UNM Biology Undergraduate Labs

Genetics

 

The Laws of Mendelian Genetics

The law of segregation states that during meiosis, each gamete receives only one member of a homologous pair.

The law of independent assortment states that it is an equal opportunity that a gamete will receive either homologue.

Definitions

• gene: a sequence of DNA that codes for a polypeptide
• allele: an alternate form of a gene
• locus: location of a gene on a chromosome
• diploid: chromosomes are in homologous pairs, one from each parent
• haploid: only one set of chromosomes, usually in gamete
• zygote: male and female gametes fused to restore the diploid number
• genotype: actual genetic component; (expressed as which alleles are present)
• phenotype: expressed trait that can be measured (expressed as trait)
• dominant: allele that can mask the presence of another allele
• recessive: allele that can be masked in the presence of another allele
• incomplete dominance: alleles blended
• homozygous: alleles are the same
• heterozygous: alleles are different (sometimes referred to as hybrid)
• pure breeding: homozygous
• epistasis: the expression of one gene affects the expression on other genes
• linkage: two very different types
  • sex linkage: allele is on the X chromosome, affecting expression in males (males having only one X)
  • autosomal linkage: loci are one the same chromosome, affecting the phenotypic (and genotypic) ratios of the offspring. Recombination affects linkage by separating the two loci. The percentage of crossover gives an estimate of how close together the two loci are, and is referred to as the number of map units between the loci.
• Punnett square: a tool to predict the genotypic and phenotypic ratios of offspring from a given mating.
• parental type: the parents in a cross (P)
• F1 generation: lit. "first filial", offspring from given cross
• F2 generation: lit. "second filial", offspring from offspring of given cross
• test cross: crossing an individual of unknown genotype to a pure breeding recessive to determine genotype of the unknown
• reciprocal cross: crossing individuals with a given phenotype, switching the sexes of the parental types to determine if the gene is sex-linked.

 

Monohybrid Cross

P: AA x aa
F1: Aa

A monohybrid cross is the cross of the F1 generation: Aa x Aa.

	A	a
A	AA	Aa
a	Aa	aa

   

 

The genotypic ratios are 1AA:2Aa:1aa.
The phenotypic ratios are 3dominant:1recessive.

Test Cross

When an individual shows the dominant phenotype, it is impossible to say whether that individual is homozygous or heterozygous for the trait.  A mechanism to determine this is a test cross. A pure breeding recessive is crossed with the unknown. The resulting offspring show whether the unknown allele is dominant or recessive.

	A	?
a	Aa	?a
a	Aa	?a

 

       

The possibilities are:       

	A	A			A	a
a	Aa	Aa		a	Aa	aa
a	Aa	Aa		a	Aa	aa
all dominant				half and half

 

 

 

Dihybrid Cross

P: AABB x aabb
F1: AaBb

A dihybrid cross is a cross between two of the F1 individuals: AaBb x AaBb

	AB	Ab	aB	ab
AB	AABB	AABb	AaBB	AaBb
Ab	AABb	AAbb	AaBb	Aabb
aB	AaBB	AaBb	aaBB	aaBb
ab	AaBb	Aabb	aaBb	aabb

 

 

 

The genotypic rations are 1 AABB:2 AABb:2 AaBB:1 AAbb:4 AaBb:1 aaBB:2 Aabb:2 aaBb:1 aabb.
The phenotypic ratios are 9 double dominant:3 first dominant, second recessive:3 first recessive, second dominant:1 double recessive.

Linkage

Sex Linkage

If a gene is located on the X chromosome, the male has the problem that he does not have a homologue for this chromosome. He has a Y chromosome that doesn't have the same information on it as the X. Therefore, whatever allele he inherits from his mother's X chromosome, he expresses that allele. (Remember he must get his X from his mother, not his father - he had to get the Y chromosome from his father.) Females will show normal mendelian genetics for this trait, as they do have homologues for the X chromosome, one from the mother and one from the father.

An example is color blindness. Males who inherit a dominant alelle from their mother show normal vision. Males who inherit a recessive allele from their mother are color blind. Females may be homozygous dominant (normal vision), heterozygous (normal vision, but carriers of the recessive allele), or homozygous recessive (color blind).

Below is a cross between a normal male and a carrier female. X^C denotes the X chromosome with the dominant allele.  X^c denotes the X chromosome with the recessive allele.  Y denotes the Y chromosome.

	X^C	Y
X^C	X^C, X^C	X^C, Y
X^c	X^C, X^c	X^c, Y

 

 

Here, the proportions are 1 normal female: 1 carrier female: 1 normal male: 1 color blind male.

Autosomal Linkage

Autosomal Linkage refers to genes being on the same chromosome.  These genes tend to show up together in the same combinations in the offspring. However, recombination, which occurs in prophase I of meiosis, can split the two alleles inherited from a parent, giving recombinant types. These tend to be in smaller proportions than the parental, linked types. The percentage of offspring showing the recombinant types can give an estimate of how close the two genes are to each other on the chromosome. The closer the genes are, the fewer recombinant types should occur. This is because the chances of recombination occurring in the right location to split the two decreases as they get closer and closer together. This estimate is referred to as the number of map units separating the two genes, which is the percentage of recombinant types seen in the offspring.

The following is a test cross for an individual with the dominant phenotype for both alleles on known linked genes.

	A-b	a-B
a-b	A-b, a-b	a-B, a-b

 

The offspring show that the parent has the dominant allele on different homologues because each offspring shows only one dominant trait.

Here I will redo problem 15 from the lab manual. In rabbits, C gives color, c is albino. The C gene is epistatic and affects expression of the B gene. If the dominant allele C is present, B gives black, and b gives brown. The problem tells us that the parental types were:

CCbb x ccBB

We know then the F1 should be: CcBb, all rabbits are black.
If we just did a test cross we get:

	CB	Cb	cB	cb
cb	CcBb	Ccbb	ccBb	ccbb
	black	brown	albino	albino
	25%	25%	25%	25%

 

 

This gives us the result of: 50% albinos, 25% black, and 25% brown.  We now can compare these predictions to some actual results: 100 albinos (50%), 25 black (12.5%), and 75 brown (37.5%). These data do not match, so we can assume linkage and redo the cross.

The parents were C-b C-b, and c-B c-B. This makes the F1: C-b c-B. The test cross gives us: C-b c-B x c-b.

We assume that recombination occurs in some cases to give us C-B and c-b.

	C-b	c-B	C-B	c-b
c-b	C-b c-b	c-B c-b	C-B c-b	c-b c-b
	brown	albino	black	albino
	37.5%	37.5%	12.5%	12.5%
	75%		25%

 

 

 

This gives us a prediction that most of the rabbits will be albinos, and that there will be more browns than blacks. If we use the data provided, we can estimate the percentage of crossover.

Working backwards from the data, we can tell that black rabbits are ONE of the recombinant types. There are TWO recombinant types (the other is recombination-linked type albino and cannot be separated from normal-linked type albinos), so the percentage of recombinants is 12.5% * 2 = 25%. This tells us that the normal types occur 75% of the time, so each type, in this case normal-albino and brown should each occur 37.5% of the time.

 

Hardy Weinberg Equation

 Basic Relations

            A = dominant allele
  a = recessive allele

            p + q = 1
      Where    p = frequency of A allele      
                    q = frequency of a allele

            p² + 2pq + q² = 1
      Where    p² = frequency of AA genotype
                   2pq = frequency of Aa genotype
                   q² = frequency of aa genotype

 Example Problem

          You are studying the gene that regulates interlocking fingers in a small isolated village with a population of 2500 individuals.  It is already known that a dominant allele (F) causes one to interlock their fingers in such a way that the left thumb is nearly always on top, while a recessive allele (f) in the homozygous conditions results in the right thumb being on top.  You have asked the entire village to gather at the town hall one Saturday afternoon so that you can run your experiment and collect the data all at once.  After everyone is seated you ask them to each clasp their hand together in front of them, interlocking their fingers.  Then you ask them to look down at their hands and note which thumb is on top.  With the help of your trusty assistant, you are able to collect all the data that afternoon.  Now your only job left is to calculate all the related frequencies and determine how many homozygous dominant and heterozygous individuals for your trait of interest are in this population.

             Data Collected:            Left thumb over right = 2275 individuals = FF, Ff genotypes
                                     Right thumb over left = 225 individuals = ff genotype

 1.         Since there are 2500 individuals in the population, it is understood that there are 5000 alleles in the population.

 2.         Given the data and basic relations, the frequencies for the recessive genotype and allele can be calculated.

             Recessive genotype frequency:

                       Right thumb over left represent 9% of the population.
                                                            9% are homozygous recessive = aa

             Recessive allele frequency:

                               The frequency of the recessive (f) allele is 30%.

 3.         Now the frequencies for the dominant allele and homozygous dominant and heterozygous genotypes can be calculated.

             Dominant allele frequency:

                               The frequency of the dominant (F) allele is 70%.

             Homozygous dominant genotype frequency:

           

             Heterozygous genotype frequency:

           

                                                              Left thumb over right represent 91% of the population.
                                                            49% are homozygous dominant = AA
                                                            42% are heterozygous = Aa

 4.         Finally the actual numbers of people that are homozygous dominant and heterozygous can be calculated.

             Known:            225 individuals (9%) are homozygous recessive = aa
                          49% are homozygous dominant = AA
                          42% are heterozygous = Aa

             Number of homozygous dominant individuals:

              

             Number of heterozygous individuals:

           

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Review Questions

- You surveyed 200 people for the presence of a widow's peak on their hairline.  Widow's peak is due to the presence of a dominant gene, W.  Suppose you obtained the following data:

        Phenotype                            No. of people
        Widow's peak                        182
        No widow's peak                    18

    1.    Write out he genotypes for each phenotype.
    2.    Calculate the frequency of each phenotype.
    3.    What is the frequency of the dominant allele?
    4.    What is the frequency of the recessive allele?

- If you had a widow's peak, how would you determine your exact genotype?