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.

DNA Replication

DNA replication is needed for the process mitosis, cell reproduction. DNA replication is a five step process that uses five different enzymes and reads from the 3 carbon end(OH-) to the 5 carbon end. The first enzyme is helicase. The helicase breaks hydrogen bonds between nitrogen bases at the origin of replication, where there are a lot of A-T repeats. The second enzyme is RNA primase, RNA primase creates a polar end for DNA Polymerase 3 (polar), and lays down RNA nucleotides. The next enzyme DNA Polymerase 3 puts in complementary DNA. DNA polymerase 1 replaces the RNA  with DNA. And the final enzyme Ligase glues the Okazaki fragments together.

                                                                                 



 

 

Survival of the Sickest: Chapter 6 and Your Inner Fish: Chapter 3

In Chapter 6 of Survival of the Sickest, it discusses "jumping genes." Nobel-prize winning scientist, Barbara McClintock found that corn genes mutated faster when the corn was under stress like drought or extreme heat. This process of mutation is called “jumping genes”, in this process the cell suppresses the proofreading cells so mutation will occur. These mutations were passed on. McClintock also found that mutations under stress happened so fast because the cells were fighting for survival. This flower is similar to the mutation in corn.

 

 

 

In Your Inner Fish, Chapter 3, the picture below is similar to the experiment of Mary Gasseling, where a patch of the tissue responsible for developing digits (ZPA: zone of polarizing activity) were planted on the opposite side of a developing limb. Which led to developing full duplicate set of digits on the opposite side.



 



Survival of the Sickest Chapter 6 Summary

Human life starts with 46 genes, 23 of them are passed from the father and the other 23 from the mother, these genes help build proteins that build the human body.  These genes act as an active network, and each gene does multiple tasks, not just one, when some genes are removed the other genes help to compensate for these removed genes.

It was believed that evolution happens when mutation in genes occurs randomly, but this phenomenon is rare because there are cells that proofread the process of copying genes, and errors rarely happen. That means that evolution has to take a long time and is not random.

Nobel-prize winning scientist, Barbara McClintock found that corn genes mutated faster when the corn was under stress like drought or extreme heat. This process of mutation is called “jumping genes”, in this process the cell suppresses the proofreading cells so mutation will occur. These mutations were passed on. McClintock also found that mutations under stress happened so fast because the cells were fighting for survival.

 

Future study of human genes found that a huge percentage of human DNA is related to viruses, it is also believed that these viruses contribute to human evolution. Additionally, jumping genes phenomenon occurred in these viruses which is called retrotransposons, there is also another kind of jumping genes called DNA transposons. Both of these tranposons contribute to the rapid evolution of humans.   

 

 

"From Atoms to Traits"

1. The significance of Gregor Mendel is that through his experiments with peas in the 1850's and 1860's, he proved that variants in species are inheritable. Mendel also proved that genetic variants do not "blend" away in future generations, they reappear. This is evidence for Darwin's theory of evolution through natural selection.  

2.  James Watson and Francis Crick discovered the double helix structure for DNA.

 

3. Homopolymers are stretches of DNA with eight or more identical letters in a row are prone to copying errors during DNA replication. An example of homopolymers is microsatellites consists sequences of two or more nucleotides in a row.  The second variation in DNA is when the bases change from G to A. An example of this is in peas when the variation occurs it shortens growth.  The third variation is “jumping” elements which changes gene activity patterns by creating new regulatory sequences. An example is the wrinkly seeds of peas.  The fourth variation is the change in regulatory genes that regulate cell division. An example is the differences between the maize plant and the tesinte plant. The fifth variation is change in pigment cells. An example is the change in human skin color.

4. Evo-devo is studying the effect of changes in important developmental genes and the role they play in evolution.

5. Migration and human reliance on other animal’s milk as a source of food was a result of the mutant form of the lactase gene which is only active in infants in other mammals, continues to be active in human populations they depend on milk as nutrition.

Founder Mutation

            Founder Mutations are changes in the DNA which are inherited future generations. These mutations first occurred in an individual, the founder and were later passed on. These mutations occurred through the process of evolution, to protect against more dangerous and deadly diseases. For example, Sickle-cell anemia originated in Africa to protect against deadly malaria. Sickle- cell anemia makes the red blood cells less hospitable, and malaria only thrives in healthy red blood cells.

            Unlike Founder mutations, hotspot mutations are spontaneous mutations of genes in individuals, individuals who have these mutations are not related to one another and do not share the same DNA. People who have founder mutations have similar DNA because they share a common ancestor.

            When an individual has one copy of the mutant gene they have a better chance of survival than those who have no copies. Individuals with two copies of the mutant gene will probably die before they could reproduce. This is called balancing selection.

 

            The chromosome region surrounding the mutant gene, the haplotype, gets shorter over generations due to recombination of chromosomes. Geneticists can use the haplotypes to trace the origin of a certain population and its migration. By tracing back the founder mutation to a certain region and time in history, geneticists can find when the mutation began and the reason the mutation started.

