Answer:
25% or 1/4
Explanation:
The gene for colour in Heliodors is controlled by two contrasting alleles that codes for Red (R) and Yellow (Y) colours. However, these two alleles exhibit incomplete dominance, which is a phenomenon whereby a combination of both alleles gives rise to a third intermediate phenotype that is a blending of the other two parental phenotypes. In this case, both colours gives rise to a heterozygous Orange coloration (RY) in Heliodors.
However, if two orange Heliodors (RY) are crossed, four possible offsprings will be produced with the genotypes: RR, RY, RY, YY. This shows a phenotypic ratio of 1 red: 2orange: 1yellow. Hence, the probability of having a child with red coloration is 1 out of 4 possible offsprings i.e. 1/4.
Expressing this in percentage, we have 1/4 × 100 = 25%.
The chemicals that helps us digest food are called enzymes.
Too bad there's not the options, but the answer must be the hydroxyl and carbonyl group.
The solubility of sucrose in water as well as organic solvents is very high. In non-aqueous solvents is generally lower. On the other hand, sucrose is not soluble in nonpolar solvents. This solubility in the polar solvent as well as the water is due to the hydroxyl (-OH) and carbonyl (-C = O) functions creating hydrogen bonds with the solvent molecules making it easier to dissolve.
<span>a.
</span>The frequency of the dominant allele: ___0.4______
<span>b.
</span>The frequency of the recessive allele: ____0.6_____
<span>c.
</span>The percentage of mice that are homozygous
dominant: __16%_______
<span>d.
</span> The
percentage of mice that are heterozygous: _48%________
<span>e.
</span>The percentage of mice that are homozygous
recessive: __36%_______
Let us assign the dominant allele (that of brown hair) letter R
while
We assign the recessive
allele (that of white hair) letter r
We then note down the
Hardy-Weinburg equation p2 + 2pq + q2 = 1
Brown fur population (p2
+ 2pq) = 64% = 0.64
White fur population (q2) = 36% = 0.36
Then we also remember that the frequencies of both allele
add up to 1 (p + q = 1);
Therefore q = ü.36
= 0.6
P = 1 – q = 1 – 0.6 = 0.4
The heterozygous population will be 2*0.6*0.4 = 0.48 = 48%
Homozygous domain population will therefore be (64% - 48%) =
16%
You should probaly put the images up, so people can help you (:
