Writing in Biology

Mates: Different investments between females (few # of costly eggs) vs males (high number of low cost sperm) naturally create both sexually dimorphism and behavioral differences between the genders. Natural selection favors those with the highest reproductive success and typically, birds form lifelong mated pairs that cooperate to raise their young as they require assistance to do so. However, as males will have higher success by finding multiple mates they often engage in cross pair copulation with other females. Furthermore, females have ways to protect their larger investments by selectivly choosing their "partner" as well as secondary partners. It is this "choosy" behavior that has driven the extravagant courtship displays seen in the wild. In addition, females will often breed with several neighboring species to ensure that at least some offspring are viable.

The TLC analysis, the solvent that was proven to be the best was 95% Hexane, 5% Ethyl acetate although it the ratio between the two compounds was not perfect for separation of the products. The solvent that was proven the best was 70% Hexane, 30% Ethyl acetate. This solvent showed a greater separation of products and distinct identity of each product in lane B compared to lanes A and C. 

In this lab, an Sn2 reaction took place with 2-naphthyl as the nucleophile which displaced the iodide to form butyl naphthyl ether and sodium iodide. The percent yield I obtained from this experiment was 13.6%. The percent yield is relatively low. The first reason my products yield was low was because the melting point was very low, I noticed that as I was weighing the product some of the product melted on the filter paper. I could have avoided the loss of product this way by being more careful when handling the solid product and making sure no heat was applied to it. Also the addition of NaOH before refluxing is important so the product is properly deprotonated so that the product does not become oil when recrystallizing. If the product is oil instead of crystals then the product would go through the filtration. The melting point obtained was 31- 33⁰C. Compared to the melting point of butyl naphthyl ether which is 33-35^oC, the product produced in this lab was close in range to the one given to us. Therefore, my product is mostly pure although the discrepancies between the temperature given to us and the one obtained in the experiment can be explained the presence of possible impurities that are left in the product.

Both 2-naphthyl and the product produced were dissolved in ethyl acetate. Then on three TLC plates put in one lane just the dissolved 2-naphthyl, the second lane is two spots of both dissolved 2-naphthyl and dissolved product, and in the third lane the dissolved product. The first plate is dissolved in the solvent of 95% ethyl acetate and 5% hexane. The second plate is dissolved in the solvent of 95% hexane and 5 % ethyl acetate. After analyzing which solvent worked better which was 95% hexane and 5% ethyl acetate, the ratios were changed to get the best results. The final plate was dissolved in 70% hexane and 30% ethyl acetate. 

Next, fill a 50 mL beaker with ice up to about 30 mL line. Then add the mixture over the ice, and rinse the round bottomed flask twice with 1 mL of ice cold water to help transfer any product left behind. Stir the mixture in the beaker until only 10% of ice is left. The product is then collected through a vacuum filtration and rinsed with ice water. Let dry under suction for at least 25 minutes. After 25 minutes, use a small amount to run a TLC analysis. After at least one night left to dry measure the melting point and percent yield. The melting point was 31- 33^oC and the percent yield was 13.6%.

In a 10 mL round bottomed flask add 0.064 g of NaOH, 0.20 g of 2-naphthyl, and boiling stones. Then add 3 ml of absolute ethanol to the round bottomed flask and attach to the air condenser. Reflux the mixture by hovering the flask over the sand bath until there are no longer any solids and continue for 25 minutes. After the 25 minutes, cool the mixture for 3 minutes. After the 3 minutes, add 0.2 mL of n-butyl iodide through the top of the air condenser and then reflux the mixture for an hour. Cool water in an ice bath for the last 10 minutes of the reflux.

After the eight groups have been organized, each will be assigned a variable to be investigated. Once they get their variable, they will plan out locations where they will be able study different degrees of the that variable. When they get to the area where they will be studying, they will mark out a one square meter area to be sampled. They will then scan the marked area, looking for different species of grass, weed, or flower, and record the species richness. After they study an area with one form of their variable, they will go on to study another area with a different form of their variable. For example, after a group finds the species richness of an area of high pedestrian traffic, they will go study an area with low pedestrian traffic. Using the data from these two areas, the group will be able to compare and analyze whether their variable has an effect on species richness. Once all of the groups are done, they can all compare their results to figure out which variable has the most significant impact on the plants.

The species of grasses and weeds that inhabit the ground and turf people walk on often go by unnoticed. In this experiment, we propose to study the diversity and richness of these species and figure out what variables play a major role in changing the diversity and richness. In order for the project to be compliant with eight different groups collecting data, we came up with eight different variables. These variables include: soil pH, amounts of sunlight, amounts of foot traffic and nearby vehicular traffic, and elevation in relation to water runoff. In addition to that, other variables groups could study are: frequency the area is mowed, how recently the grass was planted, and the grass’s proximity to a body of water.

Henrietta Lacks was a thirty-one year old African-American woman who died of complications due to cervical cancer and the treatment she received. Not too many people know of her story or even her name, but in the science community she is best known as HeLa, the immortal cell line. HeLa cells were the first ever human immortal cell line to divide infinitely, it has been used in many medical advancements and achievements such as; helping to develop drugs to treat herpes, leukemia, influenza, hemophilia, and Parkinson’s disease and the cells have been used to study sexually transmitted diseases, gene mapping and sequencing, somatic cell fusion, and cloning. HeLa cells are still being used till this very day all over the world as the cells are still dividing effortlessly.

What were the authors trying to test, and what predictions did they make?

    The authors were studying the geographically isolated Scandinavian population of grey wolves. This population was founded by only two individuals and recovered by the arrival by addition of a single immigrant. The original population suffered from inbreeding depression. The addition of an immigrant granted new genetic variation and allowed the population to thrive.

What, exactly, did the authors do?

    The authors collected tissue and blood samples from Scandinavian wolves and wolves from Finland and northwest Russia. DNA was extracted and analyzed. Twelve microsatellite markers were chosen to examine. The relationships between individuals was analyzed using Kinship.

What did the authors find (i.e, what were their data)?

    The authors found that “ Scandinavian wolves are highly differentiated both from the neighbouring eastern population and from historical Scandinavian wolves.” All of the wolf populations were found to be significantly differentiated. The genetic variability was significantly lower in the Scandinavian wolf population than other populations. A sharp decline in heterozygosity was observed. The decrease in heterozygosity was measured at 30 % per generation which matches the estimate. The addition of a new immigrant to the population increased mean heterozygosity from 0.49 to 0.62.

How did the authors interpret their findings?

    The sharp decline in heterozygosity was thought to be due to inbreeding, The expected loss of heterozygosity for a population of two individuals would be 25% per year. The observed value was 30%, which is relatively close. However the formula is likely not accurate in such small populations. As expected, the addition of the new immigrant and new alleles increased heterozygosity.