BIOL 153L General Biology II Lab Black Hills State University Lab 5: Genetics I This lab will be divided into two parts: part one will review mitosis and meiosis, while part two will focus on Mendelian genetics (monohybrid and dihybrid crosses). See textbook (Chapters 3 & 8) for more information about genetic concepts and processes! Note that this exercise will focus on human genetics and traits, which are more familiar and intuitive to many students. Basic genetic mechanisms discussed today are shared widely across eukaryote lineages. Nonetheless, important distinctions between life cycles and reproductive biology of plants and animals will be described in future labs. Before coming to lab this week, please watch the following videos about mitosis and meiosis. Mitosis: splitting up is complicated" : https://www.youtube.com/watch?v=l0k-enzoeom Meiosis: where the sex starts : https://www.youtube.com/watch?v=qclmr9-yy7o PART 1: Mitosis and meiosis (for diploid organisms) This exercise will use pipe cleaners as simple physical models of chromosomes. Let's start by reviewing important concepts and terminology used in this lab! Allele: alternate form of a gene. Homologous chromosomes: chromosomes that have the same genes in the same location however, the alleles may be different; one homologous chromosome comes from the mother and one from the father. Centromere: connects the sister chromatids during cell division. Sister chromatids: two identical copies of a chromosome that are held together by a centromere; once they separate during anaphase (mitosis) or anaphase II (meiosis) they become two separate daughter chromosomes. Synapsis: pairing of two homologous chromosomes that occurs during prophase I. Tetrad: two homologous chromosomes each of which is composed of two sister chromatids come together in prophase I; this is where crossing over occurs. (Called a tetrad because there are four chromatids.) Chiasma: two homologous, non-sister chromosomes cross over and exchange genetic material; looks like an X. Somatic: body cells do not make gametes. Gametic: reproductive cells cells that make gametes (i.e. eggs & sperm). Having reviewed the concepts above, obtain a bag of pipe cleaners and beads from your instructor. Each bag will contain the following (adjust beads or pipe cleaners if necessary to match description): 2 white long chromosomes with blue beads; 2 yellow long chromosomes with brown beads; 2 green short chromosomes with red beads; 2 blue short chromosomes with orange beads; 4 paperclips and 1 string piece. 1
Note there are other items in the bag provided (two pennies as well as ivory, yellow, blue and brown beads); these may be left in the bag until later in today's lab! Long chromosomes carry the gene for eye color (blue and brown beads represent alleles for blue and brown eyes). Long chromosomes are white and yellow (from dad), and are homologous. Short chromosomes carry the gene for cystic fibrosis (red and orange beads represent CF and non- CF alleles). Short chromosomes are green and blue (from dad), and are homologous. The paperclips represent centromeres. First exercise: pipe cleaner mitosis Here you will use pipe cleaner chromosomes to model mitosis. You should first practice with your partner until you are confident in the steps it's trickier than it looks on paper! Thereafter, you will demonstrate both mitosis and meiosis (see below) to your instructor, who will initial your handout when you have satisfactorily explained the processes. Mitosis and meiosis will be important for several labs later the semester (Chapter 3, pp. 62-71; Fig. 3-40) and in the BIOL 153 lecture. Step 1. Use string to make a circle; this represents a cell. Step 2: Place the following in the cell: 1 yellow chromosome, 1 white chromosome, 1 green chromosome, and 1 blue chromosome. This represents a diploid cell with two different types of chromosomes (i.e., 2n = 4). Step 3: Walk through the below-described stages of mitosis using the pipe cleaner models. Interphase: DNA is copied. To show this, add all four pipe cleaners to the cell, and connect identical chromosomes with paperclip (e.g. 2 yellow connected, 2 white connected, etc.). (Pretend the contents of the cell are blurry at this point because the chromosomes have not yet condensed.) Prophase: chromosomes condense. At this point, each chromosome is visible. Note that although the DNA in the cell has doubled, it is not a polyploid! Each chromosome is composed of two sister chromatids connected by the centromere. Metaphase: chromosomes align at the center of the cell. Line up all four chromosomes (with their respective chromatids) in the center of the cell. Anaphase: sister chromatids separate and become two daughter chromosomes that are pulled to opposite sides of cell. Separate the chromatids at the paperclip centromere and pull to opposite sides of cell. Do this for all four chromosomes. Telophase: chromosomes begin to uncondense, and cytokinenis begins. Move the string to divide the cell into two daughter cells. (In a plant the string represents the cell plate; in an animal it represents the cell membrane.) Demonstrate your knowledge of the above steps to an instructor and then proceed to answer the questions about mitosis on the next page. 2
1. During the process of mitosis, one cell becomes cells 2. After mitosis, each daughter cell is genetically identical to or different from the original cell? (Circle correct answer.) 3. The chromosome number for each daughter cell is 2n = 4. In plants, mitosis occurs most often in meristem. Is meristem somatic or gametic tissue? (Circle correct answer.) Second exercise: pipe cleaner meiosis Now you will use pipe cleaners to simulate meiosis (rather than mitosis)! Note that reference to Chapter 8, pp. 155-159; Fig. 8-10; and Summary Table p. 172 may be helpful here. Meiosis I: reductional division; separation of homologous chromosomes. Place the following in the cell: 1 yellow chromosome, 1 white chromosome, 1 green chromosome, 1 blue chromosome. This represents a diploid cell with two different types of chromosomes (i.e., 2n = 4). Interphase: DNA is copied. To show this, add four replicate pipe cleaners to the cell, and connect identical chromosomes with a paperclip (e.g. 2 yellow connected, 2 white connected, etc.). (Pretend the contents of the cell are blurry at this point because the chromosomes have not yet condensed.) Prophase I: chromosomes condense; homologous chromosomes pair up (synapsis); chiasma form; crossover occurs. Pair the homologous chromosomes, and make an X with non-sister chromatids to model crossover. (Note: please do not twist the pipe cleaners together, just ponder what the result of crossover would look like in the model chromosomes.) Metaphase I: homologous chromosomes line up side by side along in center of cell. Put homologous chromosomes side-by-side one on either side of metaphase plate, with the non-sister chromatids still enmeshed in crossover. Anaphase I: homologous chromosomes separate and move to opposite sides of cell. Move homologous chromosomes to opposite sides. Note, do not separate at centromeres the sister chromatids should remain attached. Telophase I: Cytokinesis. This results in two daughter cells, each with half the genetic material of the parent. However, each chromosome still consists of two chromatids. Use string to divide cell. Interkinesis (a rest period) may occur; no DNA replication takes place. Meiosis II: equational division and separation of sister chromatids. These steps are similar to mitosis (see above) but only involve one set of chromosomes in each daughter cell. Prophase II: chromatids recondense. (Chromosomes may have "loosened" during telophase I and interkinesis, and so re-condense at this stage.) 3
Metaphase II: chromosomes line up at center of cell. Line up the two chromosomes (with their respective chromatids) in the center of each daughter cell. Anaphase II: sister chromatids separate and move to opposite poles; they become separate chromosomes as soon as they move apart. Remove paperclip connecting sister chromatids of the two chromosomes and then move one of each to opposite sides of cell. Telophase II: chromosomes de-condense, cell wall or membrane reforms. Use string to divide the two cells into four cells. (Note that in females, it is common that only one of these daughter cells will develop into an actual gamete.) Demonstrate your knowledge of the above steps to an instructor and then proceed to answer the following questions about meiosis. 1. During the process of meiosis, one cell becomes cells 2. After meiosis, daughter cells are genetically identical to or potentially different from the original cell? (Circle correct answer.) 3. At the start of meiosis, how many total chromosomes were in this cell? 4. At the end of meiosis, how many total chromosomes were in each cell? 5. The somatic (body cells) chromosome number in this example: 2n = 6. The gametic (reproductive cells) chromosome number in this example: 1n = 7. During what phase does the cell(s) become haploid (i.e., when does reduction occur)? 8. Does meiosis occur in somatic or gametic tissues? (Circle correct answer.) 9. How does the "random" lining up of homologous chromosomes (in metaphase I) and their subsequent separation (in anaphase I) lead to genetic variability in offspring? 10. How does crossover (during prophase I) lead to genetic variability in offspring? 4
