TOPIC: MENDELIAN GENETICS

OBJECTIVES: After completing this exercise you will be able to:

1. Define true-breeding, hybrid, monohybrid cross, diploid, haploid, genotype, phenotype, dominant, recessive, complete dominance, homozygous, heterozygous, incomplete dominance, codominance, sex-linked, dihybrid cross, probability, multiple alleles;
2. Solve problems illustrating single-trait and double-trait crosses, including those with incomplete dominance, codominance, sex-linkage, and problems involving multiple alleles.

NOTE: IT IS IMPERATIVE THAT YOU UNDERSTAND MEIOSIS BEFORE ATTEMPTING GENETICS PROBLEMS.

DISCUSSION: In 1866 an Austrian monk, Gregor Mendel, presented the results of painstaking experiments on the inheritance of the garden pea. Those results were heard, but probably not understood, by Mendel’s audience. Now, more than 100 years later, biologists take the importance of Mendel’s work for granted. In this era of genetic engineering-the incorporation of foreign DNA into chromosomes of unrelated species, it is easy to lose sight of the basics of the process that makes it all possible.

For a better understanding of genetics it would be most helpful to understand the following terms:

Mitosis- Qualitative and quantitative cell division. Mitosis is the process by which all cells in the body are produced except the gametes (sperm and egg).

Cytokinesis- Cytoplasmic division.

Meiosis – Reductional division. Meiosis is the process by which gametes are produced. In males it is called spermatogenesis. In females it is called oogenesis.

Chromosome- The combination of DNA and histones.

Sister Chromatids- The two identical copies of DNA that occur after DNA replication. These two copies remain attached at the centromere until Anaphase of mitosis or Anaphase II of Meiosis when they separate (segregate) to become daughter chromosomes.

Chromatin- Loose, uncoiled DNA found during Interphase.

Homologous Chromosomes- Chromosomes of the same length, which carry genes for the same traits. We acquire one homologue from our fathers in the sperm and one homologue from our mothers in the egg.

Alleles- The different forms of a gene, i.e. one homologous chromosome may carry the gene (allele) for brown eye pigment and the other homologous chromosome may carry the gene (allele) for blue eyes.

Gene- A sequence of nitrogen bases on DNA which codes for the production of a polypeptide or protein.

Locus- The location of a gene (allele) on a chromosome.

Homozygous alleles- The condition which occurs when an individual has both alleles (one on each homologue) which are identical, i.e. brown eye pigment genes on both homologous chromosomes.

Pure-Breeding- A self-fertilizing parent whose offspring display the identical phenotype as their parent, generation after generation. (The parents would have to be homozygous for the trait being studied)

Homozygous dominant- Having both alleles on each homologue identical and each of these is a dominant gene, i.e. AA

Homozygous recessive- Having both alleles on each homologue identical and each of these is a recessive gene, i.e. aa

Heterozygous- Having different genes (alleles) on each homologue, i.e. Aa

Genotype- The genetic makeup of the individual regarding one or more traits. Another way to view this is ” What genes are present. ” e.g. Bb, BB, bb.

Phenotype- The physical expression of genotype. Another way to view this is ” What you see”.

Autosomes- Chromosomes that code for non-sex traits. We have 22 pair of autosomes.

Sex chromosomes- Chromosomes that code for sexual traits, i.e. reproductive organs. Males have XY genotype. Females have XX genotype.

X-Linked traits- Traits only found on the X sex chromsome. These traits are not found on the Y chromsome.

Synapsis- The joining of homologous chromosomes. This only occurs during Prophase I of Meiosis.

Crossing over- The exchange of genetic information between nonsister chromatids of a homologous pair. This only occurs during prophase I of meiosis.

Complete dominance- One allele, the dominant allele, masks the expression of the recessive allele. Individuals with the heterozygous genotype display the dominant phenotype.

Codominance- Two genes (alleles) that are equally dominant, therefore both are fully expressed, i.e. what occurs in roan cattle.

Incomplete dominance- Having two genes (alleles) neither of which dominates the other, therefore neither is fully expressed. This typically results in intermediate phenotypes, i.e.
Black + White = Gray

I. MONOHYBRID PROBLEMS WITH COMPLETE DOMINANCE

Garden peas have both male and female parts in the same flower and are able to self-fertilize. For his experiments, Mendel chose parental plants that were true-breeding, meaning that all self-fertilized offspring displayed that same form of a trait as their parent. For example, if a true-breeding purple-flowered plant is allowed to self-fertilize, all of the offspring will have purple flowers.

