LABORATORY #12: GENETICS LAB
Introduction
Genetics studies the mechanism by which traits are passed down from parent to offspring. This is
known as inheritance. The modern Chromosomal Theory of Inheritance was started by the works of
Gregor Mendel in 1866. Mendel’s work focused attention on the general patterns of inheritance and
eventually led to an understanding of the role of chromosomes and DNA in inheritance. The realm of
inheritance patterns is now known as Mendelian genetics.
Mendel’s most significant principle was the Principle of Segregation. It is the foundation of Mendelian
genetics and it states that organisms get two genes for every trait. These genes come from the parents
(one gene from each). During the production of gametes, these genes separate or segregate into
separate cells. When the gametes unite to form a zygote, each gamete supplies one parental set of
genes. The modern explanation of this principle states that genes for the same trait are on homologous
chromosomes and that homologous chromosome pairs separate during meiosis – or the formation of
gametes. The homologous chromosomes become a pair again during gamete fusion or fertilization.
Mendelian genetic problems require the knowledge of certain definitions. These definitions are listed
for you below:
1. Genotype – the genetic makeup of the trait
2. Phenotype – the physical or biochemical manifestation of the genotype
3. Allele –the different forms of a gene for a specific trait. Each parent contributes one allele. The
two alleles combine to form the genotype for that trait. Letters are given to denote the allele
e.g. E = earlobe attachment
4. Dominant and Recessive Alleles – the types of gene forms found. Alleles are described as
dominant, recessive, codominant or partial. Dominant alleles completely “overshadow” the
expression of the recessive alleles. Therefore the phenotype observed is controlled by the
dominant allele. A dominant allele is denoted with a capital letter, the recessive allele with a
lowercase letter. Codominant alleles and partially dominant alleles do not completely hide the
phenotype of the recessive allele. Another word for codominant alleles is blending
5. Homozygous – the two alleles are of the same form – e.g. EE or ee. Homozygous is another way
of reporting the genotype of an individual. You may either be homozygous dominant or
homozygous recessive
6. Heterozygous – the two alleles are of different forms – e.g. Ee. Heterozygous is another way of
reporting the genotype of an individual
Mendelian Genetics Problems
Complete all genetics problems in your lab notebook. Be sure to show your work.
I.
Probability problems:
Many genetics problems predict the likelihood that certain events will occur. This is done using
probability. Probability values range from zero (event is impossible) to one (event is certain to occur).
The probability of an event occurring is sometimes stated in advance of the nature of the event (apriori).
This includes many types of genetic problems.
To calculate the probability of an event, you must divide the expected frequency for that event by the
total frequencies for all possible events. Furthermore, there are two rules of probability. The Product
Rule states that the probability of two or more independent events occurring together (i.e. “this and
that”) is the product of their independent probabilities. The formula for the product rule is:
P(E*F) = P(E) * P(F)
where P is the probability, E and F are the events that can occur and they can occur at the same time
because the probability of E occurring has nothing to do with the probability of F occurring. For
example, when picking cards from a deck of 52 cards, the probability of getting an ace is 4/52 = 1/13,
because there are 4 aces among the 52 cards. The probability of picking a heart is 13/52 = 1/4. The
probability of picking the ace AND it being a heart is 1/4*1/13 =1/52.
The other is the Sum Rule which states that the probability of one of two mutually exclusive events (i.e.
“either or” events) occurring is the sum of their individual probabilities. The formula for the sum rule is:
P(A+B) = P(A) + P(B)
where P is the probability, A and B are the events that can occur but A and B cannot occur
simultaneously. For example, you want to calculate the probability of a person walking into a room
being a student or a teacher. The two cannot walk into the room at the same time. So if the probability
of a student walking into a room is 0.8 (P(A) = 0.8) and the probability of a teacher walking in is 0.1 (P(B)
= 0.1), then the probability of the person walking into the room being either a teacher OR a student is
0.8 + 0.1 = 0.9.
