Lab 12 Human Genetics

Note: All your answers to questions must be in Red or other color (not including blue) for easier grading. Points will be deducted if you do not distinguish your answers. Lab 12. Patterns of Inheritance Objectives:  Determine the genotype by observation of individuals with given traits and their relatives.  Determine inheritance involving autosomal dominant, autosomal recessive and x-linked recessive alleles.  Determine inheritance involving multiple allele inheritance and use blood type to help determine paternity.  Describe a normal human karyotype and discuss the various abnormalities that can be detected using this technique  Using a karyotype analyze a patient’s chromosomes to determine if inheritance is normal or a chromosomal anomaly is present.  Analyze a pedigree to determine the pattern of inheritance for an allele – autosomal dominant, autosomal recessive or x-linked recessive.  Create a pedigree to determine the probability of inheritance of a particular phenotype when given generational information only. Vocabulary:  Genotype  Phenotype  Autosomal dominant  Autosomal recessive  Heterozygous  Homozygous  Probability  X-linked recessive  X-linked dominant  Multiple alleles  Codominance  Karyotype  Pedigree

Introduction: Just like in pea plants, Mendel’s laws of inheritance also apply to humans. Humans inherit 46 chromosomes that occur in 23 pairs, 22 pairs of autosomes and one pair of sex chromosomes (either XX or XY). This means that all autosomal genes exist in two forms, called alleles. These alleles may be genetically identical, called homozygous (AA or aa) or they may be genetically different (Aa or aA). The fact that one allele may be dominant over the other determines what phenotype is exhibited. Dominant alleles are represented by capital letters (A) and recessive are represented by lower case (a). Thus, you can have homozygous dominant (AA) or homozygous recessive (aa) and of course heterozygous (Aa). When a gene follows Mendel’s laws of inheritance (MOST genes do not), it is easy to predict the ratios of potential offspring using Punnett squares, or for example to determine the genotypes of parents from the genotype or phenotype of the offspring. In this section of the lab, we will practice determining the genotype given information about the gene and inheritance of it. Part 1: Determining Genotype Table 1 shows several autosomal human traits and indicates which is dominant and which is recessive. Fill out Table 1 and answer the following questions. (If you are not sure what the trait looks like, refer to Figure 1-5 , or you can do a quick internet search for an image of that trait.) Figure 1: Left – Widow's peak (dominant) versus Right– straight hairline (recessive). Figure 2: Left – No Hitchhiker’s thumb (dominant) versus Right– Hitchhiker’s thumb (recessive).

Trait (Capital letter – dominant allele) (Lowercase letter– recessive allele) Possible Genotype s Your Phenoty pe Your Genotyp e Widow’s peak – W Straight hairline – w WW, Ww, ww Widow’s peak Ww Earlobes unattached – U Earlobes attached – u UU, Uu, uu Earlobes unattach ed Uu Skin pigmentation: Freckles – F No freckles – f FF. Ff, ff No freckles ff Hair on back of hand – H NO hair on back of hand – h HH, Hh, hh No hair on back of hand hh Dimples – D No Dimples – d DD, Dd,dd No dimples dd Polydactyly (more than 5 fingers) – P Five fingers – p PP ,Pp, pp Five fingers pp Thumb hyperextension – “hitchhiker’s thumb” Last segment cannot be bent backward – T Last segment can be bent backward - t TT, Tt, tt Last segment can’t bent backwar d Tt Questions: 1. What is the homozygous recessive genotype for dimples? What is the phenotype? The homozygous recessive is dd. The phenotype is no dimples. 2. Does an individual with Pp exhibit the symptoms of polydactyly (more than 5 fingers)? Why or why not? Yes, any person with polydactyly has the genotype of Pp, where P represents the allele that causes polydactyly and p presents the normal allele of this gene. 3. What is the genotype for an individual who is heterozygous for hitchhiker’s thumb? Will they exhibit the phenotype for hitchhiker’s thumb? The genotype for someone who is heterozygous for hitchhiker’s thumb is Tt. They will not exhibit the hitchhiker’s thumb phenotype instead they will have a straight thumb.

4. Two people who are heterozygous for earlobes unattached have a child. List the genotypes possible for their children regarding earlobe attachment. Show your work. UU (for earlobes unattached) Uu (for earlobes unattached) Uu (for earlobes unattached) uu (for earlobes attached) 5. Ryann does not have freckles, but both their parent’s do. Deduce the genotype of their parents. What is Ryann’s genotype? The parents genotype are FF and Ff. Ryann’s genotype is ff. 6. Godric is heterozygous for widow’s peak and has no hair on the back of his hand. His partner has a straight hairline and is heterozygous for hair on the back of their hand. a. What is Godric’s genotype? Ww , hh (heterozygous widow’s peak; no hair back of hand) b. What is his partner’s genotype? ww , Hh (straight hairline, heterozygous-hair back of hand) c. What possible genotypes will their children have for hair lines? WW, Ww, ww d. What possible genotypes will their children have for hair on the back of the hand? Hh, hh

which means SNPs are common. The three SNPs (see table 1) found in the TAS2R38 gene leads to changes in the amino acid sequence which can potentially change the proteins’ function. Table 1: SNPs Present in Tasters vs Non-Tasters for PTC Nucleoti de Position (bp) Nucleotide Change Codon Change Amino Acid Change Phenoty pe Non- Taster Taster Non- Taster Taster Non- taster Taster 145 G C G CA C CA Alanine Proline 785 T B G T T G C T Valine Alanine 886 A G A TC G TC Isoleucin e Valine Individuals who are tasters can be TT (homozygous dominant) or Tt (heterozygous). Individuals who are non-tasters will always be tt (homozygous recessive). To understand how the genes are inherited, examine table 2 below where the potential offspring of two heterozygous parents are analyzed. There is a 75% chance of having children that are tasters for PTC and a 25% chance of having children that are non-tasters. Table 2. Sample Inheritance Pattern for PTC Tasting Parent Alleles T T T TT (homozygous taster) Tt (heterozygous taster) t Tt (heterozygous taster) Tt (homozygous non-taster) In this section of the lab, you will use a set of data that was collected from a previous class. This data has not yet been analyzed, therefore you will have to sort through it and determine what conclusions, if any, can be made. To collect this data, 35 participants were asked to place a strip of PTC paper (paper coated with phenylthiocabamide) on their tongue. As soon as they detected a taste, they were instructed to remove the paper and report what the taste was, or if they never detected any taste. Those who reported that it was extremely bitter, “to the point of almost not being able to take it” as several commented, were also recorded. It is normal that some individuals who are tasters are also much more sensitive to PTC and thus sense the bitterness much more strongly than normal tasters. We call these individuals “Super tasters”.

Questions: 7. What is the total number of non-tasters in the class? 9 8. What is the percent of the class that are non-tasters of PTC? 9/35 x 100=25.7% 9. What is the genotype on non-tasters? tt 10. What is the total number of tasters in the class? 26 11. What is the percent of the class that are tasters of PTC? 74.2% 12. What is the genotype(s) of tasters? The genotype of tasters is TT or Tt 13. Given that some tasters are super-tasters, what is the percentage of the class that are super tasters? 22.9% 14. Individuals that are known as “Super tasters” are thought to have more of the protein which binds to and detects PTC on their tongues compared to tasters. Given this information, what genotype do you think that Supertasters are and why? Supertasters genotype would be TT. They have much more sensitivity due to the gene TAS2R38 which encodes the chemical receptors in the mouth that binds to PTC. 15. If two parents, one heterozygote and one homozygous dominant for PTC tasting were to have children, what are all the possible genotypes and phenotypes? Parents genotype: Tt (heterozygous) and TT (homozygous dominant) Genotypes: (homozygous) 50% chance of a child being TT (heterozygous) 50% chance of a child being Tt (Tasters) 100% chance all children being able to taste PTC Phenotypes: (Taster) this results from both the TT and Tt genotypes since the presence of at least one dominant allele T confers the ability to taste PTC.

