The two forms of Non-retroviral mobile DNA elements which are observable in huge numbers in the genomes of mammalian (incl. human) genomes are known as: Question 19 options: LINE & SINE/Alu sequences LINE sequences & IS elements transposons & LINE sequences transposons & retrotransposons SINE & Alu sequence

The two forms of Non-retroviral mobile DNA elements which are observable in huge numbers in the genomes of mammalian (incl. human) genomes are known as: Question 19 options: LINE & SINE/Alu sequences LINE sequences & IS elements transposons & LINE sequences transposons & retrotransposons SINE & Alu sequence

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter19: Genomes And Proteomes

Section: Chapter Questions

Problem 1ITD: Below is a sequence of 540 bases from a genome. What information would you use to find the...

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The two forms of Non-retroviral mobile DNA elements which are observable in huge numbers in the genomes of mammalian (incl. human) genomes are known as: Question 40 options: LINE & SINE/Alu sequences LINE sequences & IS elements SINE & Alu sequences transposons & retrotransposons jumping genes & deletors
It has been noted that most transposons in humans and other organisms are located in noncoding regions of the genome regions such as introns, pseudogenes, and stretches of particular types of repetitive DNA. There are several ways to interpret this observation. Describe two possible interpretations. Which interpretation do you favor? Why?
For the following short sequence of double stranded DNA and the given primers, there will be one major duplex DNA product after many cycles (imagine 10 cycles) of PCR. Provide the sequence of this one major duplex product and label the 5’ and 3’ ends of each strand. Sequence to be amplified: 5’- GGTATTGGCTACTTACTGGCATCG- 3’ 3’- CCATAACCGATGAATGACCGTAGC- 5’ Primers: 5’-TGGC-3’ and 5’-TGCC-3’
In relation to central dogma of molecular biology answer the following questions: The following segment of DNA is part of the transcription unit of a gene. You know already that RNA polymerase moves in a specific direction along this piece of DNA to convert one of the DNA strands into a single strand RNA transcript so that this entire region of DNA is made into RNA. 5′-GGCATGGCAATATTGTAGTA-3′ 3′-CCGTACCGTTATAACATCAT-5′ Given this information, a student claims that the RNA produced from this DNA is: 3′-GGCATGGCAATATTGTAGTA-5′ Give two reasons why this answer is incorrect.
Below are several DNA sequences that are mutated compared with the wild-type sequence. Eachis a section of a DNA molecule that has separated in preparation for transcription, so you are onlyseeing the template strand. For each mutated DNA sequence, translate and record the resultingamino acid sequence. What type of mutation is each? Wild-type sequence: 3’-T A C T G A C T G A C G A T C-5’ Mutated DNA Template Strand #1: 3’-T A C T G T C T G A C G A T C-5’Amino acid sequence of peptide:Type of mutation: Mutated DNA Template Strand #2: 3’-T A C G G A C T G A C G A T C-5’Amino acid sequence of peptide:Type of mutation: Mutated DNA Template Strand #3: 3’-T A C T G A C T G A C T A T C-5’Amino acid sequence of peptide:Type of mutation: Mutated DNA Template Strand #4: 3’-T A C G A C T G A C T A T C-5’Amino acid sequence of peptide:Type of mutation:
Which of the following set(s) of primers a-d could you use to amplify the following target DNA sequence, which is part of the last protein-coding exon of the CFTR gene? Explain briefly. (Note: The three dots represent the body of the region to be amplified, whose beginning and end are only being shown.) 5' GGCTAAGATCTGAATTTTCCGAG . TTGGGCAATAATGTAGCGCCTT 3' 3' CCGATTCTAGACTTAAAAGGCTC . AACCCGTTATTACATCGCGGAA 5' a. 5' GGAAAATTCAGATCTTAG 3'; 5' TGGGCAATAATGTAGCGC 3' b. 5' GCTAAGATCTGAATTTTC 3'; 3' ACCCGTTATTACATCGCG 5' c. 3' GATTCTAGACTTAAAGGC 5'; 3' АССCGTTATTАСАТСGCG 5 d. 5' GCTAAGATCTGAATTTTC 3'; 5' TGGGCAATAATGTAGCGC 3'
