1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Zielflug [23.3K]
3 years ago
12

50 points!!! Describe the process of gamete formation in sexually reproducing organisms

Biology
1 answer:
erastova [34]3 years ago
7 0

1. During sexual reproduction, a male and female gamete will merge together to form a new organism. The two haploid cells will fuse together to form a diploid cell called a zygote.

2. A large scale mutation can have a range of effects: No change occurs in phenotype. Some mutations don't have any noticeable effect on the phenotype of an organism.

You might be interested in
During the production of gametes, the chromosomes of a cell have condensed, and the chromatids of homologous chromosomes are cro
snow_tiger [21]

Answer:

The correct answer will be the cell is in prophase I of meiosis and will enter metaphase I next.

Explanation:

Meiosis is a type of cell division which is involved in the formation of haploid gametes.

During meiosis the division of the genetic material is markedly observed and based on this meiosis is divided into 4 phases: prophase, metaphase, anaphase and telophase.

During prophase: the chromatin structure of DNA starts condensing to form chromatids which attaches to homologous chromatids and are known as sister chromatids. These sister chromatids exchange their genetic material via a process called called recombination or crossing over which takes place during pachytene stage of prophase I. After crossing over it is proceeded by the next stage that is metaphase I in which chromosomes gets aligned at the center of the cell.

Thus, the cell is in prophase I of meiosis and will enter metaphase I next is the correct answer.

7 0
3 years ago
Read 2 more answers
Ancient photosynthetic prokaryotes were very important in the history of life because theya. produced the oxygen in the atmosphe
OLga [1]

Answer:

a) produced the oxygen in the atmosphere

Explanation:

  • Prokaryotes are likely to first form of life on earth and they existed billions of years ago before animals and plants.
  • As these prokaryotes can do photosynthesis so there most important function is produce oxygen in atmosphere for the evolution of aerobic life.
  • At that time the conditions were harsh on earth so these prokaryotes are used to survive under earth like in oceans. They first produce oxygen in oceans.
  • In eukaryotes photosynthesis emerge by endosymbiosis with photosynthetic prokaryotes.

7 0
3 years ago
Read 2 more answers
Explain how the original source of energy for egrets, and all the other consumers , is the sun
Maslowich
The term cell growth is used in the contexts of biological cell development and cell division (reproduction). When used in the context of cell division, it refers to growth of cell populations, where a cell, known as the "mother cell", grows and divides to produce two "daughter cells" (M phase). When used in the context of cell development, the term refers to increase in cytoplasmic and organelle volume (G1 phase), as well as increase in genetic material (G2 phase) following the replication during S phase.[1]

Contents
Cell populations Edit

Cell populations go through a particular type of exponential growth called doubling. Thus, each generation of cells should be twice as numerous as the previous generation. However, the number of generations only gives a maximum figure as not all cells survive in each generation.

Cell size Edit

Cell size is highly variable among organisms, with some algae such as Caulerpa taxifolia being a single cell several meters in length.[2] Plant cells are much larger than animal cells, and protists such as Paramecium can be 330 μm long, while a typical human cell might be 10 μm. How these cells "decide" how big they should be before dividing is an open question. Chemical gradients are known to be partly responsible, and it is hypothesized that mechanical stress detection by cytoskeletal structures is involved. Work on the topic generally requires an organism whose cell cycle is well-characterized.

Yeast cell size regulation Edit
The relationship between cell size and cell division has been extensively studied in yeast. For some cells, there is a mechanism by which cell division is not initiated until a cell has reached a certain size. If the nutrient supply is restricted (after time t = 2 in the diagram, below), and the rate of increase in cell size is slowed, the time period between cell divisions is increased.[3] Yeast cell-size mutants were isolated that begin cell division before reaching a normal/regular size (wee mutants).[4]


