The correct answer is fat-soluble vitamins.
The fat-soluble vitamins are soluble in lipids and usually absorbed as fat globules (chylomicrons). They are absorbed via the lymphatic system of the small intestines and then transported into the blood circulation within the body.
Vitamin A (retinol) has an important role in maintaining healthy vision.
Vitamin D (cholecalciferol) is important for bone health and development and it is produced naturally in the human body when the skin is exposed to the sun.
Vitamin E (tocopherol) is an antioxidant that can help the body destroy free radicals.
Vitamin K (phylloquinone) has a role in forming the blood clots.
disintegrate and one large megaspore.
O a gametophytes
O b. pollen sacs
O c pollen grains
O d. microspores
o e ovules
Meiosis and unequal cytoplasmic division in seed plant produce three small cells that disintegrate and one large megastore
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.
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:
Parental)
Double recombinant)
Comparing them we will realize that between
s u TU (parental)
s u tu (double recombinant)
and
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:
Chromosome 1:
---s---TU---u--- (Parental chromatid)
---s---tu---u--- (Double Recombinant chromatid)
Chromosome 2
---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 = (21+17+4+2)/230
P1 = 44/230
P1 = 0.191
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
0.365=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
Note:
CC= ((2 + 4)/230)/0.174x0.191
CC=(6/230)/0.0332
CC=0.7857
The coefficient of interference, I, is complementary with CC.
I = 1 - CC
I = 1 - 0.7857
I = 0.2143
Answer:
Kinds of bones, how old the earth is , and kinds of dog that like t rex bones
Explanation:
i sorrys
Answer: Need sugar for energy
Explanation: Plants exhibit an autotrophic type of nutrition that is, plants manufacture their own food. They have chlorophyll a green pigment located in the leaves to trap sunlight and convert carbon dioxide and water to glucose and oxygen in a process called photosynthesis. The glucose which is a simple sugar is then transported to other plant tissues where they are needed as source of energy for growth and other metabolic functions.
Animals exhibit a heterotrophic type of nutrition that is, animals cannot manufacture their own food rather they depend on plants and/or other animals for food. Animals use sugars derived from food they eat to obtain energy for all bodily functions.
Even though the light dependent reactions of photosynthesis are not affected by a change in temperature, the light independent reactions of photosynthesis can be dependent on temperature. They are reactions of enzymes. As the enzymes approach their optimum temperatures the rate increases.