Answer:
The concentration of glucose must be maintained within a fairly narrow range in most vertebrates. This statement is an example of — d. homeostasis.
Abnormal tissue growth on a mucous membrane.
Ribosome- both
endo reticulum- both
golgi apparatus- both
cell wall- plant only
vacuoles- plant only
lysosomes- animal only
mitochondria- both
cell membrane- both
cytoplasm- both
chloroplasts- plant only
The amount of energy that is transferred at each level of the food chain is about 10%. This is called the 10% rule (10 per cent rule).
Explanation:
The phenotypes and genotypes of the progeny can be determined by a dihybrid cross of the parents.
The heterozygous male will have the genotype 'SSww' and the heterozygous female will have the genotype 'ssWW'.
When crossed, the F1 offsprings will have a hybrid genotype of 'SsWw'. These offsprings are heterozygous with spotted skin and wooly hair.
On self-crossing of the F1 hybrids, we find four different combinations of the alleles- SW, Sw, SW and sw. The probability of getting each of these combinations is 1/4.
Hence, the probability of any dihybrid type is 1 out of the 16 possible genotypes. Using Punnet square, we find
9 SSWW: 3 SSww: 3 ssWW : 1 ssww
This is the phenotypic ratio of the offsprings.
The ratio of the possible genotypes will be 1:2:1:2:4:2:1:2:1.
