Unit 3 Biochemistry
🇬🇧
In English
In English
Practice Known Questions
Stay up to date with your due questions
Complete 5 questions to enable practice
Exams
Exam: Test your skills
Test your skills in exam mode
Learn New Questions
Popular in this course
Learn with flashcards
Manual Mode [BETA]
Select your own question and answer types
Other available modes
Complete the sentence
Listening & SpellingSpelling: Type what you hear
multiple choiceMultiple choice mode
SpeakingAnswer with voice
Speaking & ListeningPractice pronunciation
TypingTyping only mode
Unit 3 Biochemistry - Leaderboard
Unit 3 Biochemistry - Details
Levels:
Questions:
127 questions
| 🇬🇧 | 🇬🇧 |
| Inside the nucleus, the DNA is tightly wound around proteins called? | Inside the nucleus, DNA is tightly wound around proteins called histones until they become nucleosomes |
| What happens when the lac 1 protein is removed from a prokaryotic bacteria? | With the lac 1 protein removed, RNA polymerase binds to the promotor region, causing transcription to occur. While transcription occurs, the enzyme B-galactosidase can be produced, it's job being to break down lactose into simple sugars (glucose and galactose) |
| What are the basic units that make up DNA molecules? | Nucleotides are the basic unit makeup of DNA |
| What happens when the lac 1 protein is removed from a prokaryotic bacteria? | With the lac 1 protein removed, RNA polymerase binds to the promotor region, causing transcription to occur. While transcription occurs, the enzyme B-galactosidase can be produced, it's job being to break down lactose into simple sugars (glucose and galactose) |
| What are the 3 components that nucleotides are made of? | Phosphate group, nitrogenous base, and a five carbon sugar (deoxyribose) |
| How many nucleotides does DNA contain? Using this information, explain why nucleotides (basic units that makeup DNA) are different. | DNA contains 4 different nucleotides; nucleotides in of themselves are different because they have different nitrogenous bases |
| What are purines, and what are pyrimidines? | Adenine (A) and Guanine (G) are purines. Purines are doubled ringed structures. Thymine (T) and Cytosine (C) are pyrimidines. Pyrimidines are single ringed structures. |
| DNA is a helical shape, how do the strands in DNA stay linked together? List the pairs that form between nitrogenous bases. | Strands in DNA have nitrogenous bases (G, A, C, T) that pair with the bases in the other strand. Adenine is paired with Thymine (A-T) Guanine is paired with Cytosine (G-C) |
| What is Chargaff's rule? How does this help scientists? | Chargaff's rule is that if you know one strand's nucleotide (deoxyribose) sequence, then you can deduce the sequence of the other strand as well, since each nitrogenous base can only pair with one other. This helps scientists find the suitable pair to a strand of DNA. |
| What are base pairs, how are they held together in DNA? | Base pairs are 2 complementary nucleotides (deoxyribose) that pair their nitrogenous bases (GACT) together. These pairs are held together by hydrogen bonds |
| How many hydrogen bonds do the nitrogenous bases thymine and adenine have when paired? How many hydrogen bonds does cytosine have when paired with guanine? | Adenine and Thymine have 2 hydrogen bonds when paired together. Guanine and Cytosine have 3 hydrogen bonds when paired together. |
| What was the model propose by James Watson and Francis Crick about the direction nucleotides (deoxyribose) form in strands of DNA? | The model that was proposed by Hames Watson and Francis Crick proposed that nucleotides (deoxyribose) form 2 anti-parallel strands in DNA. |
| What makes strands in DNA anti parallel to each other? | Strands in DNA are anti-parallel because they're upside down in relation to each other. The 5th carbon in deoxyribose points upward in one strand, and downward in the other |
| What links the sugar in each strand (DNA) together? | In DNA, each strand's sugar(s) are linked together by a phosphate group, these strands then are held together by complimentary nitrogenous bases (GACT) through hydrogen bonds. |
| In DNA, the 2 strands of nucleotides (deoxyribose) are twisted into a right handed helix, when does it make its complete turn? What is the backbone of these 5' - 3' and 3' - 5' orientated strands? | The 2 strands of nucleotides in DNA are twisted together into a right handed helix, making its complete turn every 10 nucleotides (a distance of roughly 3.4 nm) Its backbone is made up of sugar phosphate, in both 5' - 3' orientated strands, and 3' - 5' orientated strands (which are anti-parallel) |
