Answer:
The cell interior would experience higher than normal Na+ concentrations and lower than normal K+ concentrations.
Explanation:
The Na/⁺K⁺ pump is an ATPase pump which is responsible for maintaining low Na⁺ and high K⁺ concentrations within the cytoplasm while maintaining high Na⁺ and low K⁺ concentrations in the extracellular fluid.
Since these two ions are moved against their concentration gradient, ATP hydrolysis is required to provide the energy for this process. This is done by moving in two K⁺ ions inside while moving three Na⁺ ions outside the cell for every molecule of ATP hydrolysed to ADP and Pi.
If a competitive non-hydrolyzable analog of ATP is applied on the cytoplasmic side of a plasma membrane that contained a large concentration of the Na/⁺K⁺ pump, it will act by inhibiting the action of the Na/⁺K⁺ pump. This will result in an accumulation of Na⁺ ions inside the cell and lower than normal K⁺ ions concentration.
Admitting that the "a" is a capital A for normal pigmentation and "d" is a capital D for dimpled chin, meaning that these are the dominant traits, the fraction expected to be albino with a non-dimpled chin is of 1/16.
When two heterozygous are crossed and two characteristics are being analysed, the offspring quantity that will possess the two recessive traits can be represented by 1/16. This is easily confirmed when a Punnett square is made. Considering that both parents were heterozygous, on both sides of the crossing in the Punnett square, you would have the following alleles' combination: AD, Ad, aD, and ad. The offspring that would be homozygous recessive (aadd) would correspond to only 1/16.
Answer:
DNA
Explanation:
<h2>What is DNA?</h2>
DNA is a self-replicating material that is present in nearly all living organisms as the main constituent of chromosomes. It is the carrier of genetic information.
Answer:
C low competition for glucose
Explanation:
From the given answer choices and information,
It cannot be A since there is no visual disease or any indication of disease in the experiment
Cannot be B, since space would not be an issue since it clearly hit 111 on day 3
and cannot be D since there is no indication of a change in temperature.
However, we know Petri Dishes have nutrients to stimulate cell growth, and those resources are not unlimited therefore we can only attest that a large portion of the nutrients have been consumed and starved some of the cells thus causing a population decrease
