GENETICS EXAM 2 (HW 4-6, QUIZZES 8-14)

Let us assume that hairy toes (hh) and brittle ear wax (ww) are both recessive traits in humans. In families of three children where both parents are heterozygotes at both loci (they have smooth toes and sticky ear wax) what is the probability that the family will consist of one hairy-toed, brittle ear-waxed boy and two smooth-toed, sticky ear-waxed girls? (your answer should be a decimal not a fraction, for example enter 0.25 not 1/4.)

0.0074 ANSWER: Three traits are segregating in these crosses—toe hair, ear wax, and sex; we assume that these three loci all assort independently. The cross is Hh Ww XX by Hh Ww XY. The probability of getting one hh ww XY and two H_ W_ XX in any order is the product of the probabilities: the probability of getting the hairy-toed, brittle ear-waxed boy (hh ww XY) is 1/4 ´ 1/4 ´ 1/2 = 1/32 and the probability of getting a smooth-toed, sticky ear-waxed girl (H_ W_ XX) is 3/4 ´ 3/4 ´ 1/2 = 9/32. The probability of getting a second smooth-toed, sticky ear-waxed girl (H_ W_ XX) is the same: 9/32. There are three possible orders for the three children (boy first, second or third), thus the overall probability is 3 ´ 1/32 ´ 9/32 ´ 9/32 = 243/32,768 = 0.0074.

In snapdragons, the genes A, B, C, D, and E assort independently of one another. If an Aa Bb Cc dd EE plant is crossed with an Aa bb cc Dd Ee plant, what is the probability of obtaining an offspring that is phenotypically dominant for all five traits? (Type your answer as a a decimal, for example if your answer was 1/4 you would input 0.25 as your answer)

0.09375 ANSWER: The probability of obtaining an offspring that is phenotypically dominant for all five traits is 3/4 x 1/2 x 1/2 x 1/2 x 1 = 3/32 = 0.09375

Solid black, black with tan belly, and agouti coat color in mice are all caused by alleles of the agouti gene. All three colors can be true-breeding. When using mice from true-breeding strains: (1) Black × Agouti F1s are all agouti, (2) Black × Black with Tan belly F1s are all black and tan, and (3) Black with Tan belly × Agouti F1s are all agouti. If you were to cross the F1s from (1) with the F1s from (2), what proportion of the resulting offspring would you expect to be black with a tan belly? (input your answer as a decimal not a fraction, for example if your answer was 1/8 you would input 0.125)

0.25

Suppose that a diploid cell contains 8 chromosomes (2n = 8). How many different combinations in the gametes are possible (assuming that no crossovers take place)?

16

A diploid somatic cell from a rat has a total of 42 chromosomes (2n = 42). What is the total number of telomeres in a rat cell in G2?

168

A diploid somatic cell from a rat has a total of 42 chromosomes (2n = 42). What is the total number of chromosomes in a polar body cell from a rat?

21

A diploid somatic cell from a rat has a total of 42 chromosomes (2n = 42). What is the total number of chromosomes present in the cell during metaphase I of meiosis?

42

Consider the following cross (where a is 10.6 cM from b, and b is 13.4 cM from c, and the gene order is abc) a+b+c/abc+ x abc/abc Calculate the number of individuals of a+a bb c+c genotype if 1000 progeny result from this testcross (to the nearest integer).

46

It turns out Ewok fur color is determined via basic Mendelian Patterns of Inheritance controlled by gene F, where black fur (F) is dominant to brown fur (f). A homozygous black furred Ewok has a cub(child) with a brown furred Ewok. What is the chance that the cub is brown furred? A) 0% B) 50% C) 100% D) 75% E) 25%

A) 0%

Assuming no gene linkage between genes A and B, in a cross of AABB x aabb, what will be the proportions of the gametes produced by the F1 A) 1:1:1:1 B) 4:3:2:1 C) 1:0:0:1 D) 1:2:2:1

A) 1:1:1:1

A dihybrid cross yields 320 F2 offspring. How many are expected to resemble the homozygous recessive parental (assuming the genes follow simple Mendelian Patterns of Inheritance)? A) 20 B) 10 C) 16 D) 180 E) 60

A) 20

Genes for body color (B black dominant to b yellow) and wing shape (C straight dominant to c curved) are located on the same chromosome in a species of beetle. A pure-breeding yellow beetle was crossed to a pure-breeding curved-wing beetle, and testcrossing their F1 progeny gave a second generation with these phenotypes: black, straight: 17 yellow, curved: 12 black, curved: 337 yellow, straight: 364 What is the map distance between B and C? A) 4 m.u. B) 46 m.u. C) 96 m.u D) 50 m.u. E) 2 m.u.

