What does the nucleus look like in S phase of Meiosis?

What does the nucleus look like in S phase of Meiosis?

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I was watching an animation video about Meiosis and this is what the video shows (pics attached.) It shows that before synthesis, each chromosome exists as single chromatid and then after replication, it has a sister chromatid. I am confused because germ cells are diploid, so shouldn't the S phase look like this? It should start with 23 chromosomes (46 different DNA's) and then replicated to 46 chromosomes (92 different DNA) which then get assorted into 4 different cells?

In terms of chromosome number

During meiosis, the meiocyte is a diploid cell ( 46 chromosomes for human). So during the S-phase , the meiocyte would contain 46 chromosomes.

At the end of meiosis I, the divided meiocyte contains 23 chromosomes so it's called the reductional division

At the end of meiosis II each cell contains 23 chromosomes.

In terms of chromatid number and number of DNA molecules

Meiocyte in S phase contains 2 chromatid in each chromosome, so the number of DNA molecules is doubled (if you consider that each chromatid is composed of a pair of DNA strands like in Watson Crick Model, then each chromosome of S phase contains 4 strands… So 46×4 strands of DNA is present. Often each helix / duplex is considered one DNA molecule , so in that case you have 46×2 molecules. So you have to be careful with the terminology mentioned.

After meiosis I there are 23 chromosomes in each divided cell with 2 chromatids in each chromosome… So there are 23×2 DNA molecules.

After meiosis II there are 23 chromosomes with 1 chromatid… So they contain only 23 DNA molecules.

It states mitosis here but it is the same for meiosis at this stage. What you have drawn for the inside of the first cell is already 44 chromosomes.

Hope this helps

At the beginning, germ cells are diploid and as far as I know they are the cells that give rise to gametes which are haploid.

Here how it goes, the germ cell is containing 46 chromosomes and in S phase, these chromosomes replicate so that each chromosome will be made of 2 chromatids resulting in 92 chromatid.

In meiosis I, the germ cell divides and the 46 chromosomes divide as well giving two cells each containing 23 chromosomes and 46 chromatid.

The meiosis yet hasn't finished. In meiosis II, each daughter cell containing 23 chromosomes will divide to two daughter cells. In this division, the chromatids in each chromosome will separate. In other words, the 46 chromatids in each cell will divide and each cell from the new daughter cells will be containing 23 chromatids. So now we have 4 haploid cells each one containing 23 CHROMATIDS ( a gamete ) all coming form one original diploid germ cell.

Those chromatids will then be referred to as chromosomes.

What does the nucleus look like in S phase of Meiosis? - Biology

The nucleus is a highly specialized organelle that serves as the information and administrative center of the cell. This organelle has two major functions. It stores the cell's hereditary material, or DNA, and it coordinates the cell's activities, which include intermediary metabolism, growth, protein synthesis, and reproduction (cell division).

Only the cells of advanced organisms, known as eukaryotes , have a nucleus. Generally there is only one nucleus per cell, but there are exceptions such as slime molds and the Siphonales group of algae. Simpler one-celled organisms ( prokaryotes ), like the bacteria and cyanobacteria, don't have a nucleus. In these organisms, all the cell's information and administrative functions are dispersed throughout the cytoplasm.

The spherical nucleus occupies about 10 percent of a cell's volume, making it the cell's most prominent feature. Most of the nuclear material consists of chromatin, the unstructured form of the cell's DNA that will organize to form chromosomes during mitosis or cell division. Also inside the nucleus is the nucleolus, an organelle that synthesizes protein-producing macromolecular assemblies called ribosomes.

A double-layered membrane, the nuclear envelope, separates contents of the nucleus from the cellular cytoplasm. The envelope is riddled with holes called nuclear pores that allow specific types and sizes of molecules to pass back and forth between the nucleus and the cytoplasm. It is also attached to a network of tubules, called the endoplasmic reticulum, where protein synthesis occurs. These tubules extend throughout the cell and manufacture the biochemical products that a particular cell type is genetically coded to produce.

Chromatin/Chromosomes - Packed inside the nucleus of every human cell is nearly 6 feet of DNA, which is divided into 46 individual molecules, one for each chromosome and each about 1.5 inches long. Packing all this material into a microscopic cell nucleus is an extraordinary feat of packaging. For DNA to function, it can't be crammed into the nucleus like a ball of string. Instead, it is combined with proteins and organized into a precise, compact structure, a dense string-like fiber called chromatin.

Each DNA strand wraps around groups of small protein molecules called histones, forming a series of bead-like structures, called nucleosomes, connected by the DNA strand. Under the microscope, uncondensed chromatin has a "beads on a string" appearance.

The string of nucleosomes, already compacted by a factor of six, is then coiled into an even denser structure, compacting the DNA by a factor of 40. This compression and structuring of DNA serves several functions. The overall negative charge of the DNA is neutralized by the positive charge of the histone molecules, the DNA takes up much less space, and inactive DNA can be folded into inaccessible locations until it is needed.

There are two types of chromatin. Euchromatin is the genetically active portion and is involved in transcribing RNA to produce proteins used in cell function and growth. Heterochromatin contains inactive DNA and is the portion of chromatin that is most condensed, since it not being used.

