Answer AND Explanation:
Normal: GGA CTC CTC
Abnormal: GGA CAC CTC
mRNA for normal DNA: CCU GAG GAG
mRNA for abnormal DNA: CCU GUG GAG
Amino acids for normal DNA: Proline Glutamic acid Glutamic acid
Amino acids for abnormal DNA: Proline Valine Glutamic acid
Answer:
A. DNA replication
B. Two haploid daughter cells
C. Four haploid daughter cells
D. No, they are not the same
E. Gametes
Explanation:
A) Step 1 and Step 2, according to the image attached to this question, depicts the stages of interphase where the REPLICATION of genetic material occurs i.e formation of sister chromatids.
B) Meiosis occurs in two division steps namely meiosis I and meiosis II. Meiosis I, which involves the separation of homologous chromosomes produces TWO haploid daughter cells.
C). However, in meiosis II, sister chromatids separate to produce FOUR haploid daughter cells.
D) The cells in step 4 are not all the same due to a process called CROSSING OVER, which occurs in the Prophase of meiosis I. Crossing over exchanges chromosomal segments between two non-sister chromatids of homologous chromosomes as seen in the color of the images attached.
E) The four daughter cells produced in step 4 will mature into GAMETES, which will be used in sexual reproduction.
The second graph presented in the image below shows the survivorship of whales.
As we can see, the population grows rapidly after the calves reach adulthood, but the environmental factors limit the growth of the population and the population growth reached a plateau after a period of intense growth.
0.1 M solution of a disaccharide solution will contain 2000 monosaccharide molecules.
<h3>What are monosaccharides?</h3>
Monosaccharides, also known as simple sugars are the simplest monomers of carbohydrates which may either be 3 carbon, 4 carbon, 5 carbon, 6 carbon or 7 carbo compounds.
There are two types of monosaccharides;
- aldoses sugars, e.g. glucose, and
- ketose sugars e.g. fructose.
When two monosaccharides are linked together by glycosidic bonds to form a single compound, the compound formed is called a disaccharide.
Considering the give question:
Suppose a 0.1M solution of a monosaccharide contains 1000 monosaccharide molecules. How many monosaccharide molecules would be in a 0.1 M solution of a disaccharide.
The number of monosaccharides molecules present in the 0.1 M solution of a disaccharide is determined as follows:
1 disaccharide molecule contains 2 monosaccharide molecules
0.1M solution of a monosaccharide contains 1000 monosaccharide molecules.
0.1 M solution of a disaccharide will contain 2 * 1000 monosaccharide molecules
0.1 M solution of a disaccharide will contain = 2000 monosaccharide molecules.
Learn more about monosaccharides and disaccharides at: brainly.com/question/731310
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I think it is D
Hope my answer help you?