A) 4 m.u.

How many phenotypic classes are possible for a polygenic trait that is influenced by 3 separate genes (assuming a diploid organism)? A) 7 B) 6 C) 8 D) 4 E) 9 F) 5 G) 3

A) 7 2*n+1 = phenotypic classes

Which of the following does not follow basic Mendelian Patterns of Inheritance? A) A red flowered plant crossed with a white flowered plant produce only pink flowered plants. B) Two heterozygous round pea plants crossed together produce 75% plants with round peas and 25% plants with wrinkled peas. C) A purple flowered plant crossed with a white flowered plant produces 50% white flowered plants and 50% purple flowered plants. D) A homozygous tall plant crossed to a homozygous short plant produce only tall plants.

A) A red flowered plant crossed with a white flowered plant produce only pink flowered plants.

If you want to determine recombination frequencies between two linked traits, what is the most useful type of cross that you can perform with the F1 progeny to figure out the genetic distance between the two genes? A) A test cross B) A crossover C) a dihybrid cross D) a back cross to the homozygous dominant parent

A) A test cross

In pea plants, the tall plant allele is dominant to the short plant allele and the yellow seed coat is dominant to the green seed coat. You have a tall plant with yellow seeds and want to know its genotype. What type of cross will help you determine its genotype? (CHOOSE ALL THAT APPLY) A) A test cross B) Crossing it with a short/green plant C) Crossing it with a homozygous short/yellow plant D) Crossing it with a homozygous tall/green plant E) Crossing it with a homozygous Tall/Yellow plant.

A) A test cross B) Crossing it with a short/green plant

*** ON LAPTOP *** What gamete genotypes could result from crossovers at position 2 and position 3 in the diagram below? Choose all that apply. Hint: Don't forget that crossovers could occur between different chromatids. For example, the crossover at position 2 could occur between the left blue chromatid and the left yellow chromatid, but the crossover at position 3 could occur between the left blue chromatid and the right yellow chromatid. Consider ALL possibilities and choose all answers that apply. A) C1A2B1 B) C1A1B2 C) C2A2B1 D) C2A1B1 E) C2A2B2 F) C1A1B1 G) C1A2B2 H) C2A1B2

A) C1A2B1 B) C1A1B2 C) C2A2B1 D) C2A1B1 E) C2A2B2 F) C1A1B1 G) C1A2B2 H) C2A1B2

Which of the following contributed to Mendel's success in discovering basic patterns of inheritance. (CHOOSE ALL THAT APPLY) Hint: Don't overthink this. A) His understanding and application of statistics. B) His fortunate selection of the pea plant. C) His methodical and careful approach. D) His education.

A) His understanding and application of statistics. B) His fortunate selection of the pea plant. C) His methodical and careful approach. D) His education.

*** ON LAPTOP *** If the pedigree is for an autosomal recessive characteristic, which individuals are definitely heterozygous (Choose all that apply)? A) II-5 B) III-1 C) II-4 D) I-4 E) I-1 F) I-2 G) II-2

A) II-5 B) III-1 C) II-4 E) I-1 F) I-2

Which of the following play an important role in generating genetic diversity? (CHOOSE ALL THAT APPLY) A) Random fertilization B) Sister chromatid separation C) Sexual Reproduction D) Random kinetochore orientation leading to independent assortment in meiosis E) Asexual reproduction F) Recombination in meiosis

A) Random fertilization C) Sexual Reproduction D) Random kinetochore orientation leading to independent assortment in meiosis F) Recombination in meiosis

Hair color is determined in Labrador retrievers by alleles at the B and E loci. A dominant allele B encodes black pigment, whereas a recessive allele b encodes brown pigment. Alleles at a second locus affect the deposition of the pigment in the shaft of the hair; dominant allele E allows dark pigment (black or brown) to be deposited, whereas recessive allele e prevents the deposition of dark pigment, causing the hair to be yellow. What type of gene interaction does this represent? A) Recessive epistasis B) Duplicate dominant epistasis C) Dominant and recessive epistasis D) Duplicate recessive epistasis E) Dominant epistasis

A) Recessive epistasis

Which of these genotypes could a brown pepper have? (Choose all that apply) A) Rr Cc xx B) RRCCxx C) Rr CC xx D) RR Cc XX E) rrCCxx F) Rr Cc Xx G) rr cc Xx H) Rr CC XX

A) Rr Cc xx B) RRCCxx C) Rr CC xx

Which of these statements is/are true about a dominant/recessive allele pair? (Choose all correct answers). A) The dominant allele's effect is the only one seen in the heterozygote. B) The dominant allele is translated; the recessive allele isn't. C) The dominant allele always gives the same phenotype, even when paired with other alleles (i.e. not the recessive allele mentioned above). D) The dominant allele is transcribed; the recessive allele isn't. E) The dominant allele is often represented by a capital letter.

A) The dominant allele's effect is the only one seen in the heterozygote. E) The dominant allele is often represented by a capital letter.

Two true-breeding mutant strains of Drosophila have black body color instead of the wild-type gray yellow. When the two strains are crossed, all the F1 flies have wild-type body color. How can these data be interpreted? (CHOOSE ALL THAT APPLY) A) The mutations involved are on two different genes. B) Complementation occurred. C) A new mutation occurred in all the offspring. D) Recombination occurred.

A) The mutations involved are on two different genes. B) Complementation occurred.

Mendel concluded that the two parent plants in his crosses made equal contributions to the 'genetic factors' of the offspring. All of the following statements are true; which of them explains this conclusion? (Choose all that apply.) A) The pollen and egg each have one haploid set of chromosomes. B) Some phenotypes are influenced by many genes. C) Each chromosome contains many genes. D) Sister chromatids are randomly oriented at meiosis.

A) The pollen and egg each have one haploid set of chromosomes.

You perform a chi-squared test to determine if your observed data support your null-hypothesis. Your P value = 0.1. What can you conclude? A) You cannot reject your null hypothesis B) It is inconclusive C) You reject your null hypothesis

A) You cannot reject your null hypothesis

You are analyzing recombination between the linked C and E genes in species of plant. You perform a testcross on a plant with genotype CcEe; this plant inherited the c and e alleles from the same parent. Which of the following genotypes can be conclusively considered the recombinant progeny (i.e.the result of a crossover)? (Choose all that apply.) A) ccEe B) CCEe C) Ccee D) CcEe E) ccEE F) CCee G) CCEE H) ccee I) CcEE

A) ccEe C) Ccee Make sure you understand what is meant by "testcross"

A true breeding red carnation is crossed with a true breeding white carnation. All of the progeny are pink. The pink plants are then crossed with one another and produce red:pink:white flowered plants in a 1:2:1 ratio. This is an example of A) incomplete dominance B) co-dominance C) mendelian genetics D) simple dominance

A) incomplete dominance

This is a two-part wrong-word question. In part A you'll choose the wrong word to remove from this sentence, and in part B, you'll choose the correct word to replace it with. Consider the incorrect statement: During meiosis Ieach pair of homologous chromosomes is pulled apart by the spindle fibers attached to their cohesin, and cell division produces two haploid cells. Part B Choose a word to replace the word you removed in part A that would make the statement correct. A) kinetochores B) meiosis II C) cohesin D) haploid E) sister chromatids F) telomeres G) centromeres H) mitosis I) separase

A) kinetochores

Use the information below to answer the next three questions about the genetics of peppers. The colors of commercial bell peppers are determined by alleles of three genes, R, C, and X. Gene R makes a red pigment from yellow carotenoid; the r allele makes no red pigment. Gene C stimulates pigment synthesis by R (allowing for the complete conversion of yellow carotenoid to red pigment), a cc homozygote reduces the conversion of carotenoid to red pigment (so some red and some yellow pigments are present). This affects the ability of the R protein to fully convert all of the yellow pigment to red pigment (this is an epistatic interaction). Even though the R allele is dominant to the r allele, the cc homozygote can mask the effect of the R allele. Think back to Art class, what color does mixing red and yellow make? Gene X removes the green chlorophyll from mature fruit, an xx homozygote makes green persist. A pepper with both red and green pigment is brown. (For these questions I suggest that you draw out the biochemical pathways that are outlined above, otherwise this will be very difficult) What color will a ripe Rr cc XX pepper be? A) orange B) brown C) green D) red E) yellow

A) orange

What color will a ripe rr CC Xx pepper be? A) yellow B) green C) red D) brown E) orange

A) yellow

A cell with the genotype AABB goes through meiosis. What is the genotype of one of its gametes? (Be sure to use caps and lowercase where needed)

AB

A cell in G1 of interphase has 8 chromosomes. How many chromosomes and how many DNA molecules will be found per cell as this cell progresses through the following stages: G2, metaphase of mitosis, anaphase of mitosis, after cytokinesis in mitosis, methaphase I of meiosis, metaphase II of meiosis, and after cytokinesis of meiosis II?

