Protists belong to the group eukaryotes (having their DNA enclosedinside the nucleus). They are not plants, nimals or fungi but they act likeone. They can be in general subgroups such as unicellular algae, protozoa andmolds.
Answer:
Organism cloning. Organism cloning (also called reproductive cloning) refers to the procedure of creating a new multicellular organism, genetically identical to another. In essence this form of cloning is an asexual method of reproduction, where fertilization or inter-gamete contact does not take place.
Answer:
When this reaction takes place, water dissolves ions from the mineral and carries them away. These elements have undergone leaching. Through hydrolysis, a mineral such as potassium feldspar is leached of potassium and changed into a clay mineral.
Explanation:
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:
More and more of the medium ground finch population would develop the characteristic of having a thick beak for breaking larger seeds. It would be a small amount of them having the characteristic at first, but then gradually the amount would increase until all of them keep this characteristic.
This is an example of natural selection and adaptation. The two words are related, in a way. Natural selection is when a species change over time with a characteristic that helps them adapt better to their environment. The medium ground finch population has to adapt in the change of environment through natural selection of getting the characteristic of thicker beaks for bigger seeds. Developing this characteristic will help more survive and reproduce, allowing the natural selection process to take place easier and quicker.
Explanation: