Choose from the list the molecules necessary for every living thing on earth to survive. Select one or more: a. Lipids b. Carbohydrates c. Phosphate d. Nucleic Acids e. Proteins f. Oxygen ( :
Every living thing requires lipids, carbohydrates, nucleic acids and proteins for survival and proper functioning. These are the biological macromolecules that carry out crucial functions such as energy provision, genetic instruction, and various metabolic reactions.
All living things require certain molecules to survive and function. These molecules include Lipids, Carbohydrates, Nucleic Acids, and Proteins. Lipids are fats that provide long-term energy storage and insulation. Carbohydrates are the body’s main source of energy. Nucleic Acids such as DNA and RNA contain the genetic instructions for the development, function, growth, and reproduction of living organisms. Proteins perform a vast array of functions within organisms, including catalyzing metabolic reactions, DNA replication, responding to stimuli, and transporting molecules from one location to another. Phosphate and oxygen, while also essential to life, are technically not molecules but atoms or ions involved in various biological processes.
Learn more about Essential Molecules for Life here:
Which of the following statements about primates is NOT true?a. All primates typically have five digits on each hand and foot. b. All primates rely more on vision and less on their sense of smell. c. Primates tend to have large brains in relation to their body size d. Primates usually give birth to one offspring at a time.
b. All primates rely more on vision and less on their sense of smell.
Primates make up a group of vertebrate animals and include apes (monkeys), lemurs and humans. All these beings have many similarities to each other, for example, all primates usually have five digits on each hand and foot, primates tend to have large brains in relation to body size and the larger the smarter brain the primate is, primates usually give birth one baby at a time. Not all primates rely more on sight and less on smell; humans tend to rely on sight more than on smell, but most other primates have more developed than humans, so the smell is more important to them.
Note the mechanism of absorption (passive or active transport) of the following food breakdown products, and indicate by a check mark whether the absorption would result in their movement into the blood capillaries of the lymphatic capillaries (lacteals).Substance Mechanism of absorption Blood Lymph Monosaccharides Fatty acids and glycerol Amino acids Water Ca2+ , Na+ , Ca2+
The absorption of monosaccharides takes place via facilitated and cotransport mechanism and is transported through blood capillaries present in the villi.
The absorption of glycerol and fatty acids takes place through diffusion and the majority of them are transported via lymph capillaries, while some are transported through blood capillaries.
The absorption of amino acids takes place via a cotransport mechanism with sodium ions and is transported through blood capillaries.
The absorption of water takes place by the process of osmosis via diffusion and is transported with the help of blood capillaries.
The absorption of calcium and sodium ions takes place via an active transport mechanism, while the absorption of chlorine takes place via the process of diffusion. The transportation of all these ions takes place with the help of blood capillaries.
Monosaccharides, amino acids and ions are generally absorbed through active transport into the blood capillaries, while fatty acids and glycerol are transported into the lymphatic capillaries. Water is absorbed via osmosis into both the blood and lymph capillaries.
The absorption mechanisms of the various food breakdown products are different and determine whether they move into the blood capillaries or into the lymphatic capillaries. This absorption is either through passive or active transport.
Monosaccharides like glucose are absorbed through active transport into the blood capillaries.
Fatty acids and glycerol are also absorbed into the cells lining the small intestine, they are then reassembled into fats and transported into the lymphatic capillaries.
Amino acids are absorbed through active transport into the blood capillaries.
Water is absorbed mainly via osmosis, a type of passive transport, into both the blood and lymph capillaries.
Ions (Ca2+, Na+, Ca2+) are absorbed through active transport into the blood capillaries.
Learn more about Absorption of Food Breakdown Products here:
Fine spines (s), smooth fruit (tu), and uniform fruit color (u) are three recessive traits in cucumbers whose genes are linked on the same chromosome. A cucumber plant heterozygous for all three traits is used in a testcross. The progeny from this testcross are:S U Tu 2 s u Tu 70 S u Tu 21 s u tu 4 S U tu 82 s U tu 21 s U Tu 13 S u tu 17__ Total 230 a. Determine the order of these genes on the chromosome. b. Calculate the map distances between the genes. c. Determine the coefficient of coincidence and the interference among these genes. d. Draw the chromosomes of the parents used in the testcross.
Answer and Explanation:
We have the number of descendants of each phenotype product of the tri-hybrid cross.
S U Tu 2
s u Tu 70
S u Tu 21
s u tu 4
S U tu 82
s U tu 21
s U Tu 13
S u tu 17
The total number, N, of individuals is 230.
In a tri-hybrid cross, it can occur that the three genes assort independently or that two of them are linked and the thrid not, or that the three genes are linked. In this example, in particular, the three genes are linked on the same chromosome.
Knowing that the genes are linked, we can calculate genetic distances between them. First, we need to know their order in the chromosome, and to do so, we need to compare the genotypes of the parental gametes with the ones of the double recombinants. We can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. So:
s u TU (70 individuals)
S U tu (82 individuals)
S U Tu (2 individuals)
s u tu (4 individuals)
Comparing them we will realize that between
s u TU (parental)
s u tu (double recombinant)
S U tu (Parental)
S U TU (double recombinant)
They only change in the position of the alleles TU/tu. This suggests that the position of the gene TU is in the middle of the other two genes, S and U, because in a double recombinant only the central gene changes position in the chromatid.
So, the order of the genes is:
---- S ---- TU -----U ----
In a scheme it would be like:
---s---TU---u--- (Parental chromatid)
---s---tu---u--- (Double Recombinant chromatid)
---S---tu---U--- (Parental chromatid)
---S---TU---U--- (Double Recombinant chromatid)
Now we will call Region I to the area between S and TU and Region II to the area between TU and U.
Once established the order of the genes we can calculate distances between them, and we will do it from the central gene to the genes on each side. First We will calculate the recombination frequencies, and we will do it by region. We will call P1 to the recombination frequency between S and TU genes, and P2 to the recombination frequency between TU and U.
P1 = (R + DR) / N
P2 = (R + DR)/ N
Where: R is the number of recombinants in each region, DR is the number of double recombinants in each region, and N is the total number of individuals. So:
P1 = (R + DR) / N
P1 = (21+17+4+2)/230
P1 = 44/230
P1 = 0.191
P2= = (R + DR) / N
P2 = (21+13+4+2)/230
P1 = 40/230
P1 = 0.174
Now, to calculate the recombination frequency between the two extreme genes, S and U, we can just perform addition or a sum:
P1 + P2= Pt
0.191 + 0.174 = Pt
The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).
The map unit is the distance between the pair of genes for which one of every 100 meiotic products results in a recombinant product. Now we must multiply each recombination frequency by 100 to get the genetic distance in map units:
GD1= P1 x 100 = 0.191 x 100 = 19.1 MU
GD2= P2 x 100 = 0.174 x 100 = 17.4 MU
GD3=Pt x 100 = 0.365 x 100 = 36.5 MU
To calculate the coefficient of coincidence, CC, we must use the next formula:
CC= observed double recombinant frequency/expected double recombinant frequency
observed double recombinant frequency=total number of observed double recombinant individuals/total number of individuals
expected double recombinant frequency: recombination frequency in region I x recombination frequency in region II.
CC= ((2 + 4)/230)/0.174x0.191
The coefficient of interference, I, is complementary with CC.
I = 1 - CC
I = 1 - 0.7857
I = 0.2143
What happens to a cell during INTERPHASE?
What happens to a cell during interphase is when Interphase refers to all stages of the cell cycle other than mitosis. During interphase, cellular organelles double in number, the DNA replicates, and protein synthesis occurs. The chromosomes are not visible and the DNA appears as uncoiled chromatin.
I think it is A or you can get google to help you find it by asking your teacher, parents or using google.
Explain movement of water in the root
Water initially moves into the root hair cells by osmosis, because the mineral content of the cells is higher than that of the surrounding environment. Thus, a root pressure is established and extends into the microscopic tubes of the xylem.