Mendelian Genetics Labaratory paper


Mendelian Genetics Lab





Gregor Mendel, the Austrian monk who is known as the “Father of Genetics”, first worked out the probability of offspring expressing certain traits using garden peas. It is now understood that his work can be applied to all sexually reproducing organisms.


The purpose of today’s lab is to explore various genetic traits, including obvious genetic traits that humans show in their phenotypes. Remember that you are a unique individual, genetically speaking, unless you have an identical twin!


We will first examine corn to help us understand these ratios. Then we will apply these principles to some aspects of human genetics.  Use the following definitions to help you through the first activity.


P Generation:  True-breeding parent generation.  They are either homozygous dominant or homozygous recessive.

F1 Generation:  Hybrid offspring from a parent cross.

F2 Generation:  A cross of the F1 generation.




  1. Demonstrate Mendelian ratios using plant and animal traits.
  2. Understand some genes are inherited based on more than 2 alleles.
  3. Discuss chromosomal inheritance of sex, and abnormalities associated with it.


Introduction to Mendelian Ratios


Below are images of two, true-breeding, parental corn plants used in a monohybrid, or single trait, cross.  What color are the kernels of the plants of the P (Parental) generation?  This is their phenotype.


Parent 1: ______________                                         Parent 2: _________________




Below is an image of the F1 generation.  What is the phenotype of this plant? _____________


Based on what you know so far, the dominant kernel color is: ________________, the recessive kernel color is: _____________.


When writing the genotype of an individual, be aware that the dominant trait should be represented with a capital letter, while the recessive trait is represented by the same letter, but lower-case.  If the dominant parent in this cross has a genotype of PP, the recessive parent has a genotype of: ________?


Draw a Punnett square to illustrate the Parental and F1 crosses.


Parental Cross:                                                            F1 Cross:








Looking at the F1 cross, how many F2offspring will have the dominant phenotype? ________

How many will have the recessive phenotype? __________


We express this as a ratio, the expected number of the dominant phenotype to the expected number of the recessive phenotype (#:#).  What is the expected ratio for the F2?

____ : ____




  1. Below is a photograph of an F2 ear of corn. Count the number of kernels of each color in a single row of kernels. If the rows are uneven simply do your best.  Enter your numbers in the Table 1.
  2. Count the number of kernels of each color in a second row of kernels. Add these numbers to the first row and put your two-row total in Table 1.  To clarify, this two-row total should have approximately twice as many kernels as your one-row data.
  3. The next two rows are already filled in for you. They contain kernel counts from another class for 8 rows (from a 4-person lab group) and 48 rows (6 four-person groups).
  4. Calculate the ratio of purple kernels to yellow kernels by dividing the number of purple kernels by the number of yellow kernels. Express it as “your answer”: 1.


For example, 62 purple kernels and 23 yellow kernels would be 62 ÷ 23 for a ratio of 2.7: 1


Table 1. Enumeration of corn kernel color.


  # purple kernels # yellow kernels Ratio P:Y
Single row      
Two rows      
Group data 261 92  
Class data 1353 448  


Did the ratios change as the number of kernels counted increased? _______________



How does this compare to the predicted ratios from the Punnet squares above?



What does this tell you about the importance of sample size in a statistical analysis?







Mendelian Inheritance in Humans


Mendel’s law of segregation states that allele pairs separate during the formation of gametes. Traits that are controlled by a single gene with two alleles can be used to demonstrate this pattern.  This is also what you observed in the corn kernel activity.


  1. You will be determining your phenotype for a variety of simple dominant/recessive traits such as your eye color or whether or not you have a dimple on your chin. Before you get started, do you expect to see more dominant or recessive traits?




  1. For each of the traits described below, determine your phenotype and circle orunderline it in Table 2. Do an online image search if you are unsure what you are looking for.I’ve completed this for the demo table 2 as if I have a dimpled chin.


  1. Based on your phenotype, record your genotype for each of the traits.


If you have a recessive characteristic, you must have both recessive alleles (dd).  If you have a dominant characteristic you could be either homozygous dominant (DD) or heterozygous (Dd). You likely have no way to know so we use a dash (_) or a question mark (?) to represent the unknown second allele (for example, D_or D?).


To help you with this I filled out a demo table 2 as if I have a dimpled chin.  If you have a dimpled chin this means you have the dominant trait.  Your genotype would be either DD or Dd, since you wouldn’t have a way of knowing which, write it as D_ or D?.  If you have no dimple, then you are expressing the recessive trait.  Write this as dd.


  1. Using the historical data calculate the percentage of the class with the dominant and recessive condition for each trait we looked at. Record this in Table 2.  I’ve completed this for the demo table 2.


  1. Looking at these percentages, determine, for each trait, whether or not the dominant or recessive trait was more common. Record this in Table 2.  Again, I’ve completed this for the demo table 2.


Dimpled chin A cleft, or dimple, in the chin is a dominant trait (D); the absence of a dimple is recessive (d).


Free earlobe Earlobes that hang free are due to the dominant allele (E), while an attached earlobe is a recessive trait (e).


