Personal_Knowledge_Base/Biology_and_Chemistry/20241020165306-genetics_mendelian_genetics.org

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#+title: Genetics: Mendelian Genetics
#+filetags: :genetics:lecture_notes:
#+STARTUP: latexpreview




Mendel was a monk who experimented with plants such as the pea plant.
From his experiments, he determined several characteristics of inheritance. In fact, he postulated three key principles:

*Principle 1:* /In heterozygotes, one allele may conceal the presence of another./
*Principle 2:* /Two different alleles segregate from each other during the formation of gametes./
*Principle 3:* /Alleles of different genes assort independantly of each other./

- Mendel crossed pure-breeding tall pea plants with pure-breeding dward pea plants.
- It turned out, that when he did this, /all/ of the progeny were tall. These progeny were then self-fertilized.
- The offsrping of the self-fertilization resulted in both tall and dwarf plants. The ratio of tall to dwarf was 3:1

  What Mendel did here was a *monohybrid cross*.
  The phenotypic ratio for a monohybrid cross is *3:1*

  Likewise when a *dihybrid cross* was performed, i.e. a cross that tested two genes (each being heterozygous), a ratio of 9:3:3:1 was observed.
  The phenotypic ratio for a dihybrid cross is *9:3:3:1*

  The 9 dominant phenotypes often have differing genotypes.
  The absolute frequencies of the phenotypes are found by dividing by the total (in this case 16.)

  It is important to note Mendel's assumptions while performing these experiments:
    1. Genes segregated into alleles
    2. These segregations are not linked to each other
  Mendel was wrong in the second assumption. Some genes are in fact linked. (see[[id:8c1675c6-1b10-4fb4-9d77-25cd03a373cd][Genetics: Gene Linkage]])


* Test Crosses

Test crosses are performed to test whether a dominant phenotype contained a heterozygous or homozygous genotype.
Test crosses always use an individual that is /double recessive./
If the results include a 1:1 ratio of heterozygote dominant:recessive, the individual being tested must have been a heterozygote.

For example, if an individual has the genotype G g W w and is crossed with g g w w, the resulting phenotypic ratio will be *1:1:1:1*


* Chi-Square Test

A simple test that is done to measure if the error in an experiment is expected or above expected and in that case, not accepted.
It is given by the following formula:

\begin{equation*}
X^2 = \sum \frac{(observed-expected)^2}{expected}
\end{equation*}

As long as the value of the Chi-sqaure method is less than the 5% critical value of the experiment, the error is considered acceptable.
The 5% critical value depends on the *Degrees of Freedom* of the experiment. This can be found as shown below:

\begin{equation*}
Degrees of Freedom = Number of Phenotypes - 1
\end{equation*}

If there are 4 possible phenotypes for a given procedure, then there are 3 degrees of freedom.


Degrees of Freedom|5% Critical Value
1                 | 3.841
2                 | 5.991
3                 | 7.815


* Binomial Probability

Only for *2* phenotypic classes
Takes into account birthing order/all possible patterns

Let's say that there is a family of 4 children. What is the probability that 2 kids will be dominant and 2 will be recessive?
    Do a fraction for /every kid./
    The odds of being dominant are 3/4 (because of the 3:1 ratio)
    The odds of being recessive are 1/4.
    The method of getting number of patterns is shown below:

    \begin{equation*}
    \frac{4(kids)!}{2(dominant)! 2(recessive)!}
    \end{equation*}

    This expression simplifies to 6.
    Thus,


    \begin{equation*}
    \left(\frac{3}{4}\right)^2 \times \left(\frac{1}{2}\right)^2 \times 6 = \frac{54}{256}
    \end{equation*}


If a problem deals with an "at least" type of language, do each case separately. Then, add up all the results to get the answer.