Explanation:
The problem says that the hairless phenotype never breeds true. That means that it's not the result of a homozygous genotype (H₁H₁ or H₂H₂), so it is caused by the heterozygous genotype (H₁H₂).
The <u>expected </u>offspring from the cross between two Mexican hairless would be:
<h3>P
H₁H₂ x
H₁H₂</h3><h3>F1 1/4
H₁H₁, 2/4
H₁H₂ and 1/4
H₂H₂.</h3>
And the <u>expected</u> phenotypic ratio 3:1. However, the observed offspring shows a 2:1 ratio. What's happening?
If the observed phenotypic ratio in the offspring of a monohybrid cross (a single gene with two alleles) is 2:1, we can suspect that one of the genotypes is lethal in homozygosis and therefore does not appear in the progeny (the puppies are born dead).
If we proposed that the H₂ allele is lethal in homozygosis, then:
- The H₁H₁ genotype would cause normal puppies --> 1
- The H₁H₂ genotype would cause hairless puppies --> 2
- The H₂H₂ is lethal and causes the death of puppies --> 0
The phenotypic ratios change to 2:1, as observed in the experiment.
Answer:
The solution would be HYPOTONIC. The concentration of water is higher inside the cell
Explanation:
Hypertonic: cell shrink
Isotonic: cell will stay the same
Hypotonic: cell will swell
Answer:
The correct answer is option b. "Alpha eventually replaced by theta".
Explanation:
Electroencephalogram (EEG) techniques allows researchers to monitor the phases of sleeping according of what brain waves are seen in the equipment. A normal adult connected to EEG will respond with alpha waves during a wakeful state. However, as the person becomes drowsy and enters to the first stage of sleep, the alpha waves will be eventually replaced by theta waves. Theta waves are the dominant waves during sleeping, and also could be seen during deep meditation.
As you stated before, cellulose is found throughout the cell walls of plant cells. Cellulose makes cell walls rigid, so that would indicate that cellulose is a carbohydrate.
Answer:
the role an organism plays in a community