Widow’s peak A hairline that forms a distinct point in the center of the forehead is called a widow’s peak.  This is due to the dominant gene (W).  The recessive allele (w) causes a straight hairline. Omit this tabulation if a gene for baldness has had some effect on your hairline.


Bent little finger Lay both hands flat on the table, relax your muscles, and note whether the last joint of the little finger bends sharply inward toward the fourth finger. A dominant allele (B) causes the bend, while the recessive allele (b) keeps a straight little finger.


Hitchhiker’s thumb Distal hyperextensibility of the thumb, can be determined by bending the distal joint of the thumb back as far as possible. The ability to bend it far back (sometimes until there is almost a 90-degree angle between the two joints) is an effect of two recessive alleles (h).  The inability is due to the dominant allele (H).


Pigmented irises No pigment in the front part of the eyes allows the blue layer at the back of the iris to show through resulting in blue eyes.  This is due to the recessive allele (p).  The dominant allele (P), causes pigment to be deposited in the front layer of the iris, thus masking the blue to varying degrees, resulting in amber, green, and other eye colors.


Table 2. Frequency of genetic traits. (demo)

Your possible phenotype (circle or underline) Your possible


Historical Class data Which traits were more common?
  number percentage Dominant or recessive?
Dimpled chin

Nondimpled chin

D? 2 8.3%  
  22 19.7% Recessive More Common


Table 2. Frequency of genetic traits.

Your possible phenotype (circle or underline) Your possible


Historical Class data Which traits were more common?
  number percentage Dominant or recessive?
Dimpled chin

Nondimpled chin

Free earlobes

Attached earlobes

Widow’s peak

No widow’s peak

Bent little finger

Finger not bent

Normal thumb

Hitchhiker’s thumb

Pigmented iris

Unpigmented iris



For the traits observed, did you find that the dominant or recessive allele was expressed most often in your class? ___________________

For those of you who expected to see more dominant traits, take a second to think about whether or not any of the above conditions will increase an individual’s chances of survival or reproductive success.  If the answer is no, then there is no selective advantage, and how common these traits are in a population is more a function of chance.


Please watch the accompanying video, it gives the evolutionary background of some of the previously described traits:


Chromosomal Abnormalities and Inheritance of Sex


You have two types of chromosomes in your cells: autosomes and sex chromosomes. 22 of the 23 pairs are autosomes; the 23rd pair is designated the sex chromosome (X and Y). Females have two X chromosomes; males have one X and one Y chromosome. Whereas the X chromosome carries about 1500 genes (many of which are unrelated to sex), the Y chromosome carries significantly fewer, a little over 200 genes have been identified, many involved in sex-determination and development in males.


Is it possible to live without an X chromosome? _________ Explain your answer.




Is it possible to live without a Y chromosome? _________ Why did you answer the way you did?


Draw a Punnet square to show inheritance of sex.








Mitosis and meiosis are usually very exact processes that result in the correct distribution of chromosomes to the daughter cells.  However, mistakes occasionally result in an abnormal number of chromosomes or pieces of extra chromosomes in the daughter cells.  Usually, zygotes with abnormal chromosomal compositions are spontaneously aborted.  However, if the combinations are not lethal at an early stage in development, the phenotype and viability of the resulting individual may be seriously affected.


Remember that during meiosis homologous chromosomes synapse, or come together, during prophase I and then separate from each other during anaphase I.  Sometimes a pair of chromosomes may adhere so tightly that they do not pull apart during anaphase I.  This will result in one of the cells receiving duplicate chromosomes and the other cell receiving none of that type of chromosome. This failure of chromosomes to separate is called nondisjunction.  Nondisjunction may also occur during the second meiotic division if the chromatids do not separate from each other during anaphase II (see Figure 1.)

Figure 1. Nondisjunction in Meiosis and fertilization

From left to right: the results of normal meiosis and fertilization, nondisjunction during meiosis IIand fertilization, and nondisjunction during meiosis I and fertilization


Normal meiosis in a human female results in the production of egg cells with 22 autosomes and one X chromosome.  Normal meiosis in a human male results in the production of two types of sperm: one-half with 22 autosomes and one X chromosome and one-half with 22 autosomes and one Y chromosome.


If an X-containing egg is fertilized by an X-containing sperm, an XX zygote will be formed.  Will this individual be male or female?  _____________________________________________


A Barr body is an inactivated X chromosome, resulting from a cell containing more than one X chromosome.  Cells do not require two versions of the genes on sex chromosomes.  Therefore, in females, one X chromosome is inactivated in order to prevent too many gene products.  X chromosome inactivation takes place in all somatic cells but is completely random as to which one is inactivated.  A calico cat is a great visible example of random inactivation.  The alleles for black and orange fur color reside on the X chromosome.  Some cells inactivate the orange fur allele, while others inactivate the black fur allele, resulting in tortoiseshell coloration.


How many Barr bodies will an XX zygote have?  _______________


If an X-containing egg is fertilized by a Y-containing sperm, an XY zygote will be formed.  Will it be male or female?   _______________


How many Barr bodies will it have?  _______________



If nondisjunction of the X chromosomes occurs during meiosis in a human female, some of the eggs will contain two X chromosomes and other cells will contain no X chromosome.  In a male, nondisjunction can result in four kinds of sperm:  1) containing both X and Y chromosomes, 2) containing XX chromosomes, 3) containing YY chromosomes, or 4) containing no sex chromosomes (O).


Gametes produced by nondisjunction usually fertilize, or are fertilized by, normal gametes.  Based on this information and information complete Tables 4 and 5.  Parts of these tables are already filled in, including the entire first row, and the normal gametes required to produce the given zygotes.  Start by filling out which sex chromosome(s) would be in the abnormal egg or sperm to produce the genotype in each row.


You can find a video under Lab Materials: Instructional Videos: Mendelian Genetics Table 4 to help you more clearly understand how to complete this part of the lab.


After watching the “Mendelian Genetics Table 4” video, select which random beetle appeared in the video?


Table 4.  Genotypic and phenotypic expression of nondisjunction in a human male.







Abnormal Gamete (Sperm) Expected


Number of

Barr Bodies

Name of


XXX X XX Female 2 Triple-X syndrome
XYY X       Jacob’s syndrome
XXY X       Klinefelter’s syndrome
XO X       Turner’s syndrome



Table 5.  Genotypic and phenotypic expression of nondisjunction in a human female.







Normal Gamete (Sperm) Expected


Number of

Barr Bodies

Name of


XO   X     Turner’s syndrome
XXX   X     Triple-X syndrome
XXY   Y     Klinefelter’s syndrome
OY   Y     Lethal


Triple-X Syndrome  The individual will develop into a normal-appearing female, but may be sterile.  She may also have cognitive difficulties. May be taller than average and may have minor physical abnormalities.


Klinefelter’s Syndrome  The individual will develop as a male.  During early development he appears normal, but abnormalities become apparent at puberty.  Testes do not fully develop.  The person is usually taller than average, his muscular development may be somewhat feminine, breast development may occur, and his voice may be higher-pitched than normal.  Although the individual may develop a “female” appearance, fluorescent cell staining always reveals the presence of a Y chromosome.

Turner’s Syndrome  The individual will be female.  She appears normal during early development, but at puberty does not menstruate, breasts do not develop, and no eggs are produced by the ovaries. Can have “webbing” develop on neck and can have other physical abnormalities such as low-set ears and swollen hands and feet.


 Jacob’s Syndrome  Individual appears to be sexually normal. He may have delayed learning and motor skill development.


Note that all of these syndromes, except Jacob’s syndrome (XYY), may result from nondisjunction in either the male or the female parent.



Autosomal nondisjunction


Autosomal nondisjunction involves chromosomes other than the sex chromosomes.  The most common non-lethal autosomal nondisjunction results in Down syndrome, described below.  It is caused by nondisjunction of chromosome 21, which increases with the age of the mother (although some evidence suggests that the father may sometimes be responsible for providing the extra chromosome).


When female gametes are formed (eggs), the process is complex and begins even before the individual doing the meiosis is born.  Immature sex cells begin their development around 5 months after conception.  These cells pause after prophase I of meiosis.  They remain frozen in prophase I until puberty, when, once a month, one of these cells completes meiosis I and II after ovulation in order to form an egg.  Note, this process actually produces 4 cells.  One large cell (the egg), and smaller unusable cells (polar bodies).  This is due to an unequal distribution of the cytoplasm.  The polar bodies typically die and get reabsorbed into the mother.


With this better understanding of the formation of female gametes, why do you think the incidence of Down syndrome increases with the age of the mother?  Please pull information from the above reading, and from the lecture and text to support your answer.




Down Syndrome/Trisomy 21  Three number 21 chromosomes are present.  Characteristics of individuals include a fold of the upper eyelid, short stature, broad hands, stubby feet, a wide, rounded face, a large tongue, and cognitive disabilities.


Trisomy 18  This syndrome involves an extra chromosome 18. Characteristics of individuals include a misshapen skull, eye problems, overlapping fingers, heart defects, feeding problems, and severe mental and developmental disabilities.


Why do you think that an extra chromosome leads to so many abnormalities in the child bearing it? Please pull information from the above reading, and from the lecture and text to support your answer.








Why do you think that relatively few autosomal trisomy conditions (compared to sex chromosome trisomies) are seen among people?  Please pull information from the above reading, and from the lecture and text to support your answer.  You may wish to discuss Barr bodies.












Lab modified from:

“Biology 101 Laboratory Manual” by Carol Morris, Tompkins Cortland Community College is licensed under CC BY-NC 4.0 / A derivative from the original work


Image Attributions:

“Yellow Parent” by Stephen Snyder is in the Public Domain, CC0

“Purple Parent” by Stephen Snyder is in the Public Domain, CC0

“F2 Corn” by Stephen Snyder is in the Public Domain, CC0

“File:Nondisjunction_Diagrams.svg” by Tweety207 is licensed under CC BY-SA 3.0

“File:21_trisomy_-_Down_syndrome.png” by U.S. Department of Energy Human Genome Program. is in the Public Domain, CC0

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