Part 2: Mendelian genetics (for a diploid organism) This exercise will use pipe cleaners as models of chromosomes and simulated sexual reproduction to investigate patterns of inheritance. Let's start by reviewing important concepts and terminology used! (Learn more in your textbook, Chapter 8, pp. 159-163.) Allele: alternate form of a gene. Dominant allele: this allele will be expressed if the individual is homozygous or heterozygous for this trait; a dominant allele is abbreviated with a capital letter. Recessive allele: this allele is covered up by dominant alleles; it will only be expressed if the individual is homozygous for the recessive trait; a recessive allele is abbreviated with lowercase. Homozygous: an individual has the same allele at both gene loci. Heterozygous: an individual has two different alleles at both gene loci. Monohybrid cross: mating between individuals who are heterozygous for one gene of interest. Dihybrid cross: mating between individuals who are heterozygous for two genes of interest. Genotype: the genetic makeup of an individual; the combination of alleles (e.g. AA, Aa, aa). Phenotype: the observable physical appearance of an individual (e.g., red or white flower). Pipe cleaner babies: For this exercise, you will need to modify the pipe cleaners used for modeling mitosis/meiosis by switching beads (i.e., alleles) to match the list below. Please note: these pipe cleaner/bead combinations are different from what you used above! 1 white long chromosome with blue bead, 1 white long chromosome with brown bead; 1 yellow long chromosome with blue bead, 1 yellow long chromosome with brown bead; 1 green short chromosome with red bead, 1 green short chromosome with orange bead; 1 blue short chromosome with red bead, 1 blue short chromosome with orange bead; please return paper clips to your bag, as you will not use them for this exercise. You have two long chromosomes (white & yellow) that carry the eye color gene. Beads reflect eye color (brown represents brown allele, blue represents blue allele). Brown eyes (B) are dominant to blue (b). As a point of interest, genetics of eye color is more complicated than it is presumed here and in many introductory genetics labs; learn more about inheritance of eye color online at the following link, How blue eyed parents can have brown eyed children, The Tech, http://genetics.thetech.org/how-blue-eyed-parents-can-have-brown-eyed-children If a person has the following genotypes, what eye color will they have? BB Bb bb 5
You also have two short chromosomes (blue and two green) that carry the cystic fibrosis (CF) gene. Beads reflect disease status. Orange beads are wild type, non-cf alleles and red beads are CF alleles. Non-CF alleles (N) are dominant to CF alleles (n). Cystic fibrosis is a recessive disease known for mucous secretions that damage lungs and other organs. Fifty years ago, most people with CF did not survive infancy; life expectancies are now much better. Single Gene Disorders, Learn.Genetics, Choose Cystic Fibrosis at this link http://learn.genetics.utah.edu/content/disorders/singlegene/ If a person has the following genotypes, will they be healthy or have cystic fibrosis? NN Nn nn First Exercise: monohybrid cross. A monohybrid cross is a cross between two parents that are heterozygous for one gene. In this case, parents are heterozygous for eye color (Bb). Work in pairs! Designate one partner as mom, and one partner as dad both are heterozygous for eye color (Bb). Mom will have 1 white chromosome with a brown bead and 1 white chromosome with a blue bead. Dad will have 1 yellow chromosome with a brown bead and 1 yellow chromosome with a blue bead. If the beads and chromosomes arrangements aren't right, fix them now! Mom, shuffle the white chromosomes behind your back and then hold them in separate hands. Dad, shuffle the yellow chromosomes behind your back and then hold them in separate hands. Now Dad, without looking, pick one of mom s hands mom, toss the chromosome from that hand onto table. Mom, without looking, pick one of dad s hands dad, toss the chromosome from that hand onto table. You ve just made your first pipe cleaner baby! Congratulations!!! Enter the eye-color alleles you selected into the proper columns on the table below. Also determine the genotype and phenotype of the baby. Repeat until you have made 16 babies. Table 1: Monohybrid cross (eye color) Baby no. Eye allele (from dad) Eye allele Genotype Phenotype Example B (brown) b (blue) Bb Brown Baby 1 Baby 2 Baby 3 Baby 4 Baby 5 Baby 6 Baby 7 Baby 8 Baby 9 Baby 10 Baby 11 Baby 12 Baby 13 Baby 14 Baby 15 Baby 16 6
How many babies have the following phenotypes? (Please enter these #s on the white board) No. brown-eyed = No. blue-eyed = How many babies have the following genotypes? (Please enter these #s on the white board) BB = Bb = bb = Second exercise: dihybrid cross. A dihybrid cross is a cross between parents heterozygous for two genes. In this case, each parent is heterozygous for eye color and CF (i.e., genes on different chromosomes) so the genotypes are BbNn. Mom will have white + green chromosomes, dad will have yellow + blue chromosomes. Each parent will have 1 brown, 1 blue, 1 red, and 1 orange allele. Because the genes controlling eye color and CF are on different chromosomes, it is essential that you independently and separately select the two chromosome types during baby making! Mom, shuffle the white chromosomes behind your back and then hold them in separate hands. Dad, shuffle the yellow chromosomes behind your back and then hold them in separate hands. Then Dad, without looking, pick one of mom s hands mom, toss the chromosome from that hand onto the table. Now Mom, without looking, pick one of dad s hands dad, toss the chromosome from that hand onto the table. Now, repeat the process with the other chromosome! Mom, shuffle the green chromosomes and hold them in separate hands. Dad, shuffle the blue chromosomes and hold them in separate hands. Dad, pick one of mom s hands mom, toss the chromosome from that hand onto the table. Mom, without looking, pick one of dad s hands dad, now toss the chromosome from that hand onto the table. Congratulations, you ve made your first dihybrid cross baby! Now enter the eye-color and CF alleles you selected into the proper columns on the table below also determine the genotype and phenotype of the baby. Repeat until you have made 16 babies. Table 2: Eye color and CF status Baby no. Genotype (eye) Genotype (CF) Genotype Phenotype Example Bb nn Bbnn Brown, CF Baby 1 Baby 2 Baby 3 Baby 4 Baby 5 Baby 6 Baby 7 Baby 8 Baby 9 Baby 10 Baby 11 Baby 12 Baby 13 Baby 14 Baby 15 Baby 16 7
How many dyhybrid cross babies have the following phenotypes? (enter #s on board) No. with brown eyes, non-cf No. with brown eyes, CF No. with blue eyes, non-cf No. with blue eyes, CF Punnett squares: A Punnett square is a mathematical approach (with graphical interface!) that allows you to make predictions about the phenotypic and genotypic ratios of a mating. You can learn more about Punnett squares in your textbook (Fig. 8-13 p. 161 & Fig. 8-15 p. 163). Step 1: Determine what gametes each parent makes. Remember, a gamete will only have one form of an allele. Example 1: a plant that is heterozygous for stem height (L is long-stemmed, l is short-stemmed) makes two types of gametes through the process of meiosis: L & l. Example 2: A plant that is heterozygous for both height and flower shape (R is radially symmetrical, r is bilaterally symmetric) makes four types of gametes through the process of meiosis: LR, Lr, lr, & lr. Step 2: Place the gametes from each parent on the Punnett square. Place the gametes from the dad on top and the gametes from the mom on the side. Step 3: Make the cross by combining a gamete from the mom and the dad in the square where the row and column meets. Continue through all possible gamete combinations. Step 4: Count the phenotypes of the offspring and convert to a ratio (for a monohybrid cross count the phenotypes in the four squares). For example, 4:0 or 3:1 or 2:2 or 1:2:1. Step 5: Count the genotypes of the offspring and convert to a ratio. Example 1: Monohybrid cross: Ll Ll Mom s genotype is Ll, which means she makes the gametes L & l. Dad s genotype is Ll, which means he makes the gametes L & l. L (from dad) l (from dad) L LL Ll l ll ll The above Punnett Square shows the predicted phenotypic ratio for a monohybrid cross is 3 long stems to 1 short stem (3:1). If you cross parents of unknown genotypes, and get a 3:1 ratio, you d conclude it was a monohybrid cross (each parent was heterozygous). If you do not get a 3:1 ratio, you d conclude it was not a monohybrid cross. In practice, random deviations from an expected ratio are anticipated in a 'real world' data set, even when the inheritance pattern described by the Punnett Square is accurate; statistics are thus used to test whether the observed ratio is 'close-enough' to or 'significantly different' from the expected ratio (see below for further discussion). 8
Example 2: Dihybrid cross: LlRr LlRr Mom s genotype is LlRr, which means she makes the gametes LR, Lr, lr, lr. Dad s genotype is LlRr, which means he makes the gametes LR, Lr, lr, lr. LR (from dad) Lr (from dad) lr (from dad) lr (from dad) LR LR LR LR Lr LR lr LR lr Lr Lr LR Lr Lr Lr lr Lr lr lr lr LR lr Lr lr lr lr lr lr lr LR lr Lr lr lr lr lr What is the predicted phenotypic ratio for the above cross? : : : No. long, radial = No. long, irregular = No. short, radial = No. short, irregular = The Example 2 Punnett Square shows the predicted phenotypic ratio for a dihybrid cross is 9:3:3:1. If you cross parents of unknown genotypes, and get a 9:3:3:1 ratio, you d conclude it was a dihybrid cross (each parent was heterozygous for alleles at two genes). If you do not get a 9:3:3:1 ratio, you d conclude it was not a dihybrid cross. Now let's apply this logic using a Punnett Square for eye color (monohybrid cross Bb Bb) and your pipe cleaner parents and pipe cleaner babies. Parents are heterozygous (Bb) Types of gametes made by mom: Types of gametes made by dad: & & Monohybrid cross (eye color): Bb Bb (from dad) (from dad) Now summarize phenotypic and genotypic ratios on the next page. Please read the discussion about calculation and comparisons of phenotypic ratios below before proceeding! To determine the actual phenotypic ratio, first determine proportions of each phenotype that you found (e.g., # progeny with phenotype 1 / total # progeny) and then multiply this by the number of squares on the Punnett Square (in the above case, 4). For example, if you have 13 brown-eyed babies and 3 blue-eyed babies, the proportion of brown is 13/16 (=0.8125) while the proportion of blue is 3/16 (0.1875). If these proportions are then multiplied by 4, the ratio is 3.25:0.75, close to 3:1. Note that for research purposes, you would use statistics to test whether your observed phenotypic ratio is 'close-enough' to the expected ratio or differs significantly from the expected ratio given your sample size (number of progeny) and assuming random union of gametes during reproduction. We will not be using statistical tests (like the Chi-square) today in BIOL 153L but you can learn more about these important tests online, for example at the following websites. Bozeman Science 'Chi-squared test' video: https://www.youtube.com/watch?v=wxpbofdqnvk Hobart & William Smith 'Chi Square Statistic': http://math.hws.edu/javamath/ryan/chisquare.html 9
Phenotype What is the predicted phenotypic ratio for brown eyes:blue eyes (use Punnett Square)? : What is the actual phenotypic ratio from your group s data (16 babies)? : What is the actual phenotypic ratio from the class data? : Does the actual data match the predicted data? Explain. Does your group data match the data from the entire class? Explain. Genotype What is the predicted genotypic ratio for BB:Bb:bb (use Punnett Square)? : : What is the actual genotypic ratio from your group s data (16 babies)? : : What is the actual genotypic ratio from the class data (see board)? : : Do the actual ratios match the predicted ratios? Explain. Do the actual ratios match the predicted ratio? How do the ratios from the class data and your group data differ? Do you conclude that this was a monohybrid cross? Explain. Now continue to the next page, where you'll make a Punnett Square for eye color and CF status in the dihybrid cross (BbNn BbNn) used by your pipe cleaner parents. 10
Parents are heterozygous for both traits (BbNn). Types of gametes made by mom: & & & Types of gametes made by dad: & & & Dihybrid cross: BbNn BbNn (from dad) (from dad) (from dad) (from dad) What is the predicted phenotypic ratio for this cross? : : : No. brown, non-cf = No. blue, non-cf = No. brown, CF = No. blue, CF = What is the actual phenotypic ratio from your group s data (16 babies)? : : : No. brown, non-cf = No. blue, non-cf = No. brown, CF = No. blue, CF = What is the actual phenotypic ratio from the class data? : : : No. brown, non-cf = No. blue, non-cf = No. brown, CF = No. blue, CF = How closely matched are the ratios from the class data vs. your group data? What does this result tell you about the importance of progeny sample size when performing genetic experiments? Do the actual ratios for the class match the predicted ratio? Do you conclude that this was a dihybrid cross? Explain. 11