Answer:
OB. The life expectancy of similar organisms
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Answer:
The correct answer is: C. third nucleotide position.
Explanation:
- Genes located on the chromosomes encodes for proteins.
- A Gene is made up of a Deoxyribonucleotide (DNA) sequence which is transcribed into the messenger Ribonucleotide (mRNA) sequence by the help of RNA polymerase.
- This mRNA sequence is further translated into the amino acid sequence, that folds to form the functional protein, by the help of the Ribosome.
- The Ribosome reads the mRNA sequence in the form of triplets (three nucleotide together) and each such triplet nucleotide codes for an amino acid.
- Each such triplet nucleotide is known as a Codon.
- The Genetic Code is a table which represents the amino acid encoded by each codon.
- However, the Genetic Code is degenerate in nature. This means that one amino acid can be coded by more than one codon.
- This is because, among the the three nucleotide positions in a codon only the first two determine the specificity of the amino acid while the third nucleotide, also called the wobble nucleotide, is not specific. Presence of any nucleotide in the third position of the codon will not alter the amino acid encoded by the codon.
- In the given question, organisms producing homologous protein have similar amino acid sequence but they vary in the corresponding nucleotide sequence of the gene which codes for the homologous protein.
- This is because at the nucleotide level the variation lies in the wobble nucleotide position that occupies the third position in the codon.
The cell wall gives strength to a cell.
Answer:
A: Colder before the front and warmer after the front.
Explanation:
After the warm front sweeps over the area, it becomes warmer.
Answer:
1.31 cM
Explanation:
Total offspring = 2205
Since two genes are involved, F1 progeny should have four types of combination. Out of them two are 17 and 12 which definitely means they are in lesser number. Since recombinants are always less than parental progeny in linkage, the given two types are recombinants.
Recombination frequency = (Number of recombinants / Total progeny) * 100
= [ ( 17 + 12 ) / 2205 ] * 100
= ( 29 / 2205 ) * 100
= 1.31 %
Map distance = Recombination frequency
Hence, distance between two genes = 1.31 cM
