Q: WILD-TYPE MC1R GENE (LIGHT COAT-COLOR PHENOTYPE) DNA GTG TAC GAA CGT mRNA Amino Acid
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Q: GENE TABLE 1: WILD-TYPE MC1R GENE (LIGHT COAT-COLOR PHENOTYPE) DNA GAA CAG GTG GTT CCA AAG…
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Q: DNA = TAG - TAG - GAT MRNA = Amino Acids = Phenotype = %3D
A: THE ANSWER IS IN NEXT STEP :
Q: DNA 3’ AGA ACA TAA TAC CTC TTA ACA CTC TAA AGA CCA GCA ATT CGA TGA ACT GGA GCA 5’ mRNA protein
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WILD-TYPE MC1R GENE (LIGHT COAT-COLOR
DNA |
TTG |
AGG |
TGG |
GCG |
TGT |
CCG |
CAA |
GGA |
mRNA |
||||||||
Amino Acid |
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- e. four-base, not overlapping4. An example of a portion of the T4 rIIB gene in whichCrick and Brenner had recombined one + and one −mutation is shown here. (The RNA-like strand of theDNA is shown.)wild type 5′ AAA AGT CCA TCA CTT AAT GCC 3′mutant 5′ AAA GTC CAT CAC TTA ATG GCC 3′a. Where are the + and − mutations in the mutant DNA?b. The double mutant produces wild-type plaques.What alterations in amino acids occurred in thisdouble mutant?c. How can you explain the fact that amino acids aredifferent in the double mutant than in the wild-typesequence, yet the phage has a wild-type phenotype?THE MOLECULAR GENETICS OF SICKLE CELL ANEMIA The following is the base sequence of DNA that codes for first eight amino acids of the B chain of hemoglobin. The B 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? 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 this? 8. 7. What mRNA will be formed from the mutated template strand of DNA? What amino acids will this new mRNA from the mutated template strand code for? 9. Are these new amino acids the…Table I CACGT A GA CTGAGG ACTC CACGTAGACTGAG G ACAC Wild-type beta-globin gene fragment Sickle-cell beta-globin gene fragment > Circle the mutation in DNA of the sickle-cell beta-globin gene fragment Compare fragments of DNA the wild-type and mutant beta-globin genes in the Table I above, what are the similarities and differences you observe?
- #3 HaelII --- 5’ CC ↓ GG 3’ 5’ ACGCCGGCCGTATTAT CCGGATCCGCCG CCGGCTGTCCCGGATCA 3’ 3’ TGCGGCCGGCATAATAGGCCTAGGCGGCGGCCGACAGGGCCTAGT 5’ Restriction enzyme: Recognition sequence: Number of pieces of DNA: Type of cut: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#4 BamI --- 5’ CCTAG ↓G 3’ 5’ ACGCCTAGGACGTATTATCCTAGGTAT CCGCCGCCGT CATCA 3’ 3’ TGCGGATCCTGCATAATAGGATCCATAGGCGGCGGCAGTAGT 5’ Restriction enzyme: Recognition sequence: Number of pieces of DNA: Type of cut:
- Associated SNPs outside of gene no effect on protein production or function. T G Associated SNPs within gene no effect on protein production or function Regulatory sequences A Coding region с T Noncoding SNP: changes amount of protein produced www.Biolnteractive.org Causative SNPs within gene Unassociated SNP far from gene on same chromosome or different chromosome Protein Coding SNP: changes amino acid sequence b. Which types of SNPs might be identified in a GWAS? 4. Consider the different types of SNPs shown in Figure 3: associated, unassociated, and causative (including both noncoding and coding). a. Which types of SNPs affect protein production or function for the gene of interest? Figure 3. A diagram showing various ways in which a SNP could be associated with a certain gene and its trait. GWAS in the News Read the following news release, which describes a GWAS study with dogs. Note that a dog's coat refers to its fur or hair. Variants in Three Genes Account for Most Dog Coat…What would the amino acid sequence be for the following DNA Transcript? 5’AAGCCATTTAAAGGC 3’ 3’ TTCGGTAAATTTCCG 5’ Phe Gly Lys Phe Pro Phe Leu Lys Phe Val Lys Phe Phe Lys Pro Lys Pro Phe Lys Gly More information is neededWild type genomic sequence of a portion of a gene and the wild type sequence of portion of the protein gene product. The protein sequence only matches partially. TAT AAA GGG CGA CCA CCT GGT AAT GGG ACT TTG AGG Tyr Lys Gly Arg Pro Pro Ala Pro Arg Gln Tyr Trp Note: The five amino acids at the amino end of the WT protein match the coding sequence perfectly, but not the six at the carboxyl end. This is for an obvious reason that you should figure out before starting the problem. One of the difficulties with this problem is that you do not have the DNA sequence for the carboxyl end of the protein, but that should not be a problem! And, so far, there are no mutations; nothing unusual, it is just a normal eukaryotic gene!! A mutant is found that has the following protein sequence in that part of the protein: Tyr Lys Gly Arg Thr Cys Ser Premature Stop. What is the likely mutation? between A12 and C13 insert T. O between C13 and C14 insert T. A15 -> del. O C13 and C14 -> del. G8 -> T. OG20 ->…
- Wild type genomic sequence of a portion of a gene and the wild type sequence of portion of the protein gene product. The protein sequence only matches partially. ТАT AAA GGG СGA ССА ССТ GсT AАT GGG АСт тTG AGG Tyr Lys Gly Arg Pro Pro Ala Pro Arg Gln Tyr Trp Note: The five amino acids at the amino end of the WT protein match the coding sequence perfectly, but not the six at the carboxyl end. This is for an obvious reason that you should figure out before starting the problem. One of the difficulties with this problem is that you do not have the DNA sequence for the carboxyl end of the protein, but that should not be a problem! And, so far, there are no mutations; nothing unusual, it is just a normal eukaryotic gene!! A mutant is found that has the following protein sequence in that part of the protein: Tyr Lys Gly Arg Thr Cys Ser Premature Stop. What is the likely mutation? between A12 and C13 insert T. between C13 and C14 insert T. A15 -> del. C13 and C14 -> del. G8 -> T. G20 -> A. T21…The protein encoded by the cystic fibrosis gene is 1480amino acids long, yet the gene spans 250 kb. How is thisdifference possible?Based on the alignment of these protein sequences, which [pairs] of the genes appear to be most similar to each other? And give reasons why?