Throughout the life of a cell, chromatin fibers take on different forms inside the nucleus. During interphase, when the cell is carrying out its normal functions, the chromatin is dispersed throughout the nucleus in what appears to be a tangle of fibers. This exposes the euchromatin and makes it available for the transcription process.

When the cell enters metaphase and prepares to divide, the chromatin changes dramatically. First, all the chromatin strands make copies of themselves through the process of DNA replication. Then they are compressed to an even greater degree than at interphase, a 10,000-fold compaction, into specialized structures for reproduction, termed chromosomes. As the cell divides to become two cells, the chromosomes separate, giving each cell a complete copy of the genetic information contained in the chromatin.

Nucleolus - The nucleolus is a membrane-less organelle within the nucleus that manufactures ribosomes, the cell's protein-producing structures. Through the microscope, the nucleolus looks like a large dark spot within the nucleus. A nucleus may contain up to four nucleoli, but within each species the number of nucleoli is fixed. After a cell divides, a nucleolus is formed when chromosomes are brought together into nucleolar organizing regions. During cell division, the nucleolus disappears. Some studies suggest that the nucleolus may be involved with cellular aging and, therefore, may affect the aging of an organism.

Nuclear Envelope - The nuclear envelope is a double-layered membrane that encloses the contents of the nucleus during most of the cell's lifecycle. The space between the layers is called the perinuclear space and appears to connect with the rough endoplasmic reticulum. The envelope is perforated with tiny holes called nuclear pores. These pores regulate the passage of molecules between the nucleus and cytoplasm, permitting some to pass through the membrane, but not others. The inner surface has a protein lining called the nuclear lamina, which binds to chromatin and other nuclear components. During mitosis, or cell division, the nuclear envelope disintegrates, but reforms as the two cells complete their formation and the chromatin begins to unravel and disperse.

Nuclear Pores - The nuclear envelope is perforated with holes called nuclear pores. These pores regulate the passage of molecules between the nucleus and cytoplasm, permitting some to pass through the membrane, but not others. Building blocks for building DNA and RNA are allowed into the nucleus as well as molecules that provide the energy for constructing genetic material.

The pores are fully permeable to small molecules up to the size of the smallest proteins, but form a barrier keeping most large molecules out of the nucleus. Some larger proteins, such as histones, are given admittance into the nucleus. Each pore is surrounded by an elaborate protein structure called the nuclear pore complex, which probably selects large molecules for entrance into the nucleus.

Diagrammatic Representation of a Nucleus

As the organelle that contains the genetic material of a cell, the nucleus can be described as the command center. As such, the nucleus consists of a number of structured elements that allow it to perform its functions. This section gives focus to the structure of the cell.

In general, the nucleus has a spherical shape as shown in most books. However, it may appear flattened, ellipsoidal or irregular depending on the type of cell. For instance, the nucleus of columnar epithelium cells appears more elongated compared to those of other cells. The shape of a nucleus, however, may also change as the cell matures.

Dissolution of the Nuclear Envelope

In most cells, the disassembly of the nuclear envelope marks the end of the prophase of mitosis (Figure 8.29). However, this disassembly of the nucleus is not a universal feature of mitosis and does not occur in all cells. Some unicellular eukaryotes (e.g., yeasts) undergo so-called closed mitosis, in which the nuclear envelope remains intact (Figure 8.30). In closed mitosis, the daughter chromosomes migrate to opposite poles of the nucleus, which then divides in two. The cells of higher eukaryotes, however, usually undergo open mitosis, which is characterized by breakdown of the nuclear envelope. The daughter chromosomes then migrate to opposite poles of the mitotic spindle, and new nuclei reassemble around them.

Figure 8.29

The nucleus during mitosis. Micrographs illustrating the progressive stages of mitosis in a plant cell. During prophase, the chromosomes condense, the nucleolus disappears, and the nuclear envelope breaks down. At metaphase, the condensed chromosomes (more. )

Figure 8.30

Closed and open mitosis. In closed mitosis, the nuclear envelope remains intact and chromosomes migrate to opposite poles of a spindle within the nucleus. In open mitosis, the nuclear envelope breaks down and then re-forms around the two sets of separated (more. )

Disassembly of the nuclear envelope, which parallels a similar breakdown of the endoplasmic reticulum, involves changes in all three of its components: The nuclear membranes are fragmented into vesicles, the nuclear pore complexes dissociate, and the nuclear lamina depolymerizes. The best understood of these events is depolymerization of the nuclear lamina—the meshwork of filaments underlying the nuclear membrane. The nuclear lamina is composed of fibrous proteins, lamins, which associate with each other to form filaments. Disassembly of the nuclear lamina results from phosphorylation of the lamins, which causes the filaments to break down into individual lamin dimers (Figure 8.31). Phosphorylation of the lamins is catalyzed by the Cdc2 protein kinase, which was introduced in Chapter 7 (see Figure 7.40) and will be discussed in detail in Chapter 14 as a central regulator of mitosis. Cdc2 (as well as other protein kinases activated in mitotic cells) phosphorylates all the different types of lamins, and treatment of isolated nuclei with Cdc2 has been shown to be sufficient to induce depolymerization of the nuclear lamina. Moreover, the requirement for lamin phosphorylation in the breakdown of the nuclear lamina has been demonstrated directly by the construction of mutant lamins that can no longer be phosphorylated. When genes encoding these mutant lamins were introduced into cells, their expression was found to block normal breakdown of the nuclear lamina as the cells entered mitosis.

Figure 8.31

Dissolution of the nuclear lamina. The nuclear lamina consists of a meshwork of lamin filaments. At mitosis, Cdc2 and other protein kinases phosphorylate the lamins, causing the filaments to dissociate into free lamin dimers.

In concert with dissolution of the nuclear lamina, the nuclear membrane fragments into vesicles (Figure 8.32). The B-type lamins remain associated with these vesicles, but lamins A and C dissociate from the nuclear membrane and are released as free dimers in the cytosol. This difference arises because the B-type lamins are permanently modified by the addition of lipid (prenyl groups), whereas the C-terminal prenyl groups of A- and C-type lamins are removed by proteolysis following their incorporation into the lamina. The nuclear pore complexes also dissociate into subunits as a result of phosphorylation of several nuclear pore proteins. Integral nuclear membrane proteins are also phosphorylated at mitosis, and phosphorylation of these proteins may be important in vesicle formation as well as in dissociation of the nuclear membrane from both chromosomes and the nuclear lamina.

Figure 8.32

Breakdown of the nuclear membrane. As the nuclear lamina dissociates, the nuclear membrane fragments into vesicles. The B-type lamins remain bound to these vesicles, while lamins A and C are released as free dimers.

Nuclear size

The nucleus increases in size from the time of its formation, immediately following NE assembly, to when it reaches its final size in interphase. This raises the question of what controls nuclear size: is it a mere consequence of increase in volume over time, or are there factors that regulate or limit nuclear expansion? Decades of observations suggest that the latter is true, and that nuclear and cytoplasmic volumes are somehow related to each other this phenomenon is referred to as the karyoplasmic ratio (Gregory, 2005). Moreover, in both budding and fission yeasts, the ratio of nuclear to cellular volume remains constant throughout the cell cycle, even as cell volume increases (Jorgensen et al., 2007 Neumann and Nurse, 2007). This suggests the existence of a mechanism that links nuclear and cellular volumes. If indeed such a mechanism exists, does cellular volume dictate nuclear volume, or does nuclear volume determine cell volume? What cellular factors determine nuclear volume? And finally, why is nuclear volume important?

Variation in nuclear shape. The nuclei of most cells, such as those of the C. elegans embryo (A), are either oval or round. However, various cell types or conditions display non-round nuclei. Shown are the nuclei of neutrophils (B), of cells from a patient with HGPS (C) and of cells from a 96-year-old individual (D, right panel) compared with nuclei of cells from a 9-year-old individual (D, left panel). Visualization of nuclei was performed with a GFP-tagged NPC component, NPP-1 (A), an antibody specific for lamin B (B), an antibody specific for emerin (a lamina-associated protein, C) and an antibody specific for lamin A and lamin C (D). The image in B was reprinted with permission from Ada Olins and Donald Olins (Olins and Olins, 2005). The image in C was reprinted with permission from Goldman (Goldman et al., 2004). The images in D were provided by Tom Misteli and Paola Scaffdi (NCI, Bethesda MD) (see also Scaffidi and Misteli, 2006). Nuclei are not shown to scale.

Variation in nuclear shape. The nuclei of most cells, such as those of the C. elegans embryo (A), are either oval or round. However, various cell types or conditions display non-round nuclei. Shown are the nuclei of neutrophils (B), of cells from a patient with HGPS (C) and of cells from a 96-year-old individual (D, right panel) compared with nuclei of cells from a 9-year-old individual (D, left panel). Visualization of nuclei was performed with a GFP-tagged NPC component, NPP-1 (A), an antibody specific for lamin B (B), an antibody specific for emerin (a lamina-associated protein, C) and an antibody specific for lamin A and lamin C (D). The image in B was reprinted with permission from Ada Olins and Donald Olins (Olins and Olins, 2005). The image in C was reprinted with permission from Goldman (Goldman et al., 2004). The images in D were provided by Tom Misteli and Paola Scaffdi (NCI, Bethesda MD) (see also Scaffidi and Misteli, 2006). Nuclei are not shown to scale.

Factors that affect nuclear volume

There are conflicting reports regarding the dominant cellular factors that determine nuclear volume. One idea, known as the nucleoskeletal theory, is that DNA content influences the volume of the nucleus, which in turn influences the size of the cell (Cavalier-Smith, 2005 Gregory, 2005). Intuitively, DNA may affect nuclear volume, because the size of the nucleus could be directly proportional to amount of DNA it contains and the extent to which that DNA is compacted. Simply comparing genome size to nuclear and cell volume among species supports this theory, because species with larger genomes generally have larger nuclear and cellular volumes (Cavalier-Smith, 2005 Jovtchev et al., 2006). Experiments in mice also give credence to the nucleoskeletal theory: it has been shown that tetraploid mouse embryos have nuclei that are twice as large as those in a diploid control (Henery et al., 1992 Henery and Kaufman, 1992).

