Don't give AI generated solution The genomic DNA information on bovine FABP4 gene is listed below: source 1..4389 /organism="Bos taurus" /mol_type="genomic DNA" /chromosome="14" gene 1..4389 /gene="FABP4" /note="fatty acid binding protein 4, adipocyte mRNA join(1..133,2843..3015,3608..3709,4173..4389) /gene="FABP4" /product="fatty acid binding protein 4, adipocyte" CDS join(61..133,2843..3015,3608..3709,4173..4223) /gene="FABP4" Give correct typing answer with explanation

Don't give AI generated solution The genomic DNA information on bovine FABP4 gene is listed below: source 1..4389 /organism="Bos taurus" /mol_type="genomic DNA" /chromosome="14" gene 1..4389 /gene="FABP4" /note="fatty acid binding protein 4, adipocyte mRNA join(1..133,2843..3015,3608..3709,4173..4389) /gene="FABP4" /product="fatty acid binding protein 4, adipocyte" CDS join(61..133,2843..3015,3608..3709,4173..4223) /gene="FABP4" Give correct typing answer with explanation

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter10: From Proteins To Phenotypes

Section: Chapter Questions

Problem 10QP: b. Compounds A, B, C, and D are known to be intermediates in the pathway for production of protein...

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For E. coli strains with the lac genotypes show below, use a plus sign (+) to indicate the synthesis of β-galactosidase and permease and a minus sign (–) to indicate no synthesis of the proteins.
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Question 2
Proteinase K should be used as the first step of DNA purification (before applying phenol CIA mixture) to remove most proteins. Proteinase K alone cannot remove all proteins from the lysate. Explain why the Proteinase K enzyme cannot remove ALL proteins from the cell lysate.
Cell wall bullding block D-Ala-D-Alal-Lys-tail NH,Me Heptapeptide backbone R1 R. R2 ii) With the aid of the figure above showing the important features of the binding of Vancomycin (lower part) to a bacterial cell wall peptia discuss the antibiotic effect of Vancomycin. i) Vancomycin resistant strains often have a mutation in which one of the D-Ala moieties is exchanged to a D-lactate moiety. Discuss why this mutation makes the strain Vancomycin- resistant but still viable. iv) Glycoconjugate vaccines based on bacterial capsular polysaccharides have been used for 30 years without any need for changes in their structures, while the flu vaccine, based on attenuated or killed virus particles, requires many different structures and a check each time when there is an epidemic that the right vaccine is used. Discuss the reason behind this.
Peptide Sequencing Answer numbers 1 and 2. For number 2, show the OVERLAP method.
In a mixed heteropolymer experiment, messages were createdwith either 4/5C:1/5A or 4/5A:1/5C. These messages yielded proteinswith the amino acid compositions shown in the followingtable. Using these data, predict the most specific coding compositionfor each amino acid.4/5C:1/5A 4/5A:1/5CPro 63.0% Pro 3.5%His 13.0% His 3.0%Thr 16.0% Thr 16.6%Glu 3.0% Glu 13.0%Asp 3.0% Asp 13.0%Lys 0.5% Lys 50.0% 98.5% 99.1%
This answer is incorrect pls provide the right answer and explanation
e.) ( acid buffer an appropriate choice? Why or why not? If I need to perform an enzymatic reaction at pH 6.5, is a citric )Describe the process of transcription in as much detail as possible using pictures and words beginning with a paired (duplexed) strand of DNA and ending with a processed mRNA which is ready for translation. 7.
Wild-type E. coli cells can synthesize all 20 common amino acids, but some mutants, called amino acid auxotrophs, are unable to synthesize a specific amino acid and require its addition to the culture medium for optimal growth. Besides their role in protein synthesis, some amino acids are also precursors for other nitrogenous cell products. Consider the three amino acid auxotrophs that are unable to synthesize glycine, glutamine, and aspartate, respectively. Match each nitrogenous product with the mutant or mutants that would fail to synthesize it. Each mutant may fail to synthesize more than one nitrogenous product, and some nitrogenous products may fail to be synthesized by more than one mutant. glycine auxotrophs glutamine auxotrophs aspartate auxotrophs Answer Bank adenine nucleotides cytosine nucleotides guanine nucleotides uridine nucleotides
Please give me correct solution for both questions c and d.