Part 3: Determining Inheritance Figure 7: Two Punnett squares showing possible sperm on the vertical edge and possible eggs on the horizontal edge. All boxes are filled in according to the alleles contributed by the sperm and egg. This Photo by Unknown Author is licensed under CC BY 3.0 Recall that using a Punnett square allows you to determine all the possible genotype combinations that offspring could inherit from two parents. In a Punnett square all possible types of sperm are lined up on one side of the square and all types of eggs are lined up on the other. Then the squares are filled in according to what each sperm and egg would give to the resulting zygote. By analyzing all the resulting genotypes, it is possible to calculate the probability of each genotype occurring. Figure 7 shows two Punnett squares with all possible sperm displayed on the vertical edge and all possible eggs on the horizontal edge. All boxes have been filled in according to the alleles contributed by the sperm and egg for that box. We can now calculate the probabilities of each genotype and phenotype. All probabilities are out of 4, as this is a 4 square box. For the left Punnett square, the single box filled with genotype bb is highlighted in red. As this is the only bb present it has a probability of 1 in 4, ¼, or 25%. This is both the genotypic and phenotypic ratio. For the right Punnett square there are three boxes highlighted in red, indicating the dominant phenotypes. Because there are three of the four boxes the phenotypic ratio is 3 of 4, ¾, or 75%. This is not the genotypic ratio however as there are two genotypes included in these three selected squares. The genotypic ratio for that Punnett square would be a 1:2:1 ratio. Questions: 16. Fill in the Punnett square: a a

20. Punnett squares can also be used to predict genotypes of the parents. A guinea pig with short hair is crossed to one that has long hair. Over a period of years, a scientist keeps records of the offspring and finds that 44 of them have curly hair and 46 of them have straight hair. Based on this data, what are the genotypes of the parents: _CC_ x _cc_ Show the Punnett square of this cross to explain your reasoning. Genetic counselors are trained to detect inheritance patterns of genetic diseases based on information they obtain from the family. Imagine that you are a genetic counselor, and you must solve the following cases based on the information provided. Use the following steps to solve each problem:  Create a legend that indicates the gene pairs (alleles) involved. Use a capital letter to denote the dominant allele and lowercase letter to denote the recessive allele. Example: D= dimples d= no dimples C = Curly hair (dominant) c = Straight hair (recessive)  Write the genotype and phenotype of the parents. Example: DD → dimples CC or Cc  curly hair cc  straight hair  Use a Punnett Square to cross the potential gametes of the parents.  Determine the probability based on Punnett Square. 25% probability offspring having CC 50% probability offspring having Cc 25% probability offspring having cc Autosomal Disorders: 21. Autosomal Recessive Inheritance (trait only expressed if homozygous recessive) An albino man (nn) marries a normally pigmented woman (N_) who has an albino mother (nn). What is the chance that their children will be albino? _C _ _c _ c _ Cc cc _c _ CC cc

The chance that their children will be albino is 50%. The children will be carriers of the albino gene (Nn). 22. Autosomal Dominant Inheritance (trait expressed as long as one dominant allele is present) A daughter wants to know what the chances are that she will develop Huntington’s disease, a degenerative disorder of the nervous system that appears during the ages of 30s to 40s. Her mother has Huntington’s while her father does not have the disease. Try to determine the possibilities from the information you have at hand. What further information do you need in order to more accurately determine the probability? The daughter’s genotype could be one of the following: Hh (inherited the mutated gene from her mother) or hh (inherited the normal gene from both parents). By inheriting the mutated gene (H) from her mother, she is at risk of developing Huntington’s disease. If she inherits the normal gene (h) from both parents, then she would not develop the disease. However, we would need to know her specific genotype (Hh or hh), to be able to determine the exact probability. 23. Autosomal Recessive Inheritance (trait only expressed if homozygous recessive) Cystic fibrosis (CF) is a recessive genetic disease that leads to persistent lung infections reducing the ability to breathe over time. Patients with CF have a mutation in the CFTR gene which prevents the movement of chloride – a component of salt – across cell membranes. This makes mucus in the body very thick and sticky so that it clogs airways and traps bacteria. It also prevents digestive enzymes from releasing and causes poor nutrition. CF is very prevalent in Caucasians of northern European ancestry. A carrier is a person who appears normal but carries the recessive allele for a disease. If two parents are both carriers (Cc ) for CF, what is the chance they will have a child with CF? 25% or 1 in 4 chances The chances are: CC (normal), Cc (carrier), cc (affected with cystic fibrosis) 24. Von Willebrand Disease is an autosomal dominant disorder (not located on the sex chromosomes) where blood will not clot properly. What would be the two possible genotypes of a person who has the disorder? (VV) , (Vv) 25. If a person is heterozygous for the trait (having the disease) is married to a normal spouse (dd), what is the chance that their children will have the disorder? 50% X-Linked Disorders:

