Mendel’s Laws: Their Application to Solving Genetics Problem
adventure·@bilalshoona·
0.000 HBDMendel’s Laws: Their Application to Solving Genetics Problem
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)
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
