Need a study break? Check out this fun 5-minute NY Times video about some recent research investigating the genetic and other biological differences between dogs and wolves. You can find the video here.
Answers to Example Midterm Questions from 2016
#1: C
#6: A
#16: A) black scales, bubble eyes; B) 1/4 x 1/4 = 1/16; C) (1/2 x 1/2) x 100 = 25 fish
#20: A) two (from genes F and G); B) 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1 = 1/64
Week 3 Reflections
This week we expanded our knowledge of different variations on dominance concepts (for example, incomplete dominance and codominance) and entered the world of pedigree analysis. Pedigrees are particularly useful in species (such as humans) where it is impossible or highly impractical to get the very large numbers of offspring required for the analytical approaches of Mendel. The patterns by which affected individuals appear on the pedigree provide ‘tell tale’ signs indicative of the dominance relationships of the alleles. We also entered the fun (IMO) world of epistasis where the interactions between gene products, for example in biochemical pathway contexts, can perturb the F2 dihybrid ratios away from the usual expected 9:3:3:1 for genes whose products do not interact. There are lots of different possible modifications of the 9:3:3:1 that result from epistasis, such as 12:3:1, 9:7, and 15:1. These different ratios result from different particular underlying biochemical explanations. You don’t need to memorize these, but do need to be able to reason your way to these modified ratios (and their corresponding phenotypes, genotypes) when given cross data. After our epistasis adventure, we moved into a review of mitosis and meiosis, emphasizing the dynamics of chromosomes during these cell division processes. We concluded Thursday’s lecture with an overview of sex chromosome transmission processes, and learned about nematode sex 🙂
Week 4 Sneak Peek: Of course, the main event will be the Tuesday Midterm during Week 4. Study hard! Bring your calculators! Know your probabilities! After the midterm, we will delve further into chromosome biology in Chapter 5.
Answers to Quiz #2
Correct Answers are in bold; note that there are two possible correct answers for #4.
#1. (5 points) An allele is:
a) the homozygous genotype
b) the heterozygous genotype
c) one of many possible forms of a gene
d) another word for a gene
#2. (5 points) You perform a testcross involving two genes, D and E, each with two alleles showing simple dominant/recessive relationships. D is dominant to d, and E is dominant to e. Which of the following crosses accurately describes a testcross involving these genes?
a) An individual with both dominant phenotypes, crossed with a DD, EE
b) An individual with one dominant and one recessive phenotype, crossed with a DD, EE
c) An individual with both dominant phenotypes, crossed with a dd, ee
d) An individual with one dominant and one recessive phenotype, crossed with a dd, ee individual
#3. (5 points) You are studying Mendel’s peas and have set up a cross between a true-breeding plant with round peas and a true-breeding plant with wrinkled peas. The round phenotype is dominant to the wrinkled phenotype. You then perform a F1 x F1 cross and analyze 400 F2 How many F2 plants do you expect to have the round pea phenotype?
a) 400
b) 300
c) 100
d) 0
#4. (5 points) Two genetic mutants of Neurospora crassa are defective in the Alanine biosynthesis pathway. After complementation testing, the two mutants were placed into different complementation groups. Which of the following statements most accurately describes the two mutants:
a) they must have mutations in different specific genes
b) they must have mutations at different exact nucleotide positions
c) they must have mutations in the same specific gene
d) they must have mutations at the same exact nucleotide positions
e) both A and B are true
f) both C and D are true
Answers to HW #3
Chapter 3
3.4. b: 0, c: ½
3.8. Panel A is anaphase I of meiosis, because the homologous chromosomes are paired. Panel B is anaphase II of meiosis, because the chromosome number has been reduced by half. Panel C is anaphase of mitosis, because the homologous chromosomes are not paired.
3.10: 5/6 expected to survive.
