Answer:
B. Two functions of carbohydrates are structural support and transferring genetic information
Explanation:
Carbohydrates are synthesized in green plants, thanks to photosynthesis, in which carbon dioxide (CO2) and water are converted into glycides, with energy provided by light. The energy stored in the glucose molecule during photosynthesis is released in the degradation (catabolism) of glucose, producing water and CO2.
From the point of view of metabolism, the main function of carbohydrates in plant organisms is to provide energy and provide structural support. This can be seen in the question above, because we can see that in spring, when stem growth is at its fastest pace, cellulose production rates increase in some plants, while in months with less sunlight, the level of starch decreases in some plants.
Answer:
According to the result recorded, there was no cholesterol in the albumin solution. This is probably because the cholesterol is in the egg yolk and not the egg white. In order to get a positive response on the presence of cholesterol in eggs, an egg yolk albumin stock needs to be prepared. So using the same procedure as before and only replacing the egg yolk for the egg white, prepare a 0.5 mL of egg yolk with 4.5 mL distilled water mixture in a test tube. Follow the steps given in the lab manual to check for presence of lipids in this albumin stock. Take a filter paper and add one drop of Sudan III solution onto it, let it dry and then place a drop of the stock prepared and analyze if the color transfer happens.
Hope that answers the question, have a great day!
Answer:
a. DNA polymerase proofreading: consequence of its absence is the DNA mutation
b. Mismatch repair enzymes
: consequence of its absence impedes homologous recombination resulting in the final mutation
c. Nucleotide excision repair enzymes
: the absence of nucleotide cleavage repair enzymes would impede the functioning of damaged DNA repair mechanisms
Explanation:
a. DNA polymerases are the enzymes that form the DNA in cells. During DNA replication (copying), most DNA polymerases can "check their work" with each base they add. This process is called review. If the polymerase detects that you have added a wrong nucleotide (incorrectly paired), remove it and replace it immediately, before continuing with DNA synthesis
b. In homologous recombination, the information from the homologous chromosome that matches that of the damaged one (or from a sister chromatid if the DNA has been copied) is used to repair the fragmentation. In this process the two homologous chromosomes are approached and the undamaged region of the homologue or the chromatide is used as a template to replace the damaged region of the broken chromosome. Homologous recombination is "cleaner" than the union of non-homologous ends and does not usually cause 11 mutations
c. Excision repair: damage to one or a few DNA bases is usually fixed by removing (excising) and replacing the damaged region. In repair by base cleavage, only the damaged base is removed. In nucleotide excision repair, as in the mating repair we saw earlier, a nucleotide section is removed
Answer: The answer is that the phenotypic ratio among phenotypes produced from an F1 X F1 dihybrid cross is 9:3:3:1.
Explanation:
Independent assortment of genes explains how alleles on different chromosomes arrange independently of one another during gamete formation.
So, a dihybrid cross involving TWO characters (e.g Seed color & seed shape) would have its respective alleles DISTRIBUTED whether dominant or recessive, for crossing to occur and yield varying proportion of offspring in the well spread ratio of 9:3:3:1; making it a consequence of independent assortment of genes
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%.