Figure 9: Some types of rickets may follow an XD mode of inheritance. (Wikipedia- Mrish-CC:AS) (middle and left) Two pedigrees consistent with XD inheritance. (Original-Deyholos_CC:AN) Keep in mind that in autosomal dominant and recessive traits, the sex of the individual does not change the probability of being affected, because all individuals have two copies of autosomes. Questions: 26. In humans, hemophilia is a sex-linked trait. Females can be normal, carriers, or have the disease. Males will either have the disease or not (but they won’t ever be carriers). X H X H = female, non-hemophilic X H Y = male, non-hemophilic X H X h = female, carrier X h X h = female, hemophiliac X h Y= male, hemophiliac Show the cross of a man who has hemophilia with a woman who is a carrier.

What is the probability that their children will have the disease? 2 out of 4, 50% 27. A woman who is a carrier marries a non-hemophilic man. Show the cross. What is the probability that their children will have hemophilia? 50% What sex will a child in the family with hemophilia be? Male (XhY), hemophilia sex linked trait 28. A woman who has hemophilia marries a non-hemophilic man. How many of their children will have hemophilia, and what is their sex? 50% of their children. Male (XhY) and female (XhXh) possible genotypes: XHY (male non-hemophilic), XhY (male, hemophilic) XhXH (female, carrier), XhXh (female, hemophilic) 29. The possible genotypes and phenotypes for an x-linked recessive disorder color-blindness is as follows: X B X B = female, normal vision X B X b = female, carrier X b X b = female, color blindness X B Y = male, normal vision X b Y= male, color blindness Show the cross of a female carrier with a normal vision male. What is the probability that their children will have color blindness? 50%

34. A black male is crossed with a calico female. Show the Punnett square and list the phenotypes and proportions. Phenotype= 3:1 Part 4: Human Traits not Governed by Mendelian Genetics - Multiple Alleles and Codominance There are many exceptions to Mendel’s Rules. For example, blood types in humans exhibit two exceptions: codominance and multiple alleles. When a trait is controlled by multiple alleles , the gene will have more than two possible alleles. However, each individual can still only ever have two alleles, one from each chromosome given by their parents. The classic example of multiple alleles is human blood types – A, B and O. The alleles are denoted as: I A , I B , i . We do not use O to denote the allele, as O and o are too similar and mistaking blood types is almost always fatal. Red blood cells present antigens on their surface which act as passes allowing the red blood cells to move through the body without being mistaken for an intruders or pathogens. Individuals with the I A allele will have the A antigen present on their red blood cells. The same for the I B allele, which causes the B antigen to be presented. If a person has both the I A and I B alleles their red blood cells will present both A and B antigens, this is called codominance . Conversely, if an individual has the i allele their red blood cells will not present any antigen. See Figure 10 below.

Figure 10: Representation of red blood cells and the antigens they carry. " Human Inheritance" by LibreTexts is licensed under CC BY-NC-SA 3.0. Table 4: Basic Genotypes, Antigens and Blood types of Humans Genotypes Antigens on Blood Cells Blood Type I A i , I A I A A A I B i, I B I B B B I A I B A and B AB ii None O Questions: 35. If two individuals with blood type AB marry, what are the odds that each of their children will have the AB blood type? IAIAType A: 25%; IAIB Type B: 25%; IBIB Type AB: 50% 36. A disputed paternity case! Hermione’s new baby has a blood type of O. Her blood type is B and Ron Weasley’s is A. Harry Potter (blood type AB) insists he is the child’s father. CAN THIS BE TRUE? No, it is not true because Hermione Type B (IBIB or IBi) and Ron Type A (IAIA or IAi) could both pass O alleles to their child, resulting in a type O (ii) blood child. It is impossible for Harry blood type AB (IAIB) to be the father.