1. A card is drawn from a deck of 52 cards. What is the probability the card:
a. will be the ace of hearts? 1/52
b. will be a king? 4/52 or 1/13
c. will be a king or a seven? 1/13 + 1/13 = 2/13
2. In a single throw of two dice, what is the probability that:
a. a seven will appear? 1/6
what are the ways to roll a seven?
= 6&1 OR 1&6 OR 2&5 OR 5&2 OR 3&4 OR 4&3
the chances of rolling a 1 AND a 6 = 1/6 x 1/6 = 1/36
the ways to roll a seven = 1/36 + 1/36+ 1/36 + 1/36 + 1/36 + 1/36 = 6/36 or 1/6
b. a doublet (two of the same number) will appear? 1/6
-say you roll a 1 and a 1, the chances of rolling a 1 AND another 1 = 1/6 X 1/6 = 1/36
-to roll a 2 and a 2 = 1/36
-to roll a 3 and a 3 = 1/36
-add them all up = 6/36 or 1/6
c. a doublet or an 8 will appear? 11/36
-how can you roll an 8?
2&6 OR 6&2, 3&5 OR 5&3, 4&4 = 5/36
-chances of rolling a doublet = 6/36
=6/36 + 5/36 = 11/36
d. an even number will appear? ½
-you can roll an even number by rolling 2 even die OR 2 odd die
(e.g. a 2&2 or a 1&3)
-chances of rolling 2 evens = ½ x ½ = ¼
-chances of rolling 2 odds = ½ x ½ = ¼
-chances of rolling 2 evens OR 2 odds = 1/4+ 1/4 = 2/4 or 1/2
3. A bag contains 6 blue marbles, 4 red marbles and 2 green marbles (i.e. 12 marbles). If you select
a single marble at random from the bag, what is the probability that it will be:
a. red = 4/12 or 1/3
b. blue = 6/12 or 1/2
c. green = 2/12 or 1/6
d. white = 0
e. pink = 0
f. red or white = 1/3 + 0 = 1/3
g. green or white = 1/6 + 0 = 1/6
4. If you toss a coin six times, what is the probability you will:
a. get six heads = 1/64 = ½ x ½ x ½ x ½ x ½ x ½
b. get five heads and one tails – to do this problem, you need to determine all the way
that you can get five heads and one tail
e.g. HHHHHT or HHHHTH or HHHTHH or HHTHHH or HTHHHH ior THHHHH
each of these combinations is 1/64
-so you need to add these up to get 6/64 or 3/32
c. get three tails and three heads – same thing as above
d. get four heads and two tails
II. Monohybrid problems: Monohybrid problems follow one trait between two generations. These traits
are dominant, recessive or codominant. Parental genotypes, gamete genes and possible offspring
genotypes usually need to be calculated
a. The following steps are often useful in monohybrid crosses
i. list the symbols for each trait
ii. write down the parental genotypes
iii. write down the possible gametes of each parent – i.e a Punnett square
iv. determine all of the possible offspring genotypes – complete the Punnett
square
The following is an example of a Monohybrid cross problem. Give it a try and record your answers in
your lab notebook
QUESTION: If two heterozygous brown eyed parents have five children, what is the probability that
three of them will have blue eyes? The allele for brown eyes can be denoted with the letter “B”, while
blue eyes can be denoted with the letter “b”.
Do the following with a Punnett square in your lab notebook:
1. In humans, six fingers (F) is the dominant trait and five fingers (f) is the recessive trait. Both
parents are heterozygous for six fingers.
