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Lecture 2-Ch. 3

by: Hannah Kennedy

Lecture 2-Ch. 3 30156

Hannah Kennedy
GPA 3.98

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These notes cover everything discussed in lecture (Mendelian genetics, law of segregation, law of independent assortment, etc.) as well as complementary material from chapter 3.
  Dr. Helen Piontkivska
Class Notes
Genetics, Biology, mendelian genetics
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This 9 page Class Notes was uploaded by Hannah Kennedy on Saturday July 23, 2016. The Class Notes belongs to 30156 at Kent State University taught by   Dr. Helen Piontkivska in Spring 2016. Since its upload, it has received 32 views. For similar materials see ELEMENTS OF GENETICS in Biological Sciences at Kent State University.


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Date Created: 07/23/16
© Hannah Kennedy, Kent State University July 19 thLecture Notes: Ch. 3 Mendelian Genetics 1. Problem solving a. Q: In an animal zygote with a complement of 2 homologous chromosome pairs, symbolized as A, a, and B, b, which of the following is expected in its somatic cells during growth? Explain i. None of the above (AaBB, AABb, AABB, aabb) because this zygote will result in a heterozygous AaBb somatic cell b. Q: Determine the diploid (2n) numbers of the following organisms from the per cell observations given below: i. 16 tetrads at metaphase I 1. 2n = 32 a. this is because a tetrad = 2 chromosomes (i.e. 4 homologous pairs) ii. 60 chromatids at metaphase I 1. 2n = 30 a. this is because 60 chromatids = 30 chromosomes iii. 12 centromeres at anaphase I 1. 2n = 6 a. this is because we are at a stage prior to division. So, 4n = 12, n = 3, 2n = 6 iv. 12 centromeres at anaphase II 1. 2n = 12 a. this is because we are at a point where we have 2n c. Q: The haploid amount of DNA in humans is 1.91 x 10^12 daltons. (will refer to this value as “a”) How much DNA would be found in a male cell… i. During G1? 1. 2a ii. After the S phase? 1. 4a iii. After mitotic telophase and cytokinesis? 1. 2a iv. After telophase and cytokinesis of meiosis I? 1. 2a v. After telophase and cytokinesis of meiosis II during spermatogenesis? 1. A d. Q: In the pea, tall plant height (T) is dominant over short (t). Pure-breeding tall and short plants are crossed i. What will be the genotype and phenotype of the F1 1. Genotype: Tt 2. Phenotype: heterozygous tall ii. If the F1 is selfed and 400 F2 plants are raised, how many would be expected in each phenotypic class? 1. 100 in the TT 2. 100 in the tt 3. 200 in the Tt iii. How many of the F2 would be expected to be pure breeding when selfed? 1. 200 (TT and tt) 1 © Hannah Kennedy, Kent State University e. Q: consider a tall pea plant: i. If nothing is known of the breeding history of the plant, how should its genotype be written? 1. T? x tt ii. If the plant were to be testcrossed, what would be the phenotype of the other parent? 1. Homozygous recessive iii. If all of the progeny from the testcross were tall, what would be the genotype of the tall plant? 1. TT iv. If the tall plant were heterozygous, what would be the phenotypic ratio of the testcross progeny? 1. 1:1 2. Mendelian genetics: Gregor Mendel (1822-1884) a. Key terminology: i. Crossed = a mating between 2 distinct individuals. An analysis of their offspring may be conducted to understand how traits are passed from parent to offspring ii. Hybrids = offspring obtained from a hybridization iii. Hybridization = the mating of 2 organisms of the same species with different characteristics (e.g. between the purple flower and the white flower) iv. Character = in genetics, this word refers to a general characteristic of an organism v. Trait = variant = any characteristic that an organism displays; morphological traits affect the appearance of an organism. Physiological traits affect the ability of an organism to function. A 3d category of traits are those that affect an organism’s behavior (i.e. behavioral traits) vi. True-breeding line = strain = a strain of a particular species that continues to produce the same trait after several generations of self- fertilization or inbreeding vii. P-generation = parental generation = the first generation in an experiment/cross (usually true-breeding) st viii. F1 generation = 1 filial generation = offspring produced because of the cross of the parental generation; all plants show phenotype of 1 but not the other ix. F2 generation = 2 nd filial generation = offspring produced because of the cross of the F1 generation b. Basics of transmission genetics: Mendelian laws that describe how hereditary characteristics are transmitted from parents to offspring c. Overview of Mendel’s experiments i. Goal: Mendel stated that the inheritance pattern for a single character may