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.