Misr University for Science and Technology

Misr University for Science and Technology Faculty of Biotechnology Principles of Genetics 102

"Laboratory manual"

Instructor Dr. Mona Ibrahim

Demonstrators Nermin Gamal Heba Radwan

1 | P a g e Contents  Chapter 1: Introduction to Drosophila melanogaster

 Introduction of Drosophila melanogaster o Development of Flies. o Culturing. o Handling of the Flies o Distinguish sex o Phenotype characterization. o Securing virgin females.

 Chapter 2: 1st Mendelian law

 Drosophila and maize experiments in genetics :Monohybrid crosses o The first experiment in Drosophila( Monohybrid cross) o ♂ EE x ♀ ee o ♀ EE x ♂ ee o The first experiment in Maize (Monohybrid cross)

 Chapter 3: Chi square test

 The Chi Square Test

o Analysis of monohybrid's offspring in Drosophila. o Analysis of monohybrid's offspring crosses in maize.

 Chapter 4: 2nd Mendelian law

 Drosophila and maize experiments in genetics: Dihybrid crosses o The second experiment of drosophila (dihybrids crosses) o ♂eevgvg x ♀EEVgVg o ♀ eevgvg x ♂EEVgVg o Analyzing the offspring (F2) of the dihybrid cross in drosophila(Chi-square test)

o The second experiment of maize ( Dihybrid cross ) o Analyzing the offspring (F2) of the dihybrid cross in maiz(Chi- square test)

 Chapter 5: Gene interaction and epistasis

 Gene interaction and epistasis

2 | P a g e o Gene interaction in maize o Analysis of gene interaction ratios in maize by (Chi-square test )

 Chapter 6: sex linkage

 Drosophila experiments in genetics: sex - linkage o Introduction of sex chromosomes and gene transition o Sex linkage in Drosophila o ♂RГx♀ww o ♀RRx♂wГ. o Analysing the offspring (F2) of the sex linkage experiment in drosophila

 Chapter 7: Pedigree Analysis

 Glossary

3 | P a g e Chapter one: Introduction to Drosophila

4 | P a g e Introduction to Drosophila melanogaster The purpose of this chapter is to introduce students to the equipment and techniques used to handle “fruit flies" in the laboratory. You will learn about the life cycle of the Dipteran Drosophila melanogaster. You will also learn to differentiate between male and female flies and how to identify aberrant phenotypes that are attributable to specific mutations, many of which will be studied during the semester.

 Objectives of this investigation Upon completion of this exercise you should be able to

. Distinguish between male and female D. melanogaster . Categorize mutant flies based on aberrant phenotypes . Prepare controlled genetic crosses of D. melanogaster

Background The fruit fly Drosophila melanogaster is a very useful organism for genetic research and has probably been used to define more fundamental genetic principles than any other ulticellular eukaryote. One person responsible for development of D. melanogaster into a model genetic system was named Thomas Hunt Morgan who, in 1910, published one of the first descriptions of sex linkage

Morgan and his students elucidated many basic principles of heredity, including sex-linked inheritance, epistasis, multiple alleles, and gene mapping. Thomas Hunt Morgan and colleagues extended Mendel's work by describing X-linked inheritance and by showing that genes located on the same chromosome do not show independent assortment. Studies of X-linked traits

5 | P a g e helped confirm that genes are found on chromosomes, while studies of linked traits led to the first maps showing the locations of genetic loci on chromosomes"

Due to its simple culturing requirements, short generation time, copious offspring, and well-defined genetics, this diminutive organism has become a versatile model system that is routinely used for inquiries into the genetics of eukaryotic development, behavior, and population dynamics. Since the arrival of recombinant DNA technology, much has been learned about the biology of Drosophila melanogaster.

 Development of Flies

The Life Cycle of D. melanogaster

D. melanogaster progress through four stages during their life cycle: egg, larva, pupae, and adult. Fertilization of eggs is internal, and females deposit fertilized eggs on the surface of the culture medium. Usually within one day, the eggs develop into larvae, which burrow into the nutritive medium. Over a period of 4-7 days, the larvae pass through three stages, or instars, and eventually crawl onto a firm surface to pupate.

During the pupal stage, which usually lasts from 4-6 days, metamorphosis occurs and the adult form develops. The adult fly emerges (ecloses) as an imago, which is slender, elongated, and light in color, with rumpled and unexpanded wings. Within a few hours, the adult matures, becoming darker and more rotund, with fully expanded wings. Adult flies may live for a month or more.

The rate of D. melanogaster development is greatly influenced by temperature. When propagated at 21o C (69.8o F), the progression from egg to adult usually takes about two weeks However, if temperature is maintained at 25o C (77.7o F), this time can be shortened to about 10 days.

6 | P a g e Working with D. melanogaster

 Culturing of the flies

 Medium For our purposes, D. melanogaster will be raised in plastic culture vials, which may contain a small piece of plastic netting to increase surface area for attachment of pupae. The culture medium is a complex mixture of agar, sugars, other nutritional supplements, and mold inhibitors.

 Material needed for each student for this investigation  Medium Stereo dissecting microscope Bottles of culture Gummed labels Etherizer Re-etherizer (Petri dish) Ether in dropping bottle Dropper Teasing needle or fine camel's hair brush Morgue containing 70% ethyl alcohol

7 | P a g e Handling the Flies

 To etherize and examine adult flies, proceed as follows: 1- Place a few drop of ether on the absorbent material of etherizer

2- Strike the base of bottle lightly on the palm of the hand so the flies will drop to the bottom.

3- Remove the culture bottle, plug quickly replace it with the mouth of the etherizer, invert on the bottle over the etherizer, and shake the flies into the etherizer.

4- Subject the flies to ether for about 30seconds after they cease moving. Avoid over etherization if they are to be used in further mating the flies will die if left in etherizer too long .over etherized flies hold their wings vertically over their body (in contrast to the normal at-rest position).

5- Transfer the etherized flies to clean white card. A 3-by- 5 in .file card is ideal.

6- Examine the etherized flies with a dissecting microscope at 10 x to 25x magnification. Use a soft brush or teasing needle for moving the flies about on the stage of the microscope.

8 | P a g e 7- If the flies revive before finish examine them ,add a few drops of ether to the absorbent pad on the re-etherizer and cover flies on the microscope stage for a few seconds .

8- If the flies are not needed after observation they may be discarded in the morgue. Etherized flies to be used for further mating should be permitted to recover in a dry vial or on a dry surface in the culture bottle before they come into contact with moist medium.

 Distinguishing sex

Female Male

Size Larger than the male

The caudal extremity is Sharpe and Where as in male is round and blunt Shape protruding. The abdomen of the female is the abdomen of the male is relatively distended and appears spherical or ovate. narrow and cylindrical . Black pigment is more than male and Black pigment is more extensive on Uniformly distributed the caudal extremity of the male than on that of the female

The marking or pigment occurs only in the On male the marking extend Color dorsal region completely around the abdomen and meet on venteral side.

Sex combs Only male have small tuft of black bristles called a sex comb on the anterior margin at the basal tarsal joint of each front leg

9 | P a g e Examination of the external genitalia under magnification is the best means of distinguishing the sex of flies. Only males flies exhibit darkly colored external genitalia which are visible on the ventral side of the tip of the abdomen following characteristics, illustrated in figure (1.2) can also help to distinguish males from females

Female genitalia Male genitalia

Female flies have anal plates and very dark ovipositor Males have anal plates, a dark colored genital plates on their ventral side. arch and a penis.

10 | P a g e Scoring Phenotypes

It will often be necessary to accurately identify phenotypic characteristics each as eye color or shape, body color, and bristle or wing morphology.

In most cases, you will be using a stereo microscope (dissection microscope) to examine various phenotypes or determine the sex of an individual fly. To accurately score flies for phenotype differences that are difficult to distinguish, it is easiest to make direct comparisons by having a wild type fly and the mutant strain both in the field of view of the stereo microscope.

11 | P a g e  Securing virgin females

To begin across between two varieties of flies, you must secure a virgin female. Once inseminated; females retain viable sperm for several days. Thus, the only way to ensure a controlled mating between different genetic stocks is to use virgin females.

The most common method for obtaining virgin females is to select those that have recently emerged from their pupa cases. These flies (D.melanogaster) do not mate for at least 8 hours (probably not for 10 hours) after emerging from the pupa case. Thus, 8-hour-old females will still

12 | P a g e be virgins even if male flies are present. To secure virgin females, follow this procedure

1- Empty the appropriate stock bottle of all adult flies .Record the time on the bottle 2- Within 8 hours of removing the adult flies, etherize any newly emerged adults. Such newly emerged flies are distinguished by their pale body color and a characteristics dark spot on the ventral side of the abdomen, slightly to the left of the midline. 3- Females among those newly emerged adults will be virgins that can be used in the appropriate cross

Another method for obtaining virgin females is to isolate pupae from which the adult flies are about to emerge .Single pupa cases should be placed in small vials containing a narrow strip of moistened paper towel. The vial should be stoppered with cotton. Of necessity, any females hatching alone in a vial will be a virgin. This method of obtaining virgin females is laborious and can prove unwieldy because it requires many vials.

Exercise

You will be given flies with “unknown” mutations to identify and complete the table

Trait Wild Unknown Unknown2 Uknown3 Unknown4 Unknown5 Unknown Type 1 6

13 | P a g e Body color

Eye color

Eye shape

Wing shape and size

Notes …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… ……………………………………………………………………………………………

14 | P a g e …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… ……………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… ……………………………………………………………………………………………

15 | P a g e …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… ……

Chapter two: 1st Mendelian law

16 | P a g e Drosophila and maize experiments in genetics Monohybrid crosses

In this exercise, you will investigate the inheritance of traits in D. melanogaster, and maize using the Mendelian model to develop your hypotheses. You will work with monohybridd crosses to determine the genotypes of the F1 and F2 generations and the phenotypes.

Objectives of this investigation Upon completion of this lab, you should be able to:

. Predict the offspring of a monohybrid, cross. . Determine the parents of a designed cross from the ratios of observed offspring. . Calculate phenotypic ratios and carry out Chi-squared analyses for each cross.

Background:

Each cross begins with true breeding (homozygous) parents. A homozygote has two identical alleles for a given trait. If parents with different traits are crossed (aa x AA), the offspring will be heterozygous (Aa), receiving one allele from each parent. However, the offspring will express only one allele, which Mendel described as dominant. The masked trait is recessive; two alleles must be present for a recessive trait to be expressed. The parents in this cross are referred to as the P generation, and the hybrid offspring, as the F1 (first filial or hybrid) generation. Two F1 hybrids are crossed to produce the F2 generation.

17 | P a g e The predicted Mendelian results for the F2 generation from a monohybrid cross (Aa x Aa) is a 1:2:1 genotypic ratio and a 3:1 phenotypic ratio. Mendel formulated his first law of segregation to explain these results: allele pairs separate during gamete formation and randomly return to the paired condition during the fertilization to form a zygote

First experiment Monohybrid crosses in Drosophila

D. melanogaster has historically been the model system for research in eukaryotic genetics and plays an important role in the development of our knowledge of heredity. D. melanogaster have a low chromosome number (n= 4), referred to as X (1), 2, 3, 4 chromosomes. Chromosomes 2, 3, 4 are autosomes (same in both sexes). X and Y are sex chromosomes. Females are XX and males are XY. Chromosomes 4 and Y contain few genes and for practical purposed can be ignored. Almost the entire genetic content of the D. melanogaster genome resides on only three chromosomes: X, 2, and 3.

Making crosses: 1- When making crossing between two varieties, For example, in crossing flies having ebony body color with those having white eyes, only body color and eye color need to be observed carefully.

2- Secure etherized male flies of one variety and etherized virgin females of the other. While holding the culture bottle on its side, place these flies in the bottle. Be sure to add some dry yeast granules or yeast suspension to the medium before introducing the flies.

3- Keep the bottle on its side until the flies have recovered from etherization. This position will prevent their becoming stuck in the medium.

4- Label the bottle and record data in table 1.2 after 7 or 8 days remove

the parent flies (p1) to prevent their being confused with or mated with their offspring.

18 | P a g e 5- Flies of the first filial generation (F1) will soon being to emerge.

After several F1 flies have appeared, etherized and examine them,

especially with regard to the characters by which the P1 flies differed.

6- Record in table 1.2 the phenotypes of the F1 flies of each sex; place these F1 flies in a fresh bottle of medium. (Be sure to label the bottle) This mating will allow for a production of the second filial

generation (F2). It is not necessary that the F1 female flies are virgins for this mating.

Exercise Why the instructor may wish to have you test cross an F1 female flies? …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… ……..

In this case you should use a virgin female? …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… ………

Why a virgin female is needed in this case? …………………………………………………………………………………………… ……………………………………………………………………………………………

19 | P a g e …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… ………

What is test cross? …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… ………

Laboratory record Mush of success in elucidating the basic principle of inheritance depend on his keeping accurate records about his experiments .keeping accurate and detailed notes is also necessary to the successful completion of your laboratory work .To this end ,you are provided with Table 1.2 in which to records as you do the various steps of this investigation.

Table 1.2 Record of a drosophila Experiment (1)

Experiment number……………. .Name…………………………………………………………. 1. Cross………………………….female x …………………………………..male.

2. Date P1 mated………………………………………………………………………………………….

20 | P a g e 3. Date P1 s removed…………………………………………………………………………………..

4. Date F1 first appeared ………………………………………………………………………....

5. Phenotype of F1 males ……………………………………………………………………….....

6. Phenotype of F1 females ……………………………………………………………………….

7. Date F1 male and female placed in fresh bottle (or date F1 virgin female test crossed)……….

8. Date F1 flies removed ………………………………………………………………………….

9. Date F2 appeared……………………………………………………………………………......

10. Record F2 in the following table:

Male Female Phenotype Number Phenotype Number Total a …………………. ……………….. …………………… ……………….. …………. b. ………………… ………………… …………………….. ………………….. …………... c. …………………. …………………… ………………………. ………………….. ………….

D ……………………. ……………………… ………………… ………………… ………….. e. …………………… ……………………… ………………….. …………………… …………... f. ………………… ……………………… …………………… ………………….. …………. g. ………………… …………………….. …………………….. …………………… ………. h. ………………. ……………………… …………………… ……………… …………..

Total ……………….. ……………………….. ……………….

21 | P a g e

11. Indicate the genotypes and phenotypes of the flies in each of three generations Male Female

…………………………………… …………………………………………

Genotype of P1

Genotype of F1 ..………………………………… ……………………………………………

Phenotype of F1 .…………………………………… …………………………………………

Genotype of F2 …………………………………… ………………………………………………

What F2 phenotypic ratio do you expect to obtain? …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………

What F2 genotypic ratio is expected? …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… ………………………………………………..

Which trait is dominant in your experiment? …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… ………………………………………………… How do you know? …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………

22 | P a g e Second Experiment Monohybrid cross in Maize The extensive genetic variability in maize also obeys mendelian principles; both seedling (shape) and aleurone (endosperm) (color) characters lend themselves to classroom use. Figure (1.3) shows an ear of F2 kernels of corn that are segregating in a ratio of 3 colored: 1colorless aleuronic.

figure 1.3 You probably will not be able to conduct actual experimental mating with it, Because maize has relatively long life cycle (3 months or more to complete)

Your instructor might provide you with ears of corn or flats of F 2 seedlings and request that you determine the number of individuals having the various phenotypes and then to interpret the data in terms of mendelian principles.

For example, suppose you counted 40 green and 12 albino seedlings in an F 2 population. How would you interpret these data? …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… ……………………......

23 | P a g e Ears bearing F2 kernels can be purchased from a number of biological supply companies. Without removing the kernels from the ears, you can count the number of kernels in the different phenotypes, record data in table 1.3, and then formulate hypotheses to explain the data collected. Two possible illustrations of endosperm traits follow.

Parents Parents 1- SuSu x susu 2- CC x cc Starchy sweet Pigmented non pigmented (smooth) (wrinkled) (Purple) (White)

F1 F1 Phenotype………………………………… Phenotype……………………………… Genotype……………………………….… Genotype……………………………….

F2 F2 Phenotype………………………………… Phenotype………………………………. Genotype…………………………………. Genotype…………………………......

What F2 phenotypic ratio do you expect to obtain? …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………

What F2 genotypic ratio is expected? …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… ………………………………………………… Which trait is dominant in your experiment? …………………………………………………………………………………………… …………………………………………………………………………………………… ……………………………………………………………………………………………

24 | P a g e …………………………………………………………………………………………… ………………………………………………… How do you know? …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………

Chapter three Chi square test

25 | P a g e Chi square test

The purpose of the chi square X2test is to determine whether experimentally obtained data constitute a good fit to a theoretical, expected ratio. In other words, the X2 test enables one to determine whether it is reasonable to attribute deviations from an expected value to chance. Obviously, if deviations are small then they can be more reasonably attributed to chance.

26 | P a g e The question is how small must deviations are in order to be attributed to chance?

The formula for X2 is as follows:

∑ (O-E)2 /E

Where O = the observed number of individuals of a particular phenotype, E = the expected number in that phenotype, and ∑ = the summation of all possible values of (O-E)2 /E for the various phenotypic categories. The following is an example of how one might apply the X2 test to genetics. In a cross of tall tomato plants to dwarf ones, the F1 consisted entirely of tall plants, and the F2 consisted of 102 tall and 44 dwarf plants. Does this data fit a 3:1 ratio? To answer this question, X2 value can be calculated (see table 1.4).

The calculated X2 value is 2.0548. What does this mean? If the observed had equaled the expected, the value would have been zero. Thus, a small X2 value indicates that the observed and expected ratios are in close agreement. However, are the observed deviations within the limits expected by chance? In order to determine this, one must look up the X 2value on a chi square table (see table 1.5). Statisticians have generally agreed on the arbitrary limits of odd of 1 chance in 20 (probability = .05) for drawing the line between acceptance and rejection of the hypothesis as a satisfactory explanation of the data tested. A X2value of 3.841 for a two-term ratio corresponds to a probability of .05. That means that one would obtain X 2 value of 3.841 due to chance alone on only about 5% of similar trials.

27 | P a g e When X2exceeds 3.841 for a two term ratio, the probability that the deviations can be attributed to chance alone is less than 1 in 20.

Thus, the hypothesis of compatibility between the observed and expected ratios must be rejected. In the example given the X2 value is much less than 3.841. Therefore, one can attribute the deviations to chance alone.

Notice that across table 1.5 are probability (P) values and down the side are degrees of freedom (df) values. The number of degrees of freedom is one minus the number of terms in the ratio. In the example above (3:1) there are two terms. Therefore, the degrees of freedom is 2 - 1 = 1. Thus, on the 1 degree of freedom row and under the .05 column the X2value of 3.841 is found. In the example given, the X2 value obtained is 2.0548. Looking on the 1 degree of freedom row, one finds that value between the .05 (X2= 3.841) and the .20 (X2= 1.642) probability columns. This means that the probability that the deviations occurred by chance alone is between 5% and 20%. In other words, if you were to repeat this experiment 100 times, you would expect to observe deviations at least as large as you obtained in between 5 and 20 of the 100 experiments due to chance alone.

Important points of chi-square It must be used only on the numerical data itself, never on percentage or ratios derived from the data. Two mathematical chick points are always percent in chi-square calculations:

28 | P a g e (1) The total of the expected column must equal the total observations and (2) The sum of the deviations should equal 0 the squaring of negative deviations convert all values to a positive scale.

Application of the Chi Square Test  First experiment

Transfer the F2 data obtained in the first experiment (see Table 1.2) to table 1.6.Calculate X2for the total of males and females based on the

.hypothesis that the classical monohybride F2 ratio has been obtained

.Table 1.6 Calculation of x2 on Data from experiment one Phenotypes %Ratio O E O-E O-E)2) O-E)2/E)

29 | P a g e Totals ……=X2

1. How many degrees, of freedom do you have in the interruption of the X2value? …………………………………………………………………………………………………………………………… .…..………………………………………………………………………………………………………………………

Using Table 1.5 determine what X2values lie on either side of the X2 .2 ...……………………………………………calculated in Table 1.6 record these values here

What probability values are associated with these table values of X2 .3 …………………………………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………………………………

In this case, do you accept or reject the hypothesis that these data .4 ?approximate a 3:1 ratio …………………………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………………………………

 Second experiment Transfer the F2 data obtained in the second experiment to table 2 1.3.Calculate X for the hypothesis that the classical monohybride F2 ratio .has been obtained .Table 1.3 Calculation of x2 on Data from experiment two Phenotypes % Ratio O E O-E O-E)2) O-E)2/E)

Totals .………=X2

30 | P a g e How many degrees, of freedom do you have in the interruption of the .1 ?X2value ………………………………………………………………………………………………………………………………………

Using Table 1.5 determine what X2values lie on either side of the X2 .2 ………………………………………………calculated in Table 1.3 record these values here

In this case, do you accept or reject the hypothesis that these data .4 ?approximate a 3:1 ratio …………………………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………………………………

Chapter four: 2nd Mendelian law

31 | P a g e Drosophila and Maize Experiments in Genetics: Dihybrid Crosses

In this exercise, you will investigate the inheritance of traits in D. melanogaster, and maize using the Mendelian model to develop your hypotheses. You will work with dihybridd crosses to determine the genotypes of the F1 and F2 generations and the phenotypes.

32 | P a g e  Objectives of this investigation Upon completion of this lab, you should be able to:

. Predict the offspring of a dihybrid, cross. . Determine the parents of a designed cross from the ratios of observed offspring. . Calculate phenotypic ratios and carry out Chi-squared analyses for each cross.

Background

In considering two traits simultaneously, Mendel postulated second law of independent assortment of the genes. Mendel studied seven different traits in the garden pea, and, in the entire situation investigated, he obtained dihybride F2 ratio of 9:3:3:3:1and dihybride test cross ratio of 1:1:1:1:1. This law states that allele pairs separate independently during the formation of gametes. Therefore, traits are transmitted to offspring independently of one another.

What is the mechanism of independent assortment? Genes located on separate pairs of chromosomes assort independently of one another .Thus ,one might assume that the seven different traits studied by Mendel were controlled by seven gene loci located –one locus per chromosome pair-on the seven chromosome pairs of the garden pea .This ,in fact, is not the case ,as we documented by Novitski and Blixt (1978) .Because of recombination .genes located on the same chromosome pair but at great distance from one another ,will also assort indecently of one another . Investigation of other organism other than garden pea revealed that independent assortment, occurs in them also,Figure (2.1) shows an ear of F2 Kernels of corn :the Kernel are distributed according to the law of independent assortment ,with (A)9/16 of the Kernels being colored with starchy (smooth)endosperm, (B)3/16 being colored with sugary (wrinkled) endosperm .(C)3/16 being colorless starchy and,(D)1/16 being colorless sugary.

33 | P a g e Figure 2.1

Third Experiment: Dihydrid cross in Drosophila

1.The gene for vestigial wings (vg) is located on the second chromosome, and the gene for ebony body color (e) is located on the third chromosome. Mating may be made between flies from this two stocks .Be certain to use virgin females flies for this cross. Reciprocal crosses ,that is , after about 8 days remove and discard the parent flies record all data relative to thisexperiment

2. Predict the results in the F1 and F2generations from the mating prepared

F1 phenotype(s): ………………………………………………………………………………………………...

