<span><u>The answer is A. 72.25 percent.</u>
The Hardy-Weinberg principle can be used:</span>
<em>p² + 2pq + q² = 1</em> and <em>p + q = 1</em>
where <em>p</em> and <em>q</em> are the frequencies of
the alleles, and <em>p²</em>, <em>q²</em> and <em>2pq </em>are the
frequencies of the genotypes.
<span>The <em>r</em> allele (<em>q</em>) is found in 15% of the population:
q = 15% = 15/100
Thus, q = </span><span>0.15
To calculate the <em>R</em> allele frequency (<em>p</em>), the formula p + q = 1 is
used:
If p + q = 1, then p = 1 - q
p = 1 - 0.15
Thus, </span><span>p = 0.85
Knowing the frequency of the <em>R</em> allele (<em>p</em>), it is easy to determine the
frequency of the RR genotype (p²):
p² = 0.85² = 0.7225
Expressed in percentage, p² = 72.25%.</span>
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(sry have to make an answer 20 characters /!:!27837)
Answer:
Co-dominance because he expresses both of the alleles simultaneously
Explanation:
The ABO blood group system is used by humans. This blood group type is controlled by multiple alleles. Alleles A and B are both dominant over allele O but are co-dominant. Co-dominance is a type of inheritance pattern in which two alleles of a gene both express themselves i.e. neither is recessive.
This is the case of this family whose parents have a genotype of AO (blood type A) and BO (blood type B) respectively. The children have blood types A, B, and AB. However, the child with genotype AB possesses both the A and B allele, which are both expressed in his blood group (phenotype), hence, it can be said that the child is exhibiting CO-DOMINANCE for the blood group trait.
Answer:
The greatest in the first blank
the least in the second blank
Explanation:
