Mendel’s Laws: Their Application to Solving Genetics Problem

Steps to Solving Genetics Problems

1. READ the problem

2. Write down what you know

3. Assign letters for the alleles (traditionally, you should use the letter of the recessive allele)

a. Use a capital letter for the dominant trait

b. Use a lower case letter for the recessive trait

4. Determine the genotypes involved

5. Make gametes (sex cells – each gamete will carry only ONE allele for a trait, not both)

6. Solve using a Punnett Square

7. Reread the question & make sure that you have answered it

Single Trait Problems (Monohybrid Crosses)

Red pigeon is in front of the more commonly-colored brown pigeon. Photo source: naturetales.blogspot.com

SAMPLE PROBLEM: The allele for red feather color in pigeons, B, is dominant to the allele for brown feathers, b. A red pigeon who had a red parent and a brown parent is mated with a brown pigeon. What are the genotypes of the two pigeons being mated? Identify the gametes produced by each of the pigeons being mated. What proportion of the F1 progeny would be expected to have brown feathers?

STEPS TO THE SOLUTION:

1. Write down what you know

Grandparent Pigeons – Red X Brown

(Genotypes) __ __ __ __

Parent Pigeons - Red X Brown

(Genotypes) __ __ __ __

F₁ (first filial or offspring or babies) --

2. Assign letters for the alleles (traditionally, you should use the letter of the recessive allele)

a. Use a capital letter for the dominant trait

b. Use a lower case letter for the recessive trait

Since we know that red color is dominant to brown, we will use b for the alleles. Red is dominant, so it should be B. Brown is recessive, so it should be b.

3. Determine the genotypes involved

We know that the brown parent must be homozygous, or bb. Otherwise, it would appear red in color. The red parent is a little trickier. That parent could be either homozygous or heterozygous. Rereading the problem, we see that this parent was produced from the crossing of a red and a brown pigeon. In other words we are at the very least crossing B_ X bb to get the red parent. In order for the offspring of this cross to be red it must have one dominant allele that it will inherit from its red parent. We know that the brown pigeon must give all of its offspring the recessive allele. Thus, the red parent pigeon in this problem must have a heterozygous genotype for color, or Bb.

Now we can record the parent’s genotypes.

Grandparent Pigeons – Red X Brown

(Genotypes) B __ bb

Parent Pigeons - Red X Brown

(Genotypes) Bb bb

F₁ (first filial or offspring or babies) --

4. Make gametes (sex cells – each gamete will carry only ONE allele for a trait, not both)

The brown pigeon has a genotype of bb. Thus all of the gametes it will produce will have the b allele.

The red pigeon has a genotype of Bb. Thus it will produce gametes with B alleles and gametes b alleles in equal proportions.

Parent Pigeons	Red X	Brown
(Parent Genotypes)	Bb	bb
Gametes Produced

5. Solve using a Punnett Square

Gametes

Genotypic ratio of the F₁ generation = 2 Bb : 2 bb. This can be simplified as 1 Bb : 1 bb because both numbers in the ratio are divisible by 2.

Phenotypic ratio of the F₁ generation = 2 red : 2 brown. Again, this can be simplified as 1 red : 1 brown.

6. Reread the question & make sure that you have answered it

a. What are the genotypes of the two pigeons being mated?

b. Identify the gametes produced by each of the pigeons being mated.

c. What proportion of the F₁ progeny would be expected to have brown feathers?

a. The red parent’s genotype is Bb. The brown parent’s genotype is bb.

b. The red parent produces 2 types of gametes. Half carry the allele of red & half carry the allele for brown.

The brown parent produces only one type of gamete. All of it’s gametes carry the allele for brown.

c. We would expect half of the offspring to have brown feathers.

