B- indicates dominant phenotype

Transcription

1 BIO 208 Genetics Applied Human Genetics Pedigree Analysis Monohybrid Cross Dihybrid Cross Chi Square Analysis Probability Epistasis I. Applied Human Genetics/Single Gene Traits The classical study of single gene traits is by pedigree analysis. Over 7000 human single gene traits have been recognized, many also assigned to the particular chromosome on which they reside. The Human Genome Project will uncover many more genes and further study in proteomics will assign function to these genes. For more information on the mechanism of inheritance of single gene traits, allelic variations, and chromosome mapping, consult Online Mendelian in Man The genotype of a homozygous recessive individual is usually known by the phenotype exhibited. For example, individuals with blue eyes are bb. However, the dominant phenotype could arise from two genotypes. A person with the dominant brown eye phenotype could have the BB or Bb genotype. Therefore the genotype is represented as B-. The dash indicates that while a second allele is present, the nature of that allele is unknown BB = homozygous dominant Bb = heterozygous bb = homozygous recessive B- indicates dominant phenotype Single gene traits Interlocking fingers (I) Interlock your fingers and observe which thumb is on top (right or left). The tendency to place the left thumb on top is due to a dominant allele (I). Those who place the left thumb on top have the genotype I- (either II or I-). The right thumb on top is determined by the ii genotype Ear lobes (E) Examine your partner s earlobes. The dominant allele (E) results in the phenotype of free earlobes. The recessive allele (e) is for attached earlobes. Earlobes are attached if the bottom lobe is attached directly to the head. Earlobes are free if the lobe hangs free. Widow s peak (W) Widow s peak occurs when the hairline forms a distinct point in the center of the forehead. Lift up the hair of your forehead to score this trait. People without widow's peak have a smooth hairline with no dip. Men starting to go bald (or already bald) may be unable to score this trait. Widow's peak is controlled by a dominant allele (W) Tongue curling (T) A dominant allele (T) results in the ability to curl the tongue in a U-shape Hitch hiker s thumb (H) A person homozygous recessive for this trait (hh) can bend the last (distal) thumb joint back to about a 90 degree angle. Those with the H allele cannot Pigmented iris (M) A person with the dominant allele has brown or brownish eyes. The recessive allele encodes blue eyes

2 BIO 208 Genetics Polydactyly (P) A person with polydactyly has >5 fingers on 1 or more hands and feet. It is dominant. Mid-digital finger hair Examine middle segment of your fingers. If hair is present, even one, you have mid-digital hair. PTC tasting (T) If you can taste PTC, you have a dominant allele for this trait (T). Place the strip of PTC paper on your tongue for a few seconds. If you cannot taste anything, you do not possess the dominant allele. Fill in the following table Gene Describe phenotype Genotype Earlobes (E) Widows peak (W) Tongue roll (R) Hitchhikers thumb (H) Pigmented iris (M) Polydactyly (P) PTC taster (T) Interlocking fingers (I) Class phenotypic frequencies. Gene f (E-) dominant phenotype f(ee) recessive phenotype Tongue roll (R) Pigmented iris (P) Polydactyly (PD) PTC taster (T) Interlocking fingers (I)

3 BIO 208 Genetics II. Pedigree Analysis Symbols 1. A man who has pointed ears came to the attention of a geneticist. The geneticist finds the following: Pointy ears prove to be an inherited trait due to a single genetic locus. The man's sister has pointed ears, but his mother, father, his brother, and other sister have normal ears. The man and his normal-eared wife are first cousins and have seven children, including four boys and three girls. Two girls and two boys have pointy ears, the two boys are identical twins. (a) Draw the pedigree. Include appropriate symbols and shading for the affected individuals. Fill in the genotype of each individual in the pedigree. If the second allele cannot be determined, use a -. (b) Which one of the following best describes the inheritance of pointy ears in this family? Autosomal recessive Autosomal dominant Sex-linked

4 BIO 208 Genetics Asparagus officinalis (asparagus) is a member of the lily family and a popular vegetable to grow and eat. A pungent urinary odor produced within a few hours of eating asparagus is due to sulfurcontaining metabolic breakdown products, or S-methyl thioesters. The smell cannot be noticed in raw or cooked asparagus. The odor is described as ammonia-like. About 40% of the population can detect S- methyl thioesters in urine. All people who eat asparagus produce the thioesters, it is the ability to smell them that is inherited. Pedigree of the ability to detect S-methyl thioesters I II III IV (a) Is the ability to detect S-methyl thioesters inherited as a dominant, or as a recessive, trait? How do you know (which individuals inform you about the inheritance)? (b) Label the genotypes of all individuals on the pedigree. If the second allele cannot be determined, use a dash to represent it. (c) Explain (or diagram) how individuals III,1 and III, 6 could have a non-smeller child

5 BIO 208 Genetics Alkaptonuria is a rare genetic disease in which the body does not have enough of the enzyme homogentisic acid oxidase (HGAO). Because the HGAO enzyme is low or absent, homogentisic acid (HGA) builds up in the body. Some HGA is eliminated in the urine which turns brownish black. The rest of the pigmented material is deposited in body tissues where it is toxic. Patients with alkaptonuria are usually not aware of the disease until about age 40 when symptoms are present. Dark staining of the diapers sometimes can indicate the disease in infants, but usually no symptoms are present until much later in life. Alkaptonuria can affect many body systems. Diets low in protein--especially in phe and tyr- -help reduce levels of HGA, thereby lessening amount of pigment deposited in body tissues. There is no cure. Skeletal knees and hips most. Deposits of pigment cause brittle cartilage Cardiovascular - The aortic and mitral heart valves are most affected. Genitourinary - Pigment deposits can form stones in the prostate. Respiratory - Pigment deposits in the cartilage of the larynx, trachea, and bronchi. Ocular - Vision is not usually affected, but there is pigmentation in the white part of the eye. Cutaneous (skin)-skin takes on blue-black speckled discoloration. Sweat stains clothes brown. The gene for alkaptonuria (ALK) is on human chromosome 9 and is linked to gene encoding the ABO blood group. A pedigree of a family with the disease is shown below, affected individuals indicated in black. In addition, the blood type of family members is given. The 2 alleles at the ALK locus are denoted ALK+ and ALK-. The 3 alleles at the ABO blood group locus will be denoted I A, I B (co-dominant) and i (recessive to I A and I B ). (a) What is the genotype of individual 1 at the ALK and ABO loci? (b) What is the genotype of individual 2 at the ALK and ABO loci? (c) What is the genotype of individual 3 at the ALK and ABO loci? (d) Individuals 3 and 4 are expecting their fifth child. A physician draws a prenatal blood sample. What is the probability that the child will have alkaptonuria AND type B blood? Explain your answer.

6 BIO 208 Genetics III. Corn Genetics Zea mays is a model organism in genetics. All kernels on one ear represent the progeny (offspring) of a single cross. The effects of many genes can be observed in the kernels, eliminating the need to grow plants. Many genes determine the phenotypes of the 3 tissues that control the color of a corn kernel. Endosperm (kernel) traits Color gene P allele = purple (color) p allele = yellow (no color) Kernel shape gene R allele= smooth r allele= wrinkled A. Collect Data STATION 1 1. What are the two color phenotypes observed? 2. Count 50 kernels. Compute the ratio of phenotypes: 3. Which phenotype is caused by a dominant allele? 4. Identify the parental plants that produced the progeny. STATION 2 Parental genotypes and Parental phenotypes 1. What are the two kernel shape phenotypes observed? 2. Count 50 kernels. Compute the ratio of phenotypes: 3. Which phenotype is caused by a dominant allele? 4. Identify the parental plants. Parental genotypes and Parental phenotypes STATION 3: The ear of corn represents the progeny of a cross between two purple, smooth parents. Genotype of parents: PpRr X PpRr Phenotype: 1. What are the 4 allelic combinations of gametes (eggs and sperm) each parent can produce? Remember, a gamete can only and must contain one allele of each gene.

7 BIO 208 Genetics Construct a Punnett square showing the expected genotypes and phenotypes of the offspring 3. What are the 4 phenotypes (with respect to both kernel shape and color) observed in the offspring of this cross? 4. Count 100 kernels. Compute the ratio of phenotypes (divide each number by smallest number) 5. The expected ratio of phenotypes is 9/16: 3/16: 3/16:1/16 (9:3:3:1) Does the observed ratio approach the expected ratio? B. Chi Square Statistical Analysis The purpose of the Chi Square (X 2 ) test is to determine if data obtained experimentally (observed data) fit into a model of expected data. If data are shown to fit into the model, then any variations from the model are due to chance. If the observed data do not approximate the model, then variations are said to be significant, or not due to chance. If deviations are small they are more easily attributed to chance than are large ones. X 2 = (O E) 2 E O observed data E expected data Examine the Chi Square table below. Across the top of the table are probability values (P). A P-value >0.05 indicates that in over 5% of trials, similar data would be obtained. A p >0.05 leads the researcher to accept the null hypothesis, that the differences between observed and expected data are due to chance. In other words, the observed data fits the expected data. A P value <0.05 corresponds to a 1 chance in 20 (5%) that deviations are due to significant variations. A p < 0.05 indicates that observed variations cannot be explained by the expected model. In other words differences in the observed data are not due to chance. The null hypothesis states that any

8 BIO 208 Genetics variations between observed and expected data are only due to chance. Rejecting the null hypothesis indicates that observed and expected data are significantly different. To determine which row of Chi Square values to use, first determine the degrees of freedom. The degrees of freedom are one less than the number of phenotypic classes. For example if white, yellow, purple, and red corn kernels are observed, the degrees of freedom is 3. Critical Values of the Chi Square Distribution Enter the data obtained from a single gene cross (maize) Phenotype Observed O Expected E Deviations (O-E) 2 (O-E) 2 E TOTAL X 2 = 1. Degrees of freedom (df) = 2. Locate the two Chi Square values on either side of the calculated X 2. What is the lowest probability value assigned to these Chi squares? p value = 3. Do you accept the null hypothesis (variations between observed and expected data are only due to chance) or do you reject the null hypothesis (variations between observed and expected data are significantly different? Explain.

9 BIO 208 Genetics Enter the data obtained from a 2 gene cross (maize) Phenotype Observed O Expected E Deviations (O-E) 2 (O-E) 2 E TOTAL X 2 = 1. How many degrees of freedom (df) are there? 2. What is the lowest probability value assigned to these X 2 s? p = 3. Do you accept the null hypothesis (variations between observed and expected data are only due to chance) or do you reject the null hypothesis (variations between observed and expected data are significantly different? Explain. IV. Probability The coin toss is a convenient tool to demonstrate principles of probability. Probability in coin tossing involves chance. There is a ½ chance that a head will face up and a ½ chance that tails will occur. No matter how many times a coin is tossed, the probability of heads or tails remains the same. If two coins are tossed at the same time, there are 4 combinations that can occur: 2 heads HH 1 head, one tail HT, TH 2 tails TT The chance of obtaining two heads is determined by employing the product rule: the probability of two events occurring simultaneously is the product of their individual probabilities. 1. What is the probability of obtaining two heads? The chance of obtaining a two heads OR a tail and a head is determined by using the sum rule: The chance of either of two events occurring is the sum of their individual probabilities. First use the product rule to determine the chance of HT and TH, then the sum rule to determine the chance of HT OR TH. 2. The probability of obtaining HH or TH is 3. Data collection Obtain 3 coins.

10 BIO 208 Genetics a. What are the possible combinations of each class? Enter this information in the table b. What is the probability of each class? Enter this information in the table c. Toss the 3 coins 12 times. Combine your data with that of another group so that you have a total of 24 throws. Enter in the observed numbers column Classes Combinations Probability Observed #s 3 heads 2 heads, 1 tail HHT, HTH, THH 1/8+ 1/8+ 1/8 = 3/8 1 head, 2 tails 3 tails Totals Binomial expansion Expectations for combinations of outcomes in groups can be obtained mathematically by using the following binomial formula: n! P = p x q n-x x! (n-x)! Note: 0! = babies are born. What is the probability that 2 are boys and 3 are girls? n! = 5! = 5 X 4 X 3 X 2 X 1 n = number of events x! = 2! = 2 X 1 x = number of event 1 (n-x)! = 3 X 2 X 1 n-x = number of event 2 p = probability of a boy = ½ q = probability of a girl = ½ Determine P (2 boys, 3 girls) 2. Albinism is a recessive trait. Those with the albino phenotype are aa. Heterozygotes are Aa, they are not albino, but are carriers for the trait. Two black mice, heterozygous for albinism were mated. Aa X Aa 1. What is the probability of obtaining a white mouse? 2. A black mouse? 3. In a litter of 6 mice, determine P (3 white, 3 black)? 4. In a litter of 5 mice, determine P (all black)?

11 BIO 208 Genetics V. Dominance, Allelic Series, and Epistasis COAT COLORATION AND PATTERNING GENES IN CATS LOCUS GENOTYPES NOTES W W- ww The whole coat is white. all white not all white White is not the same as albino which is recessive B B- bb black brown Eumelanin pigment (seal) (chocolate) C C- c s c s c s c s = Coat color with points full color Siamese Mask on face, legs, tail D D- dd with B- = blue (gray) intense color dilute color with bb = lilac with O o = cream A A- aa agouti (tabby) non-agouti (solid color) S S- ss White spotting no white spotting Cat has patches of white (absence of color) masking true color Challenge 1 Complete the following table. Genotype Phenotype ww BB CC DD aa Black cat ww bb c s c s DD aa ww BB CC DD AA ww bb CC dd aa

12 BIO 208 Genetics Challenge 2 Mate a chocolate, (Dd), Siamese, tabby (AA) female with a lilac, solid color, male homozygous for all trait. What are the possible genotypes and phenotypes of the progeny and what proportions are expected? Challenge 3 A female lilac of genotype ww bb CC dd aa produced a litter of four kittens: 1 black kitten, 1 blue kitten, 1 brown tabby, 1 lilac tabby a. What were the genotype and the phenotype of the father? b. Could these parents produce a lilac (same phenotype as the mother?) Show your work.