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# Tutor profile: Chika O.

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Chika O.
Senior at Yale University studying Neuroscience
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## Questions

### Subject:Pre-Algebra

TutorMe
Question:

Solve this problem for x: 4x - 17 = 15

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Chika O.

x = 8 The first step to solving this problem is to add 17 to both sides of the equation so that we have the variable by itself on one side: 4x = 32 ( -17 + 17 = 0 and 15 + 17 = 32) Lastly, divide both sides by four to isolate x: x = 8 (4x/4 = x and 32/4 = 8)

TutorMe
Question:

Do colleges like Yale only care about student grades and test scores?

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Chika O.

No, colleges look at students holistically, meaning they want to see students excelling academically but also involving themselves in activities that they're passionate about as well as taking up leadership roles in these extracurriculars. They want to see that you can handle academics while being an active part of your community so that you can bring those abilities and unique interests to their college campus. This isn't to say that numbers don't matter, but they're not enough to get a student into a top college nor are they enough to prevent someone from being accepted into a great school.

### Subject:Biology

TutorMe
Question:

A mother and father both express the phenotype for Huntington's Disease, an autosomal dominant disease that causes nerve cells in the brain to gradually break down. They give birth to four children, all of whom also end up having Huntington's Disease. Can we conclude from this information that at least one parent was homozygous dominant for Huntington's disease?

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Chika O.

No, we cannot conclusively say that at least one parent was homozygous dominant. For an autosomal dominant genetic disorder, possessing one mutant (i.e. disease-causing) allele will result in that individual displaying the mutant phenotype (i.e. the disease). Therefore, it is possible that both parents had the disease but were heterozygotes meaning they each only had one mutant allele. As heterozygotes, they each would have a 50% chance of passing on their mutant allele to each child. Based on probability, 75% of offspring should have Huntington's Disease, but given the small sample size, it is very possible and reasonable that all four children had Huntington's Disease. (We would better see a 3:1 ratio of diseased to non-diseased offspring if there were many offspring, but humans do not give rise to enough children to see these ratios clearly happen.) I would suggest drawing a Punnet Square to see how the 3:1 ratio is produced in the case of both parents being heterozygotes (eg. Hh x Hh).

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