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Human Fbh1 helicase is important in the process of DNA replication. When a muta occurs during the production of Fbh1, the result is a mutant Fbh1 that binds at the replication fork and prevents any helicase protein from attaching to the strand. DNA-polymerase RNA primer DNA-ligase DNA primase Lagging 3 strand 3' 5' 5' 3' Leading strand Helicase Topoisomerase DNA-polymerase Single-strand binding proteins Based on this information and the image shown, what would happen during DNA replication if this mutant helicase were present? Topoisomerase would unwind the DNA and an RNA primer would attach to the DNA molecule and initiate replication. The process would then stop at the blue triang because helicase is needed to separate the strands of DNA. Topoisomerase would unwind the DNA, but then the process would stop at the blue because helicase, the RNA primer, would not be able to attach to the DNA mole and initiate replication. The process would begin at the blue triangle when topoisomerase unwinds the DNA RNA primer attaches to the DNA molecule and initiates replication. DNA polyme would begin the synthesis of each side, but the process would stop because hel would not be present to rejoin the replicated strands. The process could not begin at the blue triangle because helicase is needed to sepa the strands of DNA and initiate the process.

Human Fbh1 helicase is important in the process of DNA replication. When a muta occurs during the production of Fbh1, the result is a mutant Fbh1 that binds at the replication fork and prevents any helicase protein from attaching to the strand. DNA-polymerase RNA primer DNA-ligase DNA primase Lagging 3 strand 3' 5' 5' 3' Leading strand Helicase Topoisomerase DNA-polymerase Single-strand binding proteins Based on this information and the image shown, what would happen during DNA replication if this mutant helicase were present? Topoisomerase would unwind the DNA and an RNA primer would attach to the DNA molecule and initiate replication. The process would then stop at the blue triang because helicase is needed to separate the strands of DNA. Topoisomerase would unwind the DNA, but then the process would stop at the blue because helicase, the RNA primer, would not be able to attach to the DNA mole and initiate replication. The process would begin at the blue triangle when topoisomerase unwinds the DNA RNA primer attaches to the DNA molecule and initiates replication. DNA polyme would begin the synthesis of each side, but the process would stop because hel would not be present to rejoin the replicated strands. The process could not begin at the blue triangle because helicase is needed to sepa the strands of DNA and initiate the process.

Human Anatomy & Physiology (11th Edition)
Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:Elaine N. Marieb, Katja N. Hoehn
Chapter1: The Human Body: An Orientation
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Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...
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Human Fbh1 helicase is important in the process of DNA replication. When a mutation occurs during the production of Fbh1, the result is a mutant Fbh1 that binds at the replication fork and prevents any helicase protein from attaching to the strand. DNA-polymerase RNA primer DNA-ligase DNA primase Lagging 3' strand 3' 5 5' 5' 3' Leading strand Helicase DNA-polymerase Topoisomerase Single-strand binding proteins Based on this information and the image shown, what would happen during DNA replication if this mutant helicase were present? Topoisomerase would unwind the DNA and an RNA primer would attach to the DNA molecule and initiate replication. The process would then stop at the blue triangle because helicase is needed to separate the strands of DNA. Topoisomerase would unwind the DNA, but then the process would stop at the blue triangle because helicase, the RNA primer, would not be able to attach to the DNA molecule and initiate replication. The process would begin at the blue triangle when topoisomerase unwinds the DNA and an RNA primer attaches to the DNA molecule and initiates replication. DNA polymerase would begin the synthesis of each side, but the process would stop because helicase would not be present to rejoin the replicated strands. The process could not begin at the blue triangle because helicase is needed to separate the strands of DNA and initiate the process.
Transcribed Image Text:Human Fbh1 helicase is important in the process of DNA replication. When a mutation occurs during the production of Fbh1, the result is a mutant Fbh1 that binds at the replication fork and prevents any helicase protein from attaching to the strand. DNA-polymerase RNA primer DNA-ligase DNA primase Lagging 3' strand 3' 5 5' 5' 3' Leading strand Helicase DNA-polymerase Topoisomerase Single-strand binding proteins Based on this information and the image shown, what would happen during DNA replication if this mutant helicase were present? Topoisomerase would unwind the DNA and an RNA primer would attach to the DNA molecule and initiate replication. The process would then stop at the blue triangle because helicase is needed to separate the strands of DNA. Topoisomerase would unwind the DNA, but then the process would stop at the blue triangle because helicase, the RNA primer, would not be able to attach to the DNA molecule and initiate replication. The process would begin at the blue triangle when topoisomerase unwinds the DNA and an RNA primer attaches to the DNA molecule and initiates replication. DNA polymerase would begin the synthesis of each side, but the process would stop because helicase would not be present to rejoin the replicated strands. The process could not begin at the blue triangle because helicase is needed to separate the strands of DNA and initiate the process.
The two-dimensional structures of mRNA, tRNA, and rRNA are shown in the diagrams. TRNA MRNA FRNA Which of the following might be one contributing factor to the function of rRNA based on comparisons between the structures of the three molecules? rRNA molecules are sufficiently flexible because they contain more RNA loops. rRNA molecules are more reactive because they have longer chains of unmatched bases. rRNA molecules are sufficiently stable because the sequence of bases allows for more internal hydrogen bonds. FRNA molecules are more rigid because they have a more regular sequence of bases than other RNA molecules.
Transcribed Image Text:The two-dimensional structures of mRNA, tRNA, and rRNA are shown in the diagrams. TRNA MRNA FRNA Which of the following might be one contributing factor to the function of rRNA based on comparisons between the structures of the three molecules? rRNA molecules are sufficiently flexible because they contain more RNA loops. rRNA molecules are more reactive because they have longer chains of unmatched bases. rRNA molecules are sufficiently stable because the sequence of bases allows for more internal hydrogen bonds. FRNA molecules are more rigid because they have a more regular sequence of bases than other RNA molecules.
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