What is the phenotype of the father and mother? Six fingered
What is the genotype for each parent? Ff and Ff
What is the probability of them having six-fingered children? 3/4
Five fingered children? ¼
2. If the father is heterozygous for six fingers and the mother has five fingers, what is the
probability that their offspring will be six fingered? 1/2
Five fingered? 1/2
3. In certain flowers, the genes for color inherited have incomplete dominance. In such
flowers, a cross between a homozygous red and homozygous white flowering plants will
always result in pink flowers. What is the probability of pink flowers appearing in the
offspring of two pink parents? Rr = 1/2
4. Using the information above, a cross is made between a red flower and a pink flower.
What are the genotypes and probabilities of the results offspring flowers?
RR = ½
Rr = ½
What is the probability of a pink flower appearing in the offspring of a pink and white
parent?
Rr = ½
rr = ½
5. The genes for freckles (F) is dominant over the gene for no freckles (f). Two freckled
individuals produced six offspring. Four of them have freckles while the other two do not.
What are the genotypes of the parents? Ff and Ff
The phenotypes? Freckled
What are the genotypes of the children? FF, Ff and ff (3:1 ratio of freckled to non-freckled)
6. In humans, true brown eyes, B, are dominant over true blue eyes, b. A brown-eyed man has
8 children with a blue eyed woman. They are all brown eyed. What are the possible
genotypes of all members of this family?
Dad = BB
Mom = bb
Kids = Bb
*** but it is possible that the Dad is Bb and just didn’t contribute a “b” sperm to these kids
7. The parents of a blue-eyed man were brown eyed. He has children with a woman who had
brown eyes. Her father was brown eyed and her mother was blue eyed. Their one child has
blue eyes. What are the most probably genotypes of all the individuals mentioned?
Dad’s dad = Bb
Dad’s mom = Bb
Mom’s dad = BB or Bb
Mom’s mom = bb
Kid = bb
8. A man and woman are both heterozygous for brown eyes (i.e. Bb). They have six children
all of whom have blue eyes (i.e. bb).
How is this explained? Just dumn luck
What is the probability that their 7th child will have brown eyes? 3/4
Or blue eyes? 1/4
9. In humans, the allele for tongue rolling (R) is dominant over the inability (r). If a man and
woman are both heterozygous for tongue rolling (i.e. Rr), what is the probability that their
child can roll his/her tongue?
Heterozygous cross  3:1 ratio
¾ for tongue rolling – i.e. RR and Rr kids
¼ for non-tongue rolling – i.e. rr
10. If a man and woman are tongue rollers (i.e, have at least one dominant R allele) but each
have one parent who couldn’t roll their tongue (i.e. rr), what is the probability that they will
have a child who can roll their tongue?
Because they can roll their tongues but had one parent that couldn’t – these parents are
heterozygous or Rr
In a heterozygous cross, the chances of a homozygous recessive (i.e. can’t roll the tongue)
is ¼
11. What is the genotypic ratio expected in the offspring of a cross between a male
heterozygous for tongue rolling and a female homozygous for this trait?
Rr x RR cross = ½ kids are RR; ½ kids are Rr
12. George and his father have brown eyes (i.e. at least one B allele), while his mother and
sister have blue eyes (i.e. bb). George’s paternal grandparents both had brown eyes (i.e. at
least one B allele), while his maternal grandparents had blue eyes (i.e. bb). George has
children with Helen. Helen has brown eyes (i.e. at least one B allele). Her father had blue
eyes (i.e. bb), while her mother had brown eyes (i.e. at least one B allele). Helen’s paternal
grandparents had brown eyes, while one of her maternal grandparents had brown eyes.
Draw a pedigree chart for this family. Determine all genotypes of all individuals.
George = Bb
Helen = Bb
George’s sister = bb
Helen’s dad = bb
George’s dad = Bb
Helen’s mom = Bb
George’s mom = bb
Helen’s dad’s dad = Bb
George’s dad’s dad = B?
Helen’s dad’s mom = Bb
George’s dad’s mom = B?
