It is known as Homo Erectus. It’s a Latin word.
The Smooth (ER) endoplasmic reticulum
Answer:
The mentioned parental types are c+m- and c-m+. Thus, the recombinants will be c+m+ and c-m-.
Now, the given distance between c and m is 8 map units. Thus, the recombinant frequency is 8% or 0.08.
The total recombinants from 1000 plaques will come out to be 80,
Thus, the recombinants of each type will be 40.
Total parental type will be 920, and therefore, each parental type count will be 460.
Thus, expected c+m- = 460, expected c-m+ = 460, expected c+m+ = 40 and expected c-m- = 40.
Answer:
Since, the original DNA sequence has not been provided, the mutation can be an insertion/deletion or a frameshift mutation.
- Mutated DNA
- Frameshift mutation/ insertion or deletion
- All the amino acids changed after the point mutation
Explanation:
Frameshift Mutation:
- A frameshift mutation is the alteration in the reading frame of the DNA due to the addition/deletion of one or two nucleotides.
- This type of mutation moves the mRNA sequence one or two bases forwards or backwards which disrupts the three base codons sequence required for translation into proteins.
- The CT at the end of the sequence is indicative of a frameshift in the DNA reading frame.
- Frameshift mutation affect all amino acids in a polypeptide chain as all codons are moved one or two steps forwards or backwards.
