BIOL 101
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Transmission Genetics 

Fullerton College
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(These are notes from the Bio 109 online course)
Basic Genetics Terminology

Most of us have seen that many features of our bodies, talents and personalities seem to be resemble a relative more than someone unrelated. We are going to first look at some body characteristics and human diseases that are affected/caused by a single gene on an autosomal chromosome. For the exam, you will need to know the definitions of basic genetic terms, Mendel's discoveries, how to use Punnett squares, and be able to solve autosomal trait problems.There are lots of good genetics glossaries available (see the related links below). But I will use some simple definitions while giving you an example of family inheritance.  Suppose there was a family in which the father Sam and daughter Susan had dimples, but the mother Sally did not show dimples.

 Sam (father)  DD  homozygous dominant  Dimples
 Sally (mother)  dd  homozygous recessive  no dimples
 Susan (child)  Dd  heterozygous  Dimples

The TRAIT (physical or measurable characteristic) we are focusing on in this family is dimples. Sam and Susan have the PHENOTYPE (visible physical makeup of this trait) of presence of dimples, whereas Sally has the phenotype of no dimples.

Each person has genetic information on their chromosomes that determine whether or not they will have dimples. The dimples GENE is that portion (DNA sequence) of a specific chromosome that controls the expression of the Dimples trait. But there appears to be two different phenotypes in our scenario, as there are two different ALLELES (alternative forms of one gene). One allele of the Dimple gene has information (a certain DNA sequence) that tells the body to make dimples. The other allele of the Dimple gene has information (a different DNA sequence) that tells the body not to make dimples. To simplify, we will use the letter "dee" for the dimple gene with the big "D" letter for the allele which specifies dimples and the letter "d" for the allele which specifies no dimples.

In a nutshell, the dimples gene gives information for the dimples trait. Allele D = dimples and allele d = no dimples.

Suppose the gene for dimples is on chromosome #5. Remember that each person is diploid or has two sets of each type of the autosomal chromosomes (the non-sex chromosomes, types 1-22). That means each person has two chromosome #5, and on each chromosome #5, there sits an allele for the dimples gene. Therefore, you have a total of two alleles for this gene.

Suppose that the GENOTYPE (the genetic makeup for this trait) of Sam is "DD". He is HOMOZYGOUS (has 2 identical alleles for that gene) for the dimples gene. Sally's genotype is "dd", so she is also considered homozygous. Susan's genotype is "Dd", so she is considered HETEROZYGOUS (has two different alleles for a specific gene). Although Susan has both the allele "D" to make dimples and the allele "d" in her genotype, her phenotype is the presence of dimples. Thus the allele to make dimples is DOMINANT (the one that is expressed even if only copy is present) and is always represented with a capital letter. The little "d" allele is RECESSIVE as the allele will not be expressed unless there are two copies present.

Remember that during the meiotic cell division process, one set of the parental chromosomes are passed to each of the gametes. So one human egg cell will contain a total of 23 chromosomes (one of each type of chromosome) and the human sperm cell also contains 23 chromosomes. If we just focus on chromosome 5 where the dimples gene sits, then we see that

Sam (who has DD on his diploid chromsomes) can either pass down a "D" allele to all of his gamete cells. Sally who has the genotype "dd" will pass down a "d" allele to each of her gametes. In the following diagram, Sam and Sally's genotypes are the boxes at top. Through meiosis, each produce gametes (in ovals). Because of random fertilization between the gametes (there is only one possibility in this example), the genotype of the offspring(s) is "Dd" and the phenotype must be "Dimpled".

flow.gif flow0.gif

If the dominant allele of a gene carried on an autosome (types 1-22) causes a particular phenotype or disease, then it is an autosomal dominant trait (disease). Examples of autosomal dominant diseases in humans include Huntington's disease, achondroplasia (dwarfism), hypercholesteremia. More common autosomal recessive disease that you may have heard about include: cystic fibrosis, Tay-sach's, albinism, phenylketouria. There are traits encoded by genes on the X-chromosome, which are called X-linked (dominant or recessive) traits . Hemophilia, muscular dystrophy, red-green color blindness are X-linked traits. There are also some traits and diseases encoded by genes on the Y-chromosome and are Y-linked. Which one do you know about - that half of our population possesses? A few genes are also found on the mitochondrial genome.

MENDEL: considered the Father of Genetics but died unrecognized for his important discoveries

Why was Mendel's work so wonderful and successful?

