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2 years ago

which action is caused by gravity? glaciation glaciation frost wedging frost wedging mass wasting mass wasting aeolian bands

Biology

1 answer:

Rom4ik [11]2 years ago

3 0

Mass wasting is caused by gravitational force.

The motion of soil and rock down the slope under the action of gravity is called mass wasting.

The most common examples of mass wasting are the downward flow of debris from a mountain, slubs, and rock falls.

Usually, these events occur as a result of an earthquake or seismic activity.

However, such events are initiated by frequent rainfall, volcanic activity, shaking, land sliding, weak slope surfaces, and storm waves. etc. Excessive precipitation is also one of the primary reasons behind the mass wasting process.

If you need to learn more about transcription gravitational force click here:

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Identify the pregnancy prevention method(s) that prevent the release of an egg, thickens the cervical mucus and reduces the buil

Darya [45]

The correct answers are a. vaginal ring and c. oral contraceptives.

The vaginal ring is a soft, plastic ring that releases a continuous dose of the hormones oestrogen and progestogen into the bloodstream and prevents pregnancy. Oral contraceptives are also drugs for preventing pregnacy and also contain combinations of the hormones estrogen and progestin )or progestin alone=. Estrogen and progestin inhibit the release of the hormones luteinizing hormone (LH) and follicle stimulating hormone (FSH) which have a role in the ovulation.

8 0

3 years ago

Devise a method for determining the molecular weight of the fragment you amplified and describe that method below. make sure to

11111nata11111 [884]

The method used here is agarose gel elecrtrophoresis in which DNA ladder is used which is a set of standard used to determine the size of DNA used .The DNA ladder is put adjacent to the DNA fragments to be visualized Suppose if the ladder shows 300 basepair so the DNA adjacent to this number will be of this size.These are the set standards used to determine the size of DNA or RNA. For a range we can take 2% of agrarose to run DNA fragments upto 500 Basepair. Most of the amplified DNA falls into this category size.

7 0

4 years ago

In Drosophila melanogaster, vestigial wings (vg) is recessive to normal wings (vg+), black body (b) is recessive to gray body (b

yan [13]

Correct progeny phenotype:

1779 vestigial wings, black body and purple eyes, vg b pr
1665 normal wings, a grey body and red eyes, vg+ b+ pr+
252 normal wings, an black body and purple eyes, vg+ b pr
241 vestigial wings, a gray body and red eyes, vg b+ pr+
131 normal wings, an black body and red eyes, vg +b pr+
118 vestigial wings, a gray body and purple eyes, vg b+ pr
13 vestigial wings, an black body and red eyes, vg b pr+
9 normal wings, a gray body and purple eyes, vg+ b+ pr

Answer:

The order of these genes is vg --- pr --- b
Map distances between the genes vg/pr = 12.2 MU
Map distance between the genes pr/b = 6.4 MU
Map distances between the genes vg/b = 18.6 MU

Explanation:

We know that

•Normal wings expressed by vg+ is dominant over vestigial wings, vg

•Gray body b+ is dominant over black body

•Red eyes, pr+, is dominant over purple ayes, pr

We have the number of descendants of each phenotype product of the tri-hybrid cross.

•1779 vestigial wings, black body and purple eyes vg b pr

•1665 normal wings, a grey body and red eyes vg+ b+ pr+

•252 normal wings, an black body and purple eyes vg+ b pr

•241 vestigial wings, a gray body and red eyes vg b+ pr+

•131 normal wings, an black body and red eyes vg +b pr+

• 118 vestigial wings, a gray body and purple eyes vg b+ pr

•13 vestigial wings, an black body and red eyes vg b pr+

• 9 normal wings, a gray body and purple eyes vg+ b+ pr

The total number of individuals is 4208.

In a tri-hybrid cross, it can occur that the three genes assort independently or that two of them are linked and the third 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)

1779 vestigial wings, black body and purple eyes vg b pr

1665 normal wings, a grey body and red eyes vg+ b+ pr+

Double recombinant)

13 vestigial wings, an black body and red eyes vg b pr+
9 normal wings, a gray body and purple eyes vg+ b+ pr

Simple recombinant)

252 normal wings, an black body and purple eyes vg+ b pr
241 vestigial wings, a gray body and red eyes vg b+ pr+
131 normal wings, an black body and red eyes vg +b pr+
118 vestigial wings, a gray body and purple eyes vg b+ pr

Comparing parental with the double recombinants we will realize that between

vg b pr (parental)

vg b pr+ (double recombinant)

and

vg+ b+ pr+ (Parental)

vg+ b+ pr (double recombinant)

They only change in the position of the alleles pr/pr+. This suggests that the position of the gene pr is in the middle of the other two genes, vg and b, because in a double recombinant only the central gene changes position in the chromatid.

So, the order of the genes is:

---- vg ---- pr -----b ----

Now we will call Region I to the area between vg and pr and Region II to the area between pr and b.

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 vg and pr genes, and P2 to the recombination frequency between pr and b.

P1 = (R + DR) / N

P2 = (R + DR)/ N

Where: R is the number of simple recombinants in each region, DR is the number of double recombinants in each region, and N is the total number of individuals.

So:

Parental)

• 1779 vestigial wings, black body and purple eyes vg pr b

• 1665 normal wings, a grey body and red eyes vg+ pr+ b+

Double recombinant)

• 13 vg pr+ b

• 9 vg+ pr b+

Simple recombinant)

• 252 vg+ pr b

• 241 vg pr+ b+

• 131 vg+ pr+ b

• 118 vg pr b+

P1 = (R + DR) / N

P1 = (252+241+13+9)/4208

P1 = 515/4208

P1 = 0.122

P2= (R + DR) / N

P2 = (131+118+13+9)/4208

P2 = 271/4208

P2 = 0.064

Now, to calculate the recombination frequency between the two extreme genes, vg and b, we can just perform addition or a sum:

P1 + P2= Pt

0.122 + 0.064 = Pt

0.186=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. Every 100 meiotic products, one of them 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 = 0.122 x 100 = 12.2 MU

GD2= P2 x 100 = 0.064 x 100 = 6.4 MU

GD3=Pt x 100 = 0.186 x 100 = 18.6 MU

---- vg ---------------------- pr ---------------------b ----

R1 R2

-----vg----12.2MU---------pr—

----pr--------6.4 MU----b—

-----vg ----------------18.6 MU--------------------b----

3 0

3 years ago

Site-specific recombination systems _________. Multiple Choice do not depend on extensive nucleotide sequence homology depend on

professor190 [17]

Site-specific recombination systems all of the choices are correct i.e.

A. do not depend on extensive nucleotide sequence homology.

B. depend on enzymes that are often specific for sequences within the host.

C. are features of some viruses.

An exchange between two specified sequences (target sites), either on the same DNA molecule or on two separate DNA molecules, is known as site-specific recombination.
DNA sequences may be integrated, excised, or inverted as a result of the exchange.
A site-specific recombinase that can work by itself or with the aid of additional components or enzymes shapes the DNA target during recombination.
The recombinase is chemically bonded to the ends of the intermediate DNA after DNA breakage at the recombination site; when this process is reversed, the intermediate DNA is resealed to form the recombinant and the recombinase is released.
During this recombination process, neither replication nor repair are necessary.

learn more about Site-specific recombination here: brainly.com/question/11458760

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2 years ago

Which human activity is most affected by the increasing number of dead zones in the ocean?

Oxana [17]

Over use of plastic have an amazing day!

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3 years ago