ANSWER: G1: 8 chromosomes and 8 DNA molecules per cell G2: 8 chromosomes and 16 DNA molecules per cell Metaphase of Mitosis: 8 chromosomes and 16 DNA molecules per cell Anaphase of Mitosis: 16 chromosomes and 16 DNA molecules per cell After Cytokinesis in Mitosis: 8 chromosomes and 8 DNA molecules per cell Metaphase I of Meiosis: 8 chromosomes and 16 DNA molecules per cell Metaphase II of Meiosis: 4 chromosomes and 8 DNA molecules per cell After Cytokinesis of Meiosis II: 4 chromosomes and 4 DNA molecules per cell

A man with albinism and a woman with a normal phenotype have several children, one of whom displays albinism. a. What can you conclude about the genotype of the mother? b. What is the probability that the children who do not display albinism are heterozygous?

ANSWER: (a) The mother must be heterozygous (a carrier) in order to have even one offspring with albinism. (b) With parents of genotype Aa × aa, any offspring that do not display albinism in their phenotype must be heterozygous, so the probability is 100%.

A deletion mutation in the CFTR gene results in the removal of a single amino acid at position 508 (phenylalanine) in the polypedtide chain. A person homozygous for this mutation will be afflicted with the severe symptoms associated with cystic fibrosis. A point mutation in the CFTR gene results in the altering of the amino-acid at position 508, changing the amino-acid from phenylalanine to tryptophan. An individual homozygous for this mutation has symptoms, which are much less severe to the individual above. Give an explanation as to why the first mutation is much more severe than the second mutation.

ANSWER: (The following explanation is actually the research verified reason for the effect of the mutation on the expressed protein) Amino-acid 508 in the CFTR gene is a very important amino-acid in the proper function of the CFTR protein. This single amino-acid deletion prevents the protein from leaving the endoplasmic reticulum to arrive at its functional location in the cell membrane. In a human with 2 copies of the deletion mutation, there will be no functional protein formed and thus proper ion transport in the epithelial cells for which the CMTR protein is expressed would be altered leading to the serious symptoms of cystic fibrosis. In the mutation where amino acid 508 is changed from Phe to Trp, the localization of CMTR to its proper site on the cell membrane is unaffected. Functional Phe and Trp are very similar so the functionality of the protein may be altered slightly but not significantly, as indicated by the much less severe phenotype. (The following explanation and other explanations that use sound logic, while not actually the correct answer, would receive full credit because it utilizes an understanding of genetics and cell biology to effectively hypothesize what might be happening). Amino-acid 508 in the CFTR gene is a very important amino-acid in the proper function of the CFTR protein as it is located in the ATP binding site of the transmembrane protein. Without this ATP binding site, it is hypothetically possible that the protein cannot actively transport ions across the cell membrane as the energy required for this process cannot be transferred from ATP. By deleting the AA 508 (as in the first case) the ATP binding site is rendered useless, whereas in the neutral mutation (case 2) the tryptophan replacement amino-acid is sufficient for moderate ATP binding efficiency (though not as good as the phenylalanine would be in the wild-type case).

Describe the purpose of mitosis and meiosis. In your description, be very clear to highlight the main objective of each process and the main differences in the processes and WHY those differences exist.

ANSWER: Both mitosis and meiosis are processes of cell division. The objective in mitosis is to ensure that each daughter cell gets an exact copy of the genetic information contained in chromosomes as was present in the mother cell. The objective of meiosis is to reduce the genetic content in half so that each resulting gamete has reduced its ploidy from a diploid (2n) to a haploid (1n) state. The reduction of genetic content into its haploid state is critical for genetic variation in sexual organisms. To facilitate mitosis, chromosomes must align in the center of the cell and sister chromatids must then separate to opposite poles. This is facilitated by the pulling forces generated by the spindle microtubules. The process of meiosis II looks very similar to the process of mitosis, although the chromosome content in meiosis II has been reduced in half due to the process of meiosis I preceding it. In meiosis I homologous chromosomes align in the center of the cell as opposed to individual chromosomes. During separation homologous chromosomes separate to opposite poles and sister chromatids remain intact. It is in meiosis I that the ploidy gets reduced from a diploid to a haploid state.

Explain and differentiate among the molecular basis of codominance, incomplete dominance, and complete dominance.