            The knowledge of founder mutations is valuable to physicians in identifying the types of diseases to test for depending on the ethnicity of an individual. For example, African Americans are more susceptible to Sickle- cell anemia, since the population has become more mixed the study of genes became very important for physicians to establish proper treatment depending on every individual’s DNA. The studies of founder mutations help geneticists find the origins of humans and its migration. In addition, it helps doctors and physician diagnose and treat diseases.

Hardy-Weinberg Problems

In class, 10/11/13, we learned the Hardy- Weinberg formula, which determines the frequency of alleles(genes) in a population. Hardy and Weinberg argued that if five conditions are met, the population's allele and genotype frequencies will remain constant from generation to generation. The five conditions are:

The Hardy-Weinberg formula is:



Here is a Hardy- Weinberg Problem:

 In certain African countries 4% of the newborn babies have sickle-cell anemia, which is a recessive trait. Out of a random population 1,000 newborn babies, how many would you expect of the three possible genotypes?

The first thing you look for is a q(recessive allele frequency) or q^2 (homozygous recessive). The q^2 in this problem is 4% of newborn babies. In order to find q you mush square root the 4%. So the value of q is 0.2. To find p you can just subtract 1-0.2, since you know that p+q=1. p=0.8, and q=0.2.

The problem also asks for the three possible genotypes (q^2, p^2, and 2pq). In order to find these values, you must multiply  your values for q^2, p^2 and 2pq by the total population, 1000.

p=0.8

q=0.2

2pq=0.32

population=1000



Now we use these values to find the number of individuals in each genotype for this population

q^2*population=number of individuals with recessive trait(sickle-cell anemia)

(0.04)^2*1000=40 individuals

p^2*population=number of individuals with the dominant trait

(0.8)^2*1000=640 individuals

2pq*population= number of individuals with the heterozygous trait

(0.32)*1000=320 individuals

Brine Shrimp Lab and Recessive Genes

 

In class, 10/3/13, we collected the second set of data for the brine shrimp lab. In this lab we placed different concentrations of salt water (0.0%, 0.5%, 1.0%, 1.5%, and 2.0%) into Petri dishes. Then we put around 20 brine shrimp eggs and covered them. We collected data after 24 hours and saw that the Petri dish that contained 1.5%   NaCl concentration had the highest number of shrimp swimming( 10 shrimps) , the second highest was the dish containing 0.5% NaCl concentration ( 6 shrimps), and the third was the 1.0% with 4 shrimp swimming. The other two containers did not have any shrimp swimming. After 48 hours we collected more data and saw that there were less shrimp swimming than before, 8 shrimps in the 0.5% and 1 in the 1.5%. The other three containers did not have any shrimp swimming. This lab shows that some shrimp developed variations which made them more adapted to higher concentration salt water than others, and the others could not survive it. This proves the theory of natural selection because only the shrimp that had these variations were more adapted (fit) to the changing environment survived and were able to pass on their genes.

 

 

 

 

            Also in class we did an activity about recessive genes.  We had two types of beads, red beads and green beads. We assigned the red beads to be the dominant gene. We place the beads into a paper bag and selected two beads at a time. If there was a pair of green beads they would die off because it was not the dominant gene. We repeated this process several times and each time we excluded the pairs of not dominant beads. Eventually we found that the dominant gene, the red beads, became more prevalent. This activity represents an isolated community and how the genes over time become similar and recessive.

 

Fava Beans and Paleontology Project

In class 9/26/13, we reviewed chapter 4 of Survival of the Sickest. Chapter 4 was about Favism, the most common enzyme deficiency in the world, 400 million people are affected by it. Favism is most common and deadly in North Africa and Southern Europe, around the Mediterranean Sea. Favism is carried on the X chromosome, making it more common in men because they have XY chromosome, while women have XX chromosome. If Favism affects one X chromosome, women still have another X chromosome, while men only have one X chromosome.

 

            Favism is a G6PD deficiency, caused by eating fava beans. Fava beans are mad of two sugar compounds: Convicine and Vicine. Which produces free radicals(unpaired electrons) that attack red blood cells and burst them, and produces hydrogen peroxide. This can lead to hemolytic anemia and sometimes death.

 

 

            The G6PD deficiency  carried in people with Favism, is due to evolution to fight Malaria which is prevalent around the Mediterranean Sea. Malaria is a deadly disease transmitted through mosquitos with parasites. The G6PD deficiency is an advantage because it makes red blood cells less hospitable for Malaria, and reduces their chances of getting Malaria and Malaria seeks healthy red blood cells.

 

 

Malaria and red blood cells

            Plants develop toxins to defend against predators, so they can survive and reproduce. For example, the Cassava plant, when eaten raw produces cyanide which can be fatal. Another example are Nightshades ( spicy peppers), Nightshades produce Capsaicin, a sticky poison which cannot be dissolved in water. The last example is the Jimsonweed, the Jimson weed produces chemicals which can cause hallucinations.

