BIO207H5 Lecture Notes - Wild Type, Penetrance, Statistical Hypothesis Testing

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redswan498

22 Mar 2014

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From the last lecture, we followed gene segregation in a cross of a true breeding shibire fly with a wild type fly. This is the segregation pattern expected for a single gene. There is no logical way to prove that we have a 3 :1 ratio. Nevertheless, we can think of an alternative hypothesis then show that the alternative hypothesis does not fit the data. Usually, we then adopt the simplest hypothesis that still fits the data. A possible alternative hypothesis is that recessive mutations in two different genes are needed to get a paralyzed fly. In this case a true breeding paralyzed fly would have genotype: aaaaa/aaaaa , bbbbb/bbbbb. Whereas wild type would have genotype: aaaaa/aaaaa , bbbbb/bbbbb. F2: p(aaaaa/aaaaa and bbbbb/bbbbb) = (1/4 )2 = 1/16 p(aaaaa/aaaaa and bbbbb/ ) = 1/4 x 3/4 = 3/16 p(bbbbb/bbbbb and aaaaa/ ) = 3/16 p(aaaaa/ and bbbbb/ ) = the rest = 9/16.

Related Questions

Introduction: A Chi-square test is used to compare observed data with expected data according to a hypothesis. For instance, if you were crossbreeding 2 heterozygous pea plants, you would expect to see a 3:1 phenotypic ratio in the offspring. In this case, if you were to breed 400 pea plants, you would expect to see 300 plants showing the dominant trait and 100 showing the recessive trait. But what happens if you observe only 260 plants with the dominant trait and 140 plants with the recessive trait? Does this mean something is wrong with Mendelian genetics or is this difference in expected results just due to chance (random sampling error)? These are the questions that can be answered using Chi-square statistics. The results of this statistical test is used to either reject or accept (fail to reject) the null hypothesis. The null hypothesis states there is no significant difference between the observed results and the expected results. This means that if the null hypothesis is accepted, the difference in observed and expected results was just a matter of chance and so the observed results basically "fit" with what was expected. Degrees of freedom (df) = number of independent outcomes (Y) being compared less 1 df = Y-1 At the 95% confidence interval we are 95% confident that there is a significant difference between the observed and expected results, therefore rejecting the null hypothesis. Probability Value - Is the decimal value determined from the X2 table and is the probability of accepting the null hypothesis. A 0.05 probability value equates to a 95% confidence interval.

The Chi-squared test formula is: Example: If we cross two pea plants that are heterozygous yellow pods, we would expect a 3:1 phenotypic ratio. So let's say we actually did the cross and got 280 plants with green pods and 120 plants with yellow pods. Question: Is this a 3:1 phenotypic ratio? This is the value of Chi-squared Test. We have a total of 400 plants and we expect a 300 green:100 yellow phenotypic ratio If the calculated Chi-squared value is less than the critical value listed in the Chi-squared table, then we accept the null hypothesis. This means that there is no significant difference between the observed and the expected values. Our degrees of freedom (df) = 2 outcomes - 1, or df = 1. Now we go the X2 table below and using the df = 1 and probability value of 0.05, our critical value is 3.84. Since our calculated X2 value is 5.33, and is larger than the critical value, we reject the null hypothesis and can say (at 95% confidence) that there is a significant difference between our observed and expected values.

The parent generation is yellowed podded and green podded pea plants. You cross a yellow podded pea plant with a green podded pea plant and you get 100% yellow podded plants in the F1 Generation (Phenotypic ratio 4 : 0, yellow to green). What will be the expected phenotypic ratio when you allow the F1 generation to reproduce?

Fill out the Punnett square.

If we actually did the cross and got 1150 yellow and 350 green. Would this be a consistent with what was expected?

Learning Outcomes Questions

1. Why would you run a Chi-squared test?

To determine if our data is consistent with expected results.
		a To determine if our data is consistent with expected results. b To determine if our data exactly matches the expected results.
		c To determine the expected results.
	d	To compare the phenotypic ratios to the genotypic ratios.

2. Determine the degrees of Freedom of the phenotypic ratio for this genetic cross.

a. 1

b. 2

c. 3

d. 4

e. 5

3. Using the data given, what is the result of your Chi-squared analysis? x2= ___.

	a.	2.22
	b	2.71
	c	4.36
	d	187.78
	e	448.27

4. Using the results of your Chi-squared analysis, do we fail to reject or reject the null hypothesis?

a.		Fail to reject the null
b.		Reject the null
c.		It cannot be determined from the data given

silversalmon27

A three point cross was performed using fruit flies. The three loci are physically linked on the same autosome. The d allele, in the homozygous state, results in the recessive âdouble dorsalâ phenotype. Similarly, the r allele results in the recessive âround earsâ phenotype and the e allele in the recessive âendless eyeâ phenotype. In contrast, the D, R, and E alleles encode for the dominant wild type phenotypes at each of these loci.