F2 phenotypes and frequencies ……………………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………

3. When the F1 flies appear in the culture bottles, check to see if your

expectations have been obtained .Record data relative to the F1 flies in

Table 2.1. Mate F1 males and F1 females in new culture bottles. After

about 8 days remove and discard the F1 flies

34 | P a g e 4. When the F2 adult flies emerge in the culture bottle, etherize and classify them for the various characteristics involved in this cross .Record data in Table 2.1

Table 2.1 Record of drosophila Experiment

Record both the phenotype and stock number of the parent flies.

Experiment number …………………………….. Name……………..

1. Cross …………………………… Female x ………………………

2. Date P1 s mated …………………………………………………..

3. Date P1 s removed…………………………………………………

4. Date F1 s first appeared ……………………………………………

5. Phenotype of F1 males …………………………………………….

6. Phenotype of F1 females …………………………………………..

7. Date of male and female placed in fresh bottle ……………………..

8. Date F1 flies removed …………………………………………

9. Date F2 progeny appeared …………………………………

10. Record F2 in the following table

35 | P a g e Male Female

Phenotype Number Phenotype Number Total a. …………. …………….. ……………. …………… ………… b...... c...... d...... e...... f...... g...... ……………. …………. ………. h. ………….. ……………. ……………. …………… …….

Totals ………………. ………….. ………..

4.-Indicate the genotype and phenotype of the flies in each of the three generations.

P1…………………………………………………………………………

36 | P a g e F1…………………………………………………………………………

F2………………………………………………………………………..

5-Calculate the expected number of individuals in each of the phenotypic categories and calculate the deviation (differences) between observed (o) and expected (E) numbers. Record this data in Table 2.2

Table 2.2 Summary of Data and Calculation for the second drosophila Experiment

Phenotypes %Ratio O E O-E O-E)2) O-E)2/E)

Totals ……=X2

How many degrees, of freedom do you have in the interruption of the ?X2value

…………………………………………………………………………………………………………………………………………… ..…………………………………………………………………………………………………………………………………………

Using Table 1.5 determine what X2values lie on either side of the X2 .2 ………………………………………………calculated in Table 1.3 record these values here

In this case, do you accept or reject the hypothesis that these data .4 ?approximate a 3:1 ratio

37 | P a g e ………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………… …………

Fourth Experiment: Dihydrid cross in maize

38 | P a g e You will be given an envelope of F2 maize kernels or an intact ear selected by your instructor from one of the alternative crosses listed in the instructor's manual.

1-Classify the kernels into four different phenotypes.

2-Select gene symbol to represent the two alleles of each gene studied in this example

Gene symbol Phenotype ………………………………………….. ……………………………………………………

…………………………………………. ……………………………………………………

………………………………………….. …………………………………………………… ………………………………………….. ……………………………………………………

A. Which characteristics are dependents on dominant alleles? …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………….. 3. Now, using the gene symbol you have chosen, give the possible genotype of the original homozygous parents (P1) of this F2 generation. …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… ………………………………………………… a. What are the phenotypes of these parents' individuals? …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… ………………………………………………… b. Assuming the parents to be homozygous, could more than one original cross have been used to produce this F2 generation? If so, what crosses could have been made? (Give genotype and phenotype) …………………………………………………………………………………………… …………………………………………………………………………………………… ……………………………………………………………………………………………

39 | P a g e …………………………………………………………………………………………… …………………………………………………

4. In any case, what must have been the genotype and the phenotype of the

F1 generation that subsequently was self-pollinated to produce the F2 kernels with you have been working? …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …….

5. Calculate the number of kernels expected in each phenotypic category and the deviation between observed and expected (O-E). Record this data in Table 2.3

Table 2.3 Summary of Data and Calculation for the second Maize Experiment

Phenotypes Ratio% O E O-E (O-E)2 (O-E)2/E

Totals X2=………

In this case, do you accept or reject the hypothesis that these data approximate a 9:3:3:1 ratio? …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… ……

40 | P a g e Chapter five: Gene interaction and epistasis

41 | P a g e Gene interaction in maize In this exercise, you will investigate the genes interaction in maize. You will

work with dihybrid crosses to determine the genotypes of the F 1 and F2 generations and the phenotypes Objectives of this investigation Upon completion of this lab, you should be able to:  Determine how some of these genes interact to determine endosperm color.  Calculate phenotypic ratios and carry out Chi-squared analyses for each cross.

Background Numerous gene loci scattered over the 10 pairs of chromosomes of maize determine endosperm color as well as other endosperm characterics . The interaction between two or more genes to control a single phenotype in which one allele of the first gene becomes (epistatic) (i.e. the gene or locus that suppressed or masked the action of a gene at the second locus) the gene or locus suppressed was called (hypostatic) The classical phenotype ratio of 9:3:3:1 observed in progeny of dihybride parents becomes modified by epistasis into ratio that are various combination of the 9:3:3:1 groupings s seen in the table below

Type of Genotype Types of epitasis Ratio epistasis aabb -aaB A-bb -A-B None 1 3 3 9 9:3:3:1

42 | P a g e Recessive 4 3 9 9:3:4 Dominant 1 3 12 12:3:1 Duplicate 7 9 9:7 recessive Duplicate 1 15 15:1 dominant Incomplete 1 6 9 9:6:1 duplicate dominant Dominant 3 13 13:3 and recessive

43 | P a g e Fifth Experiment: Gene interaction in maize

You will be given an ear of corn containing F2 kernels of two or more colors.

Count the number of F2 kernels of each color. From the F2 data and from the information provided about the P1 and F1 phenotypes, determine the mechanisms of inheritance that account for the data. Among the possible kinds of ears of corn your instructor may ask you to study the following characters

Phenotype of the true- Phenotype of the F1 Phenotype of the F2 breeding parents kernels kernels

44 | P a g e Purple x White purple Purple, red, white Yellow x Yellow yellow Yellow, purple White x white purple Purple, white Purple x yellow purple Purple, yellow ,whit ………………………… ……………………… …………………………… …

Your instructor will give you an ear of corn and tell you which of the four kinds it is. 1. Record the number of the ear here …………………………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………………………………

2. Record the phenotypes of the true-breeding parental generation here.

…………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… ………………………………………………… 3. Record the phenotype of the F1 kernels here.

…………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …

4. Count the F2 kernels on the ear and record their phenotypes, numbers, and ratios in the following spaces

Phenotype Number Ratio suggested by F2 data

45 | P a g e Total

5. How do genes interact to produce such epistasic ratio as 9:3:4? The following account explains the inheritance of mouse coat color: a true- breeding agouti (wild-type) mouse genotype CCAA, and a true breeding albino may have the genotype ccaa. If mated, two such animals would produce F1 offspring having the genotype CcAa and the agouti phenotype .In the F2 generation produced from crossing the F1 among themselves, one would expect 9 C - A- : 3C- aa : 3cc A- : 1ccaa. The 9/16 that are C-A- will be agouti, and the 1/16 that is ccaa will be albino, but what about 3/16 C-aa and the 3/16 ccA-? Thos having the ccA- genotype will also be albino because they are homozygous recessive cc, a genotype that prevents them from producing any pigment whatsoever .Those individuals constituting the 3/16 C-aa will be able to make pigment because they have the dominant C allele ,but the pigment that they produce will be black because they are homozygous recessive aa .In summary ,9/ 16 C-A- can make agouti pigment,3/16 C-aa make black pigment ,and 4/16 (=3/16ccA-plus 1/16 ccaa)can make no pigment at all and are thus albinos.

6. Now ,attempt to devise a hypothesis to explain the mechanism of inheritance of endosperm color for the ear of corn you have been studying .In doing so ,answer the following questions

a) What ratio do the F2 data approximate?

…......

46 | P a g e b) How many gene loci appear to determine the F2 phenotypes?

…………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… ……

c) Suggest gene symbols to be used in explaining the data .Indicate how each gene and its alleles are functioning to regulate endosperm color by completing the following table:

Gene symbol Gene effect

………………………. …………………………..

………………………. ……………..……………

………………………. …………….…………….

……………….……… …………………………..

d) Now, using the Parental gene symbols you have devised, write the following genotypes

Parental generation Phenotype Genotype

Parent no.1 ……………………… ………………….

Parent no .2 ……………………… …………………..

F1 generation Phenotype Genotype

47 | P a g e ……………………… …………………..

F2 generation Phenotype Genotype

………………………… …………………..

e) You have already suggested a ratio based on your F2 data. How closely do the data approximate the ratio you have hypothesized? To answer the

question, calculate the expected number (E) for each F2 phenotype based

on the ratio you have hypothesized and the total number of F2 kernels you have counted. Record your calculations in the following table.

F2Phenotypes Observed nos. (O) Expected nos.(E)

………………. ………………………. ……………………….

………………. ……………………….. ……………………….

………………. ……………………….. ………………………..

Totals ……………………….. ………………………….

If the expected numbers closely approximate the observed numbers, we conclude that the data to be satisfactorily explained by the proposed hypothesis.

48 | P a g e Phenotypes Ratio% O E O-E (O-E)2 (O-E)2/E

Totals X2=……….

Your comment

…………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… …………….

Chapter sex: sex linkage

49 | P a g e Sex chromosomes and Gene transmission

Cytology came to be better understood in the late nineteenth and early twentieth centuries, the role that chromosomes play in sex determination became more apparent .The presence of a particular nuclear structure (the x body) was detected in half the sperms of certain insects as early as 1891.

50 | P a g e Later (1902), C.E. McClung studied grasshopper species and demonstrated the difference between chromosome constitution of males and females: males – always have one less chromosome than females. Grasshopper sperm cells were shown to be of two types, those having darkly staining (heterochromatic) chromosome (the X body) and those not having such a chromosome. This heterochromatic chromosome came to be known as the X (sex-determining) chromosome.

Finally, it became clear that female grasshoppers have two X chromosomes per somatic cell (XX), whereas males have only one (XO). The sex of the offspring is determined by the kind of the sperm that fertilizes the ovum; X-bearing sperm cell produce female offspring, and the sperm lacking X chromosome result in male progeny. Subsequent research revealed that this XO system of sex determination is, in fact, less common than the XY method of sex determination. The latter process involves two heteromorphic, not completely homologous, sex chromosomes. Half the sperm cells produced per the X chromosome are female determining and half bear the Y and are male determining. Both Drosphila and humans have the XY method of determination. Research, however, has shown some fundamental differences in the basic genetics of sex determination in these two species despite their similar chromosomal basis for sex differences. In 1916, C.B Bridges reported that sex in Drosophila is determined by a balance between female-determining genes located on the X chromosome and male determining gene located on the various autosomes. In humans and perhaps in mammals generally, the male-determining genes are located on the Y chromosome.

In 1910, Thomas Hunt Morgan studied the white eye mutant gene in Drosophila in the first investigation to provide extensive experimental evidence for X-linked inheritance. Prior to that time, relationships between sex and gene transmission (e.g., the relationship noted since biblical times for hemophilia) had been observed, but no specific mechanism explaining the relationship had been set forth. Classical hemophilia, of course, is now understood to be due to an X-linked recessive gene. X-linked gene exhibit a

51 | P a g e crisscross pattern of inheritance in which the gene is transmitted by a male to all of his daughters and from them to approximately on-half of his grandsons.

Objectives of the investigation Open completion of this investigation, the student is able to:

 Diagram how an x-linked is transmitted from parents to F1 and F2 generations in an experimental mating of Drosophila.  Trace the transmission of an x-linked gene in a human pedigree.

…………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………… …………….

52 | P a g e Sixth Experiment: Sex linkage in Drosophila

For this part of the investigation you may use any one of many sex-linked mutants of Drosophila crosses so following discussion is based on the assumption that you will use the x-linked white eye (w) mutant.

Different members of the class will be asked to do reciprocal crosses so that the results can be compared .Review the procedures for making Drosophila mating (see investigation1)

A. Cross

Some of you will mate a white-eyed virgin female fly (Xw Xw ) to wild type male (X+Y).Produce F1 and F2 generations .Record data in Table 9.1 B. Reciprocal cross

Others will mate awild-type virgin female (X+X+) To white eyed male w (X Y).Produce F1 and F2 generations. Record data in Table 9.1.

