The digestion of the food in humans begins in the mouth. In the mouth, the salivary amylase enzyme digests and breaks down a small amount of the starch present in the food. The digestion process begins right in the mouth. and continues till the intestine. The carbohydrates (starch) are the ones whose digestion starts in the mouth.
Hence, the blanks can be filled with 'mouth and amylase' respectively.
Answer: Yes
Explanation:
Carrying capacity can be defined as the total number of members of the population of a species that an ecosystem can sustain in terms of providing resources in the form of food, shelter and others. When the resources are available in surplus then the population of a species increases exponentially but declines when resources become scarce. The human population is increasing tremendously all over the world this is supported by the resources like food, water, fossil fuels, air, minerals, and others. But some of these resources are decreasing due to overuse and may not be available in future to sustain the future generation.
Answer:
Viruses rely on the cells of other organisms to survive and reproduce, because they can't capture or store energy themselves
Answer:
True The human body contains 8-10 liters of blood.
Complete question:
Suppose "A" is a dominant gene for the ability to taste phenylthiocarbamide and "a" is a recessive gene for the inability to taste it. Which couples could possibly have both a child who tastes it and a child who does not?
a. father AA, mother aa
b. father Aa, mother AA
c. father Aa, mother Aa
d. father AA, mother AA
Answer:
c. father Aa, mother Aa
Explanation:
According to the given information, the ability to taste phenylthiocarbamide is a dominant trait and is imparted by the allele "A". This phenotype would be expressed in both homozygous and heterozygous conditions. The non-taster phenotype would be expressed in the homozygous recessive genotypes only.
To have both taster and non-taster children, both the parents should have at least one copy of the recessive allele. Among the given options, the father with genotype Aa and the mother with genotype Aa have the possibility to have both taster and non-taster children.
Aa x Aa= 3/4 taster (1/4 AA and 1/2 Aa): 1/4 non-taster (1/4 aa)
