The following is the base sequence of DNA that codes for first eight amino acids of the β chain of hemoglobin. The β chain of hemoglobin contains a total of 147 amino acids so this is a small part of the entire gene. DNA Template Strand: TACCACGTGGACTGAGGACTCCTC 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?

The following is the base sequence of DNA that codes for first eight amino acids of the β chain of hemoglobin. The β chain of hemoglobin contains a total of 147 amino acids so this is a small part of the entire gene. DNA Template Strand: TACCACGTGGACTGAGGACTCCTC 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?

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...

See similar textbooks

Similar questions

The following is the base sequence of DNA that codes for first eight amino acids of the β chain of hemoglobin. The β chain of hemoglobin contains a total of 147 amino acids so this is a small part of the entire gene. DNA Template Strand: TACCACGTGGACTGAGGACTCCTC 1. What is the minimum number of DNA nucleotides in this whole gene? 3 2. What is the sequence of bases on the strand of DNA that is complementary to the template strand? 3' TACCACGTGGACTGAGGACTCCTC 5' . 5' ATGGTGCACCTGACTCCTGAGGAG 3' 3. What mRNA will be formed from the template strand of DNA? Sequence of mRNA formed from DNA template strand is shown below: 3' TACCACGTGGACTGAGGACTCCTC 5' . 5'AUGGUGCACCUGACUCCUGAGGAG 3 4. What amino acids will this mRNA code for? 5. If the 20th base in the template strand of the DNA is changed from T to A, 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…
If one DNA segment has the following base composition, 5'-CAGTTAGTCA-3', which of the following sequences is complementary? 5'-TGACTAACTG-3' 3'-TGACTAACTG-5' 3'-CAGTTAGTCA-5' 3'-TGACTAACTG-5'
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: TAGTAGAAACCACAA
The following are DNA fragments containing a small gene. The top strand is the coding strand. Transcribe all 5 groups and translate. Group A 5’-GGCAATGGGTTTGTGCAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTTTCAAAAATTAAG-5’ Group B 5’-GGCAATGGGTTTGTGAAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACTTTAAGATTTTCAAAAATTAAG-5’ Group C 5’-GGCAATGGGTTTGTGCAATTCTAAGAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTCTCAAAAATTAAG-5’ Group D 5’-GGCAATGGGTTTGTGCAATTCTAACAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTGTCAAAAATTAAG-5’ Group E 5’-GGCAATGGGTTTTGCAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAAACGTTAAGATTTTCAAAAATTAAG
You have discovered a family with a new genetic form of anemia. The DNA sequences at the 5’ end of the non-template strand of the normal and mutant DNA encoding the alpha subunit of hemoglobin are given below: Normal 5’-ACGTTATGCCGTACTGCCAGCTAACTGCTAAAGAACAATTA…..-3’ Mutant 5’-ACGTTATGCCCGTACTGCCAGCTAACTGCTAAAGAACAATTA….-3’ What are the amino acid sequences of the normal and mutantpolypeptides? What are the codons in the translated portion of the mRNA transcribed from the normal and mutant genes? What type of mutation is present in the mutant hemoglobin gene?
You have discovered a family with a new genetic form of anemia. The DNA sequences at the 5’ end of the non-template strand of the normal and mutant DNA encoding the alpha subunit of hemoglobin are given below: Normal 5’-ACGTTATGCCGTACTGCCAGCTAACTGCTAAAGAACAATTA…..-3’ Mutant 5’-ACGTTATGCCCGTACTGCCAGCTAACTGCTAAAGAACAATTA….-3’ What are the amino acid sequences of the normal and mutantpolypeptides?
Below is a sequence of DNA. 5'-ttaccgataattctctctcccctcttccatgattctgattaaagaaggcgagaacgaaactatttgttaatacc-3' How many "reading frames" can be identified for this sequence? How many "open reading frames" can be identified for this sequence? What is the frame of the longest ORF?
Below is a sample of a segment of DNA…(copy from left to right) 3’ TACAATGGGCGACGCGCTTCGTTTCAGATT 5’ 5’ ATGTTACCCGCTGCGCGAAGCAAAGTCTAA 3’ 1.Assume the 6th amino acid is changed from T to G on the DNA template strand. What type of mutation is this? What effect would this have on the protein? Look up an example for this type of mutation. 2, Assume the 5th and 6th amino acids are removed from the DNA template strand. What type of mutation is this? How would this affect the protein? Look up an example of this type of mutation. 3.Which mutation changes the protein more...a point mutation or a frameshift mutation. Explain your reasoning. 4.What would be the problem if ATT was inserted into the DNA template strand after the second codon? (Be sure to consult the coding chart for amino acids). 5. What if the second amino acid was repeated over 5Ox. What amino acid is repeated? What type of mutation is this? If this is on chromosome 4, what genetic disorder is this?…
By average, how many Sau3A (5’GATC3’) sites are there in a 10 kd DNA molecule? (1/4)^6 * 10,000 = 2.44140625 => 2.4 For the above 10 kb DNA, if you use Sau3A to do a complete digestion, you will only get a lot of small fragments ( 200~500 base pairs). What should I do if I want to get larger digested fragments (say 3 to 5 kb) Please answer asap and type your answer and do not copy from anywhere please
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’
The following fragment of DNA is from the template strand. First determine the amino acids of the protein encoded by this sequence by using the table at the end of this document. Then give the altered amino acid sequence of the proteins that will be found in each of the following mutations: 3’ – TAC AAG GCT CTA TTT GCC ACA ATC – 5’ The nucleotides are numbered 1-24 from left to right. Mutant 1: A transition at nucleotide 9 Mutant 2: A transition at nucleotide 11 Mutant 3: A T to A transversion at nucleotide 15 Mutant 4: A one-nucleotide deletion at nucleotide 7 Mutant 5: A transition at nucleotide 13 Mutant 6: An addition of GGA after nucleotide 6
In the human genome for the beta chain of haemoglobin (the oxygen-carrying protein in the red blood cells), the first 30 nucleotide in the amino acid coding region is represented by the sequence 3’-TACCACHTGGACTGAGGACTCCTCTTCAGA-5' What is the sequence for the partner strand?