinetochore, telosome, Presence of Introns? Viral: none Bacterial: none Eukaryote: yes Presence of Repetitive DNA? Viral: none Bacterial: very little Eukaryote: abundant Chromosome Structure? Viral: circular or linear Bacterial: linear Eukaryote: circular Location of Genome? Viral: in the capsid (shell) Bacterial: in the nucleoid region of the cytoplasm Eukaryote: nucleus Other? Viral: N/A Bacterial: supercoiling, plasmids Eukaryote: mtDNA, clDNA, chromatin (euchromatin and heterochromatin)

Euchromatin vs Heterochromatin

Euchromatin: chromatin that undergoes condensation and decondensation in the course of the cell cycle, gene rich = transcriptionally active & high concentration of genes expressed, "relaxes" when cell comes out of cell division (interphase) Heterochromatin: chromatin that remains in a highly condensed state throughout the cell cycle; found at the centromeres and telomeres of most chromosomes, mostly devoid of transcription (not gene rich), maintains structural integrity/allows movement (centromeres and telomeres)

How does the different genome size (viral, bacterial, eukaryotic) affect replication?

Eukaryotic: linear (problem w/telomeres) multiple replication bubbles (large genome), several Ori to complete DNA replication, Telomeres: upon removal of RNA primer, there is nowhere for DNA polymerase I to bind upstream to replace those nucleotides (linear chromosome become shorter) > we need telomeres! In prokaryotes, this problem is solved by having circular DNA molecules as chromosomes. Prokaryotic: SINGLE Ori from small genome, use supercoiling instead of histones to compact DNA, have NAPs to compact DNA (nucleoid associated proteins) Eukaryotic DNA is complexed with nucleosomes: A structural unit of a eukaryotic chromosome, consisting of a length of DNA coiled around a core of histones

Bacterial and eukaryotic genomes are housed in the nucleus of the cell.

False, ONLY eukaryotic genomes. Bacterial in the nucleoid region (cytoplasm)

Viral genomes are housed in the nucleoid, a region that exists in the capsid.

False, bacterial genomes. Capsid is the the protein shell of a virus. It consists of several oligomeric structural subunits made of protein called protomers.

How did Hershey and Chase show that DNA is passed to new phages in phage reproduction?

First clue that DNA was the carrier of hereditary information came with the demonstration that DNA was responsible for transformation. Avery and his colleagues succeeded in isolating and partially purifying the transforming substance. They showed that it had a chemical composition closely matching that of DNA and quite different from that of proteins. Enzymes, such as trypsin and chymotrypsin (known to break down proteins) had no effect on the transforming substance. Ribonuclease, capable of destroying DNA, however, eliminated the biological activity of the transforming substance. They showed that the transforming substance precipitated at about the same rate as purified DNA and that it absorbs ultraviolet light at the same wavelengths as DNA. These results showed that the transforming principle--and therefore genetic information--resides in DNA.

Explain the cell cycle in detail. (Excluding details of Mitosis)

G1: cell grows and develops with nutrients, prepares for cell division S phase: replication of DNA G2: preparation for division, organelles replicate, cell keeps growing G0: rest phase, cell exits the cell cycle because it does not need to prepare for cell division

Haploid vs. Diploid

Haploid: reproductive cells (eggs, sperm, and spores), and even nonreproductive cells in some eukaryotic organisms, contain a single set of chromosomes and are haploid, gametic cells Diploid: cells that carry two sets of genetic information are diploid (somatic cells)

5 Types of Histones. What charge?

Histones have a + charge! H1 > Outside of core, after DNA is wound, histone H 1 attaches to keep everything in place, helps to create 30nm fiber 2 x (H2A, H2B, H3, H4) > make up the core!