 

 

Cassava Plant

            In class, we also chose the organism for our Paleontology project. My partner and I chose Hippocampus Sarmaticus( early seahorse) and for the rest of the class period we researched the organism.

 

Evidence of Evolution Quiz Answers

1. Fossils show that many evolutionary changes are gradual. But the incomplete fossil record can falsely suggest or conceal times of rapid change. The terrestrial animal began developing aquatic features and gained more bone structure. But as the evolution process continued certain body parts (ex: limbs) began to shrink in size due to lack of use (vestigial structures).

 

2. Marsupials began in North America (E.) Then they traveled to Australia

3. The bat, bird, and dragonfly show convergent evolution, since they are independently evolved in similar features in species of different lineages.

4. In the common descent lab, we compared the Cytochrome- C (amino acid) sequences of different organisms. If there were few differences in the sequences which means that the species are closely related and evolved from a common ancestor. If there were many differences in the sequences this means that the organisms are not closely related and evolved from different ancestors. For example, primates and humans have no differences in the Cytochrome-C sequences, which means that both evolved from the same ancestor.

5. Homology is the study of anatomical features, of different organisms, that have a similar appearance or function because they were inherited from a common ancestor that also had them. For example the forelimbs of dolphins and humans are different in appearance but they have similar bone structure and they come from a common ancestor.

 

 

Your Inner Fish

In class 9/20/13, we discussed the chapters we read for homework the night before, in Your Inner Fish. The chapters we read were about how Neil Shubin and his colleagues process of finding Tiktaalik. An important topic discussed in these chapters was where to find fossils. The way to find a good fossil is to look at rocks of the right age (time period), right type of rock (sedimentary rocks), and exposure (fossils towards the surface). Also, the chapters discussed layers of rocks, upper layers are younger and contain more complex and younger fossils, and lower layers contain more simple and primitive fossils. Fossils are formed through a slow process of layer and compacting of sediment.

The discovery of Tiktaalik in Greenland by Neil Shubin and his colleagues gives evidence of evolution. Tiktaalik has a neck which was used for a similar function as humans use necks today; structures which are used for similar functions and have similar structures are known as homologous structures, which further proves that humans and Tiktaalik have the same common ancestor. Tiktaalik is also the missing link between amphibians and fish because it has limbs and can breathe on land and in water.

 

Biochemistry Unit Review

In class 9/13/13, we reviewed for the unit test. We went over the properties of water: cohesive behavior, ability to moderate temperature, freezing upon expansion, and versatility as a solvent. We also went over basic chemistry, covalent and ionic bonding, number of protons, neutrons, and electrons, and valence electrons.

We reviewed polymers which are made through condensation processes (monomers + monomers= polymers and release of water) and we reviewed how monomers get released (polymers + water=release of monomers). In addition we went over different kinds of isomers: structural isomers (differ in covalent partners), geometric isomers (differ in arrangement of double bonds), and enantiomers (differ in spatial arrangement around an asymmetric carbon). There are also different groups of isomers: hydroxyl (ex: ethanol), carbonyl (ex: acetone), carboxyl (ex: vinegar), amino (ex: glycine), sulfhydryl (ex: ethane thiol), and phosphate (ex: glycerol phosphate).

    

  

Hydroxyl group: ethanol

Structural Isomer

After we reviewed these topics, we took a quiz, which helped us prepare for the test.

Testing Food for Biological Macromolecules/ Nutrients

In class 9/11/13, we did a lab to test food for biological macromolecules / nutrients. In the part 1 we had to observe the reactions of starch, protein, lipids, and sugars with benedict’s test, iodine test, Biuret test, and Sudan III test. For the benedict’s test we had to put each test tube in hot water in order to see a reaction. We found that sugars react to Benedict’s solution which is blue and changes to orange. Starches interact with Iodine which is brown and changes to a dark blue. Proteins react with Biuret which is lavender and changes to blue. Lipids react with Sudan III which is red and changes to bright red.

 

 

In part two, we tested 10 unknown food substances with each test, to identify the type of food, depending on the reactions to the Bendict’s, Sudan III, Biuret, and Iodine tests. We found that in the Benedict’s test six solutions reacted with food substances in tubes A, B, E, G, H, and J which indicates that these food substances contain sugars.

 

We found that in the iodine test it reacted to the food substances in tubes C, D, E, F, and G, which indicates that these substances contain starch. In the Biuret test it reacted to substances in tubes B, C, D, and E, which indicates that these food substances contain proteins.  In the Sudan III test it reacted to food substances in tubes A, F, I, and J, which indicates that these substances contain lipids.

 

 

                By testing food substances with Benedict’s, iodine, Biuret, and Sudan III tests, there reaction will indicate if they contain proteins, starch, sugars, and lipids.