Mitosis and Meiosis and Cell Cycle

1.      The centromere is a region in which…

The answer is chromatids are attached to one another

2.      What is a chromatid?

A replicated chromosome

3.      During which phase do centioles begin to move apart in animal cells

Prometaphase

4.      Which phase is the longest phase of the mitotic stages?

Prophase

5.      During which phase do centromeres uncouple, sister chromatids are separated, and the two new chromosomes move to opposite poles of the cell?

Anaphase

6.      A cell containing 92 chromatids at metaphase of mitosis would, at its completion, produce two nuclei containing how many chromosomes?

46

7.      The formation of a cell plate is beginning across the middle of a cell and nuclei are re-forming at opposite ends of the cell. What kind of cell is this?

A plant cell undergoing cytokinesis

8.      Movement of the chromosomes during anaphase would be most affected by a drug that…

Prevents shortening of microtubules

Genotype of the Father

The mother is eebb since she has yellow fur and brown nose and brown eyes. The puppies have black fur and brown fur so the dad has to be at least EE. Also, the picture shows that the puppies have brown ears, black nose, or brown nose. So based on the picture the fathers genotype would be EEBb.

 

 

 

 

Your Inner Fish: Chapter 6 The Best-Laid (Body) Plans

Biologists study embryos as the first stage of life. They found that embryos of species like fish, amphibians, birds, and mammals look very similar in the beginning stages of embryonic development. The study of similarities, differences, and development in embryos is called embryology.

Biologists found that there are three layers of tissues in  developing embryos that are responsible for developing all body parts which vary from one species to another. The outer layer is called the ectoderm, the inner layer is called the endoderm, and the middle layer is called the mesoderm. The ectoderm forms much of the body (skins and nervous system). The endoderm forms the inner structures of the body (digestive tract). The mesoderm forms tissues between guts and skin (skeletons and muscles).



DNA control controls how the body is formed. Hilde Mangold discovered a small patch of tissues that was able to direct other cells to form an entire body plan this patch of tissues is known as the Organizer.

 

 

Dihybrid Cross

Unlike a monohybrid cross, a dihybrid cross deals with two traits. Here are a couple problems.

1. In mice, the ability to run normally is a dominant trait. Mice with this trait are called running mice (R). The recessive trait causes mice to run in circles only. Mice with this trait are called waltzing mice (r). Hair color is also inherited in mice. Black hair (B) is dominant over brown hair (b). Cross a heterozygous running, heterozygous black mouse with a homozygous running, homozygous black mouse. What is the offspring phenotypic ratio?

        First we have to cross a  heterozygous running, heterozygous black mouse (RrBb) with a homozygous running, homozygous black(RRBB). The results would be 4 homozygous running, homozygous black (RRBB), 4 homozygous black, heterozygous running(RrBB), 4 heterozygous black, heterozygous running (RrBb), and 4 heterozygous black and homozygous running(BbRR). So the answer would be all running, black. 

2. Cross a homozygous running, heterozygous black mouse with a waltzing brown mouse. What are the parental genotypes?

          The genotypes would be RRBb x rrbb

3. Cross a heterozygous running, brown mouse with a heterozygous running, homozygous black mouse. What are the genotypes for the offspring at the running allele?

           So you would just focus on the crossing the running gametes so you would cross Rr x Rr. and the running genotypes would be RR, Rr, rr, because when you cross Rr x Rr you get 1 RR: 2 Rr: 1 rr.

4. Cross a heterozygous running, heterozygous black mouse with a heterozygous black mouse. What is the offspring phenotypic ratio?

           So you would cross RrBb x RrBb and you would get 9 running black : 3 running brown: 3 waltzing, black: 1 waltzing, brown.

Monohybrid Problems

1. A true- breeding plant that produces yellow seeds is crossed with a true-breeding plant that produces green seeds. The seeds of all the offspring are yellow. Why?

     The answer is the yellow allele is dominant to the green allele. Since the yellow allele is dominant all the offspring will be yellow as well.

2. A true-breeding plant that produces yellow seeds is crossed with a true-breeding plant that produces green seeds The F1 plants have yellow seeds. What is the expected phenotypic ratio of seed color of the offspring of an F1xF1 cross?

     F1 means first generation offspring so the F1 offspring would be heterozygous Aa. So we would cross AaxAa. The offspring would be AA, Aa, Aa, and aa. The question is asking for phenotypic ratio which means the physical traits ratio. Since three of the offspring have a capital A which stands for yellow seed and only one has aa, green seed. the ratio would be 3 yellow seed to 1 green seed.

3. Attached earlobes are recessive to free earlobes. What is the probability of having a child with attached earlobes when an individual with attached earlobes mates with an individual heterozygous for free earlobes?

   In this problem you would cross a recessive individual (aa) and heterozygous individual (Aa). The offspring would be Aa, Aa, aa, and aa. Since half of the offspring are aa (attached earlobes) so the answer would be 50%.

4. To determine the phenotype of an individual who expresses a dominant trait, you would cross that individual with an individual who_____.

   The answer would be homozygous recessive for that trait.

5. In garden peas, tallness is dominant and dwarfness is recessive. A heterozygous tall plant is crossed with a dwarf plant. If 40 offspring are produced, the number of tall individuals will be approximately?

    You would cross a heterozygous individual (Aa) and a recessive individual (aa). The offspring would be Aa, Aa, aa, and aa. Since half of the offspring are aa  and there are 40 offspring. The answer would be 20.

Here is a picture that explains monohybrid crosses and also a little of dihybrid crosses (2 traits)

Operon System and pGLO Lab

There are two kinds of operon systems, repressible and inducible. I am going to explain inducible because it is the same kind of operon system used in the pGLO lab.

In an inducible operon system it begins off and we have to turn it on. First, the promoter attracts RNA polymerase since the operator already has a repressor(protein) the RNA polymerase is not able to read genes anymore. For example, arabinose(sugar) would come from an outside system like agar and go into the repressor to inactivate it. Once the operator is open, the RNA polymerase is no longer locked and can continue to read genes. When RNA polymerase is reading genes it creates arabinose (enzyme) which can digest the arabinose. Once all the arabinose is all digested the protein/ repressor goes back into the operator and is locked again. The operon system only uses necessary energy, not extra.

In the pGLO lab, the bacteria (E. coli) transformed with pGLO will not glow in two weeks because in order for it to glow all the time there must be a constant supply of arabinose. But after two weeks the supply of arabinose will decrease and the bacteria will not be able to glow. Through GFP/ arabinose operon system, if arabinose is no longer supplied the repressor will go back into the operator and the RNA polymerase will no longer be able to read the GFP (green fluorescent protein) gene, and will not fluoresce anymore.

 

 

Protein Synthesis

The first step in protein synthesis of a polypeptide strand is to change the DNA into mRNA. This is called transcription, RNA Polymerase creates mRNA because DNA has a double helix and is too big to fit through a pore in the nuclear membrane. Protein synthesis can only occur in Eukaryotic cells, cells that have a nucleus. The second step is RNA processing where the introns get cut out by spliceosomes and the exons stay the same. This happens in the cytoplasm, a guanine cap (3 phosphates) and poly A tails (bunch of  A nucleotides) are placed to protect the message. After RNA processing, the mRNA goes to the APE site. The "A" stands for accepting site, where the mRNA is accepted. The "P" stands for peptide site, where peptide bonds are formed. The "E" site stands for exiting site where the mRNA codon (3 pairs of RNA base)/ peptide bonds are matched with anticodons (tRNA) and leave the ribosome which reads from 5 carbon end to 3 carbon end, whereas in transcription it reads from 3 carbon end to 5 carbon end. The tRNA also buses amino acids. The last step is translation, when the mRNA is translated into proteins.