| DNA is kept in cells, however, if laid out end to end, it would form a 3 meter line, and be made up of 6 billion base pairs, how do our cells fit this massive amount of information into each cell? What are histones, and how do they help DNA fit into a cell? | DNA is extremely long, and so our body's cells choose to coil our DNA strands upon itself until it takes on the shape of a chromosome. But for our DNA to coil, it needs something to coil upon; our body uses + charged proteins called histones as "spools" for our DNA molecules to make coils. |
| What are histones and nucleosomes, how do they help our DNA coil? | Histones are + charged proteins that our bodies use to coil DNA upon; nucleosomes (coil) is what is coiled upon itself to form a chromatin fibre, which later condenses during mitosis into chromosomes |
| Compare and contrast the following pairs of terms: Purine / Pyrimidine: Ribose / Deoxyribose: Histones / Nucleosomes: | Purine / Pyrimidine: Purine - Larger, double-ring structure. Pyrimidine - Smaller, single-ring structure. Both are types of nitrogenous bases in DNA and RNA. Ribose / Deoxyribose: Ribose - Sugar in RNA, has an extra oxygen. Deoxyribose - Sugar in DNA, lacks one oxygen compared to ribose. Both are part of the nucleotide structure. Histones / Nucleosomes: Histones - Proteins that help DNA coil and pack. Nucleosomes - DNA wrapped around histones. Both involved in DNA compaction and organization in the cell. |
| If the nucleotide sequence on one side of the DNA molecule is GTCATG, what would be the sequence of bases complimentary to this? | (A-T) and (G-C) CAGTAC |
| What are the sides of the DNA “ladder” made of? | The sides of the DNA "ladder" are like a handrail made of alternating sugar and phosphate molecules. They give the DNA its structure. Imagine these sugar-phosphate handrails on a staircase, and the steps are the pairs of nitrogenous bases. |
| Inside the nucleus, the DNA is tightly wound around proteins called? | Inside the nucleus, DNA is tightly wound around proteins called histones until they become nucleosomes, and then chromatins, and then chromosomes |
| A beadlike structure formed of several histones with corresponding DNA. They pack closely together to super coil the DNA. | The beadlike structure formed of several histones with corresponding DNA is called a nucleosome. Nucleosomes pack closely together to supercoil and organize the DNA within the cell's nucleus. |
| If 20% of the nucleotides in a DNA molecule are adenine, what percentage of each of the other three bases would be found in this molecule? Explain your answer. This is known as? | If 20% of the nucleotides in a DNA molecule are adenine, according to Chargaff's rules, the complementary base-pairing in DNA dictates that the percentage of thymine (T) will also be 20%. Similarly, since adenine pairs with thymine, the percentages of cytosine (C) and guanine (G) will be equal. Therefore, both cytosine (C) and guanine (G) would each make up 30% of the nucleotides. This principle is known as Chargaff's rules, which state that in a DNA molecule, the amount of adenine is equal to the amount of thymine, and the amount of cytosine is equal to the amount of guanine. |
| T/F There are five nitrogen bases in RNA There are five nitrogen bases in the DNA molecule | Both are false; RNA has (GACU) and DNA has (GACT). Neither have (GACTU) |
| Across the center of the DNA helix, a purine must always bond to a pyrimidine. | Yes, true, a purine must always bond to a pyrimidine |
| Before mitosis, cells double their genetic information to ensure that each daughter cell has identical genetic information as the parent cell, where do the nucleotides come to assemble a copy of the parents DNA? What are the 2 models that explain how this happens? | In order for the parent cell to have a copy of its DNA, free floating nucleotides (deoxyribose) in the nucleus are assembled. 2 possible methods explain how this happens: Model #1: Conservative Replication Model #2: Semi-conservative Replication |
| What does conservative replication (for DNA) say? | Conservative replication says that one daughter DNA molecule is made entirely from free floating nucleotides, and so, the parental DNA remains unchanged/conserved |
| What does semi-conservative replication say? | Each daughter DNA molecule has one strand of the parental DNA's nucleotides (deoxyribose), and one strand made from free-floating nucleotides. Thus, conserving half of the parental DNA's material in each daughter DNA molecule, semi conserved. |
| How was semi conservative replication proven? | 2 scientists used cultures of e. coli bacteria, and were given nutrients with either heavy nitrogen (15 N) or a light nitrogen (14 N). After the bacterias used the nitrogens as nitrogenous bases, they became heavy and light; now they can be compared. The bacteria were allowed to reproduce with the heavier bacteria twice. What was discovered was that the parental bacterai was at the bottom of the tube, then gen 1 was in the middle, and by gen 2 there were 2 layers: one lighter than the other. If it was conservative replication, 1 half would have been a heavy band and another a light band since the original would've still existed. |
| What must happen before DNA replication can occur? What must be used? | Before DNA replication can occur, highly coiled and condensed DNA molecule(s) must be straightened out into a linear sequence of nucleotides. An enzyme must be used in order to uncoil/unwind the DNA. |
| [DNA replication] Step 1: | As DNA uncoils and unwinds, a class of enzymes called topoisomerases relieve tension on the molecule/DNA |
| [DNA replication] Step 2: | Hydrogen bonds that hold the 2 DNA strands together are broken by DNA helicase; producing a replication fork |
| [DNA replication] Step 3: | Single-stranded binding proteins (SSBP) attach to each DNA strand and prevent hydrogen bonds from re-forming |
| [DNA replication] Step 4: | RNA primase attaches RNA nucleotides (primers) to 3' (3 prime) end of each strand. Primers are just a base to build things off |
| [DNA replication] Step 5: | Starting at the primers, DNA polymerase 3 III adds complimentary DNA nucleotides to each strand. DNA polymerase III constructs a new strand in 5' - 3' direction only. The other strand (lagging) is built discontinuously away from replication fork; it's made up of short pieces called okazaki fragments. |
| [DNA replication] Step 6: | DNA polymerase 1 replaces RNA primers with DNA, DNA ligase joins okazaki fragments. |
| [DNA replication] Step 7: | Error correction time: Most errors occur during step 5 when DNA polymerase III is adding complimentary DNA nucleotides. Occasionally, DNA polymerase III will make a mismatch error; usually, an enzyme backs up and corrects this error before moving on. About 1 error in a million base pairs are not corrected by DNA polymerase III |
| [DNA replication] Step 7 cont | Special DNA repair complexes made of proteins and enzymes (including DNA polymerase 1 & DNA polymerase II) remove a small section of nucleotides around the error. This missing section is replaced by DNA polymerase III, and a missing phosphodiester bond of last nucleotide is joined by DNA ligase. Just to reiterate: DNA polymerase III constructs a new strand in a 5' 3' direction |
| What type of bond connects one nucleotide to the next nucleotide? | The type of bond that connects one nucleotide to the next nucleotide in a DNA strand is called a phosphodiester bond. These bonds form between the phosphate group of one nucleotide and the sugar molecule of the next nucleotide, creating a sugar-phosphate backbone in the DNA structure. |
| During replication, the two nucleotide chains .... and each chain will serve as a .... for the production of a new DNA chain. The sites where new nucleotides are added are called .... | During replication, the two nucleotide chains separate, and each chain will serve as a template for the production of a new DNA chain. The sites where new nucleotides are added are called replication forks. |
| When replication occurs, it takes place at multiple sites along the molecule called .... The first step in replication is to break the .... bonds between the .... This is accomplished by the enzyme ... | When replication occurs, it takes place at multiple sites along the molecule called origins of replication. The first step in replication is to break the hydrogen bonds between the nitrogenous base pairs. This is accomplished by the enzyme helicase. |
