TutorMe homepage

SIGN IN

Start Free Trial

Christina P.

Neurobiology Major at University of California Irvine

Tutor Satisfaction Guarantee

Biology

TutorMe

Question:

You are a geneticist doing a cross-breed between two pea plants. For this problem assume that the round gene (R) and green gene (G) are dominant and the wrinkled (r) gene and yellow (y) gene are recessive. If you cross a pure-bred round yellow pea plant with a round green hybrid pea plant what would be the ratio of phenotypes seen in the offspring?

Christina P.

Answer:

This is a relatively classic Punnett Square problem. In order to create the Punnett square, we must first figure out the genotypes of the parents given by the information from the problem. the problem states that we are crossing a pure-bred with a hybrid pea plant but what does that tell us about the genotype? It tells us that the round yellow pea plant has the same two alleles for both genes and the round green pea plant has two different alleles for both genes. That means the genotypes for the parents are going to be RRgg and RrGg respectively. Now let's make the pungent square, note that since one pea plant is a true-bred than there only has to be one column for it as demonstrated below; Rg *Note that the alleles in the brackets are from the hybrid parent* [RG] RRGg [Rg] RRgg [rG] RrGg [rg] Rrgg As indicated in the Punnett square above, there will be a roughly equal distribution between the 4 genotypes. But the problem is asking us about the phenotype ratios so let's figure out what each genotype represents; RRGg= round green RrGg= round green RRgg= round yellow Rrgg= round yellow That means there is a 1: 1 ratio of round yellow to round green offspring. Moreover, this means that for this cross there is a 50% chance of the offspring being either a round yellow or a round green pea plant.

Calculus

TutorMe

Question:

Find the derivative of the following equation; f (x)= ln (x) / (1 + ln(2x) )

Christina P.

Answer:

The first thing to do in a complicated derivative problem like this'd to identify which derivative rules you are going to need to use. In this case the rules we will use to solve this problem are U / V = (VU' - UV' ) / (V^2) and ln U = U' / U **please note that U' & V' represent the derivative for that variable** Now that we know which rules are going to be applied for f(x) lets start taking the derivative; f'(x) = [ ( 1 + ln(2x) )( 1 / x ) - ( 2 / 2x )( ln (x) ) ] / ( 1 + ln (2x)) ^ 2 Next we will simplify the answer by multiplying the appropriate variables in the numerator; ( 1 / x ) + ( ln (2x) / x ) - ( ln (x) / x ) Since all of the variables in the numerator are over x we can drop the x to be in the denominator so that f'(x) = ( 1 + ln(2x) - ln (x) ) / ( x )( 1 + ln (2x) )^2 Finally to solve for the answer we will simply the numerator one last time using one of the logarithm rule ln (U) - ln (V) = ln (U / V) for our problem that means ln(2x)-ln(x) = ln (2x / x)= ln(2) Plugging that simplification into our equation we get our final answer; f'(x) = ( 1 + ln (2) ) / ( x )( 1+ ln (2x) ) ^2

Algebra

TutorMe

Question:

Solve for X and Y in the following equations; (a) 2x + 4 = 3y (b) 6 + y = 8x

Christina P.

Answer:

To solve for X; First you must isolate y in equation (b) so that y = 8x - 6 Next you plug that answer in for y in equation (a) so that 2x + 4 = 3 (8x - 6) Then multiply 3 by (8x-6) so that 2x + 4 = 24x- 18 Move all of the x terms to one side of the equation so 4 + 18 = 24x - 2x After adding/subtracting like-terms ( 22 = 22x ) divide by the coefficient in front of x to isolate the variable so X= 22/22 = 1 To solve for Y; Plug the value you found for x into either equation (a) or (b) so that 6 + y = 8 (1) Isolate y just like you did above so y= 8 - 6 = 2

Send a message explaining your

needs and Christina will reply soon.

needs and Christina will reply soon.

Contact Christina

Ready now? Request a lesson.

Start Session

FAQs

What is a lesson?

A lesson is virtual lesson space on our platform where you and a tutor can communicate.
You'll have the option to communicate using video/audio as well as text chat.
You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.

How do I begin a lesson?

If the tutor is currently online, you can click the "Start Session" button above.
If they are offline, you can always send them a message to schedule a lesson.

Who are TutorMe tutors?

Many of our tutors are current college students or recent graduates of top-tier universities
like MIT, Harvard and USC.
TutorMe has thousands of top-quality tutors available to work with you.

Made in California

© 2019 TutorMe.com, Inc.