For each question you will have to either make a piece of DNA to complement theDNA template or synthesize a piece of mRNA and give the amino acid sequence ofthe protein. Highlight the mutation on the mRNA sequence. You can use question1 as a reference point. Be specific when about the type of mutation example:missense mutation, frameshift mutation...Question 1DNA TEMPLATE:3 'GCA TTT GAT AAA TAC CTG AGA TGA CTG ATT GGG GGC AAA 5'5' CGT AAA CTA TTT ATG GAC TCT ACG GẠC TẠA CCC CCG TTT 3'1. Synthesize a piece of DNA to complement the template:- Previousort scdelete40PPA1&backspace

Answered: Image /qna-images/answer/a05e769f-8871-43d6-9582-a57eb9317dfb.jpg

Answered: DNA TEMPLATE: 3 'GCA TTT GAT AAA TAC…

Biology Today and Tomorrow without Physiology (MindTap Course List)

5th Edition

ISBN:9781305117396

Author:Cecie Starr, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Christine Evers, Lisa Starr

Chapter7: Gene Expression And Control

Section: Chapter Questions

Problem 9SQ

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Below is a DNA strand. We will call this the original strand, and you will transcribe and translate it first. Each question thereafter is a variation of the original strand with a mutation. Transcribe and translate the DNA strand for each number, and identify the type of mutation. It is important to determine where the change in DNA occurred in each mutated strand to help you identify the mutation type. Original: DNA TAC CCA GAG ATG CAT mRNA Amino Acid Mutation Type: ______________________________ Mutation 2 DNA TAC CCA GAT ATG CAT mRNA Amino Acid Mutation Type: ______________________________ Mutation 3 DNA TAC CCA GAG ATT CAT mRNA Amino Acid Mutation Type: ______________________________ Mutation 4 DNA TAC CAG AGA TGC AT… mRNA…
Look at the amino acid chart below that the proctor placed on a board at the front of the classroom before asking the students questions. Help Arcel characterize the DNA mutation (HbS) according to its amino acid and position. a The disease is caused by GAG20→GTG20. b The disease is caused by val6→gln6. c The disease is caused by glu6→ val6. d The disease is caused by glu20→ val20.
In the following drawing, the top strand is the template DNA, andthe bottom strand shows the lagging strand prior to the action ofDNA polymerase I. The lagging strand contains three Okazakifragments. The RNA primers, which are shown in red, have not yetbeen removed. A. Which Okazaki fragment was made first, the one on the left orthe one on the right?B. Which RNA primer will be the first one to be removed by DNApolymerase I, the primer on the left or the primer on the right?For this primer to be removed by DNA polymerase I and for thegap to be filled in, is it necessary for the Okazaki fragment inthe middle to have already been synthesized? Explain.C. Let’s consider how DNA ligase connects the left Okazaki fragmentwith the middle Okazaki fragment. After DNA polymerase Iremoves the middle RNA primer and fills in the gap with DNA,where does DNA ligase function? See the arrows on either sideof the middle RNA primer. Is ligase needed at the left arrow, atthe right arrow, or both?D. When…
The illumina method of sequencing uses a unique type of nucleotide building block. What is the specific characteristic of this type of nucleotide that is important for this method of sequencing? How is the sequence of a fragment of DNA determined using this method? (USE THIS LINK AND WRITE ANSWERS IN YOUR LANGUAGE PLEASE DON'T COPY SAME AS GIVEN IN SITE https://www.mybiosource.com/learn/testing-procedures/dna-sequencing/
Using the straight-chain sugar convention where the vertical lines represent sugars and the diagonals are phosphodiester bonds, choose the correct structure of the DNADNA strand that encoded this short stretch of RNA. pppApCpUpGpGpCpApA−OH
What type of mutation is represented in the second strand of DNA? TACGGCACT TACGGCCACT answer choices A. Deletion B. Substitution C. No mutation D. Insertion
The following are DNA fragments containing a small gene. The top strand is the coding strand. Transcribe all groups and translate. FIND THE POSSIBLE MUTATIONS Group D 5’-GGCAATGGGTTTGTGCAATTCTAACAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTGTCAAAAATTAAG-5’ Group E 5’-GGCAATGGGTTTTGCAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAAACGTTAAGATTTTCAAAAATTAAG
You digest a piece of DNA that is 98 000 base pairs long, with a restriction endonuclease having a recognition site four base pairs long. Approximately how many fragments would you expect to be produced? Question 18 options: 512 662 196 40 384
The following are DNA fragments containing a small gene. The top strand is the coding strand. Transcribe all groups and translate. FIND THE POSSIBLE MUTATIONS Group B - MUTATION 5’-GGCAATGGGTTTGTGAAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACTTTAAGATTTTCAAAAATTAAG-5’ Group C- 5’-GGCAATGGGTTTGTGCAATTCTAAGAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTCTCAAAAATTAAG-5’
Select the sequences that would be recognized by a restriction enzyme? (There can be more than one answer) A) 3'-TAGCTA-5' B) 5'CGATTC-3' C) 5'GAATTC- 3' D) 5'GAGCTC-3'
Which is the BEST technique to use to answer this question? You wish to determine whether a point mutation is present in the sunscreen gene. a. Westem blot. b. Northern blot . c. Sanger DNA sequencing d. In situ hybridization. Part B. Which is the BEST technique to use to answer this question? You wish to determine where Sunshine protein is translated in a ladybug embryo. a. Immunofluorescencez b. Western blot . c. RT-PCR. d. Northern blot. e. in situ hybridization.
Can you please check my answer and make sure it is correct. Question: Describe the two roles primers play in PCR. Answer: Primers are short sequences of single stranded DNA that mark both the ends of the target sequence to be amplified during PCR. One primer attaches to the top strand at one end (forward primer) and the other one attaches to the bottom strand at the other end of the segment (reverse primer). They are made specifically to attach to these sections of DNA because the nucleotides of the primers are complementary to the target sequence. Other than marking the starting and end points for PCR, primers are also necessary for the process because without them, DNA polymerase wouldn’t be able to synthesize the complementary strand of DNA because it can only add nucleotides to an existing piece.

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