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BIOLOGY 21 ON-LINE LAB 11: GENETIC VARIATION & MEIOSIS Our biological legacy is contained in our genes

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BIOLOGY 21 ON-LINE LAB 11: GENETIC VARIATION & MEIOSIS Our biological legacy is contained in our genes. That is, we owe our traits and characters to our parents, grandparents, and so on. The study of genetics and inheritance provides fascinating insight into why we look and, perhaps, behave, the way we do. For example, we can use pedigree charts to trace certain traits back several generations. We can also use principles of inheritance to predict what our children will look like. Materials needed for this week’s lab: From Bio 21 packet: PTC taster paper for Activity 4. ACTIVITY 1: MEIOSIS ON THE WEB Go to www.pbs.org/wgbh/nova/baby/divide.html (also on Canvas under Lab 11). Read through “How Cells Divide: Mitosis vs. Meiosis” to answer the following: 1. What is the main difference between mitosis and meiosis? 2. What “problem” does meiosis solve? (Hint: why is meiosis called “reduction division”?) Scroll down toward the bottom of the page and click on “Mitosis vs. Meiosis (non-Flash version).Using information from this website, identify each step (the first one is listed for you) and describe what occurs at that step for mitosis and meiosis. Diagramming may be useful here. MITOSIS MEIOSIS Step 1 Name: Before We Split Step 2 Name: Step 3 Name: MORE? Human Biology: Lab 11 11.1 MITOSIS MEIOSIS Step 4 Name: Step 5 Name: Step 6 Name: Step 7 Name: Step 8 Name: MORE? 11.2 Human Biology: Lab 11 MITOSIS Step 9 Name: MEIOSIS These steps are not found in mitosis Step 10 Name: Step 11 Name: Step 12 Name: Step 13 Name: MORE? Human Biology: Lab 11 11.3 MITOSIS MEIOSIS Step 14 Name: Step 15 Name: ACTIVITY 2: TERMINOLOGY The key to understanding inheritance and genetics is knowing the “language.” Before you proceed with this lab, define the following terms (see Chapter 19 and the Glossary in your textbook): Term Definition allele dominant (allele or trait) gene genotype heterozygous homozygous phenotype recessive (allele or trait) 11.4 Human Biology: Lab 11 ACTIVITY 3: THE PUNNETT SQUARE EXPLAINED Reginald C. Punnett developed a special chart to show the possible combinations of traits that can arise when two organisms are bred or crossed. The chart helps us to determine the probability of having a particular characteristic. The most basic Punnett Square contains four squares. One parent’s gene combination is put across the top, while the other parent’s is put down the side (see below). In this example, we will find out how many black haired children a man with black hair and a woman with blonde hair could have. The gene combination, or genotype, for a person with black hair is BB. The genotype for a person with blonde hair is bb. This information is indicated in this Punnett square: Father B B b Mother b Next, carry the mother’s alleles across the rows and the father’s alleles down the columns. Father B? B? b? Bb Bb b? Bb Bb Mother The big “B” is called the dominant gene, or the gene that will be expressed in the child. The little “b” is called the recessive gene, and will only be expressed if a dominant gene is not present. So, if a child inherits a “B” from his father, and a “b” from his mother, the “B” gene will be expressed and he will have black hair, like his father. In this case, the chances of having a child with black hair (Bb) are 4 out of 4 or 100%. It is important to point out that Punnett Squares give us only the probability of certain combinations occurring. As every gambler knows, the outcome sometimes can be wildly different! There are no questions to answer for this activity. MORE? Human Biology: Lab 11 11.5 ACTIVITY 4: ARE YOU A TASTER? About 70% of the population gets a bitter taste from a substance called phenylthiocarbamide (PTC). It is tasteless to the rest. The "taster” allele is dominant to the non-taster allele. To determine if you are a taster, touch a piece of taste-test paper to the tip of your tongue. If you are a taster, you will detect it immediately. Otherwise, the paper will be tasteless. 1. Are you a taster? __________ You can dispose of the PTC paper in the regular trash, but keep out of reach of children and/or pets to be safe. 2. What are the possible outcomes of mating a male who is heterozygous for the PTC tasting trait and a female who cannot taste PTC? (Use the letter T for taster, and t for non-taster). Do the cross, filling out the Punnett square to indicate the genotypes. Then in the space below, write out in words the phenotype results. Father __ __ __ Mother __ ACTIVITY 5: GENETIC DISORDERS There are over 6000 known genetic disorders. Here we have you investigate two. Fragile X Syndrome: Go to https://www.fraxa.org/fragile-x-syndrome/ (also on Canvas under Lab 11). 1. What is Fragile X Syndrome? How prevalent is it (according to the website)? 2. What are some symptoms? 3. How is Fragile X inherited? Why can’t a father pass Fragile X syndrome to his sons? MORE? 11.6 Human Biology: Lab 11 ACTIVITY 5: GENETIC DISORDERS (continued) 4. Why is Fragile X more common in boys? 5. When Fragile X is present in girls, why aren’t girls usually as severely affected (you’ll have to do some critical thinking here to figure this one out). Klinefelter Syndrome: men.webmd.com/tc/klinefelter-syndrome-topic-overview (also on Canvas under Lab 11). 6. What is Klinefelter Syndrome? 7. What are some of the characteristics (or symptoms) of Klinefelter Syndrome? 8. What are some of the treatments for Klinefelter Syndrome? Human Biology: Lab 11 11.7 ACTIVITY 6: GENETICS & INHERITANCE PROBLEMS On the next several pages, you’ll find many different genetics and inheritance problems. Yes, some of it may seem repetitive, but the more you practice these problems, the better you’ll grasp this week’s topic. COMPLETE DOMINANCE A completely dominant gene will totally mask its recessive allele. Example: What type of offspring would be expected from the mating of a female homozygous for hanging ear lobes (EE) with a male with attached lobes (ee)? Parental Genotypes EE ee Gametes F1 Genotype E e Ee Phenotype hanging lobes Note that the dominant gene is always written first. Since the offspring are all the same, there are no genotypic or phenotypic ratios. Example: What type of offspring would be expected from the mating of a female heterozygous for hanging lobes with a male heterozygous for hanging lobes? This problem may be solved in the form of a “Punnett Square,” in which the gametes are placed in the form of a multiplication table. Ova Sperm E e E e EE Ee Ee ee Phenotypic ratio: 3 hanging lobes to 1 attached lobe Genotypic ratio: 1 EE to 2Ee to 1 ee Problems: Note: empty Punnett Squares can be found on the last page of this lab. For most of these problems, you need to use the Punnett Squares in order to solve the problem. Fill them out first with the information given in each question, and then answer the question. MORE? 11.8 Human Biology: Lab 11 1. Some people have freckles on their face. Some freckled individuals, when mated to other freckled individuals, have only freckled children. Other freckled people, when mated to freckled persons, have some children with freckles, some without. When both parents are without freckles, no offspring have freckles. Explain the results. Write the genotypes for these crosses using the symbols F and f. a. Genotype for no freckles: ___________________________________________________ b. Genotypes for freckles: _____________________________________________________ c. What kind of crosses would produce only freckled children? ________________________ d. What kind of crosses would produce some freckled children and some children without freckles? __________________________________________________________________ _________________________________________________________________________ INCOMPLETE DOMINANCE In some instances, the heterozygous form is phenotypically unique, displaying a condition intermediate between the two homozygous conditions. Since neither gene dominates in the heterozygous condition, this is referred to as incomplete dominance. Example: Let N represent the allele allowing normal metabolism of the sugar galactose. Let n represent the gene for lack of ability to metabolize galactose due to the lack of an enzyme. The genotypes and phenotypes possible are as follows: NN normal Nn reduced ability to metabolize galactose (low level of enzyme) nn inability to metabolize galactose (lack of the enzyme; disease = galactosemia) Problems: 2. A normal woman (NN) reproduces with a man with the reduced ability to metabolize galactose (Nn). What are the expected phenotypic and genotypic ratios of their offspring? What is expected when two persons with reduced metabolic ability mate? a. Genotypic ratios:________________________________________________________ b. Phenotypic ratios: _______________________________________________________ c. Expected outcome when two people with reduced metabolic ability mate ____________ 3. Glycogen-storage disease (excessive accumulation of glycogen in the kidneys and liver) results from the homozygous condition gg. Gg results in abnormally high level of some sugars in the blood. GG is normal. A woman with high sugar levels, and whose father had the full disease, reproduces with a man with high sugar levels, although his mother was normal. Work this cross out in a Punnett square and answer the following: a. Genotype of the woman_____________ b. Genotype of the man _______________ MORE? Human Biology: Lab 11 11.9 c. Genotype of woman’s father__________ d. Genotype of man’s mother___________ e. Possible genotypes of the woman’s mother _____________________________________ f. Possible genotypes of the man’s father ________________________________________ g. Could the woman have any brothers with the disease, or any who are normal? _________ h. Could the man have any sisters with the disease, or any who are normal? ____________ i. Expected genotypic ratios of the younger couple’s offspring: ________________________ _________________________________________________________________________ j. Expected phenotypic ratios of the younger couple’s offspring: _______________________ _________________________________________________________________________ SEX DETERMINATION The primary determinants of the sex of an individual are specialized sex chromosomes. In humans, these sex chromosomes are X and Y. The X chromosome promotes femaleness, and the Y chromosome promotes maleness. An individual with two X chromosomes (XX) is a female and an individual with one X and one Y (XY) is a male. The Y chromosome possesses very few known genes. In contrast, the X chromosome has over 800 genes, most of which do not have homologous alleles on the Y chromosome (which again, has very few genes). Because of this, the genes on the X chromosome are said to be “sex linked”. Female Male XX XY Parents Gametes X X X Y Offspring X X X XX XX Y XY XY 4. Which parent determines the sex of the offspring? ________________________________ 5. ________________ percent of the sons receive a Y chromosome from the father. MORE? 11.10 Human Biology: Lab 11 6. If a gene is on the X chromosome of the father, ______________ percent of his daughters will inherit the gene and ________________ percent of his sons will inherit the gene. SEX-LINKED FACTORS Any gene on an X chromosome is said to be sex-linked. The Y chromosome, incidentally, has very few genes on it, something that has interesting consequences as you will learn. In the following description and problems, we will use special notation for alleles of genes that occur on X chromosomes. The alleles are shown in superscript (for example, XN and Xn, where N and n are alleles for the gene under question that is carried by the X chromosome. Hint: look at the superscripts – this tells you whether the allele is dominant or recessive! Example: Let N represent the allele for normal vision. Let n represent the allele for red-green color blindness. These genes are sex linked. The possible combinations are as follows: Female (XX) Male (XY) POSSIBLE POSSIBLE GENOTYPES PHENOTYPE GENOTYPES PHENOTYPE XNXN Normal XNY Normal Normal (but a XNXn X nY Color-blind carrier) X nX n Color-blind Unlike females, males only have two genotypes since the Y chromosome does not carry alleles for these classically sex-linked genes. What would be the expected types of offspring from the mating of a homozygous normal woman with a color-blind man? Female Male Xn Y XN XN XNXn XNXn XNY XNY All offspring would have normal vision. However, the females would be carriers for colorblindness. Problems: 7. The gene for the ability to smell cyanide (C) is dominant over the gene for the lack of such smelling ability (c). The genes are sex-linked. A woman who cannot smell cyanide mates with a man who can. (Hint: C and c are superscripts in the genotypes for males and females; see colorblindness example above.) a. Will their sons be able to work as safety inspectors in the cyanide factory? __________ MORE? Human Biology: Lab 11 11.11 b. Will their daughters be able to work as safety inspectors in the cyanide factory? ________ 8. Hemophilia is a malfunction in which blood fails to clot. The gene producing this (h) is recessive to the gene for normal clotting (H). The genes are sex linked. If a normal woman mates with a normal man, and they have a hemophilic son, which parent is responsible for the malady? a. Responsible parent ________________________________________________________ b. If this son later reproduces with a normal female whose father was hemophilic, what type of offspring might be expected? (Hint: since this is a sex-linked trait, you have to state the outcome for girls and the outcome for boys): _________________________________________________________________________ _________________________________________________________________________ PUNNETT SQUARES (FOR PRACTICE) 11.12 Human Biology: Lab 11