Genetics and DNA (More about Probability)

At the beginning of the lesson, the writer told the class a story about the inheritance. To arouse students’ interests, students first were attracted by a legacy of $1 000 000 000 000 000 000 000 that will be inherited to a rich man’s children only. They were then given a checklist on the man’s characteristics, like

“Curly tongue” and “Thumb bending”. They were asked if they fitted all the characteristics and thus they might be the child of this rich man. Students were eager to check the characteristics and compared which one got the most items.

The teacher introduced the meaning of dominant gene and recessive gene. The teacher then told the students a news related to a couple of Fujianese carriers of the thalassemia gene.

They gave birth to his son four years ago. His son unfortunately inherited his severe illness of thalassemia and suffered from long-term blood transfusion. In order to get the cord blood

School Mathematics Newsletter  Issue No. 23

transplant to save the first child, even if only a quarter of a healthy baby, they still decided to risk a second child. They were then asked how they could get the probability of

4 1 . Using “A” stood for dominant gene and “a” for recessive gene, most of them could get the value by either using the tree diagram and tabulation method they learnt in junior secondary, or by using the probability rules they learned in the senior secondary curriculum.

Tabulation method M F

A a

A AA Aa

a Aa aa

Method of using the multiplication rule in the senior secondary

P(healthy baby) =

4 1 2 1 2 1 

Students were also assigned to complete the 2×2 table for other combinations.

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There should be 6 combinations.

M F

A A M F

A a

A AA AA A AA AA a Aa Aa A AA Aa A AA AA a Aa Aa a Aa Aa a Aa aa

M F

A a M F

a a

a Aa aa a aa aa a Aa aa a aa aa

However, some students claimed that they found 9 combinations. After checking, they found out that three of them were redundant.

M F

A a M F

a a M F

a a

A AA Aa A Aa Aa A Aa Aa A AA Aa A Aa Aa a aa aa

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Student’s work

Would the population with characteristics represented by the dominant gene increased after several generations, and the population with characteristics represented by the recessive gene became zero after hundred years? Hardy, a British Mathematician in year 1908, showed that this phenomenon would NOT occur. Hardy with Weinberg, who was a biologist, published this result in a magazine. This was called the

“Hardy–Weinberg principle”.

The climax was when students were going to verify the

“Hardy–Weinberg principle” by using the multiplication rule in probability. The activity was divided by two parts: one by using numerical value, and one by using algebra.

For the numerical value part, we first assumed that in parental genotypes, P(AA) =

1and P(Aa) = 3

1. Then students found out

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the probabilities of AA in child’s genotype, Aa in child’s genotype and aa in child’s genotype.

Student’s script

With the help of the spreadsheet program, students could easily got the results for different values of P(AA) and P(Aa).

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Most students were able to verify, by numerical values, that ) genotype s

child' in ( ) genotype s

child' in

(AA P aa

P  = 1.

However, was the result true for ALL possible values of P(AA) and P(Aa)? Algebra would be the solution to this part.

Assume that in parental genotypes, P(AA) = p1, and P(Aa) = p2. Then, P(aa) = p3, where p3 = 1 – p1 – p2.

Therefore, we only needed to know the values of p1 and p2 in order to perform our calculations.

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Parental genotype

Probability of such

combination of parental genotypes

Probability of child’s genotype for such combination

AA Aa aa

AA / AA

(p1)² 1 0 0

AA / Aa p

p

2 + p2

p

1 = 2p1

p

2 2

1 0

AA / aa p

p

3 + p3

p

1 = 2p1

p

0 1 0

Aa / Aa

(p2)²

4 1

2 1

4 1

Aa / aa p

p

3 + p3

p

2 = 2p2

p

3 0

2 1

aa / aa

2 2 p  p

 

  

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P(Aa in child’s genotype)



1 2



1 3 2²



2 3



1 1 1

2 2 2

p p p p

p

p p

   

1 2 1 3 2 2 3

2 1

p p p p 2 p p p

   

3²

1 1

p

2 2

p p p

  

2 2

2 3 2 3

4p p p p

  

3 2

p

 

  

3 2

2 2 p p

 

  

In what condition would give rise to a stable genotype distribution?

A stable genotype distribution meant that: the ratio of the population with characteristics represented by the dominant gene, to the population with characteristics represented by the recessive gene, would become constant, i.e. the ratio of the child’s genotype would be the same as the parental genotype.

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Since P(AA in child’s genotype) and P(aa in child’s genotype) were

1 2

2 2 p p

 

 

  and

3 2

2 2 p p

 

 

  respectively, so, ) genotype s

child' in ( ) genotype s

child' in

(AA P aa

P 

would be equal to

   

¹ ² ³ ²

1 2 3

2 2

2 2 2

p p p p

P AA P aa

p p p



   

 



  



Therefore,

) genotype s

child' in ( ) genotype s

child' in

(AA P aa

P  = 1.

A very interesting result was that:

Conversely, if

p

₁

p

₃  was true, then by squaring both 1 sides of the equation,

1 1 3 3

1 3 1 3

1 3 2

2 1

2 4

p p p p

p p p p p p

  

  



 

²²

1 1 1 2

4 p p p  p

2 2

1 1 1 2

4 p p p p  p

2 2

1 1 1 2 4

p  p p p  p

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2 2 2

1 1 2 1

2 2

p p

p  p  p      

2 2

1 1

p p  p  , by completing square



1 2

2 2 p p

p   

   and the case for another formula

3 2

2 2 p p

p   

   was similar.

This condition stated that whatever the value of p1 and p3 at the beginning, the second generation of the child’s genotype became a stable genotype distribution.

This activity was conducted in two Secondary 5 classes with a diversity in their mathematical abilities. Students could fill in the table to construct the possible combinations of genotypes.

They could also (although take time) calculate the probabilities when a concrete numerical value was given. However, some of them were scared of algebra. They were unable to complete the squares and factorising by grouping terms.

A questionnaire was conducted after the tryout lesson. 81% of students agreed that this STEM activity was challenging. Thus, the writer suggested that teachers who wanted to adopt this teaching activity might consider trimming some parts of the activity.

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Teacher spirit in the promotion of STEM was more

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