Online Lab 5 Human genetics

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Dec 6, 2023

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Lab 5: Human Genetics, single gene trait, pedigree Objectives : - Interpret male and female karyotype - Define, explain, and properly use genetic terms in context - Determine the genotypes and phenotypes of family members and use them to solve genetic problems - Use a Punnett Square to determine the probability of transmission of specific trait to offspring - Determine the most probable mode of inheritance for a trait based on experimental data. Materials: - No extra material needed for this lab. Activity 1: Understanding genetics through chromosome analysis Humans have 46 chromosomes in their body cells (with the exception of mature red blood cells). This is referred to as the diploid (2n) number of chromosomes. The diploid number of chromosomes actually consists of two pairs of 23 chromosomes . One set of 23 was inherited from your father (the paternal chromosomes) and the other 23 from your mother (maternal chromosomes). Each maternal chromosome has a matching paternal partner. The two chromosomes that form a pair are called homologous chromosomes. So, in a human cell, there are 23 homologous pairs of chromosomes. A cell preparing to carry out meiosis duplicates its chromosomes during interphase. As a result of this duplication, each maternal and paternal chromosome produces an extra copy. The duplicates are known as sister chromatids. You can see chromosomes through a light microscope, but they’re only visible during the process of cell division, It’s possible to “freeze” cell division at a particular stage so that the chromosomes can be photographed and counted. The scanned images can be manipulated by computer into an orderly set showering each pair of homologous chromosomes. This arrangement is called a Karyotype The chromosome pairs are assigned numbers (1-22) and are referred to as autosomes . The last pair is referred to as the sex chromosomes. These are designated with letters of the alphabet, X and Y. A female has two X chromosomes, and is designated XX. A male has on X and Y chromosome, and is designated XY. Questions: 1. Figure 8-1 shows an example of a female karyotype. a. Are there 46 total chromosomes?____ b. How many pairs of autosomes are in the karyotype? _______ c. How many pairs of sex chromosomes ? ______ 2. Examine the homologous pair of sex chromosomes (figure 8-1). Is there a visible difference between the maternal and paternal chromosomes? Explain your answer Fundamentals of Biology. Online Labs. Lab5

3. Figure 8-2 shows an example of a male karyotype. a. Are there 46 total chromosomes?_____ b. How many parts of autosomes are in the karyotype? c. How many pairs of sex chromosomes ? 4. Is there a visible difference between the X and Y chromosomes? Explain your answer. Fundamentals of Biology. Online Labs. Lab5

Activity 2: Single Gene trait Research by the Human Genome Project has determined that, although tongue rolling is an inherited trait, it’s not a simple matter of dominant and recessive alleles. If you do a little searching through the Internet, you’ll find that other factors may be involved. For example, some studies showed cases of identical twins in which one twin was a tongue roller and the other wasn’t. Clearly more research is needed to determine all the factors in the inheritance of this trait. One way scientists are investigating this question is by examining the frequency of this trait in the general population. To examine the frequency of this trait on a smaller scale, you’ll sample family, friends, and acquaintances. The following is a list of some autosomal human traits that have been attributed to a single gene. Although clearly other genes are involved, the inheritance of each of these phenotypic traits acts as if it were governed by a single gene. First, indicate your expression of the trait ( i.e ., circle the appropriate phenotype). Then, indicate if you exhibit the dominant or recessive trait (D or R?). Finally, using any letters/symbols you want, write your genotype for the trait (hint: if you show the recessive trait, your genotype will be homozygous recessive ( i.e ., "aa"). If you exhibit the dominant trait your genotype could be homozygous dominant ( i.e ., "AA") or heterozygous ( i.e ., "Aa"). A shorthand symbol for this is "A_"). How could you distinguish between these possibilities? Complete Table 8-A Fundamentals of Biology. Online Labs. Lab5

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Trait Phenotyp e Dominant or Recessive Genotyp e (write down your own genotype) Picture (find, copy and paste a picture of the dominant trait) Tongue Rolling (Dominant) roll/not roll Widow's Peak (D) - just like Eddie Munster yes/no Wet ear wax (D) - stick your finger in to check wet/dry L/R interlocking finger (D) - without thinking, clasp your hands together, is the right thumb over the left, or vice versa? L/R or R/L Attached earlobes (D) - ask a neighbor or check out the mirror yes/no Hitchhiker Thumb (r) - does it bend back at a 90 angle yes/no Chin fissure (D) - like actor Michael Douglas yes/no Darwin tubercle (D) - little bump on the inside of the ear yes/no S-methylthioester detection (Recessive) - can you smell asparagus odor in urine? yes/no Pigmented iris (D) - any color but blue yes/no Freckles (D) yes/no Dimples (D) yes/no Wooly hair (R) yes/no Long eyelashes (> 1cm; D) yes/no Short big toe (D) - the big toe is shorter than your second one yes/no TABLE 8-A Ask 20 people the following question: “Can you roll your tongue into a tube?” Ask them to show you. Record the number of people who can and the number of people who can’t roll their tongues in Table 8-1 Fundamentals of Biology. Online Labs. Lab5

Note it’s not necessity to indicate the number of makes or females or adults of children; however, if you sample any related individuals, you should include that information when you upload your results. Table 8-1 Tongue Rolling Results Totals Number of Tongue Rollers (related to each other) Number of Tongue Rollers (NOT related to each other) Number of Non-Rollers (related to each other) Number of Non-Rollers (NOT related to each other) Activity 3: Determining inheritance by combining alleles When studying genetics, the most common question students ask is “How can I tell which trait could be present in my baby?” Suppose we had two parents with normal pigmented skin color, both of whom have the albinism trait in their families. How can we calculate the probability that this couple could have an albino child (a recessive trait)? There are three simple steps you can follow to figure out the probability for the offspring. Step 1: determine all the possible genotypes and phenotypes for the melanin trait. Since we’re working with two alleles per person, there are only three possible allele combinations. 1. Based on your knowledge of dominant and recessive alleles, fill in the phenotypes for each allele combination. Genotypes Phenotypes AA Aa aa Step 2 : determine the genotype of each parent. With the completed genotype and phenotype information, we can determine the probability of this couple having an albino child. To simplify matters, assume that both parents are heterozygous for the melanin trait. Since there is no way to know which egg and spam will combine to form a child, all possible combinations have to be considered. An easy way to visualize these combinations is through the use of a Punnett Square . Step 3 : set up a Punnett Square for these two parents. The alleles for one of the parents are entered along the side of the square and the other parent on top. Fundamentals of Biology. Online Labs. Lab5

It doesn’t matter which parent is on which side of the square, as long as you line up the alleles with the proper rows and columns (as demonstrated in Figure 8-6) Paternal alleles Aa A a A a Figure 8-6 Constructing a Punnett Square To complete the square (which will show you all the possible genotypes for children of this couple), you must combine the maternal and paternal alleles. If you combine one dominant maternal allele (A) with one of the paternal alleles (A), The resulting combination would be one possible genotype for their child (AA). Each square in the Punnett Square represents a 25% probability that this couple could have a child with that genotype. The four squares add up to 100% (all the possible genetic combinations for this trait with this set of parents. Each time a pregnancy occurs, the probability remains the same as shown in the square. To fill in the squares, combine the allele in the shaded box at the top of the column with the allele from the shaded box to the left of the row. The process is demonstrated in the first square above. A a A a 2. Combine the appropriate alleles to complete the other three boxes in the Punnett square. a. What is the probability that this couple could have an albino child? 3. If this couple does, have an albino child, what is the probability they could have a second albino child? Explain your answer. 4. If this couple has a child with regular pigmented skin color, is there any possibility that this normal child could have an albino baby? Explain your answer. 5. Steve has normal skin color. His Mom also has normal skin color but his Dad is an albino. Steve has a child with Marilyn, who is an albino. What are the genotypes of all the individuals mentioned in the problem? Genotypes: Steve: Steve’s Mom: Steve’s Dad: Marilyn: 6. What is the probability that Steve and Marilyn could have an albino child? Maternal alleles Fundamentals of Biology. Online Labs. Lab5

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Activity 4: Generate a Pedigree In the space below, use colored pencil to create a pedigree with the following information. Ray and Elaine became a couple in 1970. They both had normal vision. They had two daughters and then a son. Both daughters, Alicia and Candace, had normal vision and never had any children of their own. The son, Mike, was colorblind. The son married Beth who also had normal vision and they had two children of their own, first Greg, then Victoria. Victoria was colorblind, but Greg was not. Color blindness is a sex-linked recessive trait. Color your individuals as follow: Red: color blind White: regular vision Blue: regular vision (carriers) Green: unknown genotype. Don’t forget what shapes are male and female. Place the names and genotypes of the people under their shape Activity 5: Linked genes 1. For many genes, such as the mentioned above, the dominant allele codes for the presence of some characteristic (like, “B” codes for “make brown pigment” in someone’s eyes), and the recessive allele codes for something along the lines of, “I don’t know how to make that,” (like “b” codes for the absence of brown pigment in someone’s eyes, so by “default,” the eyes turn out blue). If someone is a heterozygote (Bb), that person has one set of instructions for “make brown” and one set of instructions for, “I don’t know how to make brown,” with the result that the person ends up with brown eyes. There are, however, some genes where both alleles code for “something.” One classic example is that in many flowering plants such as roses, snapdragons, and hibiscus, there is a gene for flower color with two alleles: red and white. However, in that case, white is Fundamentals of Biology. Online Labs. Lab5

not merely the absence of red, but that allele actually codes for, “make white pigment.” Thus the flowers on a plant that is heterozygous have two sets of instructions: “make red,” and “make white,” with the result that the flowers turn out mid-way in between; they’re pink. In humans, there is a gene that controls formation of hemoglobin, the protein in the red blood cells which carries oxygen to the body tissue. The “normal” allele of this gene codes for “normal” hemoglobin. However, there is another allele for this gene that has one different nitrogenous base in its DNA sequence, and thus, one codon in the middle of the gene codes for a different amino acid in an important place in the hemoglobin molecule. A red blood cell (RBC) that contains this altered hemoglobin will, under stress, crinkle up into a shape that reminded someone of the shape of an old-fashioned sickle. While the letters “S” and “s” are often used to represent these alleles, since both of them code for “make hemoglobin”, in reality, neither is dominant over the other. Someone who is SS makes all normal hemoglobin, someone who is ss makes all abnormal hemoglobin (and we say that person has sickle-cell anemia), and someone who is Ss essentially has two sets of instructions, and so, makes some of each kind of hemoglobin (often referred to as sickle-cell trait). Because the RBCs of a person who is ss contain all abnormal hemoglobin, they will “sickle” very easily, with very little stress required to provoke that reaction. All those sickled cells tend to get stuck as they try to go through capillaries, and cause things like strokes, heart attacks, pulmonary embolisms, etc. that lead to death. Because only some of the RBCs of a person who is Ss contain abnormal hemoglobin, that person usually only has trouble with a lot of cells getting sick if they’re under a lot of stress trying to meet a higher-than-normal oxygen demand, and so the chances of a person dying from sickle-cell trait are much lower than for full-blown sickle-cell anemia. Malaria is a parasitic disease that’s prevalent in tropical areas. When a mosquito that’s carrying the parasites bites someone, the parasites enter the person’s bloodstream, and invades and lives in the person’s RBCs. However, if a person has sickle-cell anemia (ss), the presence of a parasite in a RBC is so stressful, it causes the RBC to sickle (crinkle up), and when that happens, that kills the parasite before it can multiply and spread to other RBCs. Thus, coincidentally, a person who is ss is also “immune” to malaria. If a person is Ss and a malaria parasite tries to invade a RBC with abnormal hemoglobin, again, the RBC will sickle, killing the parasite before it has a chance to reproduce. If a parasite invades a RBC with normal hemoglobin, it will be able to live and multiply, but if its offspring invade other RBCs with abnormal hemoglobin, they, too, will be killed. Thus, a person who is Ss is “resistant” (though not totally immune) to malaria. If a person is SS and has all normal hemoglobin, the malaria parasites do just fine, invading RBCs, growing and multiplying, and invading more RBCs. Thus, an SS person usually dies, eventually, from causes tied to the malaria. A man and woman living in a tropical area where malaria is prevalent and health care is not accessible have seven children. The genotypes of these children are ss, Ss, SS, ss, Ss, Ss, and SS. a. What must the genotype of both parents be? Hint: what would be needed to have those kinds of children? b. What kind of gametes (egg/sperm) each can produce? Mother: Fundamentals of Biology. Online Labs. Lab5

Father: c. Which of their children would you expect to live to adulthood and reproduce? 2. In humans, there is a gene that controls formation (or lack thereof) of muscles in the tongue that allow people with those muscles to roll their tongues, while people who lack those muscles cannot roll their tongues. The ability to roll one’s tongue is dominant over non-rolling. The ability to taste certain substances is also genetically controlled. For example, there is a substance called phenylthiocarbamate (PTC for short), which some people can taste (the dominant trait), while others cannot (the recessive trait). The biological supply companies actually sell a special kind of tissue paper impregnated with PTC so students studying genetics can try tasting it to see if they are tasters or non-tasters. To people who are tasters, the paper tastes very bitter, but to non-tasters, it just tastes like paper. Let’s let R represent tongue- rolling, r represent a non-roller, T represent ability to taste PTC, and t represent non-tasting. Suppose a woman who is both a homozygous tongue-roller and a non-PTC-taster marries a man who is a heterozygous tongue-roller and is a PTC taster, and they have three children: a homozygous tongue-roller who is also a PTC taster, a heterozygous tongue-roller who is also a taster, and a heterozygous tongue-roller who is a non-taster. Write the representation of the whole family: Mother: Father: Offspring #1: Offspring #2: Offspring #3: Fundamentals of Biology. Online Labs. Lab5

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Activity 6: Who’s baby is it? Ms. Johnston, Ms. Johnson, and Ms. Johnstone all entered the same hospital and gave birth to baby girls on the same day, and all three babies were taken to the nursery to receive care, there. Someone later claimed that the hospital mixed up the babies. As a hospital administrator, it is your job to make sure that each pair of parents has the correct baby, so you order blood typing to be done on all the parents and all the babies. Here are the results: Person Blood type (phenotype ) Probable Genotype ABO Probable genotype Rh Possible gametes Ms. Johnston A+ AA , AO RR, Rr AOR r AR, Ar, OR, Or Mr. Johnston B+ Ms Johnson B- Mr. Johnson O+ Ms. Johstone A+ Mr. Johstone A- Baby A O+ Fundamentals of Biology. Online Labs. Lab5

Baby B AB- Baby C B- First, for each parent, think about what possible genotype(s) could give that phenotype. If there’s more than one possible genotype, then which of the possible genotypes would give the most variation in terms of possible children? Put that genotype in the appropriate box above, for each person. Ms. Johnston is an example. a. Baby A parents are: b. Baby B parents are: c. Baby C parents are: Fundamentals of Biology. Online Labs. Lab5

Support your answers with Punnett square tool. (Watch video for further explanation) Johnston couple: Johnson couple: Johnstone couple: Fundamentals of Biology. Online Labs. Lab5

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Online Lab 5 Human genetics

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