Carbohydrates,lipids,proteins,or nucleic acids which one contains DNA and RNA

When plants grow towards sunlight, they __________.

andreev551 [17]

When plants grow towards sunlight, they __________.

A. Get smaller

B. Get more light 

C. Die faster

D. Live longer

4 0

4 years ago

Read 2 more answers

Why do our bodies rearrange the food we eat and the oxygen we breathe (O2) into carbon dioxide (CO2) and water (H2O)?

Tema [17]

Answer:

A. in order to release the energy found in food.

Explanation:

Every cell in your body needs oxygen to function. You get the oxygen your cells need from the air you breathe. The air you breathe is made up of 20 percent oxygen. The rest of the air is mostly nitrogen (79%). Your body cells use the oxygen you breathe to get energy from the food you eat. This process is called cellular respiration. During cellular respiration the cell uses oxygen to break down sugar. Breaking down sugar produces the energy your body needs. This is very similar to wood burning in a fire. As the wood burns, it combines with oxygen and releases heat energy and carbon dioxide. When the cell uses oxygen to break down sugar, oxygen is used, carbon dioxide is produced, and energy is released. But instead of heat energy, much of the energy produced in cellular respiration is stored chemically for the cell to use later. Carbon dioxide is the waste product of cellular respiration that you breathe out each time you breathe. Blood picks up oxygen and releases carbon dioxide in the lungs. The opposite takes place in the cells where the blood releases oxygen and picks up carbon dioxide.

3 0

3 years ago

Will our sun ever become a red giant

NemiM [27]

Answer:

Well,

Explanation:

Roughly 5 billion years from now, the Sun will exhaust the hydrogen fuel in its core and start burning helium, forcing its transition into a red giant star. During this shift, its atmosphere will expand out to somewhere around 1 astronomical unit — the current average Earth-Sun distance.

7 0

3 years ago

One of your skin cells is about to replicate its DNA. What happens first?

Ymorist [56]

Answer:

The correct option is C) The helicase enzyme unwind the DNA molecule into two strands.

Explanation:

DNA replication occurs in a semi-conservative manner. This means that each strand of the DNA can give rise to a new strand. For the process of DNA replication to begin, firstly it is necessary that the DNA unwinds. Topoisomerases are enzymes which are known to remove the super-coiling of the DNA. Helicases can be described as the enzymes which separate the two strands of DNA from the region where replication has to begin. Hence, option C is correct.

7 0

3 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