| The copying of the DNA molecule is a fairly reliable process. As new nucleotides are added, there is an error rate of about 1 in .... base pairs. The enzyme, .... proofreads the new strand and looks for errors. If a mistake is found, this same enzyme removes the erroneous .... and replaces it with another. After the proofreading, there is a final error rate of only 1 in .... base pairs. | The copying of the DNA molecule is a fairly reliable process. As new nucleotides are added, there is an error rate of about 1 in1,000,000 base pairs. The enzyme, DNA polymerase proofreads the new strand and looks for errors. If a mistake is found, this same enzyme removes the erroneous nucleotide and replaces it with another. After the proofreading, there is a final error rate of only 1 in10 billion base pairs. |
| What are two functions of DNA polymerase during replication? | DNA Strand Synthesis: DNA polymerase makes a new strand of DNA by adding complementary nucleotides to the template strand during replication. Proofreading and Repair: DNA polymerase corrects errors that occur during replication, to ensure the accuracy of the genetic code. |
| When replication is complete, two _____ copies of the DNA molecule have been produced and the cell is ready to begin _____. | When replication is complete, two identical copies of the DNA molecule have been produced, and the cell is ready to begin cell division (or mitosis). |
| What is the central dogma of genetics? | The central dogma of genetics: necessary genetic code for synthesizing different proteins in our body is stored in DNA.And each protein is coded for by a specific segment/part of the DNA called gene(s). |
| Where is DNA found? What constructs proteins, and where do they construct them? | DNA is found in the nucleus, and proteins are constructed by ribosomes in the cytoplasm |
| What is code from DNA transcribed into? What can ribosomes do? | Code from DNA is transcribed into molecules of mRNA (messenger RNA). Ribosomes can translate genetic code into a protein. |
| After code from DNA is transcribed into molecules of mRNA, where does that mRNA go? | The mRNA transcribed from a code of DNA goes into the cytoplasm to meet ribosomes. |
| Protein synthesis occurs in 2 steps, what are they? | #1: Transcription #2: Translation |
| During protein synthesis, what does RNA end up changing? | RNA ends up changing thymine for uracil base pair |
| What are the three major classes of RNA, and what is the function of each? | The three major classes of RNA are: mRNA, which carries genetic information stored in DNA out of the nucleus to be coded into proteins at a ribosome; rRNA, teams up with proteins to create the active parts of ribosomes, which play a crucial role in helping build peptides. and tRNA, which are small clover-leaf shaped RNA units that translate mRNA code into amino acids during translation at the ribosome. |
| Compare and contrast transcription and translation in terms of their purpose and location. | The purpose of transcription is to turn the genetic code found in DNA into RNA. In eukaryotes transcription takes place in the nucleus. After transcription, RNA exits the nucleus to find a ribosome in the cytosol where it codes during translation. In prokaryotes, transcription and translation both take place in the cytosol. Translation involves the formation of a peptide chain from mRNA code that is being read by tRNA molecules associated to the next amino acid in the growing peptide. |
| The sequence of a fragment of one strand of DNA is AATTGCATATACGGGAAATACGACCGG. | The transcription of the fragment sequence into mRNA is: UUAACGUAUAUGCCCUUUAUGCUGGCC. |
| Differentiate between a stop codon and a start codon. | A start codon initiates translation and codes for the amino acid methionine. A stop codon does not code for an amino acid and instead binds a factor that releases the mRNA and protein from the ribosome. |
| What is the wobble hypothesis? | The wobble hypothesis states that there is a redundancy built into the genetic code such that several different combinations of codons may code for the same amino acid. Usually the redundancy is in the third base of the codon. |
| Explain the benefits of having large non-coding regions at the ends of eukaryotic chromosomes? | Having non-coding sequences at the ends of eukaryotic chromosomes can prevent the shortening of chromosomes and the loss or damage of important genes |
