UNM Biology Undergraduate Labs



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.



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



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




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.


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.

Click here for tutorials.

Review Questions

- What is meant by the terms genotype and phenotype?
- How many copies of a given locus does a diploid organism have?
- Explain the difference between a gene and an allele.
- Using a Punnett Square, setup a test cross for an individual with the genotype AaBb.
- A sexually reproduction animal has 2 genes, one for head shape and one for tail length.  Its genotype is HhTt.  What are the possible gametes that could be produced from this animal?
- Two parents are both heterozygous for two traits, curly and dark hair (CcDd).  Do a dihybrid cross and answer the following questions.
        1.    What is the most common phenotype of the children?
        2.    What genotypes produce that phenotype?
        3.    How many children are homozygous recessive for both traits?   
- In Rabbits, black hair depends on a dominant allele, B, and brown on a recessive allele, b.  Short hair is due to a dominant allele, H, and long hair to a recessive allele, h. 
        1.    If a true-breeding black, short-haired male is mated with a brown, long-haired female, what will their offspring (F1) look like?
        2.    What will be the genotypes of the offspring (F1)?
        3.    Show the Punnett square for this cross.
        4.    If two of these F1 rabbits are mated, what phenotypes would you expect among their offspring (F2), in what proportions?
        5.     What genotypes would be produced in the offspring (F2)?
        6.    Show the Punnett square for this cross.
- How can two cats that have different genotypes for the particular inherited characteristic, “curled ears”, be identical in phenotype?
- Neither Anna or Nick have hemophilia, Z, but their first son does have it.  Hemophilia is a sex-linked trait.  What are the genotypes of the parents?