Figure 1:Cell cycle and growth
Wee1 protein is a tyrosine kinase that normally phosphorylates the Cdc2 cell cycle regulatory protein (the homolog of CDK1 in humans), a cyclin-dependent kinase, on a tyrosine residue. Cdc2 drives entry into mitosis by phosphorylating a wide range of targets. This covalent modification of the molecular structure of Cdc2 inhibits the enzymatic activity of Cdc2 and prevents cell division. Wee1 acts to keep Cdc2 inactive during early G2 when cells are still small. When cells have reached sufficient size during G2, the phosphatase Cdc25 removes the inhibitory phosphorylation, and thus activates Cdc2 to allow mitotic entry. A balance of Wee1 and Cdc25 activity with changes in cell size is coordinated by the mitotic entry control system. It has been shown in Wee1 mutants, cells with weakened Wee1 activity, that Cdc2 becomes active when the cell is smaller. Thus, mitosis occurs before the yeast reach their normal size. This suggests that cell division may be regulated in part by dilution of Wee1 protein in cells as they grow larger.

Linking Cdr2 to Wee1 Edit
The protein kinase Cdr2 (which negatively regulates Wee1) and the Cdr2-related kinase Cdr1 (which directly phosphorylates and inhibits Wee1 in vitro)[5] are localized to a band of cortical nodes in the middle of interphase cells. After entry into mitosis, cytokinesis factors such as myosin II are recruited to similar nodes; these nodes eventually condense to form the cytokinetic ring.[6] A previously uncharacterized protein, Blt1, was found to colocalize with Cdr2 in the medial interphase nodes. Blt1 knockout cells had increased length at division, which is consistent with a delay in mitotic entry. This finding connects a physical location, a band of cortical nodes, with factors that have been shown to directly regulate mitotic entry, namely Cdr1, Cdr2, and Blt1.

Further experimentation with GFP-tagged proteins and mutant proteins indicates that the medial cortical nodes are formed by the ordered, Cdr2-dependent assembly of multiple interacting proteins during interphase. Cdr2 is at the top of this hierarchy and works upstream of Cdr1 and Blt1.[7] Mitosis is promoted by the negative regulation of Wee1 by Cdr2. It has also been shown that Cdr2 recruits Wee1 to the medial cortical node. The mechanism of this recruitment has yet to be discovered. A Cdr2 kinase mutant, which is able to localize properly despite a loss of function in phosphorylation, disrupts the recruitment of Wee1 to the medial cortex and delays entry into mitosis. Thus, Wee1 localizes with its inhibitory network, which demonstrates that mitosis is controlled through Cdr2-dependent negative regulation of Wee1 at the medial cortical nodes.[7]

Cell polarity factors
4 0
3 years ago
4. In the process of precipitation, water vapor in air cools and becomes water droplets that clump around particles in air to fo
Elena-2011 [213]

Answer:

I don't know what you ask but just think yes

Explanation:

4 0
3 years ago
Solve for v<br>negative 38= v/2
Nadusha1986 [10]

- 38 =  \frac{v}{2}



\text{Multiply by 2 on both sides,  }


2 \times ( - 38) =  \frac{v}{ \cancel{2}}  \times  \cancel{2} \\  \\  \\  \\  =  >  - 76 = v \times 1 \\  \\  \\  =  >  - 76 = v


So, value of v is - 76
4 0
4 years ago
Other questions:
  • Why are dna fragments with more base pairs found near the sample wells (origin) and the lower number of bp fragments found near
    5·1 answer
  • What stars live the longest?
    11·2 answers
  • Look at the picture below.
    13·1 answer
  • in dogs there is a hereditary type of deafness caused by a recessive gene. Two dogs who carry the gene for deafness but have nor
    13·1 answer
  • 4-10 Assuming that the 30-nm chromatin fiber contains about 20 nucleosomes (200 bp/nucleosome) per 50 nm of length, calculate th
    13·1 answer
  • Identify the funciton of part F
    9·1 answer
  • How do fronts change weather?
    8·2 answers
  • Match each checkpoint with the action it checks for.
    14·1 answer
  • Can virus show response against heat, chemical and temperature?If yes then how?​
    8·1 answer
  • Tại sao lá cây có màu xanh
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!