| Predict the impact of losing DNA in the telomere region throughout the organism's life | The loss of DNA may lead to serious complications on the organism's life. The damage of DNA may cause apoptosis which could cause interference on proper bodily functions. It may also lead to life-threatening conditions like cancer if the DNA that was lost was involved with cell growth. |
| How does the length of an individual's telomeres compare with the length of the telomeres of an older individual of the same species? How does it compare with the length of telomeres of an older individual of a different species? | The telomeres of one individual will be longer than the telomeres of an older individual. The length of telomeres is different with every species, so the differences of length is undetermined until the speciation of the older individual is specified. |
| Why must telomerase activity take place in germ cells? | Germ cells must be able to continue replication to maintain genetic integrity from parent to off-spring across generations. |
| [Protein Synthesis Transcription] Step 1: | Initiation: Transcription begins "up-stream" of the gene at the promoter region on the 3' - 5' strand. The promoter is a specific sequence of DNA in nucleotides that RNA polymerase can recognize and bind to. |
| [Protein Synthesis Transcription] Step 2: | Elongation: DNA is unwound, exposing a template strand at the beginning of a gene. RNA polymerase synthesizes mRNA molecules in a 5' - 3' direction, through complementary base pairing with a DNA template (template strand is 3' - 5') |
| [Protein Synthesis Termination] Step 3: | Termination: RNA polymerase reaches the end of a gene, and encounters a termination sequence. So RNA synthesis stops, and mRNA & RNA polymerase are released |
| MRNA is modified in 3 ways before it leaves the nucleus, what are those 3 ways? | ->5' cap ->Poly-A tail ->Introns are removed |
| What happens during the 5' cap modification of mRNA? | Seven methyl guanosine ribonucleotides are attached to the 5' end of mRNA: helps the mRNA attach to the ribosome during translation |
| What happens during the Poly-A tail modification of mRNA? | A string of about 50-250 adenine ribonucleotides are added to the 3' end of mRNA: its purpose is to protect the mRNA from RNA digesting enzymes in the cytoplasm |
| What happens during the intron removal modification of mRNA? | MRNA strand is made of sections that code for specific proteins (exons) and those that don't (introns). Proteins called spliceosomes "cut out" introns, leaving only exons |
| List and describe the three stages of transcription. | 1. The three stages of transcription are: (1) Initiation: the process that allows RNA polymerase to bind to the DNA. The DNA promoter region contains a TATA box that allows the DNA to unwind and separate, allowing RNA polymerase to bind. (2) Elongation: the process in which RNA polymerase makes an RNA copy of the DNA template. (3) Termination: the process in which transcription stops and the mRNA is released. This is accomplished by various devices such as hair pin loops in the RNA and factors binding to the stop codon that prevent further RNA production. |
| What the 3 different types of RNA involved in translation? | MRNA - Messenger RNA tRNA - Transfer RNA rRNA - Ribosomal RNA |
| What does mRNA, tRNA, and rRNA do? | MRNA - copy of instructions for constructing a specific protein tRNA - delivers individual amino acids to ribosome for construction of protein rRNA - ribosomes consist of 2 subunits; each is a combination of rRNA and protein |
| What is the diagram/structure of tRNA? | -Single Stranded and Double Stranded sections -Unique anticodon (in this case AUA) -At the bottom of a tRNA and is a perfect match for the codon on the mRNA strand -Anticodon: 3 nucleotide segment that corresponds to codon on mRNA -Carries a specific amino acid on the 3' end (in this case tyrosine) |
| Picture of tRNA | TRNA picture |
| What is important to know about trNA? | ->61/64 codons specify an amino acid -> 3 are stop codons (UAG, UGA, UAA) -> Fewer than 61 tRNA required to deliver 20 amino acids (as little as 32) |
| What is the wobble hypothesis? | Pairing of the anticodon with the first 2 nucleotides of the codon is always precise, but there is flexibility in pairing with the third nucleotide of the codon |
| What are the key steps in the initiation of translation in eukaryotes and prokaryotes? | In both eukaryotes and prokaryotes, the key steps in the initiation of translation are the association an initiator methionine-tRNA with the small ribosomal subunit. The complex binds the mRNA at the 5' cap and scans for the AUG start codon. The large ribosomal subunit then binds, completing the ribosome, and translation proceeds. |
| What is the role of tRNA in translation? | 2. The role of tRNA in translation is to shuttle the appropriate amino acid into place in the growing peptide by recognizing the mRNA code. |
| Why is there not a specific tRNA molecule for each possible codon? | There is not a specific tRNA molecule for each possible codon because several codons code for the same amino acid. If one tRNA recognizes each amino acid, then it can work efficiently by recognizing all the codons that code for it. There are many more codons than there are amino acids, but a unique tRNA molecule is not needed for each one. |
| The wobble hypothesis states that there is increased flexibility in base pairing at the third nucleotide of some codons. Why does this not lead to frequent mistakes in the assembly of proteins? | This does not lead to frequent mistakes because mutations in the third nucleotide of a codon may not affect the binding of the proper tRNA anticodon. If there is an error in transcribing the third nucleotide in a codon, in most cases it will not affect the protein being produced. |
| Explain what occurs at the A, P, and E sites during the translation of mRNA into a polypeptide. | The A site is where the mRNA is recognized by the appropriate tRNA, which has an amino acid bound ready to add to the growing peptide chain. The P site is where translation is initiated by the association of the methionine tRNA with the mRNA and the ribosome. The P site is also the site of the formation of peptide bonds that are catalyzed by the polypeptide transferase, which uses the hydrolysis of GTP to drive the reaction. As the chain moves along, the empty tRNA exits the ribosome at the E site. |
| [Translation] Step 1: Initiation | TRNA with anticodon that corresponds to AUG (start codon) brings Met and forms a complex with small ribosomal subunit at P site. Complex binds to the 5' end of the mRNA and scans along until it reaches AUG start codon. Large ribosomal subunit binds to complete ribosomal. |
| [Translation] Step 2: Elongation | Ribosome moves along mRNA reading codons (3 bases). Next tRNA delivers amino acid to A site; Met cleaved from the tRNA and a peptide bond forms between Met and aa on tRNA in A site. tRNA with the polypeptide chain moves to the P site, and the empty tRNA moves to E site and is released |
| [Translation] Step 3: Termination | Occurs when A site of ribosome reaches a stop codon (UAA, UAG, UGA). A protein with a release factor will bind instead of a new tRNA, and ribosomal subunits will separate from mRNA until tRNA, release factor, and the polypeptide chain is released. |
| What is the function of a gene? | The function of a gene is to "code" for a protein(s) |
| If the function of a gene is to produce proteins, how do prokaryotic cells control when to produce and when not to? | E.Coli is an example of a prokaryote , it's a bacteria that lives in the large intestine of mammals, and it produces an enzyme called B-galactosidase (breaks down lactose into glucose and galactose). Because there won't always be milk in your large intestine, E.Coli developed a regulating expression of your gene called lac operon. Thus, prokaryotic cells use regulating expressions in gene to control the production of enzymes. |
| What does the regulating gene lac operon consist of? | - 3 genes that code for the enzyme B-galactosidase - A promotor region where the RNA polymerase must bind before transcription can occur - An operator region that overlaps with the promoter region - And a lac 1 protein that can bind to the operator and repress transcription by preventing RNA polymerase from binding with the promotor |
| If the regulating gene for prokaryotes, lac operon, is bound by a lac 1 protein to repress transcription by RNA polymerase, how does it get removed? | The lac 1 protein has a site meant to be bounded with lactose, when lactose (circle shape) comes, it bounds to the bottom of the lac 1 protein, bending the protein until it comes out of place; permitting transcription to occur. |
| What happens when the lac 1 protein is removed from a prokaryotic bacteria? | With the lac 1 protein removed, RNA polymerase binds to the promotor region, causing transcription to occur. While transcription occurs, the enzyme B-galactosidase can be produced, it's job being to break down lactose into simple sugars (glucose and galactose) After all the lactose molecules are broken down, the Lac 1 protein returns to its original shape, no longer bended, and binds back to the promoter region; preventing RNA polymerase from binding to the promoter and stopping transcription. |
| How does gene regulation in eukaryotes occur? How many levels are there? | There are 4 levels to gene regulation in eukaryotes: Transcriptional -> Controls whether or not a gene is transcribed from DNA to mRNA) Post-Transcriptional -> Changes that occur to the mRNA molecule after transcription -> ex. Introns removed and exon's spliced together Translational -> Controls how fast and often mRNA molecules are translated Post-Translational -> Affects rate at which a protein becomes "active" and how long it remains functional |
| Why do eukaryotes have a more complex system of gene regulation than prokaryotes? Use an example to explain your reasoning. | Eukaryotes have a more complex system of gene regulation than prokaryotes because eukaryotic gene expression requires more steps. Each level of gene expression—transcription, post-transcription, translation, and post-translation—has its own regulation. For example, transcription is regulated by modifications to histone proteins that remodel wrapped DNA, making it accessible. |
| How does the lac operon regulate the production of the enzymes needed to metabolize lactose? Summarize the key ideas and concepts in list format. | The lac operon regulates the production of the enzymes needed to metabolize lactose in the following ways: • In the absence of lactose, the upstream lacI gene synthesizes the lac repressor protein, which binds to the lac operator to stop RNA polymerase from binding and stop the transcription of the lacZ, lacY, and lacA genes that code for the proteins involved in lactose metabolism. • When lactose is made available to the cell, some of it binds to the lac repressor protein, inactivating it so that it cannot bind to the operator. • This allows RNA polymerase to bind to the promoter, and allows the transcription of the downstream lacZ, lacY, and lacA genes. • Translation of the mRNA produces the three lactose-metabolizing enzymes. • If lactose is removed from the cell’s environment, the repressor will become active and will bind to the operator, preventing further transcription of the lactose metabolizing genes. |
| Give an example of a eukaryotic regulatory mechanism that occurs: (a) during transcription (b) during translation (c) after translation | (a) A regulatory mechanism that occurs during eukaryotic transcription is methylation of cytosine bases in the promoter region; this silences the gene and prevents transcription. (b) A regulatory mechanism that occurs during eukaryotic translation is the addition of a variable-length adenine tail. (c) A regulatory mechanism that occurs after eukaryotic translation is the chemical modification of some of the residues on the protein to render it active. |
| What are mutagens? | Environmental agents that directly alter DNA within a cell, like how radiation, chemicals, or viruses do |
| How are disorders inheritated? | Inherited genetic diseases are recessive, so a person must have 2 copies of the mutated gene to have the disorder. |
| What're 2 types of mutations? | Small scale mutations: -> Mutations that involve changes to individual base pairs or small groups of base pairs; minor changes but can still have serious issues Large scale mutations: ->Mutations that involve changes in chromosomes and may involve many genes. |
| What are small scale mutations? What's changed? Are they harmful? | Small scale mutations are called point mutations. These point mutations target single nucleotides within a gene, and can either be either beneficial, neutral, or harmful |
| What are the 3/4 types of small scale mutations? | Substitution ->Replacement of one base pair by another in a DNA sequence Insertion -> Addition of a base pair to a DNA sequence Deletion -> Removal of a nitrogenous base pair in a DNA sequence Inversion -> Two adjacent bases trade places It is important to note that insertion and deletions cause frameshifts |
| What are frameshifts? Which types of small scale mutations cause frame shifts? | Frameshifts are a shift in the reading frame, resulting in multiple missense and/or nonsense mutations. Insertion and deletions cause frame shifts. |