What wind pattern helped carry early explorers from Europe to North and South America?

Bond [772]

Answer:

The answer would be C. This is because the nucleus of a cell is responsible for the creation of ribosomes. Hope this helps you :)

6 0

3 years ago

in lab micr the short tail phenotype is fominant to wild long assuming tail length is controlled by a single locus a likely expl

Ne4ueva [31]

Answer:

The answer is "Option e".

Explanation:

please find the complete question in the attached file.

Its long tail disregards its short tail. Let's assume that even a short neck is a as well as a tail over, which claim, though, the short tails were mixed, shorter, and longer tailed mousses are created. It may also presume that the short mouse parental is always Aa. And we get AA, Aa, Aa, Aa, and Aa situations once their matter and they fall pregnant to both high and short tail mice but we wouldn't get the fat tail mouse unless the tail-mouse were as AA.

we always get two types of lines. It demonstrates there was no uniform AA genera. It is a case of the heterozygous dangerous gene, that can cause a set of identical alleles inside an organism to always be lethal.

5 0

4 years ago

Indicate whether each of the following statements is true of depurination (DP), deamination (DA), or pyrimidine dimer formation

solniwko [45]

Answer:

- This process is caused by spontaneous hydrolysis of a glycosidic bond: depurination and deamination

- This process is induced by ultraviolet light: pyrimidine dimer formation

- This can happen to guanine but not to cytosine: depurination

- This can happen to thymine but not to adenine: pyrimidine dimer formation

- This can happen to thymine but not to cytosine: none

- Repair involves a DNA glycosylase: deamination

- Repair involves an endonuclease: depurination, deamination and pyrimidine dimer formation

- Repair involves DNA ligase: depurination, deamination and pyrimidine dimer formation

- Repair depends on the existence of separate copies of the genetic information in the two strands of the double helix: depurination, deamination and pyrimidine dimer formation

- Repair depends on cleavage of both strands of the double helix: none

Explanation:

Depurination is the loss of purine bases (either adenine or guanine), while deamination refers to the removal of an amino group. During depurination, a β-N-glycosidic bond is cleaved by hydrolysis and a nucleic base is released (either adenine or guanine). All DNA bases may undergo deamination, except thymine (since thymine does not have an amino group). The ultraviolet (UV) radiation can cause thymine or cytosine to form dimers (e.g., pyrimidine dimers), being thymine dimers the most common lesion when DNA is exposed to UV light. Pyrimidine dimers may be repaired by different excision mechanisms, e.g., nucleotide excision repair, where the recognition of the DNA damage leads to the removal of the DNA fragment containing the lesion. DNA glycosylases are enzymes involved in the mechanism of base excision, these enzymes recognize and remove damaged bases by hydrolysis of the glycosidic bond, producing an abasic (apurinic and apyrimidinic) site. A DNA ligase enzyme covalently joins two DNA molecules by forming a phosphodiester bond, which is required during these processes.

8 0

3 years ago

Based on the above image, how can a gene determine whether a person has albinism?

kondor19780726 [428]

Answer:

They can look at the parents, and see their color and compare

Explanation:

7 0

3 years ago

Read each of the sentences that describe a phase of meiosis. Place each sentence into the correct box.

Lady bird [3.3K]

Answer:

Prophase1-crossing over may occur between the non-sister chromatids

Metaphase1- Homologous chromosomes line up at the center of the cell

Anaphase and telophase 1- homologous pairs separate and move towards opposite ends of the cell, Nuclear membrane forms and cytokinesis follows.

Meiosis 2- haploid number of duplicated chromosome line up…

Sister chromatids separate and become daughter…

Four haploid daughter cells are formed….

Explanation:

Meiosis I

P-I: Chromosomes condense, nuclear membrane dissolves, homologous chromosomes form bivalents, crossing over occurs

M-I: Spindle fibres from opposing centrosomes connect to bivalents (at centromeres) and align them along the middle of the cell

A-I: Spindle fibres contract and split the bivalent, homologous chromosomes move to opposite poles of the cell

T-I: Chromosomes decondense, nuclear membrane may reform, cell divides (cytokinesis) to form two haploid daughter cells

Meiosis II

The second division separates sister chromatids (these chromatids may not be identical due to crossing over in prophase I)

P-II: Chromosomes condense, nuclear membrane dissolves, centrosomes move to opposite poles (perpendicular to before)

M-II: Spindle fibres from opposing centrosomes attach to chromosomes (at centromere) and align them along the cell equator

A-II: Spindle fibres contract and separate the sister chromatids, chromatids (now called chromosomes) move to opposite poles

T-II: Chromosomes decondense, nuclear membrane reforms, cells divide (cytokinesis) to form four haploid daughter cells

The final outcome of meiosis is the production of four haploid daughter cells

6 0

3 years ago