Why does DNA have a leading strand and a lagging strand during replication? Select the correct row below that fills in the following blanks to answer the question above. ___1st___ has to attach the new free-floating nucleotides to the ___2nd___ of the newly formed strand. The two strands of a DNA molecule are antiparallel, so in order for the lagging strand to be formed in the same direction, it has to be done in Okazaki fragments going in a ___3rd___ direction.

Why does DNA have a leading strand and a lagging strand during replication? Select the correct row below that fills in the following blanks to answer the question above. _1st_ has to attach the new free-floating nucleotides to the _2nd_ of the newly formed strand. The two strands of a DNA molecule are antiparallel, so in order for the lagging strand to be formed in the same direction, it has to be done in Okazaki fragments going in a _3rd_ direction.

Human Biology (MindTap Course List)

11th Edition

ISBN:9781305112100

Author:Cecie Starr, Beverly McMillan

Publisher:Cecie Starr, Beverly McMillan

Chapter21: Dna, Genes, And Biotechnology

Section: Chapter Questions

Problem 1RQ: Why is DNA replication called semiconservative?

See similar textbooks

1st	2nd	3rd
DNA polymerase	3' end	5' to 3'

1st	2nd	3rd
DNA polymerase	5' end	3' to 5'

1st	2nd	3rd
DNA ligase	3' end	5' to 3'

1st	2nd	3rd
DNA helicase	5' end	3' to 5'

Similar questions

Shown below is a double stranded DNA molecule.Replicate this DNA and show the products formed from this replication. In your answer, indicate which strands are the new daughter strands (mark with D and in a different color) and which strands are the already existing parent strands (mark with P as shown below). Label the 5’ and 3’ ends of all DNA. HINT: You should have 2 double-stranded DNA molecules after replication. P: 5’ CATAGGATTCCTACG 3’ P: 3’ GTATCCTAAGGATGC 5’
Below is a diagram of DNA replication as currently believed to occur in E. coli. From specific points, arrows are provided that lead to numbers. Answer the questions below relating to the locations specified by the numbers. (02) What end (5’ or 3’) of the molecule is here? (State which) What enzyme is probably functioning here to deal with supercoils in the DNA? What enzyme is probably functioning here to unwind the DNA? What nucleic acid is probably depicted here? What are these short DNA fragments usually called? What enzyme probably functions here to couple these two newly synthesized fragments of DNA? Is this strand the leading or lagging strand? What end (5’ or 3’) of the molecule is here? (State which)
Which letter represents the following structures: (0.5X6) Okazaki fragment Replication fork Leading strand Lagging strand DNA template RNA primer
Match the following descriptions with the enzymes involved in DNA replication. Adds an RNA primer to begin elongation Removes the RNA primer from the beginning of the newly constructed strands Splices lagging strand segments Cleaves the rung of the DNA double helix ladder Description: Answer Answer Answer Answer Enzyme: Helicase DNA Polymerase Primase DNA Ligase
Show the replication strands in each of these bubbles (note they have different DNA orientations). Label each end of each DNA strand and include arrows to show which direction it is extending. Show the Okazaki fragments in the correct places. 3' 5'
Describe the process of DNA replication as if explaining it to a fellow classmate. Imagine there is an unlabelled diagram in front of you to guide your explanation. (Try to include as many relevant proteins and concepts as possible to help demonstrate an understanding, eg. DNA Helicase, Okazaki Fragments, etc).
A portion of one strand of DNA has the sequence 5′ AATGGCTTA 3′. If this strand is used as a template for DNA replication, write the sequence of the newly synthesized strand in the direction from left to right in which it will be synthesized.
Escherichia coli (E. coli) has one replication origin that contains specific genomic sequences scattering over a 245 base pair region. List the mechanistic steps that can unwind a segment of DNA at the origin of replication site to initiate the DNA replication.
Each new double helix will contain ___________ strand of DNA from the parent strand and ____________ strand of newly created DNA.
Indicate the proteins involved in the following steps of DNA replication in E. coli a. ___________ initiator of replication by binding to the origin of replication of the prokaryote ___________ a dimer that unzips the DNA helix ___________ an enzyme responsible for relieving positive supercoils ahead of replication fork ___________ maintains DNA in single-stranded state ___________ primary replicating enzyme ___________ synthesizes RNA primers The resulting gap by removal of RNA primer is filled in by ___________. h. the resulting nick is sealed by _____________.
A portion of one strand of DNA has the sequence 5′ AATGGCTTA 3′. If this strand is used as a template for DNA replication, which of the following correctly depicts the sequence of the newly synthesized strand in the direction in which it will be synthesized? 5′ TAAGCCATT 3′ 5′ TTACCGAAT 3′ 3′ AATGGCTTA 5′ 3′ TTACCGAAT 5′ 5′ AATGGCTTA 3′
The template strand of a double helical segment of DNA consists of the following sequence: 5’- GTAGCCTTATCTAGCGATCACCGTCCGTATTACTAGTGGCCAGACTCTTTTCACCATGTATAGTTG-3’ Use the given data to answer the succeeding questions and always indicate (and pay attention to) the DIRECTIONALITY of the strands in your answers. Part I. What is the nucleotide order in the complementary DNA strand? Part 2. What is the nucleotide order in the complementary mRNA that can be transcribed from the given DNA strand? Part 3. What will be the overall anticodon sequence in tRNA? Part 4. Following the transitional process, what is the amino acid sequence that will be coded for? Show your answer using ONE-letter amino code starting from N-terminus to C-terminus Part 5. Following translational process, what is the amino acid sequence that will be coded for? Show your answer using THREE-letter amino acid code starting from N-terminus to C-terminus.