SOME PRACTICE MONOHYBRID PROBLEMS

1. Several plants with purple flowers were crossed to plants with white flowers. The seeds from the cross produced plants on which only purple flowers appeared. These purple-flower plants were then crossed to each other & the seeds from the cross produced 346 purple flowered plants & 128 white flowered plants. Illustrate the crosses involved & determine the phenotypic & genotypic ratios of the last generation of plants.

2. In peas, long-stem (S) is dominant over short-stem (s). Give the expected phenotypic ratios for the following four crosses:

a. homozygous long X short c. heterozygous long X homozygous long

b. heterozygous long X short d. heterozygous long X heterozygous long

3. In humans, dimples (N) are dominant to nondimples (n). A couple who both have dimples, have a child without dimples. What must be the genotypes of the two parents? What is the probability that their next child will have dimples?

Two Trait Problems (Dihybrid Crosses)

SAMPLE PROBLEM: In humans, brown eyes are dominant to blue eyes. Also brown hair (brunette) is dominant to red hair. Imagine that a man who is heterozygous for both traits marries a woman who is heterozygous for both traits.

a. What are the genotypes of the parents?

b. What would be the phenotypic ratio of their potential children?

1. Write down what you know

Parents - MAN X WOMAN

heterozygous brown eyes heterozygous brown eyes

heterozygous brunette heterozygous brunette

2. Assign letters for the alleles (traditionally, you should use the letter of the recessive allele)

a. Use a capital letter for the dominant trait

b. Use a lower case letter for the recessive trait

Since we know that brown eyes is dominant to blue eyes, we will use the letter b for these alleles. Brown is dominant, so it should be B. Blue is recessive, so it should be b.

We know that brunette hair color is dominant to red hair, we will use the letter r for these alleles. Brunette is dominant, so it should be R. Red hair is recessive, so it should be r.

3. Determine the genotypes involved

Remember that the parents are heterozygous for both traits.

Parents - MAN X WOMAN

heterozygous brown eyes heterozygous brown eyes

heterozygous brunette heterozygous brunette

(Genotypes) ___ ___ ___ ___ ___ ___ ___ ___

4. Make gametes (Sex cells – Remember that each gamete will carry only ONE allele for a trait, not both. However, since this is a two-trait or dihybrid problem, the gametes will carry ONE allele for eye color and ONE allele for hair color.)

HINT: Do an allele cross to make sure you get one of every possible type of gamete! (Remember FOIL)

Parents	Man X Heterozygous Brown Eyes Heterozygous Brunette	Woman Heterozygous Brown Eyes Heterozygous Brunette
(Parent Genotypes)	BbRr	BbRr
Gametes Produced

5. Solve using a Punnett Square

Gametes

You have just written all of the genotypes possible for their children! Let’s figure out what their phenotypes will be.

a) To have the brown eyes, brown hair phenotype, a child must have at least B__ R__. There are four ways to satisfy this minimum. Locate each of the following genotypes in the Punnett square and record the number of each type:

BBRR _____ ; BbRR _____ ; BBRr ______ ; BbRr______ ; Total _______

b) To have the brown eyes, red hair phenotype, a child must have at least B_rr. Locate and record again:

BBrr ________ ; Bbrr ________ ; Total _________

c) To have the blue eyes, brown hair phenotype, a child must have at least bbR_. Locate and record again:

bbRR ________ ; bbRr ________ ; Total _________

d) To have the blue eyes, red hair phenotype, a child must have at least bbrr. Locate and record again:

bbrr _________ ; Total _________

THUS, the phenotypic ratio of the man and woman’s potential children is:

_____ brown eyes, brown hair : _____ brown eyes, red hair :

_____ blue eyes, brown hair : _____ blue eyes, red hair

SAMPLE PROBLEM: Imagine that the same man, heterozygous for brown eyes and heterozygous for brown hair, marries a different woman. This woman is heterozygous for brown eyes, but has red hair. (Recall that in humans, brown eyes are dominant to blue eyes. Also brown hair (brunette) is dominant to red hair.)

a. What are the genotypes of the parents?

b. What would be the phenotypic ratio of their potential children?