3.12: A) [0.5 x 0.5] + [0.5 x 0.5] + [0.5 x 0.5] = 0.75. B) 0.5 x 0.5 = 0.25
3.17: Note first that there is no chance of the X-linked color-blindness allele being passed to II-4, because he inherits his Y chromosome from his father (I-2) and his X chromosome from his normal mother (I-1). Because we are told that I-2 is the only possible source of the mutant gene, this means that there is no possibility that the male III-3 carries the mutant gene. (Moreover, III-3 is not color blind, and so his X chromosome must be nonmutant.) On the other hand, the female II-2 must be a carrier for the color-blindness allele because she inherits an X chromosome from her father (male I-2), whom we know carries the allele for color blindness. Given that II-2 is heterozygous, there is a probability of 1/2 that she transmits the allele to her daughter (III-2) and another probability of 1/2 that III-2 transmits the allele to the male IV-1. Hence, the total probability that IV-1 is color blind is 1 x 1/2 x 1/2 = 1/4.
3.19: There are 1000 progeny altogether, and the expected numbers are 562.5, 187.5, 187.5, and 62.5, respectively, from top to bottom. This chi-square test has 3 degrees of freedom (four classes of data); the final chi-square value is 3.6. The P value is about 0.33. The data give no reason to reject the hypothesis of 9:3:3:1 segregation.
Chapter 5
5.12: The mom has a Robertsonian translocation joining the long arm of Chr. 21 with the long arm of another acrocentric chromosome. The child has 46 chromosomes. This karyotype differs from the usual situation for trisomy 21 because the extra chromosome 21 is not a free chromosome.
Additional Question not in your textbook:
#1. An insect from a strain that is true-breeding for white eyes was crossed to an insect from a strain that is true-breeding for red eyes. All the F1s had white eyes. The F2s consisted of:
243 white eyed, 58 red eyed, 21 cream eyed.
Provide the genotypes for the parents, F1s and F2s of the cross.
Hint: epistasis!
How to solve this problem? First, start by analyzing the F2 phenotypic ratios:
Total progeny = 322
White = 243/322 ≈ 12/16
Red = 58/322 ≈ 3/16
Cream = 21/322 ≈ 1/16
The 16 in the denominator indicates that the F1 is segregating for two gene pairs, and thus that the F1 is a dihybrid. The F1 can arbitrarily be given the genotype of Aa Bb. The F2s can now be analyzed with respect to a modified 9:3:3:1 dihybrid ratio:
9 A_ B_
3 A_ bb
3 aa B_
1 aa bb
Thus,
P aa BB red x AA bb white
F1 Aa Bb white
F2 A_ _ _ white
aa B_ red
aa bb cream
Genetics of Skin Color in Humans
This week, a study was published in Science magazine about the genetics underlying skin color in human populations. You can read a NY Times piece highlighting the research here.
Example Questions from 2016 Midterm #1
#1. (3 points) The process of replication in most cellular organisms involves:
a) synthesis of a RNA product from a DNA template
b) synthesis of a protein product from a RNA template
c) synthesis of a DNA product from a DNA template
d) synthesis of a DNA product from a RNA template
#6. (3 points) You are a geneticist studying feather color in pigeons. You cross a true-breeding white pigeon to a true-breeding black pigeon. All of the F1 offspring have black feathers. You then mate F1 males to F1 females. The F2 offspring have the following phenotypic ratio 9 black : 7 white.
What genetic phenomenon is the most likely cause of this result?
a) epistasis
b) codominance
c) incomplete dominance
d) incomplete codominance
#16. In goldfish, black scales (G) are dominant to gold scales (g). A separate, independent gene controls eye shape where bubble eyes (N) are dominant to normal eyes (n). You cross a true-breeding black, normal eyed male fish with a true-breeding gold, bubble eyed female fish to create F1 offspring.
Part A. (3 points) What is the phenotype of the F1 offspring?
Part B. (4 points) If you intercross the F1 offspring to create F2 offspring, what is the probability of finding a gold, normal eyed fish in the F2 offspring?
Part C. (4 points) If you genotype 100 F2 fish, how many do you expect to have the genotype GgNn?
#20. You are studying a species of sea cucumbers and analyzing seven unlinked genes, each having two alleles that demonstrate simple dominant/recessive relationships, with no epistasis. You set up the following cross:
AaBbCcddEeFFGG X aabbccDdeeFfGG
Part A. (4 points) How many of the seven phenotypes are shared by the two cucumbers in the above cross?