Now, compare the F1 and F2 phenotypes and genotypes of the reciprocal crosses by completing Table 9.2.

Write a brief summary paragraph describing how the trait you have been studying (white eyes) would be inherited in reciprocal crosses if it were controlled not by a gene that is X-linked but rather by one that is autosomes .

…………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… …………………………………………………………………………………………… ……………………………………………………………………………………………

53 | P a g e …………………………………………………………………………………………… …………………………………………………………………………………………… ……………………………………………………….

Table 9.1 Records of Drosophila Experiment

Record data of cross involving a sex-linked gene

Experiment number…………………..Name……………………………………

1. Cross………………….…………..Female X……………………………Male.

2. Date P1s mated………………………………………………………………..

3. Date P1 s removed…………………………………………………………….

4. Date F1 s First appeared………………………………………………………

5. Phenotype of F1 female ………………………………………………………

6. Phenotype of F1 male………………………………………………………….

7. Date F1 male and female placed in fresh bottle ……………………………….

8. Date F1 flies removed …………………………………………………………

9. Date F2 progeny appeared…………………………………………………….

10. Record F2 data in the following table:

Males Females 54 | P a g e

Phenotype Number Phenotype Number Total a……………………….. …………………….. …………… ………. ……... b……………………….. ……………………… ……………. ………. ……… c……………………….. ……………………… …………… ………. ……… d……………………….. ……………………… …………. ………. ………

Totals …………………….. ……...... ………

Table 9.2 A comparison of the results of reciprocal Crosses involving an X- linked Gene Cross (A) Reciprocal Cross (B)

Phenotype Genotype Phenotype Genotype

P1 female …………. ……………...... …………... ……………..

P1male …………. ……………...... …………... ……………..

F1 female …………. ……………...... …………... ……………..

F1 male …………. ……………...... …………... ……………..

F2 female …………. ……………...... …………... ……………..

F2 male …………. ……………...... …………... ……………..

55 | P a g e Chapter seven: Pedigree Analysis

56 | P a g e Pedigree Analysis

Introduction

A pedigree is a diagram of family relationships that uses symbols to represent people and lines to represent genetic relationships. These diagrams make it easier to visualize relationships within families, particularly large extended families. Pedigrees are often used to determine the mode of inheritance (dominant, recessive, etc.) of genetic diseases. A sample pedigree is below.

In a pedigree, squares represent males and circles represent females. Horizontal lines connecting a male and female represent mating. Vertical lines extending downward from a couple represent their children. Subsequent generations are

57 | P a g e therefore written underneath the parental generations and the oldest individuals are found at the top of the pedigree. If the purpose of a pedigree is to analyze the pattern of inheritance of a particular trait, it is customary to shade in the symbol of all individuals that possess this trait. In the pedigree above, the grandparents had two children, a son and a daughter. The son had the trait in question. One of his four children also had the trait. In the exercises below, assume that the trait in question is a genetic disease or abnormality. We will learn patterns of inheritance that have the following modes of inheritance:

 Autosomal dominant.  Autosomal recessive.  X-linked recessive.

Developing Conclusions About Different Modes of Inheritance

Autosomal Dominant

1. The pedigree below is for a genetic disease or abnormality. We do not yet know if it is dominant or recessive. We will determine if it is possible that the trait is autosomal dominant. If the trait were dominant, we would use the following designations:

A = the trait (a genetic disease or abnormality, dominant) a = normal (recessive)

If the trait were recessive, we would use the following designations:

A = normal (dominant) a = the trait (a genetic disease or abnormality, recessive) a) Assume for the moment that the trait is dominant (we don't know yet). The pedigree shows that three of the individuals have the recessive (normal) phenotype and one individual has the dominant (abnormal) phenotype. Write the genotype of the affected (abnormal) individual next to her symbol in the pedigree below. If you only know one of the genes (letters), use a ?" for the unknown letter. Write the genotype of the three recessive individuals next to their symbols. As you write the genotypes, keep in mind that the pedigree may not be possible for a dominant trait.

58 | P a g e b) Is it possible that the pedigree above is for an autosomal dominant trait? …………………………………………………………………………………………………… ……………………………………… c) Write the genotypes next to the symbol for each person in the pedigree below assuming that it is for a dominant trait. …………………………………………………………………………………………………… ……………………………………………….

d) Is it possible that this pedigree is for an autosomal dominant trait? …………………………………………………………………………………………………… …………………………………………….. e) What can you conclude from these two examples about the parents of a child that has a dominant characteristic? ...... 2. We will determine if the pedigree below can be for a trait that is autosomal dominant. Use "A" and "a" as you did for the pedigrees above. a) Write the genotype of each individual next to the symbol. …………………………………………………………………………………………………… …………………………………………………..

b) Is it possible that this pedigree is for an autosomal dominant trait?

59 | P a g e …………………………………………………………………………………………………… ……………………………………………. c) In conclusion, can two individuals that have an autosomal dominant trait have unaffected children? ……………………………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………………………………

Autosomal Recessive

3. We will determine if the pedigree below can be for a trait that is autosomal recessive. Use the following designations:

A = normal a = the trait (a genetic disease or abnormality) a) Assuming that the trait is recessive, write the genotype of each individual next to the symbol. …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………….

b) Is it possible that the pedigree above is for an autosomal recessive trait? …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… c) Assuming that the pedigree below is for a recessive trait, write the genotype next to the symbol for each person. …………………………………………………………………………………………………......