True breeding (homozygous) flies were crossed to produce F1 offspring. Due to poor book-keeping, the genotypes of the parent flies were not recorded â but one parent fly had one mutant phenotype, while the other parent had the other two mutant phenotypes.

Phenotype Number of F2 Flies

Ears Dorsal Eyes

1) Round ears Double Dorsal Endless Eyes 230

2) Wild-type Wild-type Wild-type 260

3) Round ears Wild-type Wild-type 4,390

4) Wild-type Double Dorsal Endless Eyes 4,100

5) Round ears Double Dorsal Wild-type 15

6) Wild-type Wild-type Endless Eyes 15

7) Round ears Wild-type Endless Eyes 510

8) Wild-type Double Dorsal Wild-type 480

10,000 total

a) What is the phenotype of the F1 flies?

b) What is the genotype of the F1 flies? Your answer should indicate the phase of linkage.

c) What are the genotypes and phenotypes of the original parent flies (previously not recorded)? Your answer should indicate the phase of linkage.

d) Calculate the map distance in cM between each of the three loci and determine the gene order. Show all work for full credit. (6 points)

(tautomerization) The most common form of guanine is the ketoform, not the enol form.

True

False

What term describes the allowed degree of base pair matchingduring annealing, varying from lots of base-pair mismatches allowedto allowing only perfect base-pair matching.

stringency

aneuploidy

polyploidy

quantitative PCR

restriction fragment length polymorphism

I have an individual with a disease we will call SSI-itis,brought on by extreme exposure to cannabis. I do a southern andfind the gene affected. Another of his classmates exhibits theexact same symptoms, but that gene in hiw is wt. A southern in thesecond student finds a mutation in another gene. This is an exampleof

pleitropy

allelic heterogeneity

locus heterogeneity

dominant traits

recessive traits

penetrance

If a person carrying a dominant allele does not demonstrate thephenotype strongly, but still differs from wt, thisdemonstrates?

haploinsufficiency

penetrance

expressivity

monogenic traits

polygenic traits

I find three alleles for a gene in a fruit fly. One allelecreates the ability for the fly to eat graphite. Another allelecreates the ability for the fly to eat coal. The third alleleallows the fly to eat diamond. These alleles are all dominant, andwt cannot eat any of these substances. Which statement below is themost true?

Upon mating a bunch of the flies carrying these alleles, Ishould not find a fly that can eat graphite, coal, and diamond.

I would consider the ability to eat these substances an exampleof pleitropy because they are all carbon.

I would expect the ability to eat diamond to be dominant to theability to eat graphite.

These flies are demonstrating locusheterogeneity.

Expression vectors in prokaryotes do not make functionaleukaryotic gene products in bacteria very well because

the codon sequence for prokaryotes is different than the codonsequence in eukaryotes

there are no disulfide bridges formed in proteins normally madein prokaryotes

prokaryotic expression vectors cannot translate eukaryticsequences

eukaryotic genes have introns, and prokaryotes don't

eukaryotic genes have exons and prokaryotes don't

Which of the following is not used in PCR amplification ofDNA?

Double-stranded DNA template

Oligonucleotide primers

DNA polymerase

DNA ligase

ddNTPs

I create a knockout mouse using the agouti/black fur systemdiscussed in the online lecture and in the book. When I cross theagouti offspring of the originally engineered mouse, I find a ratioof 2 agouti mice to 1 black furred mouse. What is the bestexplanation?

The gene knocked out was recessive.

The gene knocked out was dominant.

The gene knocked out was a lethal gene.

The knockput was integrated into a random spot, and did notknockput the original gene.

A loss of function mutation is usually recessive, but canactually be a dominant mutation because of what effect?

haploinsufficiency

dominant negative mutations

allelic heterogeneity

locus heterogeneity

pleitropy

A genetic carrier of a disease can have a form of the diseasethat only shows intermittent symptoms, while the full homozygousrecessiuve individual would show all of the symptoms of thedisease.

True

False

Wild type, or wt, refers to the genotype that is the typicalgenotype found in nature. Now read that again. Once more. Nowanswer.

Answer

True

False

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BIO207H5 Lecture Notes - Wild Type, Penetrance, Statistical Hypothesis Testing

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