The "scissors" of the molecular biology are: DNA Restriction enzymes.
A restriction enzyme is a protein capable of cleaving a DNA fragment at a characteristic nucleotide sequence called a restriction site. Each restriction enzyme thus recognizes a specific site. Several hundred restriction enzymes are currently known.
Naturally present in a large number of species of bacteria, these enzymes have become important tools in genetic engineering.
The "glue" of the molecular biology are: DNA ligase
In molecular biology, DNA ligases are ligase-class enzymes that catalyze the formation of a phosphodiester bond between two segments of DNA. DNA ligases are involved in several essential cellular processes of DNA metabolism: in DNA replication, suture of Okazaki fragments, and in DNA repair and homologous recombination.
The use of these tools in molecular biology: Cloning
Molecular cloning is one of the bases of genetic engineering. It consists of inserting a DNA fragment (called insert) in an appropriate vector such as a plasmid for example. The new plasmid thus created will then be introduced into a host cell, generally the Escherichia coli bacterium. This will then be selected and multiplied to obtain a large amount of the plasmid of interest. Cloning a gene involves inserting it into a plasmid. A clone will be the bacterial transformant that contains this particular plasmid. In this case we speak of clone because all the individuals of the bacterial colony are genetically identical. Molecular cloning is thus different from reproductive cloning (creating an individual genetically identical to another but of a different age) or therapeutic cloning (making tissues from stem cells to perform transplants compatible with the recipient).
Molecular cloning requires restriction enzymes capable of cleaving the DNA, and DNA ligase capable of re-gluing the DNA fragments. Ligase was isolated for the first time from T4 bacteriophage. This enzyme is involved in the repair and replication of DNA. It can bind DNA fragments with compatible sticky ends. At higher concentration, this enzyme is also able to bind two ends of DNA as shown here. T4 DNA ligase works using ATP and Mg ++. It has an activity optimum of 16 ° C, but remains active at room temperature.
Answer:
look at the pictures
Explanation:
1. If we cross the parental generation which is RR x rr, the probability of the F1 generation to get rough coat is 100% and they will be a carrier of the recessive trait smooth coat.
Since the phenotypic ratio is 100% heterozygous Rr, in crossing the F1 to get the F2, we will use the genotype Rr. To get the F2, use the cross Rr x Rr.
The phenotypic ration for F2 is 3:1. There is 75% to get rough coat and 25% smooth. The answer is based on the result on the Punnett square. On the other hand, the genotypic ratio is 1:2:1. There is 25% probability to get RR genotype, 50% Rr, and 25% rr.
2. Since the two parental mice got 6 albino offspring and 5 brown mice offspring, it is approximately 50%. it takes a parental mice who is Brown that is carrying an albino trait crossed with an albino to get offspring with almost the same number. Therefore, the genotype of the brown mice is Aa.
Cell membrane is a thin line.it controls the movement of materials into and out of the cell. it is partially permeable,that is,it allows only certain materials to pass through.
The correct answer is:
DNA template 3'– ……… –5'
RNA transcript 5'– ……… –3'
Protein product H2N– ……. –COOH
In nucleic acids, DNA and RNA 5’ end refers to phosphate group attached to the 5′ carbon of the ribose ring, while 3’ end refers to the ribose -OH substituent. Nucleic acids can only be synthesized in the 5′-to-3′ direction, which means that the DNA template is 3’—5’. Translation of the protein by the ribosome will also proceed in a 5′ to 3′ direction, which means that the mRNA template is 3’—5’.
The protein during the synthesis is extend from its N terminus toward its C terminus.
Entropy, is mainly responsible for the resting state of this protein.
Hydrogen bonds, disulfide bonds, and dipole-dipole interactions produce the complicated folding styles seen in secondary and tertiary structure. however, if a protein does not have these elements of structure, it'll undertake a kingdom in which its entropy is maximized.
Secondary shape is determined through hydrogen bonding within the amino acid chain backbone. Tertiary structure is the entire protein's form, decided Entropy Interplay and hydrophobic forces. The tertiary shape of a protein is the 3 dimensional shape of the protein. Disulfide bonds, hydrogen bonds, ionic bonds, and hydrophobic interactions all affect the form a protein takes.
The nonpolar amino acids have formed the nonpolar center of the protein, weak van der Waals forces stabilize the protein. furthermore, hydrogen bonds and ionic interactions between the polar, charged amino acids make a contribution to the tertiary structure.
Disclaimer: your question is incomplete, please see below for complete question
A. Entropy
B. Hydrogen bonds between amino acid residues
C. Dipole-dipole interactions
D. None of the above
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