An anticodon is 5' CGA 3'. The codon that pairs with the anticodon will be:- a) 3' UCG 5'- b) 5' GCT 3'- c) 5' UCG 3'- d) 5' TCG 3'An anticodon is 5' CGA 3'. This amino acid will be put into the protein when the RNA molecule with this anticodon is used.**Use the codon table below:**| Codons | Amino Acids ||--------|-------------|| UUU, UUC | Phe || UUA, UUG, CUU, CUC, CUA, CUG | Leu || UCU, UCC, UCA, UCG | Ser || UAU, UAC | Tyr || UAA, UAG | Stop || UGU, UGC | Cys || UGA | Stop || UGG | Trp || CCU, CCC, CCA, CCG | Pro || CAU, CAC | His || CAA, CAG | Gln || CGU, CGC, CGA, CGG | Arg || AUU, AUC, AUA | Ile || AUG | Met || ACU, ACC, ACA, ACG | Thr || AAU, AAC | Asn || AAA, AAG | Lys || AGU, AGC | Ser || AGA, AGG | Arg || GUU, GUC, GUA, GUG | Val || GCU, GCC, GCA, GCG | Ala || GAU, GAC | Asp || GAA, GAG | Glu || GGU, GGC, GGA, GGG | Gly |- a) LEU- b) ARG- c) SER- d) ALA **Question 1:**An organism has 23% G in its genome. What percent A does it have?- a) 23- b) 50- c) 27- d) 46**Question 2:**A gene has 60 introns. How many exons does it have?- a) 60- b) 180- c) 59- d) 61**Question 3: Splicing**- a) Connects proteins together- b) Removes Protein sequences- c) Removes intron RNA- d) Removes Exon RNAThe option "c) Removes intron RNA" is selected.

Transcription: "It is the process through which the enzyme RNA Polymerase transfers information from one strand of DNA to another."The strand of DNA that synthesizes the RNA is known as the template strand, while the other strand is known as the coding strand. The promoter is bound by the DNA-dependent RNA polymerase, which catalyzes polymerization in the 3′ to 5′ direction. As per our policy, we are answering the first 3 questions only. For the rest of the questions, kindly repost.Answer 1: d). 5'TCG 3' Explanation: A three-base sequence in the transfer RNA's anticodon loop base pairs with the codons in mRNA. "The mRNA codon is translated in the 5'—>3 orientation, and its first base pairs with the third base of the anticodon." Consequently, 5' TCG 3' is the codon for the anticodon 5' CGA 3'. The other option, 3' UGC 5', 5' GCT 3', and 5'UCG 3', has no reading and changeover of bases and, therefore, cannot be the answers.Answer 2: c). SER Explanation: As mentioned in answer 1 we have codon TCG for that we will have AGC mRNA and SER amino acid sequence. The rest of the options are not possible because AGC mRNA can code to only SER amino acids. Answer 3: C). A = 27% Given: If G is 23% Explanation: We know for the double-stranded, there must be base pair equality: A percent = T percent and G percent = C percent. If G is 23%, then cytosine must also be 23 %. The total amount of bases = 100% A + T + G + C = 100% As we know, A percent = T percent and G percent = C percent. Let's consider T = A then, 2A + 23 + 23 = 100% 2A + 46 = 100 2A = 100 - 46 A = 27% And thus A= 27% Answer: A = 27%, T = 27% , C = 23% and G = 23% The rest of the options can not be obtained from the above calculations and explanations hence not possible.

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Biology

An anticodon is 5' CGA 3'. The codon that pairs with the anticodon will be a) 3'UGC5' b) 5'GCT3' c) 5' UCG3' d) 5'TCG3 4 Listen An anticodon is 5'CGA3'. This amino acid will be put into the protein when the RNA molecule with this anticodon is used. Use the codon table below UUU UUC UUA C CUU CUC CUA CUG AUU AUC AUA AUG GUU GUC GUA GUG } a) LEU b) ARG c) SER d) ALA Phe Leu Leu lle Met Val UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA ACG GCU GCC GCA GCG Ser Pro Thr Ala UAU UAC UAA UAG CAU CAC CAA CAG AAU AAC AAA ;} AAG GAU GAC GAG} Tyr Stop His Gln Asn Lys Asp Glu UGU UGC UGA UGG CGU CGC CGA CGG. AGU AGC AGA AGG GGU GGC GGA GGG Cys Stop Trp Arg Ser Arg Gly

An anticodon is 5' CGA 3'. The codon that pairs with the anticodon will be a) 3'UGC5' b) 5'GCT3' c) 5' UCG3' d) 5'TCG3 4 Listen An anticodon is 5'CGA3'. This amino acid will be put into the protein when the RNA molecule with this anticodon is used. Use the codon table below UUU UUC UUA C CUU CUC CUA CUG AUU AUC AUA AUG GUU GUC GUA GUG } a) LEU b) ARG c) SER d) ALA Phe Leu Leu lle Met Val UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA ACG GCU GCC GCA GCG Ser Pro Thr Ala UAU UAC UAA UAG CAU CAC CAA CAG AAU AAC AAA ;} AAG GAU GAC GAG} Tyr Stop His Gln Asn Lys Asp Glu UGU UGC UGA UGG CGU CGC CGA CGG. AGU AGC AGA AGG GGU GGC GGA GGG Cys Stop Trp Arg Ser Arg Gly

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 20QP: If an extra nucleotide is inserted in the first exon of the beta globin gene, what effect will it...

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