When parents that are true-breeding for different forms of a trait are crossed-for example, purple flowers and white flowers-the offspring are called hybrids. When only one trait is being studied, the cross is called a single-trait cross.
We’ll look first at single-trait problems. After completing the following problems, you will understand the basis for the outcome you are observing.

1. Suppose one gene has alleles A and a. Remember that most organisms are diploid; i.e., they contain homologous chromosomes with identical genes, although the alleles may be different on each homologue. Gametes, on the other hand, are haploid, containing only one of the two homologues, and thus only one of the two alleles.
The genotype of an organism is its genetic constitution, i.e., the alleles present.
For each of the following diploid genotypes, indicate the possible genotypes of the gametes.

Diploid genotype Gamete genotype (four cells are produced)

AA _____, ______, _____, _____

aa _____, ______, _____, _____

Aa _____, ______, _____, _____

2. During fertilization, two gamete nuclei fuse, and the diploid condition is restored. Give the diploid genotype produced by fusion of the following gamete genotypes:

Gamete Gamete Diploid Genotype

a a _____

A a _____

A A _____

3. Now let’s attach some meaning to genotypes. As you see from the previous problems, the genotype is an expression of the actual genetic makeup of the organism. The phenotype is the observable result of the genotype, i.e., what the organism looks like because of its genotype. Although phenotype is determined primarily by genotype, in many instances environmental factors can modify phenotype.

Human earlobes are either attached or free. This trait is determined by a single gene consisting of two alleles, F and f. An individual whose genotype is FF or Ff has free earlobes. This is the dominant condition. Note that the presence of one or two F alleles results in the dominant phenotype, free earlobes. The allele F is said to be dominant over its allelic partner, f. The recessive phenotype, attached earlobes, occurs only when the genotype is ff. In the case of complete dominance, the dominant allele completely masks the expression or effect of the recessive allele.

Suppose a man has the genotype FF. What is the genotype of his gamete (sperm) nuclei?

When both alleles in the nucleus are identical, the condition is homozygous. Those having both dominant alleles are homozygous dominant.
Suppose a woman has attached earlobes. What is her genotype?

Her gametes (ova) carry what allele?

When both recessives are present in the same nucleus, the individual is said to be homozygous recessive for the trait.

Suppose these two individuals produce a child. Show the genotype of the child by doing the cross:

Sperm Genotype X Ovum Genotype

_____________ ____________

____________

Child’s Genotype

When both the dominant and recessive alleles are present within a single nucleus, the individual is heterozygous for that trait.

What is the phenotype of the child? (i.e., does this child have attached or free earlobes?)

4. In garden peas, purple flowers are dominant over white flowers. Let P represent the allele for purple flowers, p the allele for white flowers.

a. What is the phenotype (color) of the flowers with the following genotypes:

Genotype Phenotype

pp _________

PP _________

Pp _________

A white-flowered garden pea is crossed with a homozygous dominant purple-flowered plant.

b. What is the genotype of the gametes of the white-flowered plant?

c. What is the genotype of the gametes of the purple-flowered plant?

d. What is the genotype of the plant produced by the cross?

e. What is the phenotype of the plant produced by the cross?

A convenient method of performing the mechanics of a cross is to use a Punnett square.

The circles along the top and side of the Punnett square represent the gamete nuclei.

Insert the proper letters indicating the genotypes of the two possible gamete nuclei for the above cross in the circles and then fill in the Punnett square.

Gametes of White-Flowered Plant

Gametes of Purple-Flowered Plants

A heterozygous plant is crossed with a white-flowered plant. Fill in the Punnett square and give the genotypic and phenotypic ratios of the offspring.

Gametes of White-Flowered Plant

Gametes of Heterozygotes

Genotypic Ratio: _______________________________________

Phenotypic Ratio: ______________________________________

4. In mice, black fur (B) is dominant over brown fur (b). Breeding a brown mouse and a homozygous black mouse produces all black offspring.

a. What is the genotype of the gametes produced by the brown-furred parent?

b. What is the genotype of the brown-furred parent?

c. What is the genotype of the black furred parent?

d. What is the genotype of the black-furred offspring?

By convention, P stands for the parental generation. The offspring are called the “first filial generation,” abbreviated F1. If these F1 offspring are crossed, their offspring are called the “second filial generation,” designated F2. Note the following diagram.

P X P and F1 X F1

F1 F2

e. If two of the F1 mice are bred with one another, what will be the genotypic and phenotypic ratios of the F2?

