I have a test in a couple days and I still don’t know how to do this stuffsomeone teach mee

Many of the neurodegenerative disorders are the result of the flaws in the folding of the protein, which comprises Parkinson, Alzheimer, prion, and Huntington diseases. Once the protein specific to disease misfolds, it can then produce toxic masses that gather in the brain, resulting in cell death, neuronal dysfunction, and clinical symptoms.

As misfolding of proteins and the gathering of masses are the most upstream occurrences in the pathological cascade, thus, stabilizing or rescuing the native conformations of proteins is a better therapeutic approach.

In the present times, small molecules called chaperones have been demonstrated to be effective in minimizing the levels of misfolded proteins, thus reducing the gathering of masses and their downstream pathological outcomes. Chaperones are categorized as pharmacological, molecular, or chemical.

6 0

3 years ago

In the garden pea, several different genes affect pod characteristics. A gene affecting pod color (green is dominant to yellow)

Arlecino [84]

Answer:

Green/Wide/Long, G-W-L- = 186 individuals
Yellow/narrow/ long, gg ww L- = 186 individuals
Green/wide/ short, G-W-ll = 186 individuals
Yellow/narrow/ short, gg ww ll = 186 individuals
Green/narrow/ long, G-wwL- = 14 individuals
Yellow/wide/ long, gg W-L- = 14 individuals
Green/narrow/short, G-wwll = 14 individuals
Yellow/wide/ short, ggW- ll = 14 individuals

Explanation:

Available data:

• One gene affecting pod color (green is dominant to yellow)

• One gene affecting pod width (wide is dominant to narrow)

• Both genes are located on chromosome 5 and approximately 7 mu away from each other.

• A third gene, located on chromosome 4, affects pod length (long is dominant to short).

• Cross: A true-breeding wild-type plant (green, wide, long pods) was crossed to a plant with yellow, narrow, short pods.

• The F1offspring were then test-crossed to plants with yellow, narrow, short pods.

• The testcross produced 800 offspring.

Cross 1:

Parental) GGWWLL x ggwwll

F1) GgWwLl

Cross 2:

Parental) GgWwLl x ggwwll

F2) N=800

To calculate the numbers of the F2 generation, we first need to calculate the frequencies of recombination and parental of the linked genes. We know that they are 7MU apart from each other. The map unit is the distance between a pair of genes for which every 100 meiotic products one results in a recombinant one.

MU = 7 means that there is 7% of recombination for genes that express color and width. This is:

MU = 7 = 7% recombination

We have two possibilities of recombination: green/narrow and yellow/wide. Each of these recombinants has half of the possibilities of occurring, so:

7 map units = 7 % of recombination in total

= % Gw + % gW

= 3.5 % + 3.5 %

Gw = 3.5% = 0.035

gW = 7%/2 = 0.035

For the parentals, we can calculate

100% - 7% = 93% of parental in total

= % of GW + % gw

= 46.5% + 46.5%

GW = 46.5% = 0.465

Gw = 46.5% = 0.465

The length gene is located in a different chromosome so it assorts independently. This is, of the gametes will be long, L, and the other 50% will be short, l.

L = 50% = 0.5

Now we need to figure out how to relate these frequencies. All we need to do is to multiply the frequencies of occurrence obtained previously for linked and independent genes, for each possible phenotype. This is:

Green/Wide, GW = 0.465

Yellow/narrow, gw = 0.465

Green/Narrow, Gw = 0.035

Yellow/Wide, gW = 0.035

Long, L = 0.5

Short, l = 0.5

Phenotypic frequencies:

Green/Wide/Long, G-W-L- = GW x L = 0.465 x 0.5 = 0. 2325

Yellow/narrow/ long, gg ww L- = gw x L = 0.465 x 0.5 = 0. 2325

Green/wide/ short, G-W-ll = 0.465 x 0.5 = 0. 2325

Yellow/narrow/ short, gg ww ll = 0.465 x 0.5 = 0. 2325

Green/narrow/ long, G-wwL- = 0.035 x 0.5 = 0.0175

Yellow/wide/ long, gg W-L- = 0.035 x 0.5 = 0.0175

Green/narrow/short, G-wwll = 0.035 x 0.5 = 0.0175

Yellow/wide/ short, ggW- ll= 0.035 x 0.5 = 0.0175

Finally, as we need to obtain the numbers of the individuals with those phenotypes, we need to multiply each frequency by N, which is the total number of individuals in the F2 (N = 800).

Green/Wide/Long, G-W-L- = 0. 2325 x 800 = 186 individuals

Yellow/narrow/ long, gg ww L- = 0. 2325 x 800 = 186 individuals

Green/wide/ short, G-W-ll = 0. 2325 x 800 = 186 individuals

Yellow/narrow/ short, gg ww ll = 0. 2325 x 800 = 186 individuals

Green/narrow/ long, G-wwL- = 0.0175 x 800 = 14 individuals

Yellow/wide/ long, gg W-L- = 0.0175 x 800 = 14 individuals

Green/narrow/short, G-wwll = 0.0175 x 800 = 14 individuals

Yellow/wide/ short, ggW- ll = 0.0175 x 800 = 14 individuals

3 0

3 years ago