the diagram below, label the: DNA double helix RNA transcript Template strand • Coding strand • 5' and 3' ends of each DNA and RNA strand • GODT K Unwinding Rewinding G BARA RNA-DNA hybrid helix RNA polymerase
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Gene Interactions
When the expression of a single trait is influenced by two or more different non-allelic genes, it is termed as genetic interaction. According to Mendel's law of inheritance, each gene functions in its own way and does not depend on the function of another gene, i.e., a single gene controls each of seven characteristics considered, but the complex contribution of many different genes determine many traits of an organism.
Gene Expression
Gene expression is a process by which the instructions present in deoxyribonucleic acid (DNA) are converted into useful molecules such as proteins, and functional messenger ribonucleic (mRNA) molecules in the case of non-protein-coding genes.
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- Alternative splicing Template strand S F yIGUide1,mq00:c-00: Replisome Transforming principle Origin of replication (or)eleb al msxS ain Coding strand Transcription factors Leading strand Single nucleotide polymorphism Okazaki fragment Telomerase M Nucleoside RNA Polymerase I RNA Polymerase II RNA Polymerase IIIon & of qu 9ven UoY Insertion mutagenesis Spliceosome Transcription Unit SNP Reverse transcriptase 1 Seminal work by Oswald Avery and colleagues demonstrated that DNA is what Frederick Griffiths called this etniog OS dotsM bioW 1-2kb of newly synthesized DNA strands are called this ainiog PS Assembly of the replisome is an orderly process that begins at these precise sites Snoiteeu 4 Transcribes ribosomal RNA genes in eukaryotes 5. A large nucleoprotein complex that coordinates activity at the replication fork Single base pair differences between homologous genomic regions isolated from different members of a population Complex of proteins and snRNAs catalyzing the removal of…The BNA sequence below is transcribed from left to right (the partner/coding strand is shown). Using this sequence, write the sequence of the polypeptide that results from this gene. Be sure to appropriately label the ends of the molecule. 5'-ATGCACGGCGACTAG-3' Second letter A UAU Tyr UAC First letter U P с > < A G U UUU UUC Phe UUA UUG CUU CUC CUA CUG L GUU GUC GUA GUG Leu Leu AUU AUC lle AUA AUG Met Val C UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA ACG GCU GCC GCA GCG Ser Pro Thr Ala Cys UAA Stop UGA Stop A Trp UAG Stop UGG CAC His CGU J CGC CAA I CGA Gin CAGG CGG AAA 1 AAG Lys UGU UGC AAU Asn AGC} AAC GAC Asp GAA GAGGIU For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). BIUS Paragraph V Arial G 1 AGA 1 AGG GGU GGC GGA GGG Arg Ser Arg Gly V DCAG DCA DOA UCAG Third letter 10pt < Av V IX Q ... O WORDS POWERED BY TINYWhat 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 needed
- 3’-TCTTCGTGAGATGATATAAGAGTTATCCAGGTACCGGTAAACTGG-5’ 5’-AGAAGCACTCTACTATATTCTCAATAGGTCCATGGCCATTTGACC-3’ Write down the mRNA transcript from DNA above.What is the RNA transcript that would result from the double stranded DNA segment shown below? 5'-GCACGTTGGTCGATCACGTAATATACGCATCGACTCCCGATCGA-3' 3'-CGTGCAACCAGCTAGTGCATTATATGCGTAGCTGAGGGCTAGCT-5' 5'- ينGiven the following tRNA anticodon sequence, derive the mRNA and the DNA template strand. Also, write out the amino acid sequence of the protein encoded by this message. tRNA: UAC UCU CGA GGC mRNA: protein: How many hydrogen bonds would be present in the DNA segment?
- Arg-ser-ser-ala-pro Possibilities mRNA 3’ AGG UCA UCU GCU CCC 5’ 5’ ACC ACG CCU CCU GGC 3’ 3’ UCC ACG CCU ACU GGA 5’ 5’ CGC UCC CCU GCC CCC 3’ Possibilities coding strand 5’ TCC TCG ACT GCT GGA 3’ 3’ TCC TCG TGA CGA CGC 5’ 5’ CGG ACT ACT GCA CCA 3’ 3’ CCC ACG ACT CCT CGC 5’ Possibilities non- coding strand 3’ GGG TCA TCA CGG GGG 5’ 5’ TCC AGC AGC CGC GGC 3’ 3’ GCC TCA AGC CGA GGA 5’ 5’ TGG TGC TGA AGA TCA 3’pcc300ATAAADATATAOOTTAA 1. Use the genetic code table and the information in the diagram below to determine the amino acids that would make up the portion of the polypeptide shown. Include information for a key as well. DNA template 3' G CATA ACAGAGGATT-5' al bnsua AMAm pniwollot erfT E transcription s yd bnsita ebitgeqylog s sidmeaze of beae RNA strandUU UAOUOUU A-emoaodin 5'-CGUA AUUGUC UCCUUA- 3' J J JL erit o elinW (s) translation bluow terdt aspnso sigootiwsone polypeptide viemetis ns ebivo19 (d) ent ot etslanT Key:Give the RNA molecule sequence transcribed from the following DNA sequence of a eukaryotic gene and with the correct 5' and 3' ends. DNA: 5'-ATAGGGCATGT-3' 3'-TATCCCGTACA-5' <--- template strand Group of answer choices 5'-ATAGGGCATGT-3' 3'-UAUCCCGUACA-5' 5'-AUAGGGCAUGU-3' 3'-TATCCCGTACA-5'
- The schematic dlagram below shows the functional organization of transcribing RNA polymerase. Match parts of the diagram (labeled A-G) with the corresponding term from the answer list (design ated 1-12). Polymerase movement ANA POLYMERASE A Rowinding of DNA Unwinding of DNA E NTPs Note that some of the items from the answer list should NOT be used. Labels E and F point to the specific ends of DNA/RNA molecules. 1. Template strand 2. RNA transcript 3. Replication origin 4 Non-Template strand 5. Leading strand 6. 3 end 7. Primer 8. Replication fork 9. Transcription bubble 10. 5' end 11. DNA RNA nybrid 12. Lagging strandCoding DNA 5’- GTG ACT CGT TGT GCC ATT GCA GCT AAA CAC TTC GAG CCC TGT- 3’ mRNA 5’- GUG ACU CGU UGU GCC AUU GCA GCU AAA CAC UUC GAG CCC UGU- 3’ What is the polypeptide sequence?BM4_DNA AND PROTEIN S X /1FAIPQLSDP_g5B-629FSHNpGnTMIEppLS4A71zBd4vcUBqNUILubXONw/formResponse 4. What is the nitrogen base pair of Adenine in transcription? O Cytosine O Uracil O guanine O thymine 5. The central dogma of Molecular Biology states that There are four nitrogen bases in DNA, two purines (adenine and guanine) and two pyrimidines (cytosine and thymine). Which process is not included in the central dogma? duplication transcription translation O translocation Leadple