“His third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A. Notice that the forces are exerted on different objects.”
Answer: Natural Selection
Explanation:
Natural Selection is the process by which the chances of the fittest organism to survive increases and rest of the organisms has less chances of survival.
The diversity of the organisms increases the survival rate of the organism. The natural selection shows that the fittest organism will have a greater probability to survive.
The more stable the organism more will be the probability of the organism to resist the change.
Answer:
Please find the expected genotypes and phenotypes of the progenies of each cross below.
Explanation:
In the MN blood-group locus of the gene, alleles LM and LN exhibit codominance i.e one allele is not dominant or recessive to the other, hence, they are both expressed when they occur in a heterozygous state (MN).
Considering the following crosses (find the punnet square attached);
a)LMLM x LMLN - The progeny are LMLM and LMLN in the genotypic ratio 1:1. Phenotypic ratio is Blood type M (1) : blood type MN (1)
b) LNLN x LNLN - The progeny are all LNLN offsprings with a phenotypic and genotypic ratio 4:0. All offsprings will have a blood type N (4)
c) LMLN x LMLN - The progenies are LMLM, LMLN and LNLN in the genotypic ratio 1:2:1 respectively. The phenotypic ratio is Blood type M (1) : L
Blood type MN (2) : Blood type N (1)
d) LMLN x LNLN - The progeny are LMLN and LNLN with genotypic ratio 1:1 and phenotypic ratio blood type MN (1) : blood type N (1)
e) LMLM x LNLN - The progeny are all LMLN offsprings with penotypic ratio blood type MN (4)
Answer:
Statement C is the only one that is necessarily true for exons 2 and 3. It is also true for exons 7 and 8. While statements A and B could be true, they don’thave to be. Because the protein sequence is the same in segments of the mRNA that correspond to exons 1 and 10, neither choice of alternative exons (2 versus 3, or 7 versus 8) can alter the reading frame. To maintain the normal reading frame—whatever that is—the alternative exons must have a number of nucleotides that when divided by 3 (the number of nucleotides in a codon) give the same remainder. Since the sequence of the a-tropomyosin gene is known, it is possible to check to see the actual state of affairs. Exons 2 and 3 both contain the same number of nucleotides, 126, which is divisible by 3 with no remainder.