Genotypic Ratio: _______________________________________

Phenotypic Ratio: _______________________________________

6. The presence of horns on Hereford cattle is controlled by a single gene. The hornless (H) condition is dominant over the horned (h) condition. A hornless cow was crossed repeatedly with the same horned bull. The following results were obtained in the F1 offspring:
8 hornless cattle
7 horned cattle
What are the genotypes of the parents?

cow ____________________________________________________

bull ___________________________________________________

7. In fruit flies, red eyes (R) are dominant over purple eyes (r). Two red-eyed fruit flies were crossed, producing the following offspring:

76 red-eyed flies
24 purple-eyed flies

a. What is the approximate ratio of red-eyed to purple-eyed flies?

b. Based upon your experience with previous problems, what two genotypes give rise to this ratio?

c. What are the genotypes of the parents?

d. What is the genotypic ratio of the F1?

e. What is the phenotypic ratio of the F1?

II. MONOHYBRID PROBLEMS WITH INCOMPLETE DOMINANCE

8. Petunia flower color is governed by two alleles, but neither allele is truly dominant over the other. Petunias with the genotype R R are red-flowered, those that are heterozygous (R W ) are pink, while those with the (WW ) genotype have white flowers. This is an example of incomplete dominance.

a. If a white-flowered plant is crossed with a red-flowered plant, what is the genotypic ratio of the F1?

b. What is the phenotypic ratio of the F1?

c. If two of the F1 offspring were crossed, what phenotypes would appear in the F2?

d. What would be the genotypic ratio in the F2 generation?

III. MONOHYBRID PROBLEMS ILLUSTRATING CODOMINANCE

9. Another type of monohybrid inheritance involves the expression of both phenotypes in the heterozygous situation. This is called codominance.

One of the best known examples of codominance occurs in the blue Andalusian chicken. “Blue” birds are heterozygous (BW) and result from the mating between a black bird (BB) and a white bird (WW). Blue birds do not really have blue feathers, instead having a mixture of black and white feathers that reflects light to appear blue. Thus, the “blue” coloration is not a consequence of blending of the pigments (after all, the mix is not gray) but rather the result of both colors existing on the same bird. That is both phenotypes occur on the same individual.

a. If a blue Andalusian hen is mated with a white rooster, what will be the genotypic and phenotypic ratios in the F1 generation?

Genotypic ratio ________________________________________

Phenotypic ratio _______________________________________

b. List the parental genotypes of crosses that could produce at least some:

white offspring _______________________________

black offspring ________________________________________

IV. MONOHYBRID, SEX-LINKED PROBLEMS

10. In humans, as well as in other primates, sex is determined by special sex chromosomes. An individual containing two X chromosomes is a female, while an individual possessing an X and a Y chromosome is a male.

a. What is the genotype of your sex chromosomes?

b. In terms of sex chromosomes, what is the sex chromosomal genotype of the (ova)?

c. What are the possible sex chromosomal genotypes of a male’s sperm cells?

d. The gametes of which parent will always determine the sex of the offspring?

11. The sex chromosomes bear alleles for traits, just like the other chromosomes in our bodies. Genes that occur on the sex chromosomes are said to be sex-linked.

The Y chromosome is smaller than its homologue, the X chromosome. Consequently, some of the loci present on the X chromosome are absent on the Y chromosome.

In humans, color vision is a sex-linked trait; the gene for color vision is located on the X chromosome but is absent from the Y chromosome.
Normal color vision (XC) is dominant over color blindness (Xc). Suppose a color-blind man fathers children of a woman with the genotype XCXC?

a. What is the genotype of the father?

b. Would any of the daughters be color blind?

c. Would any of the sons be color blind?

12. One of the daughters from the above problem marries a color-blind man.

a. What proportion of their sons will be color blind? (Another way to think of this is to ask what are the chances that their sons will be color blind).

b. Could a color-blind daughter result from this marriage? How?

V. DIHYBRID PROBLEMS

All the problems so far have involved the inheritance of only one trait, i.e., they were single-trait problems. We will now examine cases in which two traits are involved: two-trait problems.

NOTE: We will assume that the genes for these traits are carried on different pairs of homologous chromosomes.

13. In humans, a pigment in the front part of the eye masks a blue layer at the back of the iris. The dominant allele P causes production of this pigment. Those who are homozygous recessive (pp) lack the pigment, and the back of the iris shows through, resulting in blue eyes. (Other genes determine the color of the pigment, but in this problem we’ll consider only the presence or absence of any pigment at the front of the eye).

