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dexar [7]
3 years ago
9

A segement that contains the information for making protein

Biology
1 answer:
bezimeni [28]3 years ago
6 0
In terms of the genetic sequences found on a mature strand of RNA, messenger RNA, the actual coding sequence of nucleotides that codes for protein specifically would be the exons. Introns, found in eukaryotes are found in between exons and do not code for proteins.
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Which of the following accurately describe characteristics of the cell membrane. Choose all that apply. *
artcher [175]

Answer:

selectively permeable

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3 years ago
What occurs when two species have the<br> need for the same resource?
Lady bird [3.3K]

Answer:

Competition. Additionally, with competition comes <u>survival of the fittest</u>.

Explanation:

Competition is formed when more than one species must fight for the same resource/commodity.

6 0
4 years ago
1. Select the statement that pertains to an issue with the Green Revolution.
BARSIC [14]

Answer:

O The Green revolution has caused a decrease in energy production.

O The Green revolution has led to many people dying of hunger

Explanation:

8 0
3 years ago
The artificial cell in the picture below is permeable to monosaccharides but not polysaccharides. Initially, the artificial cell
Ne4ueva [31]

Answer:

Osmosis.

Explanation:

Osmosis is the process which is responsible for the transportation of glucose inside the cell. Osmosis is a process in which micro-molecules such as glucose, ions and gases etc go inside the cell through a semi-permeable membrane. This semi-permeable membrane of the cell only allow monosaccharides but not polysaccharides because monosaccharides is small in size while polysaccharides are larger in size.

8 0
3 years ago
22. A plant geneticist is investigating the inheritance of genes for bitter taste (Su) and explosive rind (e) in watermelon (Cit
mylen [45]

Answer:

See the answer below

Explanation:

Using the formula for calculating Chi square (X^2):

X^2  =  \frac{(O - E)^2}{E} where O = observed frequency and E = expected frequency.

The observed frequencies for the four phenotypes are 88, 62, 62, and 81 respectively.

For the expected frequency, the phenotypes are supposed to assort in 9:3:3:1 according to Mendel ratio.

Hence, expected frequencies are calculated as:

phenotype (1) = 9/16 x 293 = 164.81

phenotype (2) = 3/16 x 293 = 54.94

phenotype (3) = 3/16 x 293 = 54.94

phenotype (4) = 1/16 x 293 = 18.31

The X^2 is calculated thus:

Phenotype            O               E                             X^2

  1                          88            164.81            \frac{(88 - 164.81)^2}{164.81} = 35.80

  2                          62             54.94            \frac{(62 - 54.94)^2}{54.94} = 0.91

  3                           62             54.94              \frac{(62 - 54.94)^2}{54.94} = 0.91

  4                            81              18.31               \frac{(81 - 18.31)^2}{18.31} = 214.64

Total X^2 = 35.80 + 0.91 + 0.91 + 214.64 = 252.26

To the nearest tenth = 252

Degree of freedom = n - 1 = 4- 1 = 3

X^2 tabulated  = 7.815

<em>The calculated </em>X^2<em> exceeds the critical value, hence, the null hypothesis is rejected. The two genes did not assort independently.        </em>

4 0
3 years ago
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