Differences between Meiosis and Mitosis

Meiosis: 1. Occurs in gametes (haploid) 2. CANNOT cycle. Leads to terminally differentiated cells. 3. two cell divisions 4. causes chromosome # in the newly formed cells to be reduced by half (2n to n) 5. produces genetically variable cells due to recombination and random alignment of chromosomes during M I and II defines segregation of chromsomes in anaphase I and II 6. product is 4 cells (some exceptions) Mitosis: 1. consists of a single nuclear division and is usuallt accompanied by a single cell division 2. Occurs in somatic cells (diploid), for the body to repair damaged tissue and grow 3. chromosome # in the newly formed cells is the same as the original cell 4. produces genetically identical cells

When do chromosomes condense in the cell cycle?

Mitosis, more specifically prophase

Octomer Nucleosome Chromatosome

Octomer: 8 histone units and protein together Nucleosome: core particle consiting of DNA wrapped about two times around an octamer of 8 histone proteins (two copies of each H2A, H2B, HJ3, and H4), 146 bp DNA Chromatosome: nucleosome + Histone H1

Draw and identify the three parts of a DNA nucleotide.

Phosphate Group: consists of a phosphorus atom bonded to 4 oxygen atoms found in every nucleotide and carry a negative charge (making DNA acidic) always bonded to the 5' carbon atom of the sugar Pentose Sugar (5 carbon atoms) Ribose: hydroxyl group attached to the 2' carbon atom Deoxyribose: hydrogen atom attached to the 2' carbon atom (one oxygen fewer overall) Nitrogenous Base (Nitrogen-containing base) RNA: Uracil, Guanine, Adenine, and Cytosine DNA: Guanine, Adenine, Cytosine, and Thymine Purines: Guanine and Adenine, 2 rings Pyrimidines: Uracil, Thymine, and Cytosine, 1 ring

List the major characteristics of prokaryotic and eukaryotic chromosomes.

Prokaryotic Chromosomes: 1. single circular DNA molecule 2. NO histone proteins 3. exists as a series of large loops, not as an open and relaxed circle 4. found in the nucleoid (cytoplasm) 5. proteins help to compact it (NAPS > nucleoid associated proteins) Eukaryotic Chromosomes: 1. single, long molecule of DNA (linear) 2. bound to histone and non-histone chromosome proteins 3. euchromatin and heterochromatin

Name the Pyrimidines and Purines. How many rings?

Pyrimidines: 1 ring, Uracil, Thymine, Cytosine Purines: 2 rings, Adenine and Guanine

How does Meiosis produce genetically different cells?

Recombination: (crossing over) DNA strands of NON SISTER CHROMATIDS break and are then repaired. Allows allele "shuffling" Random alignment of chromosomes during metaphase I and II defines segregation of chromosomes in anaphase I and II (random distribution of maternal and paternal chromosomes into daughter cells)

Is DNA replication necessary for gene expression (transcription and translation)?

Replication is the process by which DNA is copied. At the start of replication you have one string of DNA, at the end of replication you have two strings of DNA. Nothing else is produced. In protein synthesis, you have multiple steps and different intermediates. You start with DNA. The first step is transcription, which copies the code from the DNA onto mRNA. You now have your starting DNA molecule (which you don't need any more, and gets put back in the box or wherever it came from) and an mRNA molecule. The mRNA molecule is then used to produce protein in a process called translation. So when you've finished, you still have the DNA, you have a used mRNA transcript (which is broken down to reuse the nucleotides), and your finished protein. Again, the process is: DNA > Transcription > mRNA > Translation > Protein. There is a bit of a wrinkle when you ask the question "does translation need to occur before replication?" Well, the processes aren't strictly connected, but the process of replication is performed by enzymes, which are proteins. These proteins had to come from somewhere - so there must have been *some* translation happen at some point before transcription can happen!

When do chromosomes duplicate? Why?

S phase to being mitosis or meiosis

Explain how the telomeres are protected from degradation.

Telomeres need to be protected from a cell's DNA repair systems because they have single-stranded overhangs, which "look like" damaged DNA. The overhang at the lagging strand end of the chromosome is due to incomplete end replication (see figure above). The overhang at the leading strand end of the chromosome is actually generated by enzymes that cut away part of the DNA^1 . In some species (including humans), the single-stranded overhangs bind to complementary repeats in the nearby double-stranded DNA, causing the telomere ends to form protective loops. Each telomere sequence is always oriented with the string of Gs and Cs toward the end of the chromosome. The G-rich strand at the end of the chromosome, called the 3' overhang. Shelterin binds to telomeres and protects the ends of the DNA from being "reapireD" as a double strand break in the DNA. Some cells, the 3' overhang folds over and pairs with a short stretch of DNA to form a "t-loop", which also protects the end fo the telomere from degradation. Proteins associated with the telomere ends also help protect them and prevent them from triggering DNA repair pathways.

What are the three major pathways of information flow within the cell?

Transcription: DNA expresses its genetic instructions by first transferring its information to an RNA molecule in this process. Although information is transferred from DNA to RNA, the information remains in the language of nucleic acids. Translation: RNA molecule transfers the genetic information to a protein by specifying its amino acid sequence (information must be translated from the language of nucleotides into the language of amino acids) Replication: information passes from one DNA molecule to other DNA molecules

Bacterial genomes are haploid.

True

Histones are genome-associated proteins that can only be found in eukaryotes.

True

In a single nucleotide, the phosphate group is attached to the 5' carbon.

True

Viral genomes can be made of any single or double stranded nucleic acid

True

3. Low Copy Repeats or Segmental Duplications

Very high sequence identity (>95%) Can be on the same or different chromosomes Mediate genomic rearrangements (human diseases): can mediate non-allelic crossing over

Ploidy Number Viral: E. Coli: Eukaryotes:

Viral: 1n E. Coli: 1n Eukaryotes: 2n

DNA or RNA? ss or ds? Viral: E. Coli (Bacteria): Eukaryotes:

Viral: DNA/RNA, ss or ds E. Coli: DNA, ds Eukaryotes: DNA, ds

Other? Viral: E. Coli: Eukaryotes:

Viral: N/A E. Coli: Plasmids, Supercoiling Eukaryotes: mtDNA, clDNA, Highly Dynamic Condensation

Location of Genome? Viral: E. Coli: Eukaryotes:

Viral: inside capsid E. Coli: cytoplasm (nucleoid region) Eukaryotes: nucleus

Chromosome Structure Viral: E. Coli: Eukaryotes:

Viral: linear or circular E. Coli: circular Eukaryotes: linear

Genome associated proteins Viral: E. Coli: Eukaryotes:

Viral: no E. Coli: NAPs Eukaryotes: Histones, scaffold proteins, kinetochores, telosome, chromatin (euchromatin, heterochromatin)

Presence of introns Viral: E. Coli: Eukaryotes:

Viral: no E. Coli: no Eukaryotes: yes

Presence of repetitive DNA Viral: E. Coli: Eukaryotes:

Viral: no E. Coli: very little Eukaryotes: abundant

Telomere

a region of repetitive nucleotide sequences at each end of a chromosome, which protects the end of the chromosome from deterioration or from fusion with neighboring chromosomes. provide stability (prevent attachment and degradation) has a specific repetitive sequence that is highly conserved human telomeres (10-15Kb), (T loop: 3' G rich single strand segment about 200bp, 3' overhang, that rethreads itself back into the double helix on 3' end, proteins attach (shelterin/telosome) to protect telomeres from DNA repair mechanisms Protect and stabilize the chromosome ends.

2. Blocks of Tandemly Repeated Sequences

a. large segments on chromosomes Are substrates: 1. Centromeres (can be heterochromatic) 2. Telomeres (can be heterochromatic) 3. rRNA genes (rDNA clusters) 4. Repetitive DNA sequence (fish) 5. short arms of acrocentric chromosomes and ribosomal gene clusters

A diploid plant cell contains 2 billion base pairs of DNA. a. How many nucleosomes are present in the cell? b. Give the number of molecules of each type of histone protein associated with the genomic DNA.

a. one nucleosome = 8 histones w/146 bp DNA + 20-22 bp of DNA associated w/H 1 protein + 30-40bp of linker DNA 2 billion/200 = ten million nucleosomes Each nucleosome has 2 X (H2A, H2B, H3, and H4) ten million molecules of each SINGLE H2A, etc. Therefore, ten million X 2 for answer: 20 million molecules each of H2A, etc. Each nucleosome has one histone H1 with it so ten million times 1 = ten million molecules of H1

A DNA molecule 300bp long has 20 complete rotations. This DNA molecule is a. positively supercoiled b. negatively supercoiled c. relaxed

b. negatively supercoiled relaxed would have about 30 rotations (10 bp/rotation), so only 20 rotations means it is under-rotated and negatively supercoiled DNA has approx. 10 bp per turn so 100 BP would have 10 turns

Unique sequence DNA (repeats)

consists of sequences that are present only once or, at most, a few times in the genome (includes sequences that encode proteins as well as a great deal of DNA whose function is unknown). genes that are present in a single copy constitute roughly 25% to 50% of the protein encoding genes in most multi cellular eukaryotes

What is a phosphodiester linkage?

covalent bond that links the 5'phosphate group of one nucletodie to the 3'-hydroxyl group of the next nucleotide (series of nucleotides linked this way is called a polynucleotide strand

Simple Sequence Repeats

direct replication of short k-mers Tandem repeats: minisatellites (VNTRs): 15-100bp each unit microsatellites (STRs): 2-6bp each unit DNA fingerprinting, paternity tests, human identification, population genetics

Centromere

essential for disjunction primary constriction on chromosome (DNA thats methylated becomes tight) specific repetitive sequence per species attachment point for spindle MT, which are the filaments responsible for moving chromosomes during cell division joins the sister chromatids attaching the microtubules in the mitotic spindle. In this function, the centromere directs the formation of the kinetochore, which is a special protein structure that attaches to the microtubules in the mitotic spindle.

growth and devlelopment of the cell, G1/S checkpoint

Chromosome bi-orientation

occurs in prometaphase for each chromosome, a MT from one of the centrosomes anchors to the kinetochore of one of the SC, and a MT from the opposite centrosome then attaches to the other SC, anchoring the chromosome to both of the centrosomes arrangement is known as chromosome bi-orientation

Draw the cell cycle. Label where chromosomes condense. X and Y chromosomes DO form ____, but NOT ____. Spindle fibers push chromosomes _______. This ________.

ok they do pair with each other, homologous pairs to center on metaphase plate, resolves synaptonemal complex.(A protein structure that forms between homologous chromosomes (two pairs of sister chromatids) during meiosis and is thought to mediate chromosome pairing, synapsis, and recombination.

Replication and Division go hand in hand. Why? Connection between DNA replication and the cell cycle.

purpose of DNA replication is so cells can be divided. 2 daughter cells > need a set of instructions no division? no purpose for replication

S phase

synthesis of DNA

Recombinant DNA

the exchange of genetic material between different organisms which leads to production of offspring with combinations of traits that differ from those found in either parent. In eukaryotes, genetic recombination during meiosis can lead to a novel set of genetic information that can be passed on from the parents to the offspring.

Are SC identical? X and Y chromosomes pair so they can what?

yes i swear if you miss this on the exam its over for you segregate appropriately during meiosis. Facilitated by identical sequence in the telomeric pseudoautosomal regions

Genetics Final

Kaugnay na mga set ng pag-aaral

Ace world history pace 103 self test

Speech final thing

Milk

True False Math Review

Chapter 35 Integrative Care

CIS 150 chapter 1

ACIS

BIO 1201 - MORONEY - EXAM 1 SAMPLE/QUIZ 1

EGEE 102 Lesson 6

NUR220 - Mental Health Questions

Equal Credit Opportunity Act (ECOA)

smbook 6

Episode 1- The First 8 Weeks

Patho E2

info security ch 1

Microbiology- Chapter 13

CMST204D

Chemistry: 14 T

ECON 347

The Cardiovascular System: The Heart