Part B. (5 points) What is the probability of an F1 progeny of the following genotype?:
aaBbCcDdEeFFGG
Week 2 Reflections
At the beginning of the week, we went through mutation screening methods with Neurospora and associated complementation testing. These studies provide great examples of the type of logic that geneticists use – make sure you understand the logic that determines whether different particular genetic mutants go into the same versus different complementation groups. Next, we entered into classical Mendelian (aka transmission) genetics – how the results of thousands of crosses between pea plants of varying phenotypic traits and combinations of traits led to our current understanding of how different versions (alleles) of genes segregate, and how different alleles, and allele combinations, lead to different phenotypic outcomes. After revisiting the basic concepts and Punnett square approaches to understanding how alleles are transmitted, probability concept approaches were introduced that offer much faster and practical ways to solve these problems (especially when it comes to many genes). Mendel was EXTREMELY lucky to have worked with seven allele pairs (each pair for one of seven different genes) that each affect different “easy to score” phenotypic traits where there are very simple “full dominance” relationships between each of the allele pairs. We also entered into variations on dominance concepts such as incomplete dominance and codominance. Most phenotypic traits of interest to the health sciences (e.g. susceptibilities to genetic diseases and cancers, ability to defend against infectious microbes and parasites, aging) and agricultural sciences (e.g. plant drought tolerance, resistance to pests, livestock growth rates) are influenced by many genes. We will study these more complicated scenarios later in the term.
Week 3 Sneak Peek: Next week we will finish up Ch. 2, discussing more about variations on dominance concepts and epistasis. Ch. 3 will follow where discussion will turn toward the biology of chromosomes and the genetics of sex chromosomes. We will also cover the chi-square test.
Answers to Homework Assignment #2
ANSWERS: BI311 Homework Assignment #2
Chapter 2
2.2: 2/3. Because the seed is round it cannot be ww. Ww seeds give 2 bands in the gel. Of the round seeds, 1/3 are WW and 2/3 are Ww.
2.4. Note that you assume ‘rare’ alleles to exist in heterozygous state.
a) parents must be Aa x aa so the risk of having an Aa child is 1/2
b) parents must be Aa x Aa so the risk of having an aa child is 1/4
c) parents must be Aa x aa so the risk of having an aa child is 1/2
2.8.
a) Individual II-2 is aa
b) Both individuals I-1 and I-2 must be Aa
c) Individuals II-1 and II-3 are either AA or Aa
d) Because II-3 is phenotypically normal, the odds that they are heterozygous carrier is 2/3
2.16: Due to Mendelian segregation, the expected numbers are: a) 0; b) 100; c) 200; d) 100; e) 0; f) 0
2.21.
a) Recessive allele because the affected individual had unaffected parents. Also, the parents of the affected individual are related.
b) The double line indicates a mating between relatives.
c) III-1 and III-2 are first cousins.
d) I-1 and I-2 could be (AA) or (Aa). At least one individual must be (Aa) or both could be (Aa). Not possible for both to be (AA). II-1 and II-4 are most likely (AA) because they a allele is a rare trait. Individuals that marry into a family are assumed to be wildtype unless there is evidence to the contrary. III-1 and III-2 are both (Aa) because they have an affected offspring (IV-4).
Chapter 3
3.1. After the centromeres have split, each former sister chromatid is counted as a chromosome in its own right, so at anaphase there are 92 chromosomes.
Additional Questions not in your textbook:
#1. On the basis of Mendel’s observations, predict the results from the following crosses with peas:
a) true-breeding tall peas x true-breeding dwarf peas
all tall
b) the offspring of (a) self-fertilized
¾ tall, ¼ dwarf
c) offspring of (a) crossed with the original tall parent
all tall
d) the offspring of (a) crossed with original dwarf parent
½ tall, ½ dwarf
#2. In pigeons, a dominant allele C causes a checkered pattern in feathers, the recessive allele c produces a plain pattern. Feather color is controlled by an independently assorting gene; where the dominant B allele causes red feathers and the recessive allele b produces brown feathers. You cross a true-breeding checkered red pigeon with a true-breeding plain brown pigeon.
a) What is the phenotype of the F1 offspring?
checkered and red feathers
b) If these F1 offspring are crossed with each other, what phenotypes will appear in the F2 generation, and in what proportions?
9 checkered red : 3 checkered brown : 3 plain red : 1 plain brown
Basic Mendelian Genetics podcast
A link to a pretty good YouTube video about the basics of Mendelian genetics can be found here. Lasts about 15 minutes.