Helen’s mom’s dad = Bb or BB
George’s mom’s dad = bb
Helen’s mom’s mom = bb
George’s mom’s mom = bb
13. Charlie decides to breed pigs. In pigs, hair color is either black, white, or tan. Tan hair
coloring is a condition that results from being heterozygous. HINT: assume black is the
dominant color.
a. Isabel, Charlie’s black pig (i.e. BB) escapes from her pen one night. Sometime later,
she gives birth to a tan piglet (i.e. Bb) and two black piglets (i.e. BB). The color of
what piglet tells the father’s color? It is the Black piglets that are important. A tan
piglet could have come from Isabel (BB) mating with a white male pig (bb). But
this mating could not produce black piglets.
b. Charlies mates Matty, who is tan, with Butch, who is white? What color piglets can
he expect? He can expect tan and white piglets from the BbXbb cross.
c. Matty refuses to mate with Butch, so Charlies brings in Bas, who is tan. Matty ends
up having four piglets, two black and two white. What are the expected piglets in
this cross? The following piglets can be expected from this BBXBb cross: BB (1/4),
Bb (1/2) and bb (1/4)
III. Dihybrid and Multihybrid crosses: Dihybrid problems follow two traits that are not on the same
homologous chromosome pair. The Principle of Independent Assortment, another of Mendel’s
principles, explains the observed ratios in these problems. According to this principle, genes are certain
to separate from their allelic counterpart, but nonallelic gene associations occur randomly. It is now
known that random assortment occurs for all genes not linked on the same homologous pair.
Homologous pairs separate during meiosis, the chromosomes they assort with from the other pairs are
random.
2. Dihybrid problems:
1. In horses, black color (B), dominates chestnut (b). The trotting gait (T) dominates the pacing gate (t). A
cross is made between a horse homozygous for black and pacing gait (i.e. BBtt) and a horse homozygous
for chestnut and trotting gait (i.e. bbTT).
List the probable genotypes and phenotypes from offspring resulting from this cross. A BBttXbbTT cross
will give you 100% BbTt, black and trotting offspring
List the probable phenotypes of offspring that would result if the offspring from the original cross were
crossed amongst themselves. A BbTtXBbTt dihybrid cross will give you a 9:3:3:1 ratio. 9 Black and
trotting: 3 Black and pacing: 3 chestnut and Trotting:1 chestnut and pacing
2. Humans have Rh+ or Rh- blood. A person with Rh+ has a certain type of protein in his blood and is
considered the dominant trait. An Rh- person lacks this antigen and is recessive. Normal insulin
production is dominant abnormal production (i.e. diabetes). If both parents are heterozygous for Rh+
blood and normal insulin production, what probably phenotypes would they produce in their offspring.
An Rh+Rh-Ii X Rh+Rh-Ii cross is a dihybrid cross and will give you the standard 9:3:3:1 ratio. 9 Rh
positive, Insulin producing: 3 Rh positive, no insulin producing: 3 Rh negative, Insulin producing: 1 Rh
negative, no insulin producing
3. Using the above information, a father is homozygous for Rh+ blood but he has diabetes. The mother is
Rh- and is homozygous for normal insulin production. What probable phenotypes would they produce in
their children?
An Rh+Rh-ii X Rh-Rh-II cross will give you 100% heterozygous Rh+Rh-Ii, Rh positive, insulin producing
children
4. Using what you know about brown/blue eyes and freckles and no freckles, what is the probability of a
man with the genotype Bb FF and his wife who has blue eyes and no freckles, having a blue eyed child
with freckles?
a. 0
b. 1/16
c. 1/8
d. 3/16
e. 1/4
f. 1/2
g. 3/4
h. 1.1
5. In dogs, black coloring is dominant and red color is recessive; solid coloring is dominant while white
spotting is recessive. A homozygous black and white spotted make is crossed with a red and white
spotted female. What is the probability of them producing a solid black puppy?
There is no chance of producing a solid black puppy since the parents are BBss and bbss (i.e. both are
homozygous for the recessive spotted gene)