GENETICS AND GENE PROBLEMS

© 30 August 1999, John H. Wahlert & Mary Jean Holland
Revised 14 February 2004 (jhw)


CORN GENETICS

Inheritance obeys the same rules of probablilty that apply to tossing coins and rolling dice. Mendel’s great achievement was his recognition from experimental results that this is so.

A simple case: given a pair of alleles of a gene, one dominant and one recessive, their recombination at fertilization is like flipping two coins at the same time.

[paired coin flip outcomes]A coin has two sides, head and tail (like two alleles of a gene). If you flip two dimes and examine the paired outcomes (whatever is face up), you will observe three possible combinations: HH, HT, and TT. Since there is half a chance that either coin will come up heads, the probability of a homozygous outcome, HH (also of hh), is 1/2 x 1/2 = 1/4. The outcome, HT, can arise in two ways so its probability is 1/4 + 1/4 = 2/4. The outcome of many such trials of HT x HT (flipping 2 coins and examining which pair of faces is up) will be 1/4HH + 2/4HT + 1/4TT. The more times you flip the coins together, the closer you will come to these ideal ratios.


Corn (Zea mays) is a large, coarse grass. Male and female flowers are separate. Many male flowers make up each of the tassels, which are located at the tips of stems. The tiny pollen grains are transported by wind. Female flowers nestle lower down, close to the stout part of the stem and are combined in structures we refer to as an ear. Each kernel in an ear of corn is the swollen ovary of a flower; a single strand of silk extends from each kernel to the outside of the ensheathing husk. A kernel only matures if its silk is pollinated. Thus each kernel in an ear of corn represents a separate fertilization of an ovum by a sperm. An ear of corn holds the results of hundreds of separate genetic crosses.

The ear of corn, below, illustrates the outcome of a monohybrid cross of parent plants that were heterozygous for kernel color. P (purple) is the dominant allele, and p (yellow) is the recessive allele. Both heterozygous parents had the genotype Pp and phenotype purple kernels.

  • What phenotypic frequencies of kernel color do you expect as outcomes in the cross Pp x Pp?
  • Count at least two full rows of kernels and record the colors. Do the ratios of each color (number of purple divided by total number of kernels counted and number of yellow divided by total number of kernels counted) come close to your expected phenotypic outcomes?
  • Does this actual result support your original hypothesis?

[detail, dihybrid ear of corn]This detail from an ear of corn illustrates the outcome of a dihybrid cross of parent plants that were heterozygous for kernel color and kernel form. P (purple) and S (smooth) are dominant alleles, and p (yellow) and s (wrinkled) are recessive alleles. Both heterozygous parents had the genotype PpSs and phenotype purple smooth kernels.

  • What phenotypic frequencies of kernel color do you expect as outcomes in the cross PpSs x PpSs?
  • Count at least four full rows of kernels and record the colors. Do the ratios of each color come close to your expected phenotypic outcomes?

    purple, smooth divided by total number of kernels
    purple, wrinkled divided by total number of kernels
    yellow, smooth divided by total number of kernels
    yellow, wrinkled divided by total number of kernels
  • Does this actual result support your original hypothesis?
  • Calculte the frequency of purple and yellow kernels. Calculate the frequencies of smooth and wrinkled kernels. Is a dihybrid cross like two monohybrid crosses multiplied together?

    (3/4P_ + 1/4pp) x (3/4S_ + 1/4ss)

    Note: The _ can stand for either the dominant or recessive allele; it’s the same phenotype.


GENETICS PROBLEMS

  1. Ear lobes in people may be free hanging or completely attached to the side of the face. This is determined by a single gene locus; the free hanging allele, E, is dominant and the attached allele (e) is recessive.

    1. A person has the heterozygous genotype Ee. With respect to this gene locus, how many kinds of gametes, eggs or sperm, will this person produce? What will the percent or frequency of each kind of gamete be out of the total possible?

    2. A man and a woman who are both heterozygous for ear lobe condition have children. What is the probability that a child will have free ear lobes? What is the probability that a child will have attached ear lobes?

    3. A man with attached ear lobes and a woman with free ear lobes have three children; two have free ear lobes and one has attached ear lobes.

      What is the man’s genotype?

      What is the woman’s genotype?

      What are the genotypes of the three children?

  2. A plant has the genotype AArrSs. Cross out gene combinations that cannot be produced or are not normal in the haploid gametes of this plant:

    Aas     ArS     Ars     arS     aRs     Arr     AS     AArrSs

  3. Two plants with the following genotypes are crossed: AArrSs x aarrSS. What are not possible or normal genotypes in a plant of the next generation?

    AArrss     AarrSS     AaRRS      AarrSs     Aarrss     aaRrSs     AarrSss

  4. The parental cross is NnMm x Nnmm
    Use a branching diagram to determine the frequency of the genotype nnmm in the next generation.

