Mendelian Genetics Lab - Essay Papers Hq

Lab Template

Week 5: Mendelian Genetics

Submitted by:

As you complete the lab, record your answers in this template. Save the document as LastName_FirstName_BIO1020_W5A3, and submit it to the Dropbox. Full lab instructions and the rubric with which you will be evaluated can be found in the online classroom.

Activity

The laws of segregation, independent assortment, and dominance form the basis of all genetics. The ability to predict the results of crossing experiments and explain any variance between expected and observed results is still a vital part of our understanding of heredity. In this lab assignment you will experiment with monohybrid crosses and explore the role of chance in genetics.

Experiment 1

Questions

1. (10 points)

a. Set up and complete Punnett squares for each of the following crosses: (remember Y = yellow and y = blue)

· Y Y and Y y

Parent 1

Parent 2

· Y Y and y y

Parent 1

Parent 2

b. What are the resulting phenotypes for each cross? Are there any blue kernels?

Y Y and Y y

Y Y and y y

The resulting phenotypes is that all the offsprings are yellow because all the offspring have at least one Y (yellow, dominant) allele

All the offsprings are yellow

There are no blue kernels in either cross and all are yellow because the genotypes of all the kernels have at least one dominant (Y) gene which codes for yellow color.

2. (10 points)

a. Set up and complete a Punnett squarefor a cross of two of the F1 from the Y Y and y y cross above.

Parent 1

Parent 2

b. What are the genotypes and phenotypes of the F2 generation?

The genotypes of offsprings are Yy (heterozygous) and their proportion is 100% If Y= yellow an y= blue, then the phenotypes of the off springs would be the characteristics of Y gene which means all the off springs will have a yellow color.

Experiment 2

Questions

As you select the beads from the beaker, complete this table with each cross. You may complete the associated Punnett Squares on paper, but do not need to submit them as part of this lab.

Parents – randomly selected

F1 – determined from Punnett square

Cross

Genotype parent #1

Genotype parent #2

4 Genotypes

4 Phenotypes

1. (10 points)

a. How much genotypic variation do you find in the randomly picked parents of your crosses? How much in the oﬀspring?

Possible Genotype

Parents

Offspring

Total

b. How much phenotypic variation do you find in the parents of your crosses? How much in the offspring?

2. (10 points)

a. What is the ratio of phenotypes (yellow kernel color: blue kernel color) in the 20 offspring of your five crosses?

b. If you were to run this experiment 1000 times, rather than just 5 times, what would you expect the ratio of phenotypes to be in the offspring?

c. Is the ratio of observed phenotypes the same as the ratio of predicted phenotypes in the offspring? Why or why not?

3. Organisms heterozygous for a recessive trait are often called carriers of that trait. Explain what this means. (10 points)

Mendelian Gene cs

Lab 5

Lab 5: Mendelian Gene cs

Introduc on

In 1866, Gregor Mendel, an Austrian Monk, published a paper en tled “Experiments in plant hybridiza-

on”. It went largely unno ced un l 1900 when it was rediscovered and subsequently became the

basis for what we now refer to as Mendelian Gene cs.

Mendel was the Þrst to recognize:

Inherited characters are determined by speciÞc factors (now recognized these as genes).

These factors occur in pairs (genes).

When both alleles of a gene are the same they are said to be homozygous, while if they are di erent

they are said to be heterozygous. When gametes form, these factors segregate so that each gamete

contains only one allele for each gene. Remember, alleles reside on the chromosomes that are divid-

ing. These original observa ons lead to what we now refer to as The law of segrega on and the law of

independent assortment.

Figure 1: Law of Segrega on

Concepts to explore:

Gregor Mendel

Law of segrega on

Homozygous

Heterozygous

Independent assortment

Dominant vs. recessive

Incomplete dominance

Co-dominance

Genotype

Phenotype

Monohybrid cross

Dihybrid cross

Punne square

Lab 5: Mendelian Gene cs

The law of segrega on states that during

meiosis, homologous (paired) chromo-

somes split (Figure 1). The law of inde-

pendent assortment states that during

meiosis, each homologous chromosome

has an equal chance of ending up in ei-

ther gamete, and alleles for individual

genes segregate with the chromosomes

on which they are located (Figure 2).

Using corn as an example (Figure 2):

The large chromosome has the gene for kernel color (Y = yellow, y = blue).

The small chromosome has the gene for kernel texture (S = smooth (green); s = wrinkled (red)).

When a dominant allele is present, that characteris c is expressed, regardless of the second allele. In

this case both the Yy and YY o spring will be yellow.

