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*Biology Paper 5 tips*

J

Just visiting

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guys plz tell me anything about the statistical tests like when to use each and y
t-test
chi squared
and that z test
 
I

Irfan1995

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Just visiting said:
guys plz tell me anything about the statistical tests like when to use each and y
t-test
chi squared
and that z test
Click to expand...

t-test: comparing two sets of continuous data
chi-squared: comparing one set of (discontinuous) data to an 'expected' or 'standard'
(other) z-tests: I can explain it, but I don't think it's required in the A-level syllabus.
 
I

Ineedhelp

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Irfan1995 said:
Guys, I really don't know why you guys are over-complicating the problem. In part (a) to the question, you can clearly see that the 'without tail' allele is dominant, and that the gene is autosomal (not sex-linked).

In cross 1, you have two parents 'without a tail'. As you have some offspring that do have tails, you can be sure that both parents are heterozygous. If one of them were homozygous, then ALL the offspring would be 'without a tail'.

Now, we know that when two heterozygous parents cross, the ratio of the phenotypes of the offspring would be 3:1 (dominant allele:recessive allele).
In this case, we have two heterozygous parents, out of every 4 offspring, we expect 3 of them to be 'without a tail' and 1 to have a tail. So 3/4 of the offspring are expected to be 'without a tail' and 1/4 with a tail. This is just a simple monohybrid cross.
Now, to get the expected number of offspring:
Total number of offspring = 40 + 72 = 112 (taken from the observed results)
Expected number of offspring with a tail = (3/4) * 112 = 84
Expected number of offspring without a tail = (1/4) * 112 = 28
Click to expand...

Your values are switched up. in the marking scheme with a tail is 28. and without is 84.
=S soo yeah.
 
I

Irfan1995

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Ineedhelp said:
Your values are switched up. in the marking scheme with a tail is 28. and without is 84.
=S soo yeah.
Click to expand...

My bad, the expected number of offspring without a tail (dominant allele) = (3/4) * 112 = 84
expected number of offspring with a tail (recessive allele) = (1/4) * 112 = 28
Thanks for pointing that out :)
 
J

Just visiting

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Irfan1995 said:
t-test: comparing two sets of continuous data
chi-squared: comparing one set of (discontinuous) data to an 'expected' or 'standard'
(other) z-tests: I can explain it, but I don't think it's required in the A-level syllabus.
Click to expand...
thanks but what are continous and discontinuous data
PS. the z test i faced it in one of the old exams but it had another name
 
X

Xainab

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And for error bars, do we have to multiply the standard error by 2 before drawing it on the graph?
 
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