However, other data suggest that genome size per se is not the determining factor of nuclear size. Rather, it is likely that there is a nuclear-scaling mechanism whereby nuclear volume is proportional to, and determined by, the levels of one or more cellular factors. Indeed, nuclear transplant experiments support this claim: implanting a small hen erythrocyte nucleus into a HeLa cell results in expansion of the nucleus to the appropriate size for its new environment, without affecting DNA content (Harris, 1967). Moreover, the nucleoskeletal theory does not explain why cells from different tissues in a given organism have the same amount of DNA but varied nuclear sizes (Altman and Katz, 1976). Studies in yeast also contradict the notion that DNA content dictates nuclear and cellular volumes (Jorgensen et al., 2007 Neumann and Nurse, 2007). In neither fission yeast nor budding yeast does nuclear volume increase sharply during S phase, as would be expected if DNA content had a direct affect on nuclear size (Jorgensen et al., 2007 Neumann and Nurse, 2007). Furthermore, even a 16-fold increase in ploidy does not affect nuclear size in fission yeast (Neumann and Nurse, 2007). Instead, the displacement of nuclei by centrifugation in multi-nucleated fission yeast showed that nuclear size adjusted in proportion to the amount of surrounding cytoplasm (Neumann and Nurse, 2007). These studies support a mechanism whereby nuclear size is determined by cytoplasmic volume rather than DNA content.

Assuming that cytoplasmic factors determine nuclear size, what might these be? In cell-free extracts of Xenopus oocytes, an increase in nuclear volume after NE reassembly requires an intact ER (Anderson and Hetzer, 2007). This suggests that the membrane for the newly formed NE is supplied by the ER, and therefore membrane availability could be a limiting factor in determining nuclear size. The ER exists as a continuous meshwork of membrane sheets and membrane tubules. Proteins known as reticulons cause tubule formation in the ER (Voeltz et al., 2006), and high levels of reticulons are inhibitory to nuclear growth, which suggests that the availability of membrane in the form of sheets can put an upper limit on nuclear size (Anderson and Hetzer, 2008 Kiseleva et al., 2007). Work in the Xenopus system has demonstrated a requirement for NPCs and nuclear import in nuclear growth after NE assembly (D'Angelo et al., 2006 Newport et al., 1990), which suggests that the import of one or more nuclear proteins contributes to sizing the nucleus. Indeed, several nuclear lamina proteins that are transported into the nucleus through the NPCs have been found to affect interphase nuclear growth (e.g. Brandt et al., 2006 Dittmer et al., 2007 Newport et al., 1990). However, many questions remain. For example, how do yeast – which lack lamins and lamin-associated proteins – adjust nuclear volume in response to changes in cytoplasmic volume? Also, what is the mechanism, in any organism, that establishes the upper limit to nuclear growth?

Does size matter for nuclear function?

Although the mechanisms that control nuclear volume remain unclear, the existence of a karyoplasmic ratio suggests that nuclear size is important for cell function. Disturbance of this ratio is associated with certain types of cancers (Slater et al., 2005 Zink et al., 2004), suggesting that the ratio between nuclear and cytoplasmic volumes is crucial for cell integrity. Moreover, it has been proposed that cell-cycle progression depends on nuclear size (Roca-Cusachs et al., 2008 Yen and Pardee, 1979), and that cells monitor the ratio between cytoplasmic and nuclear volume to gauge the proper time to enter the cell cycle (Futcher, 1996). In addition, a strong correlation between nuclear size, RNA transcription levels and cell size has been found (e.g. Sato et al., 1994 Schmidt and Schibler, 1995). It is therefore possible that larger nuclei facilitate the increase in transcription that is required in larger cells. Additionally, the volume of the nucleus might be important for maintaining nuclear compartments, such as the nucleolus, and the activity of enzymes such as DNA polymerase, which are sensitive to macromolecular crowding (Hancock, 2004 Miyoshi and Sugimoto, 2008 Sasaki et al., 2006). An increasingly popular view of molecular dynamics within the nucleus favors self-organization of complex structures – a process that depends on biochemical and physical interactions between numerous proteins (Misteli, 2001). A recent example is the assembly of Cajal bodies, which are nuclear structures involved in the biogenesis of small nuclear ribonucleoproteins (snRNPs). Cajal bodies assemble by self-organization through stochastic interactions between the building blocks of which they are composed (Kaiser et al., 2008 Misteli, 2008). Because self-organization may be acutely sensitive to the concentration of the individual components, the regulation of nuclear volume might have an important role in enabling this process.

Before starting, it's always important to ensure that the working surface is clean and that you are wearing a pair of clean gloves to avoid contamination.

Cheek cells can be easily obtained by gently scraping the inside of the mouth using a clean, sterile cotton swab.

Once the cells have been obtained, the following procedure is used for cheek cell wet mount preparation:

  1. place a drop of physiological saline on a clean microscopic slide (central part of the slide)
  2. smear the cotton swab on to the center (part containing the saline drop) of the clean slide for about 4 seconds to get the cells on to the center of the slide
  3. add a drop of methylene blue solution on to the smear and gently place a cover slip on top (to cover the stain and the cells)
  4. any excess solution can be removed by touching one side of the slide with a paper towel or blotting paper.
  5. place the slide on the microscope for observation using 4 x or 10 x objective to find the cells
  6. once the cells have been found, they can then be viewed at higher magnificatio

* Note - Used cotton swabs and cotton towel should be safely discarded in the trash and not left lying on the working table.

Why do we have to Stain the Cells?

The cell has different parts, and those that can absorb stains or dyes are referred to as chromatic. Having absorbed the stain, these parts of the cell become more visible under the microscope and can therefore be easily distinguished from other parts of the same cell.

Without stains, cells would appear to be almost transparent, making it difficult to differentiate its parts.

Methylene blue has a string affinity for both DNA and RNA. When it comes in contact with the two, a darker stain is produced and can be viewed under the microscope.

The nucleus at the central part of the cheek cell contains DNA. When a drop of methylene blue is introduced, the nucleus is stained, which makes it stand out and be clearly seen under the microscope.

Although the entire cell appears light blue in color, the nucleus at the central part of the cell is much darker, which allows it to be identified.


Some species, such as lizards, moths, birds and flatworms, have different sex-determination genes than X and Y. These genes are Z and W. The ZZ genotype produces males, and the ZW produces females. Sex determination in some of these species is directed by temperature. High temperatures have been known dictate the sex of the animal. For example, high incubation temperatures for alligator eggs promotes male, ZZ, genotypes. However, in many lizards and turtles, high incubation temperatures favor the female, ZW, genotype.

How to see DNA with the Naked Eye

DNA, or deoxyribonucleic acid, is contained in all living organisms and is the set of instructions that tell a cell how to build a protein. In the human body, DNA tells the body how to build proteins that makes up hair, skin, muscles, and every organ in your body.

DNA is stored in the nucleus of cells. It is an extremely thin molecule averaging about 2 nanometers in width. A nanometer is one-billionth of a meter. To put this in perspective, a human hair is approximately 80,000 nanometers wide.

In this activity, you will extract DNA from green split peas. To do this, you will go through a series of steps that include breaking the cell apart, releasing the DNA from the nucleus, and protecting the DNA from enzymes that will shear or break it down.

As you perform this activity, think about why you are performing each of the steps. Finally, explain how you will be able to see DNA when it is 40,000 times smaller than a human hair.

In this activity we will:

  • Extract DNA from green split peas and observe with the naked eye
  • Analyze the steps taken to extract the DNA Materials
  • Pour the liquid portion of the mixture through the strainer into a medium sized bowl. Do not dump chunks of unblended green split peas into the bowl.
  • Add 2 tablespoons of dish detergent to the bowl and stir gently for 2 minutes.
  • Let mixture set for approximately 5 minutes.
  • Dump part of this solution into a small clear glass only filling the glass half full.
  • Add a pinch of meat tenderizer to the liquid in the small clear glass and stir for 15 seconds.
  • Remove as many bubbles from the solution as possible with a paper towel. The less bubbles in the cup the better the DNA will be seen.
  • Slowly pour cold isopropyl alcohol into the small clear glass until the glass is nearly full. Pour alcohol as gently as possible trying not to disturb the mixture that is already in the small clear glass.
  • Observe the white, stringy, frothy mixture in the glass- that is your DNA! You may need to let the solution set for several minutes before the DNA becomes visible.


  • 1/2 cup green split peas (not pictured)
  • measuring cup
  • pinch of table salt
  • dish detergent
  • meat tenderizer
  • blender
  • small clear glass
  • strainer
  • medium sized bowl
  • 1 cup water
  • cold 90% or higher isopropyl alcohol
  • tablespoon


This activity requires the use of a blender with sharp blades. Use caution when using a blender. Never reach into the blender and place a lid on the blender when blending. Ask an adult to help you if necessary.


  • Place isopropyl alcohol in refrigerator or freezer about 1 hour prior to performing activity.
  • Gather all other materials.


  • Place ½ cup of green split peas, 1 cup of water and pinch of table salt into blender.
  • Blend on high for approximately 20 seconds.

Extension Activity

  • Research enzymes to discover why the meat tenderizer is a necessary ingredient when extracting DNA.
  • Try extracting DNA from other fruits or vegetables.

After this activity, you should know the steps needed to extract DNA from green split peas. You should also understand the point of each step. The blending breaks the cell open, the soap and salt release the DNA from the nucleus, the meat tenderizer prevents enzymes from breaking down the DNA, and the DNA is not soluble in alcohol so it precipitates out at the water and alcohol boundary. You may also try extracting DNA from other fruits or vegetables. You will see that certain fruits and vegetables will yield more or less DNA. Finally, you should understand that the DNA you are seeing is not individual strands but a tangled mass of all the DNA that is present in a cell’s nucleus.

Good Luck writing the Mock Exam!!

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Extra Office Hours: Wednesday, December 6th 2:00-4:00 PM (402 ML)

Unless otherwise specified, assume independent assortment

What is one way that cells are able to decrease their membrane fluidity? a. By increasing the level of cholesterol in the membrane b. By increasing the number of unsaturated fats in the membrane c. By increasing the number of phospholipids in the membranes d. By increasing the number of saturated fats in the membrane e. By increasing the number of proteins in the membrane

In the following reaction, the carbon molecule of glucose is to form carbon dioxide and the oxygen molecule is to form water.

C 6 H 12 O 6 + 6 O 2 6 CO 2 + 6 H 2 O + Energy

a. oxidized, reduced b. reduced, oxidized c. oxidized, oxidized d. reduced, reduced e. This is not a REDOX reaction.

A genetically modified blue peach has the genotype BB and is autosomal dominant. The recessive peaches of this variety are purple. Another gene in the peach turns it white when autosomal recessive. What is this interaction known as? a. Pleiotropy b. Epistasis c. Aneuploidy d. Incomplete Dominance e. Codominance

Which off the following can easily diffuse across the membrane? a. Glucose b. Hydrogen ions c. Methane d. Potassium ions e. Sucrose

What is the final product of the Calvin cycle? a. Biphosphoglycerate b. Pyruvate c. Carbon dioxide d. Glyceraldehyde phosphate e. Oxaloacetate

What type of bond forms between complementary base pairs? a. Ionic bonds b. Hydrogen bonds c. Covalent bonds d. Disulfide bonds e. Peptide bonds

Rachel and Ross just got married. Ross was married once before and had albino child. Rachel has an albino sister. Neither Ross, Rachel, nor any of their parents are albinos. Calculate the probability that Ross and Rachel will have an albino child. Albinism is an autosomal recessive disorder. a. 1/ b. 1/ c. 1/ d. 2/

How many different types of gametes could be produced through independent assortment by an individual with a genotype AA Bb Cc DD Ee? a. 5 b. 6 c. 8 d. 16 e. 64

Use the following predigree to answer questions 15 and 16:

Which of the following best describes the disease in the above pedigree? a. X-linked dominant b. X-link recessive c. Autosomal dominant d. Autosomal recessive e. Y-link

In the above pedigree, what is the most likely genotype of the first generation male? a. HH b. Hh c. hh d. XhY e. Either a or b

In a cross between a white-eyed female fruit fly and red-eyed male, what percent of the female offspring will have white eyes? (White eyes are X-linked, recessive) a. 100% b. 75% c. 50% d. 25% e. 0%

A female Drosophila of unknown genotype was crossed with a white-eyed male fly, of genotype (w = white eye allele is recessive, w+= red-eye allele is dominant.) Half of the male and half of the female offspring were red-eyed, and half of the male and half of the female offspring were white-eyed. What was the genotype of the female fly? a. Xw Y b. Xw Xw c. Xw+ Xw d. Xw+ Xw+ e. Xw+ Y

Hemophilia in humans is due to an X-chromosome mutation. What will be the results of mating between a normal (non-carrier) female and a hemophiliac male? a. 50% of sons are hemophiliacs, 50% of daughters are hemophiliacs b. 100% of sons are hemophiliacs c. 100% of daughters are hemophiliacs d. 50% of sons are normal, 50% of sons are hemophiliacs and 100% of daughters are carriers e. 100% of daughters are carriers, 100% of boys are normal

  1. A reaction occurs when a disaccharide is formed from two monosaccharides. a. Dehydration b. Oxidation c. Hydrolysis d. Reduction e. Hydrophilic interaction
  1. The dominant P allele of a gene that controls colour in pea flowers produces petals with a purple

colour plants homozygous for a recessive p allele of this gene have white flowers. If a Pp plant and a Pp plant were crossed, what would be the expected phenotypic ratios of the offspring? a. 2 purple: 2 white b. 1 purple : 3 white c. 3 purple: 1 white d. 3 white: 1 purple e. 4 purple: 0 white

a. Metaphase chromosome b. Nucleosome c. DNA double helix d. Looped domains e. 30 nm chromatin fibre

  1. Which of the following statements describes the concept of “semi-conservative” DNA replication? a. The two parental strands reassociate after acting as templates for new strands, thus restoring the parental double helix. b. Each strand of both daughter molecules contains a mixture of old and newly synthesized DNA. c. The two strands of the parental molecule separate, and each functions as a template for synthesis of a new, complementary strand. d. DNA Polymerase III carries out synthesis by extending from the RNA primer (5’ to 3’) e. That the lagging strand of DNA is synthesized in short fragments called Okazaki fragments.

Use the following information to answer questions 32 through 34. The following sequence represents a non-transcribed strand of a gene. The transcription initiation sequence is at the first underlined T. There are introns present in between, but not including the first underlined C and second underlined T and another between but not including the underlined A and third underlined T.




What would the translated protein sequence of the above mRNA strand look like? (genetic code on last page) a. Met - Ser – Ser –Tyr – Gly – Val – His b. Met – Ser - Tyr – Cys – Gly – Gly – Val c. Gly – Cys – Met – Ser – Tyr – Cys – Gly – Gly – Val d. Gly – Ser – His – Val – Leu – Ile – Val e. Val – Gly – Leu – Trp – Cys – Ser – Gly – Gly – Val

If the underlined A was replaced with a C what would the resulting mutation be? a. No mutation would occur b. Frameshift c. Nonsense d. Silent e. Missense

The tortoiseshell coat present on some female cats is the result of? a. Incomplete dominance b. Polygenic inheritance c. X-inactivation d. Y-inactivation e. Codominance

In the fruit fly, the dominant and recessive traits for body colour and wing size are as follows:

Character Wild Type Mutant Body Colour Grey (b+) Black (b) Wing Size Normal (vg+) Vestigial (vg)

The genes encoding body colour and wing size are linked. A wild-type fruit fly, heterozygous for grey body colour and normal wings, was bred with a black fruit fly with vestigial wings. The distribution of the offspring is as follows: 437 black with normal wings, 752 grey with normal wings, 263 grey with vestigial wings and 548 black with vestigial wings. What is the recombination frequency between the genes for the body colour and wing size? a. 15% b. 30% c. 35% d. 70% e. There is no recombination

  1. How many centimorgans away from each other are the above loci? a. 60 b. 66 c. 35 (I think. Double-check) d. 70 e. 100

Which of the following statements is true about polyploidy? a. It alters the genetic balance. b. Includes traits such as height and skin colour. c. Occurs when an organism had more than the diploid set of chromosomes. d. It appears as a bell curve in the population. e. b and d are both correct

If you have a plant with blue flowers and round seeds, Bb Rr, crossed with a plant with white flowers and round seeds, bb Rr, what is the probability of getting a plant with blue flowers and wrinkled seeds? a. 1/ b. 1/ c. 1/ d. 3/ e. 1/

Which of the following is a source of genetic variability in meiosis? a. Independent Assortment b. Transformation c. Recombination d. Random fertilization e. a, c and d are correct

What is required to regenerate RuBP in phase 3 of the Calvin Cycle? a. 1 G3P, 9 ATP, 6 NADPH b. 5 G3P, 3 ATP c. 5 G3P, 9 ATP, 6 NADPH d. 1 G3P, 3 ATP e. 6 G3P, 3 AT

A plant that is heterozygous for the dominant “purple” colour is crossed with a plant that is recessive for the “white” colour. In the F1 generation, 150 plants are purple. Approximately how many plants in the F1 generation are expected to be white? a. 150 b. 25 c. 50 d. 0 e. More information is required to answer the question

Which of the following statements about meiosis 1 is false? a. During anaphase 1 homologous chromosomes are pulled to opposite poles b. During prophase 1 chromosomes condense c. Synapsis occurs during prophase 1 d. Separation of the cytoplasm occurs during anaphase 1 e. No replication occurs after meiosis 1

You have a purple, tall plant with yellow, round seeds, Pp TT Yr Rr crossed with a white, dwarf plant with yellow, wrinkled seeds, pp tt Yy rr. What is the probability of getting a purple tall plant with yellow wrinkled seeds? a. 1/ b. 3/ c. 3/ d. 1/ e. 0

Which of the following bridges the enhancer and the promoter? a. Activators b. Co-Activators c. Repressors d. General transcription factors e. Enhancer

Which of the following is true about recombination frequency? a. Genes that are closer together have a higher rate of recombination b. Genes that are further apart have a higher rate of recombination c. All genes recombine at the same frequency d. Genes with a recombination frequency equal to 1 assort independently e. None of the above

What did the Frye-Edidin experiment demonstrate? a. The membrane proteins are unable to move within the plane of the plasma membrane. b. That some membrane proteins move sideways within the plane of the plasma membrane. c. That phospholipids can “flip-flop” across the plasma membrane. d. That phospholipids cannot move sideways within the plane of the plasma membrane. e. None of the above.

Red-green colour blindness is a recessive, X-linked condition. A normal-vision daughter of a man with red-green colour blindness marries a man who is also normal for the trait. If the couple has two sons, what is the probability that both sons will be born with red-green colour blindness? a. 1/ b. 1/ c. 1/ d. 1/ e. 0

Which of the following statements is true of the lac operon?

a. It operates when glucose is present but only if there is an absence of lactose b. It operates when lactose is present but only if there is an absence of glucose c. Expression will occur when an inducer binds to the repressor protein d. a and c are both true e. b and c are both true

Which of the following statements regarding protein synthesis is true? a. Transcription involves the transfer of information from DNA to RNA, and translation involves the transfer of information from RNA to amino acids b. Transcription involves the transfer of information from DNA to RNA, and translation involves the transfer of information from DNA to amino acids c. Transcription uses nucleic acids exclusively to transfer information, while translation exclusively uses protein monomers. d. Transcription involves the transfer of information from DNA to RNA, while translation only uses RNA e. Options A and C are both correct

A difference between translation initiation in prokaryotes and eukaryotes is. a. In prokaryotes translation begins at the first start codon in the RNA strand, and in eukaryotes translation occurs at the first start codon that the ribosomal subunit encounters. b. In eukaryotes translation begins at the first start codon in the RNA strand, and in prokaryotes translation occurs at the first start codon that the ribosomal subunit encounters. c. Prokaryotes and Eukaryotes have unique start and stop codons. d. In prokaryotes translation is right to left and in eukaryotes translation is left to right. e. There are no differences in translation between prokaryotes and eukaryotes.