A recessive allele is only expressed when both alleles are recessive. In this case only the yy combina-

on is blue. The dominant allele is always represented by capital le ers, while the recessive is repre-

sented by lower case le ers.

Genotype refers to the combina on of alleles for a par cular trait. Phenotype refers to the appear-

ance of that combina on of alleles. In our example, the genotype of the diploid cell is Yy, Ss, while the

phenotype is Yellow and Smooth.

Figure 2: Law of Independent Assortment

Figure 3: Monohybrid Cross

Punne Square F1

Lab 5: Mendelian Gene cs

Alleles can exhibit incomplete dominance and co-dominance. An example of incomplete dominance is

the cross of two plants, one with red ßowers and one with white, whose o spring have pink ßowers.

In the case of codominance, the same cross would result in red and white striped ßowers.

If we know the genotype of two parents we can predict both the genotype and phenotype of their o –

spring using a Punne Square. A monohybrid cross is a cross between two parents (P), looking at a

single gene (Figure 3). In this example, both parents are pure breeding (homozygous); one for the yel-

low color and one for the blue color. This cross can be shown as a Punne Square (Figure 4), with each

parent (P) contribu ng a single gamete. The o spring (F1) are determined adding the gamete of each

parent (P) (Row and Column). The F1 genotypes are all Y , y; with yellow phenotypes. The cross of the

(F1) genera on, known as the F2 genera on, is shown in Figure 5. The Punne square can also predict

the F1 for mul ple genes.

Y Y

y Y y Y y

Figure 4: Punnet square of a monohybrid cross

(F1)

Parent 1

Parent 2

Figure 5: Punnet square of a monohybrid cross

(F2)

Y y

Y Y Y Y y

y Y y y y

Parent 1

Parent 2

Y s Y S y S y s

Y s Y Y s s Y Y S s y Y S s y Y s s

Y S Y Y s S Y Y S S y Y S S y Y s S

y S Y y s S Y y SS y y S S y y s S

y s Y y s s Y y S s y y S s y y s s

Figure 6: Punnet square of a dihybrid cross (F1)

Lab 5: Mendelian Gene cs

Using our corn example, let’s look at two genes (color and texture), also known as a dihybrid cross. In

this example we use two parents that are heterozygous for both traits (Figure 6), using the gametes we

already iden Þed in (Figure 2).

The F2 phenotypes are:

Yellow & Smooth: 9

Yellow & Wrinkled: 3

Blue & Smooth: 3

Blue & Wrinkled: 1

Experiment 1: Punne Square Crosses

Procedure

1. a) Set up and complete Punne squares for each of the following crosses: (remember Y = yel-

low, and y = blue)

Y Y and Y y

Y Y and y y

b) What are the resul ng phenotypes for each cross? Are there any blue kernels?

Lab 5: Mendelian Gene cs

2. a) Set up and complete a Punne square for a cross of two of the F1 from the Y Y and the y y

cross above.

b) What are the genotypes and phenotypes of the F2 genera on?

Experiment 2: Monohybrid Crosses

Procedure

1. You have been provided with two alloca ons of di erent colored beads. Pour 50 of the blue

beads and yellow beads into the beaker and mix them around.

The beaker contains beads that are either yellow or blue.

These colors correspond to the following traits for kernel color:

Yellow (Y) vs. Blue (y)

2. Simulate a monohybrid Cross. Randomly (without looking) take 2 beads out of the beaker.

This is the genotype of individual #1, record this informa on. Do not put these

beads back into the beaker.

Repeat this for individual #2. These two genotypes are your parents for the next

genera on. Set up a Punne square and determine the genotypes and pheno-

types for this cross.

Repeat this process 4 mes (5 total).

Materials

Blue beads

Yellow beads

100mL Beaker

Lab 5: Mendelian Gene cs

Ques ons

1. a) How much genotypic varia on do you Þnd in the randomly picked parents of your cross-

es? How much in the o spring?

b) How much phenotypic varia on do you Þnd in the parents of your crosses? How much

in the o spring?

2. a) What is the ra o of phenotypes (yellow kernel color : blue kernel color) in the 20 o –

spring of your Þve crosses?

b) If you were to run this experiment 1000 mes, rather than just Þve mes, what would

you expect the ra o of phenotypes to be in the o spring?

c) Is the ra o of observed phenotypes the same as the ra o of predicted phenotypes in the

o spring? Why or why not?

3. Organisms heterozygous for a recessive trait are o en called carriers of that trait. Explain

what this means.

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