Exam 3 Flashcards
(84 cards)
What is a null hypothesis?
A null hypothesis assumes data will fit a given ratio or model, and that any variation is purely by chance.
True or False:
A null hypothesis can be proven.
False:
A null hypothesis cannot be proven, but it can be either supported (failed to reject) or rejected.
What does the Chi-square statistical test accomplish?
Known as the ‘goodness of fit’ test, it evaluates the influence of chance on data and assesses whether the data is consistent with the model (expected outcome) and determines if the null hypothesis should be rejected or not.
What is p-value (Chi-square test calculation)?
P-value is a statistical measure of significant differences between datasets calculated using a Chi-square statistical test and has a cut-off (a) frequently set to 0.05.
What does the value (i.e. 0.5) of p-value mean?
The value (0.5) obtained from calculating p-value is the percent (5%) that if this experiment is repeated, there is a 5 percent chance that the same or greater variation in data would be observed again.
When should one reject or accept the null hypothesis based on the p-value?
If the p-value of the data is greater than the cut-off (a) then the null hypothesis is supported. However, if the p-value of the data is less than the (a) then the null hypothesis is rejected due to too much variation.
How do you determine how many degrees of freedom?
Degrees of freedom are determined by taking the sum of how many variants are being analysed and subtracted by one.
Example: Vestigial wings are completely recessive to wild-type wings in Drosophila melanogaster.
Monohybrid cross data:
315 F2 flies with wild-type wings
97 F2 flies with vestigial wings
Model?
Null Hypothesis?
(X^2)?
Degrees of Freedom?
p-value?
Fail to reject or reject?
Model? 3:1
Null Hypothesis? There is no difference between the monohybrid cross results and the model; the monohybrid results follow a 3:1 ratio.
(X^2)?
X^2 = sum of ((Observed - Expected)^2) / Expected
wild-type: 315 309 0.117
vestigial: 97 103 0.350
412
Sum (wild-type + vestigial X2) = 0.467 = X2
Degrees of freedom? 1
p-value? 0.50
Fail to reject or reject? Fail to reject
Gene mapping
What is linkage?
Genes (alleles) on the same chromosome are likely to be inherited together known as linked genes.
Gene mapping
What is a linkage group and provide an example?
A linkage group is a set of linked genes.
(i.e. Genes located on chromosome 4 are in linkage group IV; Genes located on the X chromosome are X-linked)
Gene mapping
What is a genetic map?
A genetic map shows the order of linked genes and distances between genes on chromosomes.
Gene mapping
What is an example of linkage?
Two genes on a single pair of homologs; exchange occurs between two nonsister chromatids.
Gene mapping
What determines the likelihood of recombination?
The distance between genes determines the likelihood of recombination.
Two genes that are far apart have many chances of crossing over while two genes that are extremely close together do not have the possibility of crossing over due to strong linkage.
Gene mapping
What are genetic and physical markers?
A genetic marker is a DNA sequence, such as an SNP or repeat.
A physical marker is a phenotype associated with a gene locus.
Gene mapping
What is a test cross and what are they used to determine?
A test cross crosses an F1 heterozygote with a homozygous recessive tester. If we are interested in two genes, a two-point test cross is used.
Gene mapping
How do we know which progeny (offspring) are recombinant and which are parental?
Recombinant progeny have a different combination of markers from the parent, indicating a crossover has occurred.
Parental progeny have the same combination of markers as the parent.
Gene mapping
What do the results of a test cross indicate if:
50% recombination between e and vg genes
F2 phenotypic classes:
e vg : 25% parental
e vg+ : 25% recombinant
e+ vg : 25% recombinant
e+ vg+ : 25% parental
The genes are very far apart on the same chromosome, OR on different chromosomes indicating independent assortment.
Gene mapping
What do the results of a test cross indicate if:
No recombination between e and vg genes.
F2 phenotypic classes:
e vg : 50% parental
e vg+ : 0% recombinant
e+ vg : 0% recombinant
e+ vg+ : 50% parental
Genes are very close on the same chromosome and have complete linkage.
Gene mapping
What do the results of this experiment indicate? And how do we determine the distance between the e and vg genes?
F2 phenotypic classes:
e vg : 1418 flies : 43.8% parental
e vg+ : 241 flies : 7.4% recombinant
e+ vg : 283 flies : 8.7% recombinant
e+ vg+ : 1294 flies : 40.0% parental
The e and vg genes are linked.
We determine the distance between genes by dividing the number of recombinants by the total number of progeny and then multiplying by 100 to get the number of map units (mu) the genes are apart.
((241 + 283) / 3236) * 100 = 16.2 mu
Genes e and vg are 16.2 map units apart.
Gene mapping
What happens during a double-crossover (DCO)?
A double crossover occurs when two chiasmata form and there is a double exchange between two nonsister chromatids.
Gene mapping
True or False:
Double-crossovers occur far less than single-crossovers.
True
Gene mapping
How do we know which progeny are parental, single-crossover, or double-crossover in a three-point cross?
Progeny in the lowest abundance are double-crossover recombinants.
Progeny in the highest abundance are parental.
Progeny in the intermediate range of abundance are single-crossover recombinants.
Gene mapping
What four steps will help in the determination of a three-test cross?
1) Determine the order of the genes
2) Reorganize the data to reflect the gene order
3) Calculate map units for the first two genes
4) Calculate map units for the second two genes
Gene mapping
Identify parental, region 1 SCO recombinants, region 2 SCO recombinants, DCO recombinants, gene order, recombination frequency, and map unit distance from the following data:
p+ j+ r+ 179
p j r 173
p+ j r 52
p j+ r+ 46
p+ j+ r 22
p j r+ 22
p+ j r+ 4
p j+ r 2
Total: 500
Parental:
p+ j+ r+ 179
p j r 173
SCO region 1:
p+ j r 52
p j+ r+ 46
SCO region 2:
p+ j+ r 22
p j r+ 22
DCO:
p+ j r+ 4
p j+ r 2
Gene order: pjr or rjp
DIstance p to j (region 1)
(SCO region 1 + DCO) / Total Progeny * 100
=20.8 mu
Distance j to r (region 2)
(SCO region 2 + DCO) / Total Progeny * 100
=10 mu
Distance p to r (region 1 + region 2)
20.8 mu + 10 mu = 30.8 mu