Answer:
The answer is not in any of the options and is the following:
to eliminate any remaining Hfr cells.
Explanation:
In a medium that has a streptomycin antibiotic it is used to kill Hfr donor cells after the conjugation is interrupted. This is entirely due to the streptomycin sensitivity allele (str-s) which are found in Hfr strains. The presence of the streptomycin resistance allele, located in the receptor, is used for the specific destruction of Hfr donor cells after conjugation occurs.
Answer:
UUU and UUC
Explanation:
Information that encodes certain functional products called PROTEIN are present in the DNA molecule. However, the information needs to be expressed via the processes of transcription and translation. Transcription is the synthesis of a mRNA using a DNA template.
The mRNA then undergoes translation, where it is read in a group of three nucleotides called CODON. Each CODON specifies a particular amino acid. Due to the degenerate nature of the genetic code, more than one codon can specify one amino acid.
Hence, in the case of phenylalanine amino acid, codons UUU and UUC both specify it. This means that whenever UUU and UUC are read during translation, a phenylalanine is added to the peptide chain.
Answer:
E. All the answer options are correct.
Explanation:
Cilia are very small hair-like, membrane-bound cell structures. They are present on the surface of many eukaryotic cells. They are made of microtubules and are continuous with the plasma membrane of a cell. On a single cell, they are present in large numbers as compared to flagella. The major function of cilia is to move the cell or to move substances such as mucous, fluid over or around the cell.
During DNA Replication, the DNA strand that has the bases 3'ATACGC5' produces a strand with the bases 5'TATGCG3'
Answer: Option B
<u>Explanation:</u>
The DNA Replication produces the newly formed two complimentary base pair strand from the two template base pair strand, so basically the already existing DNA strand is replicated, in order to produce complimentary DNA strand, with the template giving one strand to new each.
Thereby, here the given base was 3' ATACGC 5' so the first thing to produce the strand will be that the coding end will be reversed so the 3' end will have 5' end and 5' end will have 3' end. Then A bonds with T and T bonds with A and, so C bonds with G and G bonds with C, therefore, ATACGC will produce TATGCG. Thereby, answer would be 5'TATGCG3'.