follow quantitative natural laws. Wanted to uncover them ii. Process: 1. First, a P generation (true-breeding parents with the same trait) were bred a. Purple flowers x white flowers 2 © Hannah Kennedy, Kent State University 2. Next, an F1 generation (hybrid generation) were crossed a. All plants had purple flowers (white flower “disappeared”) 3. Finally, an F2 generation (generation with a 3:1 ratio) were made a. 705 plants had purple flowers and 224 plants had white flowers b. ratio of white flowers wasn’t random iii. Interpretation: 1. Mendel’s postulates (i.e. the principles of inheritance), 3 things a. Unit factors (i.e. genes) exist in pairs in individual organisms b. Dominance/recessiveness relationships among 2 diff unit factors exist c. During gamete formation, paired unit factors (genes) segregate independently: law of segregation i. Half the gametes receive dominant ii. Half the gamete receive recessive 3. Law of Segregation a. Key terminology: i. Monohybrid cross = a cross in which one gene is crossed ii. Gene = a unit of heredity that may influence the outcome of an organism’s traits 1. Eukaryotes have 2 copies bc genetic material is organized into pairs of chromosomes iii. Allele = different versions of the same gene iv. Mendel’s law of segregation = states that the 2 copies of a gene segregate/separate from each other during transmission from parent to offspring; each gamete can receive dominant or recessive allele with equal probability, 50:50 v. Genotype = the genetic composition of an individual (e.g. tt and TT) vi. Phenotype = an observable characteristic of an organism (e.g. dwarf and tall) vii. Punnett square = a method in which the genotypes of 2 parents are used to predict the genotypes of the offspring 4. Law of Independent Assortment a. Key terminology: i. Dihybrid cross = a cross in which an experimenter follows the outcome of 2 diff traits; contains a 9:3:3:1 ii. Law of independent assortment = during gamete formation, pairs of genes segregate independently of each other and each gamete receives one or the other allele of a particular gene independently of another gene iii. Non-parentals = describes categories in which the combinations of traits aren’t found in the true-breeding plants of the parental generation iv. Genetic recombination = the process in which chromosomes are broken and then rejoined to form a novel genetic combination; the 3 © Hannah Kennedy, Kent State University process in which alleles are assorted and passed to offspring in combinations that are different from parents b. Interpreting the data of a dihybrid cross/experiment: rejected the hypothesis of linked assortment and supported the hypothesis that diff characters assort themselves independently i. Mendel’s law of independent assortment = states that 2 diff genes randomly assort their alleles during the formation of haploid cells aka the allele from 1 gene is found within a resulting gamete independently of whether the allele for a diff gene is found in the same gamete 1. Via homologs randomly assorting themselves during meiosis c. Consequence of independent assortment is that a single individual can make many genetically diff gametes: genetic recombination d. Independent assortment is revealed by a 2-factor testocross i. This is when dihybrid individuals are mated to homozygous recessive individuals for all characters 5. Sources of Genetic Variation in Humans (2 main sources) a. Sexual reproduction (3 sub-sources) i. Independent assortment at meiosis I 1. Physical fitness ii. Random fusion of sperm and egg at fertilization 1. Diff set of alleles that will end up in the gamete iii. Cross-over/recombination at prophase I 1. Homologous chromosomes aligned as a tetrad exchanges genetic material during prophase I 2. Chromosome theory of inheritance = inheritance patterns of traits can be explained by the transmission patterns of chromosomes during gametogenesis and fertilization (based on 5 fundamental principles) a. chromosomes contain genetic material that is transmitted from parent to offspring and from cell to cell b. chromosomes are replicated and passed along from generation to generation and are passed from cell to cell during organismal development. Each chromosome type retains individuality during cell division and gamete formation c. nuclei of eukaryotic cells contain chromosomes that are found in homologous pairs (i.e. they are diploid). 