60 | P a g e d) Is it possible that this pedigree is for an autosomal recessive trait? …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… e) If a trait is autosomal recessive, what can you conclude about the children if both parents are affected? ……………………………………………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………………………………………… 4. We will determine if the pedigree below can be for a trait that is autosomal recessive. Use "A" and "a" as you did for the previous example. a) Write the genotype of each individual next to the symbol. …………………………………………………………………………………………………… …………………………………………………………………………………………………… ……………………………………………………………………………………………………

b) Is it possible that this pedigree is for an autosomal recessive trait? …………………………………………………………………………………………………… …………………………………………………………………………………………………… ……………………………………………………………………………………………………. c) If a trait is autosomal recessive, what can you conclude about the children of two parents that are not affected? ……………………………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………………………………. . 5. We will determine if the pedigree below can be for a trait that is autosomal recessive. a) Write the genotype of each individual next to the symbol. …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………….

61 | P a g e b) Is it possible that this pedigree is for an autosomal recessive trait? …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… c) In this pedigree, two generations have been skipped. What can you conclude about recessive traits skipping generations? …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………..

X-Linked Recessive

The conclusions that you made for autosomal recessive traits apply to X-linked traits. In this exercise, we will work on some additional conclusions because males have only one X chromosome and females have two.

6. We will determine if the pedigrees below can be for a trait that is X-linked recessive. Use the following designations:

XA = normal Xa = the trait (a genetic disease or abnormality) Y = Y chromosome (males only) a) Write the genotype of each individual next to the symbol. …………………………………………………………………………………………………… …………………………………………………………………………………………………… ……………………………………………………………………………………………………..

62 | P a g e b) Is it possible that the pedigree above is for an X-linked recessive trait? …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… c) Write the genotype next to the symbol for each person in the pedigree below. …………………………………………………………………………………………………… …………………………………………………………………………………………………… ……………………………………………………………………………………………………..

d) Is it possible that this pedigree is for an X-linked recessive trait? …………………………………………………………………………………………………… …………………………………………………………………………………………………… ……………………………………………………………………………………………………. e) Write the genotype next to the symbol for each person in the pedigree below. …………………………………………………………………………………………………… …………………………………………………………………………………………………… ……………………………………………………………………………………………………

f) Is it possible that this pedigree is for an X-linked recessive trait? …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………... g) Write the genotype next to the symbol for each person in the pedigree below. …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………….

63 | P a g e h) Is it possible that this pedigree is for an X-linked recessive trait? …………………………………………………………………………………………………… …………………………………………………………………………………………………… ……………………………………………………………………………………………………. i) What can you conclude about the children of mothers affected with an X-linked recessive characteristic? …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………… …………………….. 7. We will determine if the pedigree below can be for a trait that is X-linked recessive. We will continue to use the designations "XA and Xa". a) Write the genotype of each individual next to the symbol. …………………………………………………………………………………………………… …………………………………………………………………………………………………… ……………………………………………………………………………………………………

b) Is it possible that this pedigree is for an X-linked recessive trait? …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… c) Which parent did the son get the Xa gene from? …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………….. d) What can you conclude about father-to-son transmission of X-linked traits? …………………………………………………………………………………………………… …………………………………………………………………………………………………… ………………………………………………………………………………………………….. 8. We will determine if the pedigree below can be for a trait that is X-linked recessive. a) Write the genotype of each individual next to the symbol.

64 | P a g e …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………...

b) Is it possible that this pedigree is for an X-linked recessive trait? …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………... c) What can you conclude about the children if both parents are affected with an X-linked recessive trait? …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………... d) How does this conclusion compare with the one you made earlier if about both parents being affected by an autosomal recessive trait? …………………………………………………………………………………………………… …………………………………………………………………………………………………… …………………………………………………………………………………………………… e) Do the conclusions that you made for autosomal recessive traits apply to X- linked recessive traits? …………………………………………………………………………………………………… …………………………………………………………………………………………………… ……………………………………………………………………………………………………

65 | P a g e Glossary

A dihybrid cross -- is a breeding experiment between P generation (parental generation) organisms that differ in two traits. A monohybrid cross -- is a breeding experiment between P generation (parental generation) organisms that differ in one trait. An allele -- is an alternative form of a gene (one member of a pair) that is located at a specific position on a specific chromosome. Autosome -- a nuclear chromosome other than the X- and Y-chromosomes. Centromere -- a region of a chromosome to which spindle traction fibers attach during mitosis and meiosis; the position of the centromere determines whether the chromosome is considered an acrocentric, metacentric or telomeric chromosome. Chromosome -- in the eukaryotic nucleus, one of the threadlike structures consisting of chromatin and carry genetic information arranged in a linear sequence. Dominance -- An allele or corresponding phenotypic trait that is expressed in heterozygotes .

66 | P a g e Dominant -- alleles that determine the phenotype displayed in a heterozygote with another (recessive) allele. Gene -- The fundamental physical and functional unit of heredity, which carries information from one generation to the next; a segment of DNA, composed of a transcribed region and a regulatory sequence that makes possible transcription. Genotype -- genetic constitution of an organism. Hemizyg -- A male organism in which a mutant trait is present at the X chromosome (i.e. Drosophila sp). Heterozygote -- having two alleles that are different for a given gene. Homologous chromosomes -- chromosomes that pair during meiosis; each homologue is a duplicate of one chromosome from each parent. Homozygote -- having identical alleles at one or more loci in homologous chromosome segments. Hybrid -- An offspring resulting from mating between individuals of different genetic constitution. Incomplete dominance -- is a form of intermediate inheritance in which one allele for a specific trait is not completely dominant over the other allele. Linkage -- analysis of pedigree the tracking of a gene through a family by following the inheritance of a (closely associated) gene or trait and a DNA marker. Linkage -- the greater association in inheritance of two or more nonallelic genes than is to be expected from independent assortment; genes are linked because they reside on the same chromosome. Locus --The particular physical location on the chromosome of which the gene for a given trait occurs. Pedigree -- a diagram of the heredity of a particular trait through many generations of a family. Phenotype -- observable characteristics of an organism produced by the organism's genotype interacting with the environment. Recessive -- a gene that is phenotypically manifest in the homozygous state but is masked in the presence of a dominant allele. X-Chromosome -- The sex chromosome found in two doses in female mammals and many other species.

67 | P a g e Y-Chromosome -- The sex chromosome found in a single dose in male mammals and many other species.

68 | P a g e