Answer:
Common examples of social issues: Poverty and Homelessness, Climate Change, Civil Rights and Racial Discrimination, etc.
Explanation:
There's so much more examples of social issues down the list, but that'll take to long, these are enough for now.
 
        
             
        
        
        
Answer:
% GENOTYPE
FF = 0%
Ff = 50%
ff = 50%
% PHENOTYPE
Black fur = 50%
Grey fur = 60%
Explanation:
This question involves a single gene coding for fur color in dogs. The allele for grey fur (F) is dominant over the allele for black fur (f). This means that a heterozygous dog (Ff) will have a grey fur. 
In this question, a heterozygous female (Ff) is said to cross with a recessive male (ff) i.e. Ff × ff. The following gametes will be produced by each parent:
Ff - F and f
ff - f and f
Using these gametes in a punnet square (see attached image), the following genotypic proportion of offsprings will be produced:
Ff - 1/2 = 50%
ff - 1/2 = 50%
Ff is phenotypically GREY while ff is phenotypically BLACK. This means that 50% of the offsprings will be black and 50% will be grey. 
 
        
             
        
        
        
Answer:
the correct answer is c i just tuck the text
Explanation:
 
        
                    
             
        
        
        
Some plants also have large leaves to soak up nutrients and water, very helpful in this biome since it rains a lot. - Having some needles cased in wax, withstand cold temperatures and losing water from their leaves. - Having thin bark and shallow roots enables trees to withstand fire.
        
                    
             
        
        
        
Answer:
F (pp) = 0.04
F (Pp) = 0.32 
F (PP) = 0.64 
Explanation:
Studying population genetics, we used the Hardy-Weinberg equilibrium model. In the case of snails, the H-W principle predicts that the total number of cases in the population should follow the following equation: PP + 2.Pp + pp = 200.
If a population has 200 snails and 8 are affected by a recessive condition, we have 4% affected, or 0.04.
This means that 0.04 corresponds to p², and p corresponds to 0.2.
Therefore, P is equal to 0.8 because p + P = 1.
Knowing the allele frequencies, we go to the frequency of the genotypes.
F (pp) = p² = 0.04 = 8 snails
F (Pp) = 2.Pp = 0.32 = 64 snails
F (PP) = P² = 0.64 = 128 snails