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Main focus–

a) To discuss an alternative possible alternative Hypothesis when the Null Hypothesis is rejected in cross where Mendelian Genetics is expected

 

b) To review of Multiplication and addition rule in probabilities.  

  

Period  3,8:

1.  Complete the Statistics Quiz

2. Quick Drosophila Article Review – 

Drosophila Article Key p.pdf
View Download

3.  Non-mendelian Genetics:  If a the Null Hypothesis for a particular genetics cross is rejected then there must be a significant difference between the expected ratios (from the Punnet square) and the observed outcome.

This difference is due to some reason or driving factor that is “proposed’ in an alternative hypothesis.

So if Mendelian Genetics is not upheld by a rejected Null Hypothesis than a  plausible alternative hypothesis in this situation could be one or more of the following:

Chapter 14 – (end) topics that was in your NORT from the summer assignment!

Incomplete Dominance – 

Co-dominance – 

Multiple Alleles – 

Epistasis- One gene conrols or affects another

Multiple Alleles – 

Gene – Linkage – 

         a) Use X -linked recessive disorder of color blindness to demonstrate that Mendelian genetics is not upheld.

    

          – A greater percentage of males will have colorblindness than female. How can this be if the probability of being a male or female is 50% and probability of getting a mutated X chromosome is 50% from a heterozygote?   Shouldn’t the allele for color blindness  independently assort so that 50% of the females and 50% of the males can have color blindness?  Why is it that more males are colorblind? Why does sex makes difference in this disorder?

 

            The gene for color blindness is on the human chromosome that also has the genes for female characteristics!!!

 

           These 2 alleles (female characteristics) and color blindness are linked.  That is they are      

           located on the same chromosome and thus cannot independently assort.  This means that 

           the female characteristics and colorblindness traits travel together in heredity!

 

           This is non-Mendelian! 

 

Because the gene is linked (the allele for colorblindness is on the same chromosome as the female characteristics trait) males cannot be carriers and cannot be heterozygotes!!!  It just takes one X mutated chromosome to have the disorder.  

 

Because of this the ratios we would expect from our cross above would be STATISTICALLY Significant  from actual results. We would expect equal number of males and female to have the disorder but that is NOT the case.  This is Non-Mendelian.

 

If males could be heterozygotes for Colorblindness we would expect 1/4 of all offspring AND EQUAL number of males and females having the condition.  1/2   x  1/4 = 1/8   probability of being a male and being colorblind.

 

The above probability problem uses the multiplication rule because the each event (separating of a different chromosome independently) is separate if you assume that the 2 traits are on different chromosomes.  This is Mendelian!!! 

We would get 1/8 probability if the genes for these traits were unlinked.

 

But males are cannot be heterozygotes for colorblindness. and thus the actual probability of being a male and having the condition is equal to :   1/4 = the probability of being colorblind in cross between a colorblind male and a female carrier.

Because the the actual probability is 1/4  (NOT 1/8) , the 2 traits do not travel independently as Mendel always suggested.  We would use the addition rule here and count how many of the offspring would have the phenotype.  OUT OF THE 4 CHOICES IN THE PUNNET SQUARE THERE IS ONLY ONE!

 

1/4 + 0 = 1/4 of all offspring are colorblind but 50% of the males are also!

 

4. Lab – Complete both Excel graphs for Lab 3

– Change the value of the x – axis

– Limit the 0 mark for test groups that do not have data for some days

– Add data table and Graphs in to your lab.

– Begin Analysis 

 

 

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