A organism adapts to its environment by using stuff like a long layer of blubber on walruses and polar bears in the artic
Answer:
1. CGAGGTT → CGTT (Deletion)
2. ATTCGG → ATTCGGATTCGG (Duplication)
3. CTTAAT → TAATTC (Inversion)
4. CTTAAT → CTTAACGCT (Insertion)
5. CGAT → CTAT (Substitution)
6. CCGGTT + TTAGGC = CCGTTA + GTTGGC (Translocation)
Explanation:
1. CGAGGTT → CGTT (Deletion) ---- This is called deletion because it involves the removal of 3 base pairs (AGG) from the DNA sequence.
2. ATTCGG → ATTCGGATTCGG (Duplication) ---- In this case, the particular sequence (ATTCGG) is copied again or duplicated.
3. CTTAAT → TAATTC (Inversion)----- This is called inversion mutation because the DNA sequence breaks off and is reattached but this time in a reverse order i.e. CTT becomes TTC, placing the last base first and the first base last.
4. CTTAAT → CTTAACGCT (Insertion) ------ This is called insertion mutation because it involves the addition of extra base pairs (CGC) into the sequence. The Insertion occurs between the last A and T nucleotide.
5. CGAT → CTAT (Substitution) ----- This is called substitution because Guanine base is replaced by Thymine in the DNA sequence. It is specifically called a transversion substitution because a purine (Guanine) is replaced by a pyrimidine (Thymine). It is called a point mutation because it involves a single base.
6. CCGGTT + TTAGGC = CCGTTA + GTTGGC (Translocation) ----- in this case, CCGGTT and TTAGGC are sequences on different chromosomes. Portions of sequence on the first chromosome (GTT) and second chromosome (TTA) breaks off and gets reattached/exchanged in each other i.e. the first chromosome gets TTA while the second gets GTT. This kind of mutation is called translocation.
Anaerobes/ Anaerobic organisms
Answer:
Like all viruses, phages are simple organisms that consist of a core of genetic material (nucleic acid) surrounded by a protein capsid. The nucleic acid may be either DNA or RNA and may be double-stranded or single-stranded.
Explanation:
Answer:
148
Explanation:
According to Hardy-Weinberg equilibrium,
p + q = 1
p² + 2pq + q² = 1 where,
p = frequency of dominant allele
q = frequency of recessive allele
p² = frequency of homozygous dominant genotype
2pq = frequency of heterozygous genotype
q² = frequency of homozygous recessive genotype
Here,
Total population = 592
Number of NN people = 148
Frequency of N blood group or NN genotype (q²) = 148/592 = 0.25
q = √0.25 = 0.5
p = 1 - q
= 1 - 0.5 = 0.5
Hence, p = 0.5
Frequency of MM genotype = p² = 0.25
Number of people with MM genotype = 0.25*592 = 148
Hence, 148 people will have MM genotype or M blood group.