1 part of the pair is from mom and other part of pair is from dad. Gametes contain 1 set of chromosomes (i.e. they’re haploid) d. during formation of haploid cells, diff types (i.e. nonhomologous) chromosomes segregate independently of each other e. each parent contributes 1 set of chromosomes (23) to its offspring b. Random mutations 6. Designating alleles 4 © Hannah Kennedy, Kent State University a. 2 alleles encoding a trait are located in similar locations (i.e. loci) on 2 homologous chromosomes i. these alleles are part of the same homologous locus b. in a heterozygous individual, 1 allele is dominant and the other is recessive 7. Probability theory: allows us to determine the probability that a cross btwn 2 individuals will produce a particular outcome a. Key terminology: i. Probability = the chance that an event will occur in the future; depends on the number of possible outcomes ii. Random sampling error = the deviation between the observed and expected outcomes b. P = a/n : the probability of a random event i. A = # times event occurred ii. N = total # of possible cases c. When several events are considered, 2 rules can apply i. Sum rule = either/or rule = states that the probability that one of 2 or more mutually exclusive events will occur is equal to the sum of the individual probabilities of the events. e.g. p(2 or 3) when rolling one dice aka what is the probability of getting a 2 or a 3 1. Mouse example: sum rule will let us determine the probability we will obtain any 1 of 2+ diff types of offspring a. de = droopy ear b. De = normal ears c. ct = crinkly tail d. Ct = normal tail i. Expected ratio is 9:3:3:1 bc we are comparing 2 traits. 9 will have normal ears and normal tails bc these are both the dominant traits. 3 will have normal ears and crinkly tails bc this is heterozygous. 3 will have droopy ears and normal tails. 1 will have droopy ears and crinkly tails. 1. Cross: Dede Ctct x Dede Ctct 2. Q: what is the probability that an offspring will have normal ears and a normal tail or have droopy ears and a crinkly tail? 3. Step 1: calculate the individual probabilities of each phenotype by using a Punnett square a. 9/(9+3+3+1) = 9/16 = the probability of normal ears and a normal tail b. 1/(9+3+3+1) = 1/16 = the probability of droopy ears and crinkly tail 4. Step 2: add the individual probabilities a. 9/16 + 1/16 = 10/16 therefore 10/16 is probability that offspring will have either normal ears and normal tail or have droopy ears and crinkly tail ii. And rule = product rule = the probability that 2+ independent events will occur is equal to the products of their individual 5 © Hannah Kennedy, Kent State University probabilities. e.g. p(4 and 4) when rolling two dice aka what is the probability of getting a 4 on the 1 roll and a 4 on the 2 ndroll 1. Congenital analgesia example: a. P = normal allele b. p = recessive congenital analgesia allele i. Cross: Pp x Pp ii. Q: what is the probability that the couple’s first 3 offspring will have congenital analgesia iii. Step 1: calculate the individual probability of this phenotype by using a Punnett square 1. The probability of an affected offspring is ¼ iv. Step 2: multiply the individual probabilities st 1. Bc we’re asking about the 1 3 offspring, we multiply ¼ three times therefore ¼ x ¼ x ¼ = 0.016 so the probability that the first 3 offspring wil have this is 0.016 2. Example: what is the probability that the 1 offspring will be unrdfected, the second offspring will have the disease, and the 3 will be unaffected? (predicting the probability of a sequence of events that involves combos) a. Step 1: calculate the individual probability of each phenotype using a Punnett square i. Unaffected = ¾ ii. Congenital analgesia = ¼ b. Step 2: calculate the probability that these 3 phenotypes will occur in this specified order by multiplying their respected individual probabilities together i. ¾ x ¼ x ¾ = 0.14 = 14% 3. Example: AaBbCC x AabbCc. What is the probability that an offspring will have the genotype AAbbCc? (predict the outcome of a cross involving 2+ genes) a. Step 1: separately calculate the probability of the desired outcome for each gene via a Punnett square i. AA = ¼ ii. bb = ½ iii. Cc = ½ b. Step 2: multiply each individual probability from step 1 together to determine the probability that the offspring will be the specific one given i. P = ¼ x ½ x ½ = 0.0625 = 6.25% 8. Dihybrid cross: 9:3:3:1 a. F1xF1 selfing leads to a 9:3:3:1 ratios of phenotypes due to the law of independent assortment b. Spongebob example i. Family history 1. Spongebob’s Aunt Sylvia (squarepants) is married to Jack- Sponge (roundpants), and Sylvia and Jack’s 5 children all have squarepants 6 © Hannah Kennedy, Kent State University 2. Sylvia’s sister has a blue sponge and is married to Keith, who is purebred yellow (and comes from a family of yellow sponges). Their children all have yellow sponges. a. Family history is impt bc this tells us which traits are dominant b. Squarepants and yellow sponge are dominant c. Roundpants and blue sponge are recessive ii. Purebred squarepants and yellow sponge Bob wants to marry Ethel who comes from a family of roundpants with blue sponges 1. Their parents are shopping for children’s