Arabidopsis thaliana has among the smallest genomes in higher plants, with a haploid genome size of about 100 Mb. If this genome is digested with BbvCl, a restriction enzyme which cuts at the sequence CCTCAGC GGAGTCG 1. approximately how many DNA fragments would be produced? Assume the DNA has a random sequence with equal amounts of each base.
Below is a sequence of 540 bases from a genome. What information would you use to find the beginnings and ends of open reading frames? How many open reading frames can you find in this sequence? Which open reading frame is likely to represent a protein- coding sequence, and why? Which are probably not functioning protein-coding sequences, and why? Note: for simplicitys sake, analyze only this one strand of the DNA double helix, reading from left to right, so you will only be analyzing three of the six reading frames shown in Figure 19.4.
Based on the following wild type DNA sequence, indicate if each of the mutations should be classified as : insertion, deletion, missense, nonsense, silent (Use the provided Genetic Code table and remember you have been given DNA sequence). Wild Type: AUGAUUCUUAAAAGU Mutant 1: AUGAUUCUUUAAAGU Mutant 2: AUGAUUCUUGAAAGU Mutant 3: AUGAUCCUUAAAAGU Mutant 4: AUGAUCCUAAAAGU Mutant 5: AUGAUCCUUAAACAGU Socond letter Key: Ala = Alanine (A) Arg Arginine (R) Asn = UUU } UAU Tyr UGU UGC Cys UGA STOP UGG Trp UCU UCC UUC Phe Ser Asparagine (N) Asp = Aspartate (D) Cys Cysteine (C) Gin = Glutamine (Q) Glu = Glutamate (E) Gly = Glycine (G) His = Histidine (H) le = Isoleucine (1) Leucine (L) Lys Lysine (K) Met = Methionine (M) Phe = Phenylalanine (F) Pro Proline (P) Ser = Serine (S) Thr Threonine (T) Trp Tryptophan (W) Tyr Tyrosine (Y) - Valine (V) UCA UCG UAA STOP UAG STOP UUA Leu UUG S CCU CC CGU CUU CUC His CGC Arg Leu Pro CAA Gin CGA CCA CCG CUA CUG CGG Leu = AGU AUU AUC } lle AUA ACU ACC ACA Ser AAC…
All are correct about DNA gyrase in E. coli EXCEPT: It works to remove positive supercoiling introduced by the DnaB protein (helicase). It is a topoisomerase that hydrolyzes ATP during its reaction mechanism. Its mechanism involves the breaking of a single phosphoester bond in one strand of dsDNA. It works to relieve supercoiling in DNA to overcome the torsion stress imposed upon unwinding.
THE MOLECULAR GENETICS OF CYSTIC FIBROSIS and of The following is the base sequence of DNA that codes for amino acids 506-510 of the protein that regulates the chlorine channels in the cell membrane. This protein contains a total of 1476 amino acids so this is a small part of the entire gene. DNA Template Strand: 3'TAGTAGAAACCACAA5' 1. What is the minimum number of DNA nucleotides in this whole gene? 2. What is the sequence of bases on the strand of DNA that is complementary to the template strand? 3. What mRNA will be formed from the template strand of DNA? 4. What amino acids will this mRNA code for? 5. If the 6th, 7th and 8th bases in the template strand of the DNA are removed, rewrite the new template strand below. 6. When the template strand of the DNA is changed, this is referred to as a mutation. What kind of mutation is this? 65
LINEs and SINEs are repetitive sequences in humans that as retrotransposons, can also insert into genes and cause disease. Select one: True FalseWhat is the most reasonable explanation for the observation that transposons in many multicellular genomes are more often found in nongenic sequences (i.e.,sequences that do not code for genes) such as centromeric heterochromatin rather than in genic sequences (i.e,. gene sequences)? Select one: a. The transposons are “safe” from harm from the host when in nongenic regions. b. Reverse transcriptase promotes integration into nongenic DNA preferentially. c. Insertion into nongenic DNA is less likely to do harm to the host and thus would not be selected against. d. Genic DNA is protected from transposon insertion by all the transcription factors bound to the region. e. The AT-rich nature of centromere and other nongenic sequences makes it easier for transposon insertion.