Guide to the practical teaching
of biology and genetics

Tasks for Genetics - list

Cytogenetics - study of chromosomes, karyotypes

TASK 1: Structure of polytene (giant) chromosome

PP: polytene (giant) chromosome - "obrovské chromosomy", NP: salivary gland of chironomid larva (Chironomus sp.) or Drosophila melanogaster

Prepare salivary glands of the chironomid larva and control the presence of the giant chromosomes under a microscope. Pass the slide (sample-side up) through the flame of the burner 3 to 4 times to heat-fix the sample. Stain it with acetorcein for 5-10 min. Cover with a cover glass and prepare the compression sample. Observe polytene chromosomes with chromomeres (stripes of different width) and puffs.

Fig.: Polytene chromosome in cells of salivary gland in chironomid larva.

TASK 2: Sex chromatin in somatic cells

NP and PP: buccal smear - "pohlavní chromatin"

Take a sample from the inner side of the cheek using a cotton swab and impress the swab on a slide. Fix it above the flame and apply acetorcein on the slide for 5 min. and then cover with a cover glass. In somatic cells of female buccal mucosa, you can find sex chromatin (Barr body), a densely colored body in the periphery of the nucleus.

Fig.: Sex chromatin.

TASK 3: Mammal karyotype

PP: karyotype of rabbit, pig, sheep, horse, cattle and human - "karyotyp králík, prase, ovce, kůň, skot, člověk".

Observe permanent samples prepared from lymphocytes of different domestic animals or human (Note: In the samples, you can find round nucleus of lymphocytes and clusters of chromosomes = karyotype). Compare the numbers and types of chromosomes (rabbit 2n = 44, pig 2n = 38, sheep 2n = 54, horse 2n = 64 and human 2n = 46).

TASK 4: Karyotype of onion

NP: the rootlet of onion (Allium cepa) cultivated with 0.02 % water solution of colchicin and stained with acetorcein.

Prepare the compression sample from a stained onion rootlet and find the cluster of chromosomes (karyotype).

Fig.: Examples of karyotype: A – in cattle (Bos taurus), B – in onion (Allium cepa).

TASK 5: Karyotype records

Write following karyotypes:

1. karyotype of healthy man and woman
2. karyotype of bull (male cattle), mare (female horse), ram (male sheep)
3. karyotype of man with Down syndrome and man with Klinefelter syndrome
4. karyotype of woman with Edwards syndrome and woman with Turner syndrome

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Model organism

TASK 1: Genetic experiment

Genetic experiment - crossing of red-eyed drosophila (original wild form) with mutant wite-eyed drosophila. Experiment will proceed during three lessons:

The beginning of experiment (1^st lesson)
You will be divided into groups. Each group will receive an Erlenmeyer flask with a fresh medium (maze grit, dry yeast, sugar and agar) and two tubes with drosophila of different sexes (male, female) and different eye color (red, white). Observe them and then carefully place females and males with different eye colors together into the Erlenmeyer flask. Mark the flask with the date, sex and eye colors of crossed drosophila and identification of your group.

Removing of adult drosophila (2^nd lesson)
Drop ether on the cotton placed at the cover of the bottle and close it. Wait a few minutes and subsequently remove the adult fruit flies from the Erlenmeyer flask into the bottle with ether to narcotize them.

Evaluation of experiment (3^rd lesson)
Use ether to narcotize fruit flies (offspring of crossing) and sort them according to sex and eye color; deduce how the observed trait (eye color) is inherited.

TASK 2: Mutant strains

Observe different mutant strains of drosophila (white, yellow, ebony, curly and vestigial) - live, in alcohol or on the picture.

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Monohybridism

TASK 1:

In snapdragon plants (Antirrhinum maius), dominant homozygote has red flowers, recessive homozygote has white flowers and heterozygote has pink flowers.
1. Complete genotypes in P (parental), F₁ (first filial) and F₂ (second filial) generations. What is the phenotype and genotype ratio of F₁ and F₂ generations?
2. What type of inheritance does the observed trait show? (complete or incomplete dominance)

TASK 2:

Use χ² test to compare obtained phenotype ratio in F₂ generation (79 red, 170 pink and 95 white snapdragon plants) with expected ratio (see task 1). Table values are here.

TASK 3:

Color of pumpkin can be white or yellow. Dominant allele W determines white color, while recessive allele w determines yellow color.
1. After crossing of two white pumpkins, 3/4 of offspring were white and 1/4 of offspring were yellow. What are the genotypes of parents and offspring?
2. White phenotype of pumpkin can be encoded by two different genotypes. Which are they? How can you find the genotype of one particular white pumpkin?

TASK 4:

White pumpkin crossed with yellow one resulted in 327 white and 361 yellow offspring (see task 2).
1. What are the genotypes of parents and offspring?
2. How is this type of crossing called?
3. Compare obtained and expected ratios using χ² test (table values are here).

TASK 5:

Blue and brown color of eyes in humans is determined by two different alleles of one gene. In 337 families, the following data were found:

What color of eyes is dominant? Complete genotypes of parents and children.

TASK 6:

Two black female mice were crossed with two brown males. One female had 9 black and 7 brown offspring in several litters; the second female had 57 black offspring in several litters.
1. How is the coat color inherited in mice? What color is dominant and why? Explain.
2. What are the genotypes of parents and offspring from the aforementioned crossings?

TASK 7:

Leaves of cucumber can be palmate or flabellate (called ginkgo). After crossing of two plants (homozygote with palmate leaves and homozygote with flabellate leaves), all offspring were palmate.
1. What type of leaves is dominant (palmate or flabellate)?
2. What are genotype and phenotype ratios in F₂ and B₁ (back crossing = crossing of recessive homozygote and heterozygote) generations?

TASK 8:

In Andalusian strain of chicken, dominant allele B determines dark color of feather, while recessive allele b determines white color of feather. Heterozygote has bluish color of feather.

What will be the color of feather of offspring after crossing of bluish female with: a) dark colored fowl, b) bluish colored fowl, c) white fowl?

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Dihybridism, polyhybridism and branching method

TASK 1:

Complete Punnett square for dihybridism and deduce phenotype ratio for offspring of two guinea-pigs with the same genotype RrBb (R - rough coat, r - smooth coat, B - black coat, b - white coat). There is the relation of complete dominance between the two alleles of both genes.

What offspring results from the back crossing?

TASK 2:

Use the branching method for setting of genotypes of gametes produced by individuals with the following genotypes:
1. RrssTtUU
2. AaBBCcddEe
3. KkllmmNnOOppQq

TASK 3:

Use the branching method for setting of genotype and phenotype ratios in offspring of parents with these genotypes: RrssTtUU x RrSsTTuu (between alleles of all genes is complete dominance).

TASK 4:

Use the method of branching for setting of genotype and phenotype ratio in offspring of parents with these genotypes: AaBBCcddEe x aaBBCcDdee (There is the relation of incomplete dominance between the two alleles of all genes).

TASK 5:

Black rough and white rough guinea-pigs had 32 black rough, 33 white rough, 12 black smooth and 9 white smooth offspring (see task 1).
1. What were genotypes of both parents?
2. Use the branching method for determination of genotype frequencies of offspring.
3. Use χ² test for determination of significance of variation between obtained (empiric) and expected (theoretic) phenotype ratios (table values are here).

TASK 6:

In pumpkins, white color (W) is dominant over yellow (w) and disk shape (D) is dominant over round shape (d). What are the genotypes of parents and offspring in the following crossings?
1. White disk x yellow round (offspring: 1/2 white disk and 1/2 white round).
2. White disk x yellow round (offspring: 1/4 white disk, 1/4 white round, 1/4 yellow disk and 1/4 yellow round).
3. White disk x white disk (offspring: 28 white disk, 9 white round, 10 yellow disk and 3 yellow round).

TASK 7:

In hen, feathery legs (F) are dominant over featherless legs (f) and pea comb (P) is dominant over simple comb (p).

Two cocks (A and B) were mated with two hens (C a D). All of them had feathery legs and pea combs. Cock A mated with hens C and D had all offspring with feathery legs and pea combs. Cock B mated with hen C had offspring with feathery or featherless legs and only pea combs, while mated with hen D offspring had only feathery legs and pea or simple combs. What are the genotypes of parents (cocks and hens) and their offspring?

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Polymorphic genes

TASK 1: Genetic determination of blood groups

Diagnostic kit for determination the ABO blood group system (monoclonal)

• monoclonal Anti-A
• monoclonal Anti-B
• diagnostic cards
• stick for mixing reagents with blood samples

Fig.: Diagnostic kit for determination the ABO blood group system (monoclonal).

Principle: agglutination technique, which is based on antigen / antibody reaction

Procedure:

1. Place one drop of the monoclonal antiserum Anti-A within each of the blue circles
2. Place one drop of the monoclonal antiserum Anti-B within each of the yellow circles
3. Place one drop of blood within the each of the red circle (press on the bottom of the fingertip with the thumb of the same hand and quickly prick the fingertip with the help of a needle)
4. Use a stick to mix the blood with monoclonal antiserum. Mix each sample with new edge of stick.

Evaluation:

• The results are visible within 1 minute after mixing the diagnostic card with gentle rocking motion.
• Positive reaction (agglutination) indicates the presence of the corresponding antigen on the red blood cells.
• The negative reaction (no visible agglutination) indicates the absence of the corresponding antigen on the red blood cells.
• The corresponding blood group is determined by the following table.

A - has on red blood cells agglutinogen A and produces anti-B antibodies

B - has on red blood cells agglutinogen B and produces anti-A antibodies

AB - has on red blood cells agglutinogens A and B and produces no antibodies

0 - have no agglutinogens and produces antibodies against both agglutinogens

The representation of blood groups in Czech Republic:

A = 45%

0 = 30-35%

B = 15-20%

AB = 5-7%

TASK 2:

In rabbits, there is allele set with dominance in this order: colored fur C, Himalayan albinism c^h, and albinism c^a.
1. What color of fur can be expected in offspring after crossing of two homozygotes one with colored fur and the second with Himalayan albinism?
2. What are the genotypes of rabbits with colored fur and Himalayan albinism, if they had these offspring: 1/2 colored, 1/4 Himalayan albinism and 1/4 albinism?

Genetics of blood groups - examples

Humans:

1. What blood groups are expected in offspring of parents with genotypes I^Ai and I^Bi ?
2. What are the genotypes of parents, if father has blood group AB, mother B and their children 1/4 A, 1/4 AB and 1/2 B.
3. Which of two men can be excluded as a father? Mother has blood group B, children 0, first man has A and second has AB.
4. Parents are heterozygotes in blood groups B. What is the probability that their firstborn son will inherit the group B? And what is the probability, if the firstborn child is daughter?
5. Parents are heterozygotes in blood groups A. What is the probability that their two children will inherit the group A?
6. Mother has blood group B and her child A. Mother said, that father of child is man with blood group AB. Is this possible?
7. During one night in one hospital four children with blood groups A, B, AB and 0 were born. Due to mistake of nurse we do not know which child was born to which mother. Therefore the blood groups of all four pairs of parents were examined. First parents have blood groups B x B, second have 0 x AB, third parents have groups A x B and fourth pair have groups 0 x 0. Is it possible to determine which parents have which children? If yes, which children belong to first parents, which to second parents, to third and fourth parents?
8. What are the blood groups of men, which can't be fathers if we know blood groups of mother and child: a) mother 0, child A, b) mother 0, child 0, c) mother A, child B, d) mother AB, child B

Animals:

1. A cat with blood group B has been mating with the tomcat with an unknown blood group. Four kittens have blood group A and three blood group B. What are the genotypes of parents and kittens?
2. A cat with blood group A has been mating with the tomcat (male) with blood group B. What blood groups can have their kittens?
3. A cat with blood group AB had gradually kittens with two tomcats. All kittens from the first cast had blood group A, kittens from the second cast had blood groups A, AB or B. What blood groups had both tomcats and what are genotypes of parents and offspring?

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Gene interactions

TASK 1:

Purple color of vetch (wild pea) is caused by the presence of dominant alleles of both genes (C-P-). Others combination of genotypes encodes for white color of flowers.
1. What is the phenotype ratio in F2 generation? Use Punnett square. What type of gene interaction is it?
2. What color of flower can be expected in offspring resulting from the following crossings: 1) CcPp x CcPP     2) Ccpp x ccPp?

TASK 2:

In pumpkin, the shape is determined by two genes: A and B. The pumpkin is round if dominant allele of gene A or of gene B is present in the genotype, it is discoid if dominant alleles of both genes are present, and it is elongated in individuals who are recessive homozygotes in both genes.
1. What is the phenotype ratio in F₂ generation (crossing of two dihybrids)? Use combination square. What type of gene interaction is it?
2. The crossing of the discoid pumpkin with the round pumpkin resulted in 3/8 discoid, 1/2 round and 1/8 elongated offspring. What were the genotypes of parents and of offspring?

TASK 3:

The color of a feather of a canary is determined by genes A and B. Dominant allele of gene A encodes for red color, dominant allele of gene B for yellow one. Birds with genotypes aabb and A-B- are white.
1. What is the phenotype ratio in F2 generation? What type of gene interaction is it?
2. Third gene (C) determines, if the feather is smooth or fuzzed. Birds with dominant allele C are smooth, recessive genotype determines fuzzed feather. Use branching method to find phenotype ratio in offspring resulting from crossing: AaBbCC x AabbCc.

TASK 4:

In a pumpkin, the orange color is determined by a dominant allele W, white color by dominant allele Y. Plants with genotypes W-Y- and W-yy are orange, wwY- white and wwyy green.
1. What is the phenotype ratio in F2 generation? Use Punnett square. What type of gene interaction is it?
2. What will be the color and phenotype ratio of the offspring after crossing of parents with genotypes: WwYy and Wwyy?

TASK 5:

The color of a feather of a budgerigar (Melopsittacus undulatus) is determined by the interaction of two genes: F and O. Allele F encodes for yellow color (genotype F-oo) and allele O encodes for blue color (genotype ffO-). If both alleles are present, budgerigar is green (genotype F-O-). Recessive homozygote in both genes is white (genotype ffoo).
1. What is the phenotype ratio in F2 generation (crossing of two dihybrids)? Use combination square. What type of gene interaction is it?
2. What will be the feather color and phenotype ratios in the offspring after crossing of parents with genotypes: 1) FFOo x ffOo     2) FfOO x Ffoo?
3. The crossing of yellow and blue budgerigars resulted in 6 yellow and 5 green offspring. What are the genotypes of parents and offspring?
4. Green female budgerigar had white offspring. What was the genotype of female and offspring?

TASK 6:

In mice, the presence of dominant allele of gene C is important for production of dark pigment melanin. Dominant allele of gene A causes change of dark pigment into yellow.
1. What is the phenotype ratio in F2 generation? Use Punnett square. What type of gene interaction is it?
2. What offspring will result from crossing of black mouse (CCaa) with white mouse (ccAA)?

TASK 7:

In chicken, dominant allele of gene A determines colored feather, dominant allele of gene I suppresses effect of gene A, but has no effect on the phenotype.
1. What is the phenotype ratio in F2 generation (crossing of two dihybrids)? Use Punnett square. What type of gene interaction is it?
2. What feather can be expected in offspring resulting from crossing: AaIi x Aaii?

TASK 8:

Scaliness in carp (Cyprinus carpio) is determined by genes S and N. Between these genes, there is reciprocal interaction. The following phenotypes can exist: row (S-Nn), scaly (S-nn), smooth (ssNn), bare (ssnn). The presence of two dominant alleles N in the genotype is lethal.
1. What are the phenotypes in offspring resulting from crossing of two row carps (SsNn x SsNn)?
2. Use the branching method to find phenotype ratio in offspring resulting from crossing of row and smooth carps.

TASK 9:

In humans, the color of hair is determined by the interaction of six genes. Gene A encodes for pigment formation and is recessive epistatic against other genes (it means that man with genotype aa is albino). Gene B encodes for formation of brown pigment and is dominant epistatic against R gene (it means that man with genotype bb has fair hair). Gene R encodes for formation of red pigment (recessive allele r is inactive). Dominant alleles D, F, V influence intensity of hair color. In all six genes, there is complete dominance between alleles of individual genes.

Use the branching method to find phenotypes of offspring, if their parents had black hair and genotypes: AaBBRrDDFfVv and AaBBRrDDFfVv

TASK 10:

The coat color of some rodents is determined by the interactions of three genes: A, B and C. Dominant allele of gene C causes albinism and is recessive epistatic against genes A and B. Between genes A and B, there is reciprocal interaction. Dominant allele A determines gray color, recessive allele determines black color. Dominant allele B determines yellow pigmentation of hair ends ("wild color"); recessive allele b has no effect on the phenotype.

Use the branching method to find phenotypes of offspring, if their parents had genotypes: AabbCc x AaBbcc.

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Inheritance and sex

TASK 1:

What are phenotype ratios in F₁ generation of fruit flies when a red-eyed female is crossed with a white-eyed male and on other side, a white-eyed female is crossed with a red-eyed male?

TASK 2:

In cats, dominant allele of gene Y (in homozygote form in females and hemizygote form in males) determines black color of fur. Recessive allele determines yellow color and heterozygote females have tortoise-colored fur.
1. What type of inheritance is it? On which chromosome is gene for fur color located?
2. Black female has one tortoise-colored and four black offspring. What genotype and color of fur does their father have? What sex and color does the black offspring have?
3. Tortoise-colored female was mated with yellow male. What is probability that they will have yellow male and female offspring?

TASK 3:

Daltonism (color blindness) is disease with recessive X-linked inheritance. A couple has daughter with daltonism, but her mother is able to distinguish colors.

What genotypes do parents and their daughter have?

TASK 4:

Haemophilia is disease with a recessive gonosomal inheritance linked to chromosome X. A man with haemophilia has daughter with healthy woman with homozygote genotype.

What genotype does their daughter have?

TASK 5:

The color of Ayrshire cattle is determined by gene M. Cattle and bull with genotype MM are mahogany, cattle with recessive homozygote genotype is red spotted. The bull with genotype Mm is mahogany, while the cattle is red spotted.
1. What type of inheritance is it?
2. The red spotted cattle (her father was mahogany) was mated with the red spotted bull. What are genotypes and phenotypes of parents and their offspring?
3. What is the phenotype ratio in F2 generation?
4. The mahogany cattle gave birth to red spotted offspring. What is the sex of offspring?

TASK 6:

In humans, there is gene P for baldness (hairlessness). Men with genotype PP and Pp are bald, while women are bald only with genotype PP. Gene B determines color of eyes; brown one is dominant over blue one.
1. What type of inheritance does hairlessness show?
2. The hairless man with brown eyes (his father has hair and blue eyes) married a blond woman with blue eyes (her father and brothers were bald). What color of eyes can they expect in their children? Will they have hair or will they be bald?

TASK 7:

In Siamese fighting fish (Betta splendens), dominant allele of gene Z induces enlargement of fins, if there are male sex hormones presents.
1. What type of inheritance is it?
2. What type of fins can we expect in offspring resulted from crossing: Zz x Zz.
3. How can you identify genotype of male with enlarged fins and genotype of female, if you have recessive homozygotes for crossing?

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Genetic linkage

TASK 1:

What genotypes can we expect in gametes, if there is incomplete linkage of genes A and B in gametogony with genotype AaBb?

TASK 2:

What is the phenotype ratio in B1 generation resulting from crossing:

AB	x	ab
ab		ab

Genetic linkage p(AB) = 16.6 cM and number of offspring n = 1152.

TASK 3:

In hens, feathery legs are dominant over featherless (gene A), pea comb over simple (gene B) and white color over dark (gene C). What is the order of genes ABC and power of linkage between neighbouring genes based on genetic analysis of offspring (n = 2500) resulted from test crossing: AaBbCc x aabbcc. Draw chromosome map.

Test crossing:

ABC	x	abc
abc		abc

Phenotypes	ABC abc	ABc abC	Abc aBC	AbC aBc
Frequency in offspring (%)	80.9	3.9	14.6	0.6

TASK 4:

What is the order of the four genes K, L, N and S and power of linkage (distance) between neighbouring genes located on 5. chromosome of tomato? (K-red color, k-yellow; L-rounded shape, l-sharp; N-pilous stalk, n-bare; S-straight leaf, s-curly). Phenotype frequencies (%) of offspring resulted from test crossing are in tables. Use three-point cross and draw chromosome map of all four genes.

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Population genetics

TASK 1:

Gene I (AB0 blood system) exists in three forms - I^A, I^B and i. What are the genotype and phenotype frequencies in panmictic population, if frequency of allele I^B (q) = 0.4 and frequency of allele i (r) = 0.4?

TASK 2:

In humans, three types of earlobes can be found. In a group of 100 students, free earlobe (hanging free from the head) was found in 38 students (genotype dominant homozygotes), middle attached in 45 students (heterozygotes) and attached (joined to the head) in 17 students (recessive homozygotes).
1. What are the frequencies of dominant and recessive alleles?
2. Use χ² test to verify, if population is in HW equilibrium (table values are here).

TASK 3:

Population is in HW equilibrium. Frequency of allele (b) for blue color of eyes is 0.6. What is the frequency of blue-eyed people in the population? How many % of brown-eyed people with homozygote and heterozygote genotypes are in this population?

TASK 4:

In population, there is 36 % blue-eyed people (bb) and 64% brown-eyed people (BB, Bb). How many % of brown-eyed homozygotes and brown-eyed heterozygotes are there?

TASK 5:

In humans, the presence of RhD antigen (named after rhesus monkey), determines the blood type. If the antiserum against this antigen agglutinates your red blood cells you are Rh⁺ (Rh positive) (genotype DD or Dd), if it doesn't you are Rh^- (Rh negative) (genotype dd). In the Central European Caucasian population, approximately 84 % of people are Rh⁺ and 16 % Rh^-. What is the frequency of the dominant allele?

TASK 6:

Frequency of recessive allele (a) for myopy (short sightedness) in a population is q (a) = 0.14. What are the frequencies of short-sighted people (genotype aa) and of healthy carriers (genotype Aa) in this population?

TASK 7:

Albinism is recessively inherited disorder (inability to synthesize pigment melanin). In the population, there is one albino (aa) in 10 000 people (the frequency of genotype aa is 0.0001).
1. What is the allelic frequency of recessive allele a?
2. How many % of carriers (Aa) are present in this population?

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Quantitative genetics

TASK 1:

Evaluate the variability of weight in two groups of mice and compare the results. Use the arithmetic average and variance.

First group: 15.5 g, 10.3 g, 11.7 g, 17.9 g, 14.1 g

Second group: 20.2 g, 21.2 g, 20.4 g, 22.0 g, 19.7 g

TASK 2:

In eight ducks, width of head and length of wing was measured (table).

1. Count the arithmetic average and standard deviation for these two traits (width of head and length of wing).
2. Use the correlation coefficient to find out what correlation is between the two traits.

TASK 3:

The height of parents (average of father and mother) and their children was measured (table).

1. Count the arithmetic average and the variance for the height of children and parents.
2. Use the correlation coefficient to find out what correlation is between the height of parents and children.
3. Count the narrow-sense heritability for the height (use regression coefficient).

TASK 4:

Heritability of height in wheat h²_N = 0.7. There is a trend to reduce the height of wheat by selection. Will this goal be reached promptly or slowly?

TASK 5:

The height in humans has heritability h²_N = 0.9. How much is the phenotype influenced by the environmental conditions?

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Non-Mendelian inheritance

TASK 1:

A man affected with mitochondrial optic neuropathy married a healthy woman. What is the probability that their child will be affected with this disease?

TASK 2:

Recessive mutation in chloroplast DNA causes variegation of leaves in Mirabilis jalapa. What plants can be expected in the offspring of variegated maternal plant and green paternal plant?

TASK 3:

What will be the leaf color of the offspring resulting from pollination of green plant with the pollen of variegated plant?

TASK 4:

What will be genotypes and phenotypes in F₁, F₂ and F₃ generations, when P generation is represented by female snail with left-handed shell and male with right-handed shell.

TASK 5:

What was the genotype and phenotype of parents of an offspring with left-handed shell? What was the phenotype of this offspring?

TASK 6:

The pedigree bellow shows the inheritance of Kearns-Sayre syndrome, a hereditary genetic disease caused by mtDNA mutation. It is myopathy characterized by isolated involvement of the muscles controlling eye-lid and eye movement and by progressive ophthalmoplegia. Mark the affected individuals in the pedigree with black color.