gifts to match their future grandchildren. What should they shop for? a. This would be the F1 generation b. All yellow squarepants: because SpongeBob is a purebred, he is homozygous dominant and will mask any recessive allele in Ethel 2. Bob and Ethel want to stock up on clothes for their own grandchildren—what should they get? a. This would be the F2 generation b. Do a branch diagram, 9:3:3:1 i. 9 yellow square pants ii. 3 blue square pants iii. 3 yellow round pants iv. 1 blue roundpants 9. When the cross involves 3+ independent loci: Branch diagram a. Branch diagram = forked-line diagram = can be used to visualize results of di/tri/etc. crosses that involve multiple independent loci 10. Chi square analysis = analysis to test the statistical significance of differences between observed and expected values a. Key terminology i. Hypothesis testing = using statistical tests to determine if the data from experimentation are consistent with a hypothesis ii. Goodness of fit = the degree to which the observed data and expected data are similar to each other. If the observed and predicted data are very similar, the goodness of fit is high iii. Null hypothesis = a hypothesis that assumes there is no real difference between the observed and expected values iv. Chi square test = a commonly used statistical method for determining the goodness of fit. This method can be used to analyze population data in which the members of the population fall into different categories v. P-value = in a chi square table, the probability that the deviations between observed and expected values are due to random chance vi. 5% and 1% significance levels = common convention in which the scientist rejects the null hypothesis if the chi square value results in a probability is less than 0.05 or less than 0.01 vii. degree of freedom = the number of categories that are independent of each other b. Chi Square Test: about i. Used to test the validity of a genetic hypothesis, hypothesis testing ii. Goal: determine if the data from genetic crosses are consistent with a specific pattern of inheritance 7 © Hannah Kennedy, Kent State University iii. Rationale: evaluate the goodness of fit between the observed data and the data that are predicted from a hypothesis (null hypothesis); any differences are assumed to be due to random sampling error iv. If observed and predicted data are similar we can accept the null hypothesis but it doesn’t prove the hypothesis v. If there is a high deviation between observed and predicted then there’s a poor fir between the hypothesis and expected values and the null hypothesis is rejected c. Chi Square Test: Equa2ion 2 (O−E) i. X =Σ E ii. O=observed data∈eachcategory iii. E = expected data in each category based on the experimenter’s hypothesis iv. Σ=¿ ∑ thiscalculation for eachcategory 2 v. x = tells us how strong the deviation from expected is; we will accept the hypothesis if its less than the alpha number and reject it if its more 11.Pedigrees: to figure out inheritance mode of genes a. General info i. Organized in time 1. Trying to figure out if the trait is present in all the generations we’re considering or only some (skipping generations) a. If it doesn’t skip any generations, it is most likely dominant b. If it does skip generations, it is most likely recessive ii. Where the trait is present 1. In males only? a. X-linked 2. In females? a. Chromosomal (autosomal) b. The following pedigree, this trait is most likely recessive because it skips a few generations from their parents. There is also a gender imbalance present here because the trait only shows up in males which means the trait must be coming from an X-chromosome. i. This means that boys inherit their X from their mom c. The following pedigree shows a dominant trait because it skips no generations. The trait is also autosomal bc there is a transmission from dad to son (2 to 2) ruling out X-linked since dad doesn’t contribute an X chromosome to son. d. The following pedigree i. Dominant/recessive: dominant bc there are a lot of ppl affect 8 © Hannah Kennedy, Kent State University ii. Mode of inheritance: X- linked bc there is no transmittance from dad to son, just dad to daughter. And there is a transmittance from mom to son. 1. If mom is heterozygous it’s possible that she has some sons that are affected and some that aren’t affected. Like shown to the left e. Key: f. A recessive pattern of inheritance makes 2 impt predictions: (1): 2 heterozygous normal individuals will have ¼ of their offspring affected (2): all offspring of 2 affected individuals will be affected g. A dominant trait predicts that affected individuals will have inherited the gene from at least 1 affected parent unless a new mutation happened during gamete formation 9


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