1. Write down what you know

Parents - MAN X WOMAN

heterozygous brown eyes heterozygous brown eyes

heterozygous brunette red hair

2. Determine the genotypes involved

Remember that the parents are heterozygous for both traits.

Parents - MAN X WOMAN

heterozygous brown eyes heterozygous brown eyes

heterozygous brunette red hair

(Genotypes) ___ ___ ___ ___ ___ ___ ___ ___

4. Make gametes (Sex cells – Remember that each gamete will carry only ONE allele for a trait, not both. However, since this is a two-trait or dihybrid problem, the gametes will carry ONE allele for eye color and ONE allele for hair color.)

HINT: Do an allele cross to make sure you get one of every possible type of gamete! (Remember FOIL)

Parents	Man X Heterozygous Brown Eyes Heterozygous Brunette	Woman Heterozygous Brown Eyes Red Hair
(Parent Genotypes)	BbRr	Bbrr
Gametes Produced

5. Solve using a Punnett Square

Gametes

You have just written all of the genotypes possible for their children! Let’s figure out what their phenotypes will be.

a) To have the brown eyes, brown hair phenotype, a child must have at least B__ R__. There are four ways to satisfy this minimum. Locate each of the following genotypes in the Punnett square and record the number of each type:

BBRR _____ ; BbRR _____ ; BBRr ______ ; BbRr______ ; Total _______

b) To have the brown eyes, red hair phenotype, a child must have at least B_rr. Locate and record again:

BBrr ________ ; Bbrr ________ ; Total _________

c) To have the blue eyes, brown hair phenotype, a child must have at least bbR_. Locate and record again:

bbRR ________ ; bbRr ________ ; Total _________

d) To have the blue eyes, red hair phenotype, a child must have at least bbrr. Locate and record again:

bbrr _________ ; Total _________

THUS, the phenotypic ratio of the man and woman’s potential children is:

_____ brown eyes, brown hair : _____ brown eyes, red hair :

_____ blue eyes, brown hair : _____ blue eyes, red hair

SOME PRACTICE DIHYBRID PROBLEMS

Black, long-haired rabbit Photo source: http://upload.wikimedia.org/

1. In rabbits, black fur is due to a dominant allele B, and brown fur is due to its recessive allele b. Short hair is due to the dominant allele L, & long hair is due to the recessive allele l. A cross is done between a homozygous black, long-haired rabbit & a homozygous brown, short-haired rabbit. What would be the genotype(s) & phenotype(s) of the F1 generation? When the F1 offspring are allowed to breed, what will be the phenotypic ratio of the F2 generation?

Brown, short-haired rabbit Photo source: http://farm3.static.flickr.com

Photo source: http://media.knoxnews.com/

2. In mice the gene for coat color has two forms. The allele for colored coat (A) is dominant to the allele for albino (a). There are two forms for the gene controlling whiskers, as well, straight (B) is dominant to bent (b). Imagine that we had a female mouse whose mother was homozygous colored with bent whiskers and whose father was an albino that was homozygous for straight whiskers. We are going to cross this female on a male mouse that is albino and has bent whiskers. a. What percent of their offspring will be albino? b. What percent of their offspring will have straight whiskers? c. What would be the phenotypic ratio of their offspring?

3. In rabbit coats, spotted (S) is dominant to solid color (s) and black (B) is dominant to brown (b). A brown, spotted rabbit is mated with a solid, black one and all the offspring (the F₁ generation) are black and spotted.

a. What are the genotypes of the parents?

b. What are the genotypes of the offspring (the F₁ generation)?

b. What would be the phenotypic ratio of the F₂ generation if two of these F₁ black, spotted rabbits were mated?

