Answer:
Ss and ss
Explanation:
<em>Since the smooth trait (S) is dominant over the wrinkle trait (s), the genotype of the wrinkled pea parent is definitely </em><em>ss</em><em>.</em>
<em>Also, some of the progeny had wrinkled pea texture (ss). Each of the 2 wrinkled alleles must have been donated by each of the parent. It thus means that the smooth pea parent is heterozygous for the trait (</em><em>Ss</em><em>).</em>
Now, let us look at a cross between the two parents;
Ss x ss
Progeny: Ss, Ss, ss, ss.
Phenotypically, 50% of the progeny are smooth peas while 50% are wrinkled.
Looking at the population of each of the phenotype resulting from the cross. 252:247 is approximately a 50:50 ratio.
Hence, the genotypes of the two parents are Ss and ss.
This process is referred to as Ammonification, I believe, basically this allows Nitrogen to be in the appropriate form for plants to absorb it.
Hello TeamFlow, They are all living things, they all need some sort of nutrition, and all plants like fungi, moss, grass etc, have cells. :)
Answer:
To maintain this balance, the Hardy-Weinberg Equilibrium Principle states that a population should meet five main assumptions. There should be random mating, large population size, no mutation, no selection on the gene in question, and no gene flow in or out of the population.
Explanation:
The answer would be <span>insulin-dependent diabetes mellitus.</span>
