Answer:
A. Preventing fish from moving along a river.
Explanation:
1. Dams create barriers and trap the water that builds up behind it. This prevents fish from moving along the river as there's a barrier in place and is just overall harder for the fish to move freely along a river because of it.
2. It's the only negative option as the others are all positive outcomes, while the question asks for a negative outcome instead.
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Answer:
Plant cells are eukaryotic cells present in green plants, photosynthetic eukaryotes of the kingdom Plantae. Their distinctive features include primary cell walls containing cellulose, hemicelluloses and pectin, the presence of plastids with the capability to perform photosynthesis and store starch, a large vacuole that regulates turgor pressure, the absence of flagella or centrioles, except in the gametes, and a unique method of cell division involving the formation of a cell plate or phragmoplast that separates the new daughter cells.
Explanation:
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Answer:
Animal cells do not have a cell wall or chloroplasts but plant cells do. Animal cells are mostly round and irregular in shape while plant cells have fixed, rectangular shapes.
Answer:
global warming is a part of climate change
Explanation:
climate change refers to the way that climate is changing on earth this can be global warming or also cooling
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
O DNA comparisons
o observations
O comparison of fossils
O experimentation
Answer:
DNA comparisons
Explanation:
The molecular nature of DNA and its role in inheritance and evolution was not known during Darwin's time. Therefore, it was not possible to compare similarities in DNA sequences between different species to understand their evolutionary relationships.
This type of method was not in regular use until the 1980s/1990s, when sequencing and DNA amplification technologies were invented
o disturb the ecosystem because another species will not be able to replace it.
o not affect the ecosystem because another species will not be able to replace it.
disturb the ecosystem because another species will take its place.
Answer:
disturb the ecosystem because another species will not be able to replace it.
Explanation: