Question

Mutations in the promoter can affect termination of transcription. T or F 2. Primase has 3'...

Mutations in the promoter can affect termination of transcription.


T or F

2.

Primase has 3' to 5' exonuclease activity


T or F

3.

Always termination of transcription requires the transcription of the terminator sequence.

True

False

4.

Why do longer double-stranded molecules of DNA melt at higher temperatures than longer double-stranded molecules of DNA?

Because longer double-stranded molecules of DNA have more covalent bonds than shorter double-stranded molecules of DNA

Because longer double-stranded molecules of DNA have more hydrogen bonds than shorter double-stranded molecules of DNA

0 0
Add a comment Improve this question Transcribed image text
Answer #1

1.

Promoter helps in the binding of transcription factors and initiation of transcription at the appropriate sites of DNA. Mutation in the promoter region causes transcription inhibition. Transcription termination has to occur only after initiation. As initiation do not occur, termination will also not occur.

It is appropriate to state that mutation in promoter region causes transcription inhibition.

So, the statement is false.

2.

RNA primase catalyzes the synthesis of RNA primers, so that the growing chain continues on RNA primer. This is vital in initiation of replication. Later, the RNA primers are removed by 5’to 3’ exonuclease activity.

However, primase do not have exonuclease activity.

So, the statement is false.

3.

The termination sequence acts as signal for termination of transcription. After the transcription of the termination sequence, which is highly specific, there occurs a loop formation due to the ‘U’ and GC-rich region on the transcribed sequence.

For termination to occur, the termination sequence gets transcribed.

So, the statement is true.

4.

The melting temperature of DNA (Tm) depends on length and specific nucleotides of a DNA molecule. With increase in length, the hydrogen-bonds in the molecule increase, causing it difficult to melt the DNA.

Thus, longer ds DNA molecules melt at a higher temperature than shorter ds DNA molecules, due the high number of hydrogen bonds in the longer strand.

Add a comment
Know the answer?
Add Answer to:
Mutations in the promoter can affect termination of transcription. T or F 2. Primase has 3'...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • In rho-dependent transcription termination: the formation of a hairpin in the transcribed mRNA causes RNA polymerase...

    In rho-dependent transcription termination: the formation of a hairpin in the transcribed mRNA causes RNA polymerase to pause, facilitating termination. rho binds the mRNA, and when it makes contact with RNA polymerase, it assists with the removal of the mRNA from the DNA template. the rho factor binds to the -10 consensus sequence located in the promoter region to terminate transcription. a site within the poly(A) tail is cleaved which signals termination. the 3' untranslated region (3" UTR) is synthesized....

  • t sequence of enzymes necessary for DNA replication are he Meli Primase, Topoisomerase. b) Helicase To...

    t sequence of enzymes necessary for DNA replication are he Meli Primase, Topoisomerase. b) Helicase To c) Helicase, Ton merase, Primase, DNA D merase II, and DNA Ligase erase, DNA Polymerasei t , and DNA Ligase d) Helicase, Topoisomerase, Primase, DNA Polymerase I, and DNA Ligase dDNA Ligase 58. DNA tesized in the 31 5' direction b) False 59. Eukaryotes have a single origin or replication a) True b) False Okaz 6U- KI Fragments are synthesized in the 3'-5' direction...

  • 7.. In the controlled termination method of DNA sequencing, reading the gel from _____ gives the...

    7.. In the controlled termination method of DNA sequencing, reading the gel from _____ gives the sequence in the _____ direction; _____ fragments that were terminated _____ in polymerization move faster down the gel a. bottom to top; 5′ to 3′; shorter; early b. top to bottom; 5′ to 3′; longer; early c. bottom to top; 5′ to 3′; longer; later d. top to bottom; 5′ to 3′; shorter; early e. bottom to top; 3′ to 5′; shorter; early 8....

  • match 1. Cilia 2. Promoter 3. Flagellum 4. Microtubules 5. Codon 6. Termination signal 7. RNA...

    match 1. Cilia 2. Promoter 3. Flagellum 4. Microtubules 5. Codon 6. Termination signal 7. RNA polymerase 8. Anticodon 9. Transcription factor 10. Centrioles A. A special base sequence on DNA which signals the end of transcription B. A three-base sequence or triplet on mRNA that provides the genetic information used in protein synthesis C. Long projections formed by the centrioles D. The enzyme in protein synthesis that breaks E. Start point of a gene being transcribed F. Short, cell...

  • 13. Why are ribonucleoside triphosphates the monomers required for RNA synthesis rather than ribonucleoside monophosphates? A....

    13. Why are ribonucleoside triphosphates the monomers required for RNA synthesis rather than ribonucleoside monophosphates? A. Only ribonucleoside triphosphates contain the sugar ribose. B. Ribonucleoside triphosphates have low potential energy, making the polymerization reaction endergonic. C. Ribonucleoside triphosphates have high potential energy, making the polymerization reaction exergonic. D. Ribonucleoside monophosphates cannot form complementary base pairs with the DNA template. E. Ribonucleoside triphosphates are not used, rather all use deoxyriboside triphosphates. 14. How is a mutation in a bacterial cell that...

  • Answer just part 2! The answers to the first part are C, A, C, B, B,...

    Answer just part 2! The answers to the first part are C, A, C, B, B, A, A, C, B Thanks Instructions: Part I: For each of the statements below, indicate with an A, B, or C whether the statement is referring tox A. DNA replication B. Transcription C. Translation The sibosome recognizes a ribosome binding site on the mRNA molecule 2 The replisome assembles at a specific en sequence encoded in the DNA. J. The abosome cresates peptide bonds...

  • Why would changes in the genes for transcription factors be expected to generate major phenotypic differences?...

    Why would changes in the genes for transcription factors be expected to generate major phenotypic differences? They are extremely powerful genes. They can affect the expression of small numbers of other genes. Their gene products are remarkably stable. Their gene products normally denature more rapidly than other gene products. They can affect the expression of large numbers of other genes. Which enzyme, also responsible for siRNA formation, carves miRNAs from their double-stranded, fold- back RNA precursor (pre-miRNA)? Dicer ribonuclease RNA...

  • 1. Homologous recombination can happen between non-identical DNA sequences. T/F? 2. Homologous recombination can happen in_______...

    1. Homologous recombination can happen between non-identical DNA sequences. T/F? 2. Homologous recombination can happen in_______ a) meiosis b) mitosis c) both 3. Homologous recombination in meiosis has the main purpose of_____ a) DNA repair b) Creating new chromosomes   c) Sealing double-stranded breaks 4. Strand invasion usually happens without enzymatic assistance. T/F? 5. When replication fork runs into a nick, it results in a_______ a) single-stranded break b) double-stranded break 6. The invading end is usually a _______ a) 3'...

  • Please give me a complete solutions, don't work on it if you're not going to finished...

    Please give me a complete solutions, don't work on it if you're not going to finished this. This is an old homework, in which I am using to study for the exam. So please don't give me half ass answers. QUESTION 11 If a DNA segment has the sequence GCTAA, what RNA sequence will be made from it? a. CGATT b. CGUTT c. CGAUU d. GCTAA e. UGATT 1 points    QUESTION 12 Which of the following brings amino acids...

  • QUESTION 1: You are inserting a gene into an MCS found within the LacZ gene. Using...

    QUESTION 1: You are inserting a gene into an MCS found within the LacZ gene. Using blue/white colony selection, why could you assume that white colonies have modified plasmids? a. A blue colony means the LacZ reading-frame was disrupted b. A blue colony means your gene has mutations c. A white colony means the LacZ reading-frame is intact d. A white colony means the LacZ reading-frame was disrupted    QUESTION 2: You are performing a PCR using primers with a sequence perfectly...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT