Lab 10

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Dec 6, 2023

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Lab 10: Mendelian Genetics Group Number: Section: Student Names (First and Last) Student Panther ID #s James Cordova 6415838 Emilys Perez 6439296 Malek Barakat Melissa Trevol OBJECTIVES: ● Understand Mendel’s laws of segregation and independent assortment. ● Differentiate between an organism’s genotype and phenotype. ● Recognize different patterns of inheritance. ● Perform monohybrid and dihybrid crosses. ● Use pedigree analysis to identify inheritance patterns. INTRODUCTION: Through his studies of the inheritance patterns of the garden pea, Pisum sativum, Gregor Mendel changed our understanding of heredity. Mendel studied characters/traits that differed between plants and designed cross-fertilization experiments to understand how these characters transmit to the next generation. The results of Mendel’s work refuted the prevailing hypothesis of blending inheritance and provided a new framework for understanding genetics. Ultimately, Mendel postulated two laws to explain heredity: (1) the law of segregation and (2) the law of independent assortment . Monohybrid crosses and the law of segregation The law of segregation states that during gamete formation the alternate forms of a gene (i.e. alleles ) on a pair of chromosomes segregate randomly so that each allele in the pair is received by a different gamete. For example, if you were to examine the gene responsible for petal color, you may discover that the gene can be expressed as either yellow or white flowers. In this scenario, the gene is petal color, while the alleles are yellow and white. Depending on which allele is expressed, petal color will vary (Fig. 1). Figure 1. Schematic of Mendel’s law of segregation 1

In diploid organisms, all alleles exist in pairs; identical alleles within a pair are homozygou s, while different alleles are heterozygous . Allele forms are represented by a single letter that explains whether a particular trait is dominant or recessive . Dominant alleles are assigned an uppercase letter (E), while recessive alleles are lowercase (e). In general, a dominant trait is expressed when at least one of the alleles present in the resulting allelic pair is dominant (EE or Ee). In contrast, for a recessive trait to be expressed, both alleles within the pair must be recessive (ee). For example, when considering ear lobe shape, two forms (attached and unattached) are apparent (Fig. 2). This trait is regulated by a single gene where unattached ear lobes are dominant (E) while attached ear lobes (e) are recessive. Figure 2. (a) Unattached (EE or Ee) vs. (b) attached earlobes (ee) An organism’s genotype (EE, Ee, ee) is the combination of alleles present whereas the phenotyp e is the physical expression of the genotype. In the earlobe shape example above, an individual can have a genotype of EE, Ee or ee. People with EE or Ee genotypes have the unattached earlobe phenotype (Fig 2a), while those with an ee genotype express the attached earlobe form (Fig 2b). Note that in general, dominant traits can be either homozygous (EE) or heterozygous (Ee) while recessive traits are always homozygous (ee). Question : Given that the allele for brown eyes (B) is dominant and the allele for blue eyes (b) is recessive, which of the following genotypes would result in individuals with brown eyes? Which genotype(s) is/are homozygous and which is/are heterozygous? BB: Homozygous—Brown eyes Bb: Heterozygous—Brown eyes bb: Homozygous—blue eyes TASK 1 – Patterns of Inheritance I: Simple Dominance Simple dominance describes a common outcome of allelic combinations, where one allele, if present, will dominate over the other and will be expressed. Information about alleles present in a parental population can be used to determine the probability of different genotypic and phenotypic ratios for a variety of traits in the offspring. In instances when only 1 or 2 traits are being considered the Punnett square (Fig. 3) approach is used to predict the possible outcomes of the parental cross. When only one trait is being considered the cross is monohybrid , while a dihybrid cross involves 2 traits. 2

General instructions on how to perform a cross using the Punnett square approach: 1. Write down the genotypes of the parents 2. Note the gametes that each parent can contribute 3. Draw a Punnett Square 4. Across the top write the gametes that one parent contributes and along the side write the gametes contributed by the other parent 5. Perform the cross 6. Determine the genotypic and phenotypic ratios In the example above (Fig. 3), the genotypic ratio is 1:2:1 (1: CC, 2: Cc, 1: cc) while the phenotypic ratio is 3:1. Since C = curly hair and c = straight hair, ¾ of the possible offspring will have curly hair while only ¼ will have straight hair. Procedure: 1. You will now simulate a cross between two heterozygous individuals, Tt and Tt. Each group should obtain two coins from your TA. You will flip the coins simultaneously to represent the potential outcomes of a cross between two Tt individuals. A head represents the dominant tall allele (T) while a tail symbolizes the recessive dwarf allele (t). Before you begin flipping the coins, perform the Tt x Tt cross in the Punnett square below to estimate the expected genotypic and phenotypic ratios. Figure 3. Example of a Monohybrid cross Parent 1 T t Parent 2 T TT Tt Based on this cross, what do you anticipate the genotypic and phenotypic ratios to be? Write your hypotheses (H o and H a ) in Table 1 t Tt tt Table 1: Genotype Phenotype Expected Ratio: 1:2:1 3:1 H o : Have a dominant gene does correlate to your child having a dominant gene Having brown eyes will correlate to your child having brown eyes. H a : Having a dominant gene does not correlate to having a dominant gene Having brown eyes will not correlate to your child having brown eyes. 1. Begin flipping the two coins simultaneously for a total of 16 times. Record your results in Table 2. Table 2: Genotype Number TT 3 Tt 7 3

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tt 6 Questions : a. What ratio of allele combinations did you observe? 3:7:6 b. What genotypes and phenotypes result from these crosses? Genotypes: TT,Tt,and tt Phenotypes : Tails and heads c. What are the genotypic and phenotypic ratios? Genotypic ratio: 3:7:6 Phenotypic ratio: 5:3 d. How did your results compare to your expectations? Do your results support or reject your null hypothesis? They aligned for the most part with my expectations—therefore supporting my nul hypothesis e. Do you think your results would have been closer if you flipped the coins 1600 times instead of just 16? Why or why not? I think so—due to the law of large numbers(statistics) the more you do something—the closer it will be to the mean. 2. Albinism, a recessively inherited trait, results in organisms that lack pigment in their skin, hair or eyes. Anna is a female with normal pigmentation, but her mother, Sara, was albino. Anna’s husband, John, is albino. Anna and John have one child. Using the information you have learned so far complete Table 3. Table 3: Genotype of Anna Aa Genotype of John aa Allele(s) possible in Anna’s gametes Aa, AA Allele(s) possible in John’s gametes aa Possible genotype and phenotype of the child aA or aa—either albino or not albino Genotypic ratio of children 1:2:1 Phenotypic ratio of children 3:1 4

TASK 2 - Patterns of inheritance II: Incomplete vs. Complete Dominance & Codominance Inheritance of traits can occur in multiple forms. So far you have considered complete dominance , where a homozygous dominant or a heterozygous individual expresses the dominant phenotype, while an individual that is homozygous recessive expresses the recessive phenotype. However, in certain cases a cross between two different allele forms results in a phenotypic expression that combines the two allelic traits - incomplete dominance . For example, if an offspring resulting from a cross between a red (RR) and a white (rr) snapdragon plant receives the dominant allele for red flower color (R) from one parent and the allele for white flower color (r) from the other, the resulting genotype will be Rr. The heterozygous form (Rr) of the plant will bear pink flowers since neither allele is completely dominant over the other (Fig. 4). Figure 4. Pink snapdragons are an example of incomplete dominance 1. Determine the possible phenotypes of the F1 offspring when two pink snapdragons are crossed. Calculate the probability (percent chance) of the offspring having each phenotype. Parent 1 R r Possible phenotypes: Red or white Parent 2 R RR Rr Phenotypic ratio/ probability: 3:1 r Rr rr 2. What would be the resulting genotypes of a cross between a pink and a white snapdragon? Calculate the probability (percent chance) of the offspring having each genotype. Parent 1 R R Possible genotypes: RR and rr Parent 2 r Rr Rr Genotypic ratio/ probability: 100% probable that the offspring will have a dominant gene and a recessive gene carrier. r Rr Rr 5

Expression of both alleles of a particular gene is known as codominance . When alleles are inherited codominantly, both phenotypes are expressed at the same time in the heterozygous condition in contrast to incomplete dominance where the heterozygote is an intermediate between the two homozygotes (Fig. 5). The ABO blood type system is an excellent example of codominance. Humans have four blood types: A , B , AB and O . All individuals carry two alleles, one from each parent. In this system, both alleles inherited determine one’s blood type, where a person with Type AB blood possesses phenotypic traits of both A and B blood types (Table 4). Figure 5. Different types of inheritance Table 4. Relationship between blood type and genotype Blood type (phenotype) Genotype Type A I A I A or I A i, Type B I B I B or I B i, Type AB I A I B Type O i i For example, an individual with Type B blood can have two possible genotypes , I B I B or I B i, where I (dominant) and i (recessive) represent an allele from each parent (Table 4). The different blood types are characterized by the presence of a particular sugar molecule attached to the proteins on the surface of red blood cells (Fig 6). In Type A blood, the attached sugar molecule is galactosamine, while in Type B blood it is galactose. In contrast, individuals with Type O blood, have no sugars present on the surface of their red blood cells. These protein-sugar complexes are antigens that act as recognition markers for the immune system. The immune system is tolerant to its own antigens but produces antibodies against antigens that differ from its own. The antibodies formed bind to the antigens causing agglutination (clumping) and lysis of the foreign red blood cells. Therefore, an individual with Type A blood could not receive a blood transfusion from a Type B blood donor because the antigens on the donor’s red blood cells will trigger an immune response from the recipient’s antibodies. Thus, the Type A recipient will produce antibodies against the donor’s Type B antigens. 6

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Figure 6. Human ABO blood types When the wrong blood type is given to a patient, agglutination of the blood occurs and this can ultimately lead to death. Table 5 provides a quick overview of which blood types can donate to which, and which types can receive from which. An “X” indicates that mixing of the donor and recipient blood types results in agglutination whereas a blank cell means that no agglutination occurs when the blood types are mixed. Table 5. Effects of mixing different blood types Recipient (Antibodies) Donor (Antigens) A B AB O A X X B X X AB X X X O Questions : 1. Given the information above, why do you think that Type O is the universal donor and AB is the universal recipient? Because of the antigens. 7

2. A woman with genotype I A i and a male with genotype I B i have a child. What are the possible blood types (A, B, AB, or O) that the child could have? For each possible blood type, calculate the probability (percent chance) that the child has that blood type. Parent 1 IA i Possible blood types: AB, b, A, O Parent 2 IB IAB IB Phenotypic ratio/ probability: 1:1 i iA ii 3. There was a mix-up with children in the maternity ward of a hospital. The children in question and their blood types are listed below. Parent 1 ● Child 1: type A (genotype I A I A or I A i) ● Child 2: type B (genotype I B I B or I B i) ● Child 3: type AB (genotype I A I B ) ● Child 4: type O (genotype ii) IA iB Parent 2 i IAi IBi Which child or children could belong to a couple having AB and O blood types? Child 1 or Child 2 i IAi IBi 4. Based on the previous question, is it possible to prove paternity based on blood types? Explain. NO—many people can have the same blood type. 5. A woman with Type O blood has a child with the same blood type. Can the child’s father have Type AB blood? Why or why not? No—all the children would turn out to either be type A or Type B Another trait involved in blood typing is the Rh factor . The Rh factor works along the principle of simple dominance instead of codominance. An individual who is Rh positive possesses the Rh antigen on his/her blood cells while someone who is Rh negative lacks Rh antigens. Generally, the Rh status of an individual is always included with the blood type. For example, a person that is AB+ has Type AB blood and is Rh positive. This information is particularly important during pregnancy since Rh incompatibility can develop in women that are Rh- and who have an Rh+ baby. Mixing of maternal and fetal blood through the placenta can cause the mother to develop antibodies against the Rh antigens from the baby. This condition is usually not harmful to the first child but may cause mild to severe symptoms (depending on 8

the amount of Rh antibodies created) during subsequent pregnancies since the mother’s Rh antibodies attack the Rh antigens of the developing fetus. Determining blood type Procedure: 1. Prepare your station by obtaining the following supplies: a. 5 small plastic blood typing trays b. Toothpicks for mixing c. Five bottles of blood representing five different individuals d. One bottle representing A antibodies e. One bottle representing B antibodies f. One bottle representing Rh antibodies 2. You will use each blood typing tray to determine the blood type of a particular individual. Note that each tray contains 3 wells, labeled A, B and Rh. 3. Add 1 drop of blood from individual 1 to every well in a blood typing tray. 4. Add 3 drops from the bottle labeled A antibodies to the well labeled A. 5. Add 3 drops of B antibodies to the well labeled B. 6. Add 3 drops of Rh (D) factor solution to the well labeled Rh. 7. Mix each well with a toothpick. Note: Use a different toothpick for each well and tray. 8. After 1 min, examine the tray for the presence of crystals. Presence of crystals, indicates a positive result* for a particular blood type and Rh factor. 9. Repeat steps 3-8 for the remaining individuals. * Important note: In this particular experiment, crystallization indicates a positive result for a particular blood type. For example, if crystallization occurs in well A, then the individual has blood type A. However, when working with real blood (i.e. for the purposes of transfusions), agglutination/clumping would be a negative result because agglutination would indicate that the antibodies of one’s blood detected a foreign substance (an antigen), causing an immune response and cell lysis. 10.Record your results in Table 6. Note which wells crystallize. Based on your results, determine the blood type of the five individuals examined. Table 6: Crystallization (Yes or No) Interpretation Individual Well A Well B Rh factor (+/-) Blood type Possible Genotype(s) 1 N0 yes + B I B I B or I B i, 2 YEs yes - AB I A I B 3 YEs No - A I A I A or I A i, 4 NO No - O ii 5 NO NO - O ii Dihybrid crosses and the law of independent assortment 9

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The law of independent assortment states that alleles from different genes assort independently from one another during meiosis if they are located on separate chromosomes (Fig. 7). For instance, alleles for the attached earlobe gene will assort independently from those that determine height as long as they are located on different chromosomes. Think of it like this - if you are flipping a coin and recording the results of each toss (heads or tails), and your friend is doing the same thing, do your tosses have any effect on the results of his? No, because your two actions are not linked. The chance of getting any two events to happen at the same time is simply the product of the chance of each event happening at all. For example, the chance of getting a “heads” is 50%. However, the chance that you and your friend get heads on the same toss is .5 x .5 = .25 or 25%. Figure 7. Schematic of Mendel’s law of independent assortment _____________________________________________________________________________ TASK 3 - Double Gene Inheritance We will examine the inheritance of two genes found on different chromosomes, in Drosophila melanogaster fruit flies. The genes used in this experiment are for the wing shape (wild or vestigial) and body color (wild or ebony). Wild type and mutant phenotypes for males and female flies are shown in Figure 8 below. 10

Figure 8. Wild type and mutant phenotypes for males and female flies The allele for wild type wings (F) is dominant over the allele for vestigial wings (f). The allele for wild type body color (B) is dominant over the allele for ebony body color (b). The following flies are crossed. Each fly is heterozygous for both wing shape and body color genes. 1. Determine the phenotype of both flies. Female phenotype: wild wings, wild body color Male phenotype: wild wings, wild body color 2. Determine the genotype of both flies. Female genotype: Ff,Bb Male genotype: Ff,Bb 3. Determine the possible allele combinations of the gametes of each fly. Female gametes: FB, Fb, fB, fb Male gametes: FB, Fb, fB, fb 4. Complete the punnett square for this dihybrid cross. Parent 1 5. Using your punnett square, calculate the probability (percent chance) that the offspring of this cross have each possible phenotype combination and complete the table below: FB Fb fB fb P a r e n t 2 FB FFBB FFBb FfBB FfBb Fb FFBb FFbb FfBb Ffbb fB FfBB FfBb ffBB ffBb fb FfBb Ffbb ffBb ffbb Wild Wings and Wild Body Color Vestigial Wings and Wild Body Color Wild Wings and Ebony Body Color Vestigial Wings and Ebony Body Color 9 3 3 1 11

TASK 4 - Analyzing Pedigrees A pedigree is a map of relatives that is used to determine the inheritance pattern of a particular disease or trait. This map usually includes the gender of each family member, how each is related (through lines connecting individuals) and also provides information about genetic traits. Certain symbols are used to indicate these variables (see Fig. 8). Figure 8. Symbols used in pedigree analysis Questions : 1. What would you look for in a pedigree to determine if a trait was dominant? Black circle or square 2. What would you look for in a pedigree to determine if a trait was recessive? Half black circle or sqaure 3. The image to the right is a pedigree for albinism. Note: the half-filled circles and squares in this pedigree represent carriers. a. Using this pedigree, can you determine if this trait is dominant or recessive? Explain. 12

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Yes—you can see from the offspring b. In generation III, individual 1 mates with a female that is a carrier for albinism. Determine the probability (percent chance) that their child has albinism. Parent 1 A a Possible phenotypes: Albanism or no albinism Parent 2 a Aa aa Phenotypic ratio/ probability: 1:1 a Aa aa 13

Independent Review and Evaluation: Lab 10 Instructions: ● This section of the task sheet is to be done independently and is due before the start of next class ● In essay/paragraph form , answer and address the below questions and topics (items 1-3) Note: Essay / paragraph form means one cohesive document that flows naturally as one piece of literature. Your assignment should use multiple paragraphs. Do not submit documents in Q&A format (i.e. no bullet points, no question numbers, do not simply type your answers under items 1-3 below) ● Use the template provided on Canvas and complete the self evaluation assessment at the bottom of the template ● Format: ○ File format: Word Document or PDF ONLY ■ MAC “Pages” will not be accepted. ○ Length: Writing should take about half page (more is ok) ○ Font: Calibri, 11pt font ○ Single spaced ○ 1” Margins (except for headers) ○ Include all header information in this document (name, PID, assignment title, etc) Questions/Topics: 1. What was done in today’s lab and what were your main take-aways? 2. What are some key concepts from today’s lab that you think might be on your quiz next class? Why are these concepts important? 3. Applying knowledge: Nature rarely fits neatly into the boxes we often use to define it. In today’s lab we made broad generalizations about biological concepts that can be extremely intricate and complex. Look up the case of Lydia Fairchild, a woman with chimerism that had to prove that the children that she gave birth to were actually hers after a DNA test suggested otherwise. In your own words, explain why, although sometimes necessary, it can be dangerous to reduce complex natural systems and processes into simplified “black and white” terms. 14

Lab 10

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