Comparing genetic sequences is the most accurate way of determining evolutionary relationships. This is a quantitive measure, ( it uses numbers) we can use gene sequencing to identify how much of the genetic sequence in A is the same in organism B. The greater the similarities the more likely it is they have recently evolved from a common ancestor. The closer the ancestor, the less time has passed which means there’s been less time to develop mutations a base variation. That’s why there’s more genetic similarities between two things that are closely related.
Hope that makes sense
Answer:
The correct option is this: A specific enzyme removes the hydrogen from the monosaccharide and the hydroxide from the polysaccharide, creating a bond between the two and creating a water molecule.
Explanation:
The process by which a monosaccharide molecule is added to an existing polysaccharide in biological systems is called CONDENSATION POLYMERIZATION.
During the process of condensation, an hydrogen atom and an hydroxide molecule from the reactants come together to form a water molecule; this leads to the formation of covalent bond between the reactants.
Condensation reaction is used in living organisms to produce complex macro molecules such as carbohydrates, proteins and lipids, which are needed by the body for health growth.
Remember, condensation reaction always results in loss of water molecules and formation of more complex molecule. One example of monosaccharide is glucose while cellulose (found in plants) and starch (found in animals) are examples of polysaccharides.
Answer:
0.035
Explanation:
<u>cv+ is the wild-type dominant allele over cv, therefore:</u>
- cv+cv+ and cv+cv cause wild-type phenotype for crossveinless
- cv cv causes the crossveinless phenotype
<u>Sb is a dominant mutant allele over wild-type Sb+, therefore:</u>
- Sb Sb and Sb Sb+ cause Stubble phenotype
- Sb+ Sb+ causes wild type phenotype for Stubble
<h3><u>Test cross</u></h3>
It's the cross between the heterozygous female with a homozygous recessive male. Remember that cv and Sb+ are the recessive alleles.
X 
-The male produces only 1 type of gamete: cv Sb+
-The female produces 4 types of gametes:
- cv Sb+ ] Parental
- cv+ Sb ] Parental
- cv Sb ] Recombinant
- cv+ Sb+ ] Recombinant
The genes are linked and separated by 7 map units. A distance of 7 mu means that 7% of the resulting gametes will be recombinant. Because there are 2 possible recombinant gametes, each of them will appear in 3.5% of the cases.
The genotypes and proportions of the offspring resulting from the test cross can be seen in the Punnett Square. The phenotypically wild-type individuals will have the genotype cv+ Sb+ / cv Sb+ (heterozygous for crossveinless and homozygous recessive for Stubble) and a 0.035 proportion.
Category C animals are not
subjected to procedures that involve pain or distress or would be required to
use pain-relieving drugs. Routine procedures such as injections and blood
sampling from veins that produce only mild, transient pain or discomfort are
reported in this category.