  5. Genes often interact with one another. The term epistasis is applied to cases in which one gene alters the expression of another gene that is independently inherited. In Labrador retriever dogs one gene locus is involved with production of melanin pigment: B (black) and b (brown) are its two alleles. Another gene locus determines whether the melanin produced is actually deposited in individual hairs as they grow (E) or not deposited (e). Any dog with at least one dominant allele B and one dominant allele E will be a black Labrador. A dog with the homozygous recessive bb and at least one dominant E will be a lighter, chocolate Labrador. If a dog has the homozygous ee genotype it will be a yellow Labrador, regardless of the alleles at the pigment locus.

    1. Give possible genotypes for a dog with the black Labrador phenotype?

      With respect to locus B

      With respect to locus E

    2. Give possible genotypes for a dog with the chocolate Labrador phenotype?

      With respect to locus B

      With respect to locus E

    3. Give possible genotypes for a dog with the yellow Labrador phenotype?

      With respect to locus B

      With respect to locus E

  6. A yellow Labrador with the genotype bbee mates with a black Labrador that is homozygous for both alleles. What are the ratios of genotypes and phenotypes expected in the F1 and F2 generations?


CHI SQUARE TEST

When you examine the results of a genetic cross you may ask if the numbers you observe are in agreement with the hypothetical outcome of the cross. For example, among the progeny of a monohybrid cross Rr x Rr, you expect that 3/4 will have phenotype R_ and 1/4 rr. The phenotypes you observe and count probably won’t match these ratios exactly because chance plays a role in biological phenomena, in this case, fertilization events. Chance enters again with the corn cross, since you will not be counting all kernels on every ear, and some kernels are missing due to handling or consumption by mice.

Is the difference between your observation and the expected result small enough that it could have been produced by chance alone. This is the null hypothesis—that there is no real difference between the observed data and the predicted data.


EXAMPLE

Suppose you counted 79 R_ and 33 rr. The total number of individuals you counted, N, is 112. You expect 3/4 to be R_ (84) and 1/4 to be rr (28). Are your results close enough to these ratios for you to accept the null hypothesis—that there is no real difference? The Chi-square test is one tool for making this decision.

Phenotypes

Observed
(o)

Expected
(e)

d = o - e

d2

d2/e

R_

79

.75 x 112 = 84

-5

25

.30

rr

33

.25 x 112 = 28

5

25

.89

Total

112

112  

0

 

1.19

Χ2 = ∑ (observed-expected)2 / (expected). This means add up the values in the last column.

You can compare the chi-square sum, 1.19, with the numbers in a table of critical values to decide whether to accept the null hypothesis—that the observed results are so close to expected results that there is no difference, and our original hypothesis is accepted.

Table. Selected percentile values of the Χ2 distribution

 

Probabilities

df*

.99

.95

.50

.10

.05

.01

1

.000157

.00393

.455

2.706

3.841

6.635

2

.0201

.103

1.386

4.605

5.991

9.210

3

.115

.352

2.366

6.251

7.815

11.341

4

.297

.711

3.357

7.779

9.488

13.277

5

.554

1.145

4.351

9.236

11.070

15.086

6

.872

1.635

5.348

10.645

12.592

16.812

*df is degrees of freedom.

Degrees of freedom: In a two-phenotype system, when you know the number for one phenotype, the result for the other (the rest of the population) is automatically determined. In this kind of genetic data, the number of degrees of freedom is one less than the number of different phenotypes observed.

As a rule, if the probability of obtaining a particular Χ2 value is greater than 5 in 100 (P > 0.05), then the difference between expected and observed is not considered statistically significant, and the null hypothesis is accepted.

Since we observed 2 different phenotypes in the monohybrid cross, there is only 1 degree of freedom; numbers in only the first row of the above table are relevant. The value 1.19 falls between probabilities of .50 and .10. This is interpreted to mean that in 10 to 50 out of 100 observed samples (10 to 50 percent of the time), we could expect Χ2 values this big or bigger due to chance. That’s reasonable; the observed deviation can simply be a chance or sampling error. Note that the test does not prove that the hypothesis is true; it indicates that the observations provide no statistically compelling argument against it.


EXPERIMENTAL DATA

Fill in the tables below with your observed data, calculate the expected result according to your hypothesis. Determine the Χ2 value for each experiment, and use the table of probabilities to accept or reject the null hypotheses. For your convenience in lab, you can print the tables you need and bring them to lab.

Monohybrid cross

Phenotypes

Observed
(o)

Expected
(e)

d = o - e

d2

d2/e

 

 

 

 

 

 

 

 

 

 

 

 

Total

 

 

0

 

 

Χ2 = _____     degrees of freedom: _____

Range of probability: ______________

Accept or reject null hypothesis? __________

 

Dihybrid cross

Phenotypes

Observed
(o)

Expected
(e)

d = o - e

d2

d2/e

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Total

 

 

0

 

 

Χ2 = _____     degrees of freedom: _____

Range of probability: ______________

Accept or reject null hypothesis? __________


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Last updated 18 July 2014 (JHW)