BIOL3190 Lecture Notes - Lecture 15: Dihybrid Cross, Null Hypothesis, Centimorgan

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27 Oct 2017

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If you did in advance, you were at an advantage. Based on that prediction, you did a chi squared analysis. 9:7, a:o no just says how confident you are in a particular hypothesis. Has nothing to do with variation there was a lot of null hypothesis was that variation was due to chance p-value is the likelihood that what you actually observe varies from the null just due to chance. 14-6 is pretty different than 11-9 there really only is a 15% chance that you would see something that different. What if type o babies are more likely to die early. If (cid:455)ou sa(cid:449) so(cid:373)ethi(cid:374)g that did(cid:374)"t (cid:272)o(cid:374)fo(cid:396)(cid:373) to (cid:374)ull h(cid:455)pothesis the(cid:374) (cid:455)ou (cid:272)ould figu(cid:396)e out (cid:449)hat"s goi(cid:374)g o(cid:374) Mutation of the f 1 allele which puts the sugar on to a cell surface protein that then allows for the type a sugar to be added or the type b sugar to be added.

Related Questions

Q.In the non-9:3:3:1 ratio dihybrid cross below, there are 3 classes of phenotypes: white, pink, red which of the following is NOT true?

9 A_B_ white

3 A_bb pink

3 aaB_ red

1 aabb red

a.) phenotype is dependent on the genotype of gene B

b.) This is an example of recessive episasis

c.) aa is epistatic to B_, and bb

d.) phenotype is dependent on the genotype of gene A

Q,When choosing a selectable marker, you...

a.) choose a enhancer for your gene of interest

b.) choose a rescue gene to save your gene of interest

c.) choose a resistant gene that will allow for colony screening

d.) choose a gene that encodes for DNA replication

Q.The following steps detail an experiment resulting in transgenic mice with a knockout gene for fur color. Recall that agouti (brown) mice are A/A;M/M and black fur coats are a/a;M/M . List the steps in their correct chronological order.

1.Inject transformed agouti embryotic cells into an a/a;M/M embryo to produce A/A;M/m and a/a;M/M altered embryo

2.Self-cross hemizygous mice to produce homozygous transgenic mice (A/-;M/m)

3.Create a a targeting vector containing tk+ for ganciclovir sensitivity, neoR for neomycin resistance, and the gene encoding fur color.

4.Observe mice with chimeric phenotype and cross them with black a/a;M/M mice to produce transgenic hemizygous progeny (A/a;M/m)

5.Isolate all transformed embryonic stem cells that show homologous recombination and inject them into a embryonic stem cells of an agouti mouse.

a.) 3,5,4,1,2

b.) 1,5,3,4,2

c.) 3,5,1,4,2

d.) 5,3,2,4,2

If dog 1 has is black with alleles (A_B_) , and dog 2 is brown with (aaB_), and dog 3 is yellow with (_ _bb) alleles, what kind of epistasis is this and what color would dog 4 be if he was (AAbb).

1. Recessive epistasis, Yellow

2. Recessive epistasis, Black

3. Dominant epistasis, Yellow

4. Dominant epistasis, Black

A gene interaction in which pair of recessive genes at one locus prevents expression of dominant allele at another locus called

a. incomplete dominance

b. epistatis

c. complete dominance

d. polygenic inheritance

Problem A: Shipwreck Founder Effect Genetics

In the year 1782, a small passenger schooner (the HMS Leaky) was blown off course, crashing into the rocks of a then-uncharted mid-Atlantic island. The 30 survivors built shelter and found plentiful food on the island. Within a year, it was clear that there was little chance of rescue and no means of leaving. Soon, the warm tropical nights led couples to succumb to the romantic tendencies that moonlight walks on the beach can induce. Before long, babies were born, and the ânewâ civilization of the island now known as Gilligania was off to the races.

A genetic analysis of the remains of those original islanders showed the following ABO blood type phenotypes:

Blood Type	Number of People
A*	9
B**	7
AB	4
O	10

*Four heterozygotes (iA iO) and five homozygotes for iA
** Five heterozygotes (iB iO) and two homozygotes for iB

Keep in mind that the blood type alleles iA and iB are codominant over each other, and that the iO allele is recessive to both iA and iB. If you need to do so, review the genetics of this system in Campbellâs Biology. Based on this information, please answer the following questions (show your work):

1. What were the frequencies of the three alleles in the original survivor population on the island?

2. Determine the genetic structure (allele and genotype/phenotype frequencies) for this gene system in the island population in 1932, when the island was ârediscoveredâ by cartographers. The population at that time had grown to 185 people.

THis was all the information we were given

greyotter602

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

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BIOL3190 Lecture Notes - Lecture 15: Dihybrid Cross, Null Hypothesis, Centimorgan

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