White, sphere-shaped squash Photo source: http://www.thenibble.com/	4. In the summer squash, white fruit (Y) is dominant over yellow (y), & disk-fruit shape (S) is dominant over sphere-shaped (s). For the crosses given below, provide the following in each case: a. The phenotypes of the parents b. The phenotypic ratio of the offspring 1) YYss X yySS 2) YySs X yyss 3) YySs X YySs 4) YySs X YySS
Yellow, disk-shaped squash Photo source: http://www.photographsofaustralia.com/

Incomplete Dominance Problems

SAMPLE PROBLEM: In humans, inheritance of hair texture shows incomplete dominance. If a person inherits two alleles for curly hair, they have very curly hair (CC). If a person inherits two alleles for straight hair, they have very straight hair (C’C’). A heterozygous person (CC’), on the other hand, shows an intermediate condition, wavy hair. If a wavy-haired man married a wavy-haired woman, what percentage of their children would you expect to have curly hair?

Parents	Man X Wavy-Haired	Woman Wavy-Haired
(Parent Genotypes)	CC’	CC’
Gametes Produced

Because the couple are heterozygotes, they will produce two types of gametes; C and C’. We can use this information to complete a Punnett square.

Gametes	C	C’
C
C’

What percent of their offspring should have curly hair? _____________

SOME PRACTICE INCOMPLETE DOMINANCE PROBLEMS

Photo source: http://www.greenearthgrowers.net/

Photo source: http://www.jparkers.co.uk/

Photo source: http://www.weststarfarm.com/

1. In snapdragons, red flower color (W) is not completely dominant over white (W’); the heterozygous condition produces pink flowers. What will be the result of a cross between two pink-flowered snapdragons? Between a pink and a white one?

2. Two parents have wavy hair & dimples. They have a child with curly hair & no dimples. Identify the genotypes of the two parents and then determine all of the possible phenotypes that their children could have for the dimple & hair trait. (Remember that dimples is dominant to nondimples. Also the heterozygous condition in which a person inherits an allele for straight hair & an allele for curly hair results in an intermediate condition, wavy hair.)

Red Polled Photo source: http://www.rarebreeds.co.nz/	Roan Photo source: http://www.midcontinentfarms.com/	White Horned Photo source: http://www.glcattleco.com/
3. In shorthorn cattle, the polled (hornless) condition (H) is dominant over the horned condition (h), also the heterozygous condition of red coat (W) and white coat (W’) is roan. If a homozygous polled red animal is bred to a white horned one, what will the F1 be like? If two F1 were crossed, what would be the phenotypic ratio of the F2 generation?

Sex-Linked Trait Problems

SAMPLE PROBLEM: In humans, red-green color blindness is a sex-linked trait. Normal color vision is due to allele B & color blindness is due to allele b. The heterozygous condition results in a carrier condition in females (they see red-green normally, but can pass the trait on to their offspring). What would be the phenotypic ratio of offspring produced by a color blind male and a carrier female?

For this problem,

X^B X^b = carrier female X^b X^b = color blind female X^B X^B = normal female

X^b Y = color blind male X^B Y = normal male

The color blind male in this case will have genotype X^b Y. He will produce 2 types of gametes: X^b & Y. The carrier female will have genotype X^B X^b. She will produce 2 types of gametes: X^B & X^b . We can use this information to complete a Punnett square.

Gametes	X^B	X^b
X^b
Y

What will be the phenotypic ratio of this couple’s children?

________________________________________________________________

Here is an Ishihara chart to check if you have some difficulty seeing color. You should see the number beside the circles as a pattern of colored dots.

SOME PRACTICE SEX-LINKED TRAIT PROBLEMS

Calico Cat Photo source: http://www.hanne-mugaas.com/	Tortoise Shell Cat Photo source: http://www.catsarewonderful.com/
1. In cats, orange color is due to allele B’ & black color is due to allele B. The heterozygous condition (B’B) results in a color known as calico (calico is a coat pattern that is mottled in tones of black, orange, and white) in females. These alleles are known to be sex-linked. What coat color types would be expected from a cross between a black male & a calico female?