Why can C3 plants not survive in hot and dry climates? a. They undergo photorespiration b. They undergo phosphorespiration c. They do not have advanced control over their stomata like C4 and CAM plants d. They cannot get enough oxygen e. Their leaves are too big and would get dried out

Why is a testcross be performed? a. To determine the phenotype(s) of an organism’s offspring b. To determine the genotype(s) of an organism’s offspring c. To determine the traits expressed by an organism d. To determine the genotype of an organism e. To determine which traits are dominant and which are recessive

John has hemophilia, but his daughter Lucy does not. Lucy marries a man with hemophilia named Paul. If Lucy and Paul have two daughters and one son, what is the probability that none of their children will be affected by the disease? (Note that hemophilia is an X-linked recessive trait) a. 1/ b. 1/ c. 1/ d. 3/ e. 1

To what area of genetics did Gregor Mendel contribute the most? a. Determining the dominant and recessive traits in pea plants b. Discovering the basic patterns of allele sorting into gametes during meiosis c. Discovering the principles of incomplete dominance in flowers d. Creating the Punnett square method to calculate possible genotypes e. Generating interest in the field to entice future researchers

The allele R codes for red flowers and r for white flowers. This species displays incomplete dominance for colour resulting in pink flowers. The allele E codes for rough edges and the allele e for smooth edges. If a pink individual with smooth edges is crossed with a pink individual that is heterozygous for edge shape, what proportion of the offspring will display a pink phenotype? a. 0% b. 25 % c. 50 % d. 75 % e. 100 %

Mendel finds a stray green pea plant and performs a testcross. He discovers that all of the resulting offspring are green. What are the genotypes of the parents in this cross? a. GG x GG b. GG x Gg c. Gg x Gg d. gg x Gg e. gg x GG

What important contribution did Thomas Hunt Morgan make to genetics? a. He discovered that many traits are not segregated independently b. He determined the karyotype of drosophila c. He coined the term genetics d. He discovered that fruit flies have red and white eyes e. Options A and C are both correct

When a gene has more than two allelic forms, it can be referred to as. a. Incomplete dominance b. Codominance c. Multiallelism d. Multiple alleles e. Dominant and recessive alleles

Which of the following statements regarding polygenic inheritance is false? a. Several genes work together to influence a specific trait b. Produced phenotypes are displayed as a blend of many different alleles c. Polygenic inheritance is the same as incomplete dominance d. Produced phenotypes exist on a continuum and follow a bell-shaped curve e. All of these statements are true

What is the probability that an offspring will have the genotype: aaBbDDEe? a. 1/ b. 1/ c. 1/ d. 3/ e. 3/

What is most likely the inheritance pattern shown in the pedigree? a. Autosomal recessive b. Autosomal dominant c. X-linked dominant d. X-linked recessive e. Y-linked

What is the probability that the offspring from the individual 5IV will possess the trait? (Assuming an unaffected partner.) a. 0 b. 1/ c. 1/ d. 3/ e. Impossible to determine

In a sample of 1000 progeny, how many must vary, phenotypically, from the parental phenotypes for the alleles in question to be deemed unlinked? a. 250 b. 500 c. 600 d. 750 e. More information is needed to answer this question.

What type of inheritance pattern is shown by skin colour and height? a. Natural selection b. Incomplete dominance c. Codominance d. Polygenic inheritance e. Blending inheritance

Answer is 1/8 which was forgotten here

A certain flower variety shows phenotypes of red, white, and pink petal colours with smooth or rough petal edges. Red and rough are both dominant alleles. A cross was performed with a parent that was heterozygous for both traits and a parent that was homozygous recessive for both traits. If an offspring with pink, smooth petals is crossed with an offspring that has white, rough petals, what will the phenotype ratio be for the second-generation offspring? a. 1:2: b. 9:3:3: c. 6:3:3: d. 6:6:3: e. 1:1:1:

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What does the nucleus look like in S phase of Meiosis? - Biology

Tagging expressed proteins with the green fluorescent protein (GFP) from Aequorea victoria[1] is a highly specific and sensitive technique for studying the intracellular dynamics of proteins and organelles. We have developed, as a probe, a fusion protein of the carboxyl terminus of dynein and GFP (dynein–GFP), which fluorescently labels the astral microtubules of the budding yeast Saccharomyces cerevisiae. This paper describes the modifications to our multimode microscope imaging system [2], [3], the acquisition of three-dimensional (3-D) data sets and the computer processing methods we have developed to obtain time-lapse recordings of fluorescent astral microtubule dynamics and nuclear movements over the complete duration of the 90–120 minute yeast cell cycle. This required low excitation light intensity to prevent GFP photobleaching and phototoxicity, efficient light collection by the microscope optics, a cooled charge-coupled device (CCD) camera with high quantum efficiency, and image reconstruction from serial optical sections through the 6 μm-wide yeast cell to see most or all of the astral molecules. Methods are also described for combining fluorescent images of the microtubules labeled with dynein–GFP with high resolution differential interference contrast (DIC) images of nuclear and cellular morphology [4], and fluorescent images of the chromosomes stained with 4,6-diamidino-2-phenylindole (DAPI) [5].

SL Shaw, E Yeh, K Bloom and ED Salmon, Department of Biology, University of North Carolina at Chapel Hill, Chapel Hill, North Carolina 27599-3280, USA.

E-mail address for ED Salmon (corresponding author): [email protected] .

Watch the video: MEIOSIS - MADE SUPER EASY - ANIMATION (May 2022).