LABORATORY #15: GENETICS LAB – Part One

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

  1. 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
  2. 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
  3. 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.

  1. 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. Soif 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:
  2. will be the ace of hearts? 1/52
  3. will be a king? 4/52 or 1/13
  4. will be a king or a seven? 1/13 + 1/13 = 2/13
  1. In a single throw of two dice, what is the probability that:
  2. 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

thechances 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

  1. 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

  1. 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

  1. 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

  1. 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:
  2. red = 4/12 or 1/3
  3. blue = 6/12 or 1/2
  4. green = 2/12 or 1/6
  5. white = 0
  6. pink = 0
  7. red or white = 1/3 + 0 = 1/3
  8. green or white = 1/6 + 0 = 1/6
  1. If you toss a coin six times, what is the probability you will:
  2. get six heads = 1/64 = ½ x ½ x ½ x ½ x ½ x ½
  3. 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 HTHHHHiorTHHHHH

each of these combinations is 1/64

-so you need to add these up to get 6/64 or 3/32

  1. get three tails and three heads – same thing as above
  2. 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

  1. The following steps are often useful in monohybrid crosses
  2. list the symbols for each trait
  3. write down the parental genotypes
  4. write down the possible gametes of each parent – i.e a Punnett square
  5. 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

  1. 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

  1. 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
  1. 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

  1. 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)

  1. 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

  1. 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

  1. 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

  1. 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

  1. 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

  1. What is the genotypic ratio expected in the offspring of a cross between a male heterozygous for tongue rollingand a female homozygous for this trait?

Rr x RR cross = ½ kids are RR; ½ kids are Rr

  1. 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

George’s sister = bb

George’s dad = Bb

George’s mom = bb

George’s dad’s dad = B?

George’s dad’s mom = B?

George’s mom’s dad = bb

George’s mom’s mom = bb

Helen = Bb

Helen’s dad = bb

Helen’s mom = Bb

Helen’s dad’s dad = Bb

Helen’s dad’s mom = Bb

Helen’s mom’s dad = Bb or BB

Helen’s mom’s mom = bb

  1. 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.
  2. 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.
  1. 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.
  1. 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. Incomplete Dominance

  1. In radishes, the shape may be long, round or oval. Crosses between long and oval gives 159 oval and 156 long. Crosses between oval and round produce 203 oval and 199 round. Crosses between long and round give 576 oval. Crosses between oval and oval give 121 long, 243 oval and 119 round. Which of the shapes are defined by a homozygous dominant genotype? Homozygous recessive? Heterozygous?

For this problem, use the letters L and l

Long is LL

Round is ll

Oval is Ll – If a cross between long and round give 100% oval offspring, you know that long and round MUST be homozygous and the offspring must be heterozygous because only a cross between two homozygous genotypes will give offspring all with the same phenotype.

To check: the LL and Ll cross will give LL and Ll in a 1:1 ratio; the Ll and ll cross will give Ll and ll in a 1:1 ratio; crosses between Ll and Ll give a 1:2:1 ratio of long:oval:round

IV. Dihybrid and Multihybrid crosses:

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 BbTtXBbTtdihybrid 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.

AnRh+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-iiX 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. 0b. 1/16c. 1/8d. 3/16e. 1/4 f.1/2 g. 3/4h. 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)

6. In cattle, the gene for hornless (H) is dominant to the gene for horned (h), the gene for black (B) is dominant to that of red (b), and the gene for white face (or Hereford spotting) (S) is dominant to that for solid color (s). A cow with the genotype BbHhSs is inseminated by a bull of the genotype bbhhSs. What is the probability of the calf's being:

a. a black, hornless cow with Hereford spotting – B-H-S- and a cow = ½ x ½ x ¾ x ½ = 3/32

b. a red, horned bull with solid color - bbhhss and a bull = ½ x ½ x ¼ x ½ = 1/32

c. a red, hornless bull with Hereford spotting - bbH-S- and a bull = ½ x ½ x ¾ x ½ = 3/32

(Hint: the sex of the calf is part of the phenotype)

B- = black

bb = red

H- = hornless

hh = horned

S- Heresford spotting

ss = solid color

BbHhSs X bbhhSs

Bb X bb = ½ black; ½ red

Hh X hh = ½ hornless; ½ horned

Ss X Ss = ¾ Heresford; ¼ solid

7. Assume that D, E, F, G, H, and I are autosomal genes on different chromosomes. From the mating DdeeFfGGHhIi x DdEEFFGgHhii:

a. What is the probability that one of the offspring will have the genotype DdEeFFGghhIi?

½ X 1 X ½ X ½ X ¼ X ½ = 1/64

b. What is the probability that one of the offspring will be heterozygous for each allele?

½ X 1 X ½ X ½ X ½ X ½ = 1/32

c. What is the probability that one of the offspring will have the genotype DDEEFfGGhhii?

0 – because EE is not possible

Dd X Dd = ½ Dd; ¼ DD; ¼ dd

ee X EE = all Ee

Ff X ff = ½ Ff; ½ Ff

GG X Gg = ½ GG; ½ Gg

Hh X Hh = ½ Hh; ¼ HH; ¼ hh

Ii X ii = ½ Ii; ½ ii

8. Given the following information about the inheritance of characteristics in pea plants, answer the questions below:

Y (yellow) is dominant to y (green)

R (round) is dominant to r (wrinkled)

B (bitter) is dominant to b (sweet)

S (smooth) is dominant to s (hairy)

L (long pod) shows incomplete dominance to 1 (short pod) (Ll is medium in length)

Given this cross: (P1) YyRr Bb SS Ll (male) x yy RR Bb SsLl (female)

a. How many different gametes can be formed by the female plant?

The female is essentially BbSsLl = 8 gametes

b. How many different genotypes are possible in the F1 offspring?

Therefore, this cross results in 72 genotypes/offspring