LAW of SEGREGATION: pairs of 'characters' separate during gamete formation

Practice Problems

Given: In Mendel's Peas, the Height gene: T = allele for tall plants, t = allele for short plants

1) cross a homozygous tall plant with a homozygous short plant
parental genotypes: TT x tt
possible gametes: T ; t
do Punnett square

   t  t
 T  Tt (tall)  Tt (tall)
 T  Tt (tall)  Tt (tall)

Genotype probabilities of offspring = 100% Tt
Phenotype probabilities of offspring = 100% tall plants

2) cross a heterozygous tall plant with a short plant {Tt x tt } ----->

   t  t
 T  Tt (tall)  Tt (tall)
 t  tt (short)   tt (short)
Genotype = 50% Tt , 50% tt . Phenotype = 50% tall, 50% short

3) cross two tall plants together
Trick problem! You should ask if these plants are heterozygous or homozygous.
Suppose both are heterozygous (monohybrid cross) Tt x Tt ----->

  T  t
 T  TT (tall)  Tt (tall)
 t Tt (tall)   tt (short)
Genotype = 25% TT, 50% Tt, 25% tt. Phenotype = 75% tall, 25% short

4) If you are given a tall plant, how do you determine its genotype? It could be Tt or TT.
Do a TESTCROSS with a homozygous recessive plant and look at the offspring.
Compare the offspring results of problems (1) and (2) above.
If any short offspring are produced, then the tall parent must be heterozygous.

What if there are no short offspring out of 500 offspring plants? Homozygous
What if there are no short offspring out of 5 offspring plants? Inconclusive, since this is a very small sample size which may not be a true reflection of the theoretical probabilities.

You can test this idea of sampling error by flipping a coin. A coin has a 50% chance of landing as heads, and 50% chance of landing as tails. But if you flip only 4 times, do you always get 2 heads and 2 tails?
But if you flip 400 times, do you get pretty close to 200 heads and 200 tails?

LAW of INDEPENDENT ASSORTMENT: gene for one trait does not influence gene of another trait

What is remarkable is that Mendel did not know about genes, chromosomes or meiosis... when he performed his experiments and figured out his important contributions to the field of genetics. Mendel's Second Law of Independent Assortment is very easy to follow, if you recall how the chromosomes moved during Meiotic Division. During metaphase I, chromosomes lined up randomly at the equator, which eventually results in a mixture of paternal and maternal chromosomes in each gamete.

Suppose there is a gene A on chromosome 3 and another gene B on chromosome 7. One parent is heterozygous for both genes; or has the genotype "AaBb". How are these two genes passed onto the gametes and then to the next generation? Instead of drawing the alleles on chromosomes (as in the previous hyperlink), we will use a shortcut and just write the letters of the alleles.

Again, sometimes the A allele will be passed into a gamete cell along with the B allele. Sometimes, the A allele and b allele are passed into a gamete. There are two more possible combinations. The math formula for all the possible combinations would be 2(^n) or 2 to the nth power, with n = number of genes. {I don't know how to write superscript in HTML}. Two genes here, so there is a total of four possible combos. There are several different ways to figure out the different combinations of alleles. Remember that each gamete should always have one allele of each gene.

Here is the FOIL method you may have learned from math class too. You write out the alleles for gene A on the left and alleles for gene B on the right. My drawing of the FOIL method looks like a face.foil.gif

F: Take the FIRST allele of each gene (left eyebrow) to make one possible gamete.
O: Take the OUTER alleles (big smile) to make one possible gamete.
I: Take the INNER alleles (nose curve) to make one possible gamete.
L: Take the LAST allele of each gene (right eyebrow) to make one possible gamete.

Another method is the BRANCHING method, which is especially useful when considering even more genes. In this case, I have written the alleles from gene A separated vertically (see left column). From allele "A", it is possible to have either (branching arrows) "B" or "b" allele passing (big horizontal arrow) to the gamete combination (letters in oval). The possible gamete combinations are exactly the same ones you got with the FOIL method.

Now that we know how to get the gametes from a dihybrid parent (heterozygous for two different genes), let's try to cross two dihybrid parents together. We will need to use a big Punnett square, like this one:

Mom gamete 1 Mom gamete 2 Mom gamete 3 Mom gamete 4
Dad gamete 1 Offspring Offspring Offspring Offspring
Dad gamete 2 Offspring Offspring Offspring Offspring
Dad gamete 3 Offspring Offspring Offspring Offspring
Dad gamete 4 Offspring Offspring Offspring Offspring

As there are four possible gametes from each parent, random fertilization produces a total of 16 different possible combinations for each offspring. Let's finish our problem example.

The Punnett Square for the cross AaBb x AaBb should look like this:

There are four different phenotypes. The phenotypic ratio is 9:3:3:1 for a classic or Mendelian dihybrid cross (if the two genes are similar to the factors that Mendel worked on). Note that it is very easy to determine which phenotype class has all recessive alleles for both genes. How?

Suppose the dihybrid problem was that two spotted, long-eared dog parents had

18 spotted, long-eared puppies,
6 spotted, short-eared puppies,
6 solid, long-eared puppies, and
2 solid, short-eared puppies.

Can you tell which are the dominant and recessive traits? 


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