BIOLOGY 12E CONNECT ACCESS CARD
BIOLOGY 12E CONNECT ACCESS CARD
12th Edition
ISBN: 9781264938513
Author: Raven
Publisher: MCG
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Chapter 10, Problem 1U

Binary fission in prokaryotes does not require the

a. replication of DNA.

b. elongation of the cell.

c. separation of daughter cells by septum formation.

d. assembly of the nuclear envelope.

Expert Solution & Answer
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Summary Introduction

Introduction:

“Binary fission” in prokaryotes is a process of the division of the parent cell into two daughter cells. This leads to the division of the genetic material like chromosomes inside the cell. The nuclear envelope is absent in the prokaryotes and the chromosomes are present in a region of nucleoid inside the cytoplasm.

Answer to Problem 1U

Correct answer:

Binary fission in prokaryotes is the process of the separation of the chromosomes and their replication during the division of the parent cell into two daughter cells. The prokaryotes do not have a nucleus and therefore assembly of the nuclear envelope does not take place. Therefore, option d. is correct.

Explanation of Solution

Reason for the correct statement:

The prokaryotes do not have a nucleus the nuclear material is suspended in the cell and is termed as the nucleolus and the formation of the nuclear assembly does not take place during the process of binary fission.

Option d. is given as “assembly of the nuclear envelope”.

As, “the nucleus and nuclear envelope are absent in the prokaryotes therefore the process of binary fission involves only the replication and separation of the chromosomes after the division of the parent cell into two daughter cells” is the right answer.

Hence, option d. is correct.

Reasons for the incorrect statements:

Option a. is given as “replication of DNA ”.

Replication of DNA takes place during the process of binary fission. So, it is the wrong answer.

Option b. is given as “elongation of the cell”.

The elongation of the parent cell takes place before cell division. So, it is a wrong answer.

Option c. is given as “separation of daughter cells by septum formation”.

The septum is formed in between the two daughter cells which helps in the formation of the separate cells during binary fission. So, it is a wrong answer.

Hence, options a., b., and c. are incorrect.

Conclusion

The binary fission process involves the division of the parent prokaryotic cell into two daughter cells with replication and separation of the chromosomes. The nucleus and nuclear envelope are absent in the prokaryotes and the chromosomes are present as nucleoid in the cytoplasm.

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For short answer questions, write your answers on the line provided. To the right is the mRNA codon table to use as needed throughout the exam. First letter U บบบ U CA UUCPhe UUA UCU Phe UCC UUG Leu CUU UAU. G U UAC TV UGCys UAA Stop UGA Stop A UAG Stop UGG Trp Ser UCA UCG CCU] 0 CUC CUA CCC CAC CAU His CGU CGC Leu Pro CCA CAA Gin CGA Arg CUG CCG CAG CGG AUU ACU AAU T AUC lle A 1 ACC Thr AUA ACA AUG Mot ACG AGG Arg GUU GCU GUC GCC G Val Ala GAC Asp GGU GGC GUA GUG GCA GCG GAA GGA Gly Glu GAGJ GGG AACASH AGU Ser AAA1 AAG Lys GAU AGA CAL CALUCAO CAO G Third letter 1. (+7) Use the table below to answer the questions; use the codon table above to assist you. The promoter sequence of DNA is on the LEFT. You do not need to fill in the entire table. Assume we are in the middle of a gene sequence (no need to find a start codon). DNA 1 DNA 2 mRNA tRNA Polypeptide C Val G C. T A C a. On which strand of DNA is the template strand (DNA 1 or 2)?_ b. On which side of the mRNA is the 5' end (left or…
3. (6 pts) Fill in the boxes according to the directions on the right. Structure R-C R-COOH OH R-OH i R-CO-R' R R-PO4 R-CH3 C. 0 R' R-O-P-OH 1 OH H R-C-H R-N' I- H H R-NH₂ \H Name Properties
4. (6 pts) Use the molecule below to answer these questions and identify the side chains and ends. Please use tidy boxes to indicate parts and write the letter labels within that box. a. How many monomer subunits are shown? b. Box a Polar but non-ionizable side chain and label P c. Box a Basic Polar side chain and label BP d. Box the carboxyl group at the end of the polypeptide and label with letter C (C-terminus) H H OHHO H H 0 HHO H-N-CC-N-C-C N-C-C-N-GC-OH I H-C-H CH2 CH2 CH2 H3C-C+H CH2 CH2 OH CH CH₂ C=O OH CH2 NH2
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