Dimpled chins (D=allele for dimpling) are dominant over undimpled chins (d=allele for lack of dimple).

a. List all possible genotypes for an individual with pigmented iris and dimpled chin.

b. List the possible genotypes for an individual with pigmented iris but lacking a dimpled chin.

c. List the possible genotypes of a blue-eyed, dimple-chinned individual.

d. List the possible genotypes of a blue-eyed individual lacking a dimpled chin.

14. Suppose an individual is heterozygous for both traits (eye pigmentation and chin form).

a. What is the genotype of such an individual?

b. What are the possible genotypes of that individual’s gametes?

If determining the answer for the last question was difficult, recall that the principle of independent assortment states that genes on different chromosomes are separated out independently of one another during meiosis. That is, the occurrence of an allele for eye pigmentation in a gamete has no bearing on which allele for chin form will occur in that same gamete.

There is a useful convention for determining possible gamete genotypes produced during meiosis from a given parental genotype. Using the genotype PpDd as an example,
here’s the method:
Using PpDd, match the first allele (P) to the third allele (D) = PD
match the first allele (P) to the fourth allele (d) = Pd
match the second allele (p) to the third allele (D) = pD
match the second allele (p) to the fourth allele (d) = pd

therefore use: 1-3 (PD)
1-4 (Pd)
2-3 (pD)
2-4 (pd)

c. Suppose two individuals heterozygous for both eye pigmentation and chin form have children. What are the possible genotypes of their children?

You can set up a Punnett square to do dihybrid problems just as you did with monohybrid problems. However, depending upon the parental genotypes, the square may have as many as 16 boxes rather than just 4. Insert the possible genotypes of the gametes from one parent in the top circles and the gamete genotypes of the other parent in the circles to the left of the box.

	Gametes of First Parent
Gametes of the Second Parent

d. What is the genotypic ratio?

e. What is the phenotypic ratio?

15. You would probably agree that it is unlikely that a family will have 16 children. In fact, one of the most useful facets of problems such as these is that they allow you to predict what the chances are for a phenotype occurring. Genetics is really a matter of probability, the likelihood of the occurrence of any particular outcome.
To take a simple example, consider that the probability of coming up with heads in a single toss of a coin is one chance in two, or 1/2.
Now apply this example to the question of the probability of having a certain genotype. Look at your Punnett square in problem 15. The probability of having a genotype is the sum of all occurrences of that genotype. For example, the genotype PPDD occurs in 1 of the 16 boxes. The probability of having the genotype PPDD is 1/16.

f. What is the probability of an individual from the above problem having the genotype:

ppDD ____________________________________________________

PpDd ____________________________________________________

PPDd ____________________________________________________

To extend this idea, let’s consider the probability of flipping heads twice in a row with our coin. The chance of flipping heads the first time is 1/2. The same is true for the second flip. The chance (probability) that we will flip heads twice in a row is 1/2 X 1/2 = 1/4. The probability that we could flip heads three times in a row is 1/2 X l/2 X 1/2 = 1/8.

g. Returning to eye color and chin form, state the probability that three children born to these parents will have the genotype ppdd.

h. What is the probability that three children born to these parents will have pigmented eyes and dimpled chins?

i. What is the genotype of the F1 generation when the father is homozygous for both pigmented eyes and dimpled chin, and the mother has blue eyes and no dimple?

j. What is the phenotype of the individual(s) you determined in letter i, above?

16. A pigment-eyed, dimple-chinned man and a blue-eyed woman without a dimpled chin have a child. The child is blue-eyed and has a dimpled chin.

	Gametes of First Parent
Gametes of the Second Parent

a. What is the genotype of the father?

b. What is the genotype of the mother?

c. What alleles were carried by the father’s sperm?

17. Suppose a blue-eyed, dimpled chinned man whose father lacked a dimple and a woman who is homozygous recessive for both traits have children:

	Gametes of First Parent
Gametes of the Second Parent

a. What would be the expected genotypic ratio of children produced in this marriage?

b. What would be the expected phenotypic ratio?

18. In his original work on the genetics of garden peas, Mendel found that yellow seed color (YY,Yy) was dominant over green seeds (yy) and that round seed shape (RR,Rr) was dominant over shrunken seeds (rr). Mendel crossed pure breeding (homozygous) yellow, round-seeded plants with green, shrunken-seeded plants.

a. What would be the genotypic and phenotypic ratios of the F1 produced from such a cross?

b. If the F1 plants are crossed, what would be the expected phenotypic ratio of the F2?