Il TEW ņ 47% 2:38 PM can you please solve these three problems 30. Study the nucleotide sequences of wild type and it's mutant gene below. Wild type: Protein: DNA Met - Pro -Glu- Pro- Lys- Met ATG CCC GAA CCC AAA ATG Mutant: Protein : DNA Met - Pro-Glu-Pro - Lys- Met ATG CCT GAA CCC AAA ATG This mutation can best be called: A) Non-sense B) Silent C) Frameshift D) Mis-sense E) All of the above 31. Study the nucleotide sequences of wild type and it's mutant gene below. Wild type: Protein : DNA Met - Pro -Glu- Pro- Lys- Met ATG CCC GAA CCC AAA ATG Mutant: Met - Pro -Gly- Thr- Gln- Lys ATG CCC GGA ACC CAA A AT Protein: DNA This mutation can best be called: A) Non-sense B) Silent C) Frameshift D) Mis-sense E) All of the above 32. The following are examples of large-scale DNA changes in an organism EXCEPT: A) Transformation B) Transduction C) Point mutation D) Conjugation E) Recombination

Il TEW ņ 47% 2:38 PM can you please solve these three problems 30. Study the nucleotide sequences of wild type and it's mutant gene below. Wild type: Protein: DNA Met - Pro -Glu- Pro- Lys- Met ATG CCC GAA CCC AAA ATG Mutant: Protein : DNA Met - Pro-Glu-Pro - Lys- Met ATG CCT GAA CCC AAA ATG This mutation can best be called: A) Non-sense B) Silent C) Frameshift D) Mis-sense E) All of the above 31. Study the nucleotide sequences of wild type and it's mutant gene below. Wild type: Protein : DNA Met - Pro -Glu- Pro- Lys- Met ATG CCC GAA CCC AAA ATG Mutant: Met - Pro -Gly- Thr- Gln- Lys ATG CCC GGA ACC CAA A AT Protein: DNA This mutation can best be called: A) Non-sense B) Silent C) Frameshift D) Mis-sense E) All of the above 32. The following are examples of large-scale DNA changes in an organism EXCEPT: A) Transformation B) Transduction C) Point mutation D) Conjugation E) Recombination

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter11: Genome Alterations: Mutation And Epigenetics

Section: Chapter Questions

Problem 2QP: Achondroplasia is an autosomal dominant form of dwarfism caused by a single gene mutation. Calculate...

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Achondroplasia is an autosomal dominant form of dwarfism caused by a single gene mutation. Calculate the mutation rate of this gene given the following data: 10 achondroplastic births to unaffected parents in 245,000 births.
______________________________________________________________ 1. People who carry a theoretical genetic disorder (called B-disease) can be identified from a 2kb DNA sequence. People who carry this genetic disorder have a single nucleotide polymorphism that results in a change of GTATCC to GGATCC, a site that only occurs once at nucleotide number 750 in this DNA sequence. Answer the following questions based on the information provided. (a) How can you develop a simple molecular test with enzyme BamH to identify the genetic disorder? ( (b) If you have carried out the molecular test (based on the information above) on a 100 individual and found that 24 were healthy (BB) and 26 were carriers (bb); 1) What is the ratio of heterozygous? 2) Show how can you identify the three types from the agarose gel
The amino acid sequence of part of a protein has beendetermined:N . . . Gly Ala Pro Arg Lys . . . CA mutation has been induced in the gene encodingthis protein using the mutagen proflavin. The resultingutant protein can be purified and its amino acidsequence determined. The amino acid sequence of themutant protein is exactly the same as the amino acidsequence of the wild-type protein from the N terminus of the protein to the glycine in the preceding sequence. Starting with this glycine, the sequence ofamino acids is changed to the following:N . . . Gly His Gln Gly Lys . . . CUsing the amino acid sequences, one can determinethe sequence of 14 nucleotides from the wild-typegene encoding this protein. What is this sequence?
Shown below are two homologous lengths of the alpha and betachains of human hemoglobin. Consult a genetic code dictionary and determine how many amino acid substitutionsmay have occurred as a result of a single nucleotidesubstitution. For any that cannot occur as a result of a singlechange, determine the minimal mutational distance. Alpha: ala val ala his val asp asp met proBeta: gly leu ala his leu asp asn leu lys
To determine the reproducibility of mutation fre-quency measurements, you do the following experiment.You inoculate each of 10 cultures with a single E. coli bac-terium, allow the cultures to grow until each contains 106cells, and then measure the number of cells in each culturethat carry a mutation in your gene of interest. You were sosurprised by the initial results that you repeated the experi-ment to confirm them. Both sets of results display the sameextreme variability, as shown in Table Q5–1. Assuming thatthe rate of mutation is constant, why do you suppose thereis so much variation in the frequencies of mutant cells indifferent cultures?
Diamond–Blackfan anemia (DBA) is a rare, dominantgenetic disorder characterized by bone marrow malfunction,birth defects, and a predisposition to certaincancers. Infants with DBA usually develop anemia in the firstyear of life, have lower than normal production of red blood cellsin their bone marrow, and have a high risk of developing leukemiaand bone cancer. At the molecular level, DBA is causedby mutations in any one of 10 genes that encode ribosomalproteins. The first-line therapy for DBA is steroid treatment,but more than half of affected children develop resistance tothe drugs and in these cases, treatment is halted. DBA can betreated successfully with bone marrow or stem cell transplantsfrom donors with closely matching immune system markers.Transplants from unrelated donors have significant levels ofcomplications and mortality. While a stem cell transplant from an unaffected donor is currentlythe only cure for DBA, genome-editing technologies mayone day enable the correction of…
Diamond–Blackfan anemia (DBA) is a rare, dominantgenetic disorder characterized by bone marrow malfunction,birth defects, and a predisposition to certaincancers. Infants with DBA usually develop anemia in the firstyear of life, have lower than normal production of red blood cellsin their bone marrow, and have a high risk of developing leukemiaand bone cancer. At the molecular level, DBA is causedby mutations in any one of 10 genes that encode ribosomalproteins. The first-line therapy for DBA is steroid treatment,but more than half of affected children develop resistance tothe drugs and in these cases, treatment is halted. DBA can betreated successfully with bone marrow or stem cell transplantsfrom donors with closely matching immune system markers.Transplants from unrelated donors have significant levels ofcomplications and mortality. Given that a faulty ribosomal protein is the culprit and causesDBA, discuss the possible role of normal ribosomal proteins.Why might bone marrow cells be…
Q-)Silent mutations that occur in DNA are quite common in living cells and usually involve no effects onphenotype. In not more than 2 pages (using 1.5 line space of Arial or Times New Roman fonts) provideanswers for the following questions? A) Provide one example of a clinical implication of a “silent mutation” that proven to have an effect onthe phenotype and provide a brief description of its molecular characteristics?
54. 35delG mutation in the connexin 26 gene is an autosomal recessive condition that has been identified to cause congenital deafness. In a population of 367 Greek children, 75 of the children carried two copies of the 35delG mutation.In order to calculate the percentage of individuals who are heterozygous for the 35delG mutation, the calculation that must be performed is Select one: a. 1 – (p2 – 2pq) b. 1 – (p +q) c. 1 – (p2 + q2) d. 1 – q2
Below are 9 possible primer pairs.● Determine which primer pair is the best choice.● Explain why the other primers are not good choices.● Calculate the Tm for each primer. Underline or highlight the region of DNA for the primer pair you chose as the best.Forward 1: 5’ gaaataattttgtttaactttaag 3’ Tm =Reverse 1: 5’ gtttaagacaaaatagtctgg 3’ Tm =Forward 2: 5’ gtaactcagctttcaggtcg 3’ Tm =Reverse 2: 5’ tctcggaatgttgcaacagc 3’ Tm =Forward 3: 5’ agattagcggatcctacctg 3’ Tm =Reverse 3: 5’ atgtgtaatcccagcagcag 3’ Tm =Forward 4: 5’ cattgattatttgcacggcg 3’ Tm =Reverse 4: 5’ aaaatcttctctcatccgcc 3’ Tm =Forward 5: 5’ tccataagattagcggatcc 3’ Tm =Reverse 5: 5’ tgcaagcttggctgttttgg 3’ Tm =Forward 6: 5’ gatcctacctgacgcttttta 3’ Tm=Reverse 6: 5’ aaataatgaattcgagctcggt 3’ Tm =Forward 7: 5’ataaaaaaatcgagataaccgtt 3’ Tm =Reverse 7: 5’aggtcgactctagaggatc 3’ Tm =Forward 8: 5’ctacctgttccatggccaac 3’ Tm=Reverse 8: 5’ ttcgggcatggcactcttg 3’ Tm=Forward 9: 5’ tccataagattagcggatcc 3’ Tm =Reverse 9: 5’…
The electrophoresis gel shown below had a 4th lane (cut off and not shown) which contained 7 proteins of known molecular weight. A plot of these proteins was made with log MW on the y-axis and migration distance (cm) on the x-axis.The equation for the linear regression line is: Y = -0.1 X + 4.79If the distance between the well and the 3rd band of the Red sample is 4 cm, what is the approximate MW, expressed in daltons, of the protein at that location?
. Geneticists interested in human hemoglobins havefound a very large number of mutant forms. Some ofthese mutant proteins are of normal size (but haveamino acid substitutions) while others are short, dueto deletions or nonsense mutations. The first extralong example was named Hb Constant Spring, inwhich the β-globin has all of its normal amino acidsplus several extra amino acids attached after thenormal C-terminal end of the protein. a. What is the most plausible explanation forits origin?b. Is it possible that Hb Constant Spring arose fromfailure to splice out an intron?c. Estimate how many extra amino acids mightbe added to the C terminal end of the mutantprotein