ANSWER: In codominance, gene product from both alleles of a heterozygote is produced, so both allelic phenotypes are expressed. In incomplete dominance, a homozygous dominant produces two doses of gene product, a heterozygote produces one dose, and a homozygous recessive produces none. In complete dominance, one or both of the dominant alleles is capable of producing enough gene product for the normal dominant phenotype to be expressed.

Explain why you came to that conclusion in the question above?

ANSWER: There are 400 total offspring, so if the genes are unlinked the parentals should equal the recombinants at 200 each. The chi-square value is (228-200)2/200 + (172-200)2/200 = 7.8. With 1 degree of freedom, the P value is less than 0.05 (between 0.01 & 0.001), so it is unlikely that the results were from random assortment and likely that the genes are linked.

In chickens, there is a mutant gene called "frizzle" that results in weak, stringy, and easily broken feathers. When a frightfully frizzled fowl is bred to a normal chicken, the offspring are all mildly frizzled. If one breeds two mildly frizzled chickens to each other, the offspring have the phenotypic ratio of 1 normal: 2 mildly frizzled: 1 frightfully frizzled. What is the mode of inheritance of "frizzle"?

ANSWER: There are three phenotypes observed in the F1 offspring in the ratio of 1:2:1, and one phenotype is intermediate between the other two. This is the signature of incomplete dominant inheritance. The normal chicken represents the homozygous wild-type genotype, the mildly frizzled chickens are heterozygotes, and the frightfully frizzled chicken has homozygous mutant phenotypes.

A cell with the genotype AaBb goes through mitosis. What is the genotype of one of its daughter cells? (Be sure to use caps and lowercase where needed)

AaBb

A wild flower with the genotype W+W+ produces blue petals, whereas a wild flower with the genotype W-W- produces white petals. You are aware that the gene produces an enzyme responsible for catalyzing the biochemical reaction below. Naringenin (No Pigment) YIELDS Cyanidin (Blue Pigment) Explain why a heterozygous wild flower produces blue petals, similar to that seen in the W+W+ wild flower. (Choose ALL that apply) A) Because cyanidin is only produced in W-W- wild flowers B) Because half as much gene product is enough to produce the normal amount of pigment C) Because W+ is dominant to W- D) Because bees are attracted to blue flowers E) Because less pigment is made when one allele is defective

B) Because half as much gene product is enough to produce the normal amount of pigment C) Because W+ is dominant to W-

Rare alleles are always recessive. A) True B) False

B) False

True or False All genes on the same chromosome are genetically linked? A) True B) False

B) False

True or False A cell that just replicated its genome in S-phase, went from a haploid to a diploid state. A) True B) False

B) False Ploidy does not change when chromosomes are replicated (going from 1 chromatid to 2 sisters chromatids). Ploidy only changes when chromosome numbers are reduced (like in meiosis) or when cells fuse (like in fertilization). Or alternatively, when things go wrong (like cytokinesis failure).

True or False You can calculate the genetic distance of two genes in Map Units if you know the distance between the genes in base pairs. A) True B) False

B) False While the physical distance in base pairs does generally correlate with genetic distance, the only way to know the exact genetic distance is by determining recombination frequency via genetic crosses.

Mice of one species have two distinct tail sizes, long and short. A male with a long tail is repeatedly crossed with a female with a short tail. These matings produce: · 14 males with short tails, · 16 males with long tails, · 18 females with short tails, · 15 females with long tails. In a separate experiment, males and females with short tails are crossed. These matings produce: · 10 males with long tails, · 18 males with short tails, · 9 females with long tails, · 21 females with short tails. Which of the following patterns of inheritance is the best explanation of these results? A) Two tail-length genes; short alleles of both are needed for a short tail, and one gene is X-linked. B) One autosomal tail-length gene with two alleles; the short allele is lethal when homozygous and causes a short tail when heterozygous. C) One autosomal tail-length gene with two alleles; short > long. D) One X-linked tail-length gene with three alleles. E) One X-linked tail-length gene with two alleles; long > short.

B) One autosomal tail-length gene with two alleles; the short allele is lethal when homozygous and causes a short tail when heterozygous.

Consider a gene encoding for a transcription factor important in promoting gene expression. Would you expect a loss-of-function allele of this gene to be dominant, recessive or neither to the normal allele? (Choose all that apply) A) Recessive, because the regulated gene would never come on. B) Recessive, because the regulated gene would still be turned on by the product of the normal allele. C) Dominant, because the regulated gene would never come on. D) Dominant, because the regulated gene would stay on. . E) Neither dominant nor recessive, because half as much of a transcription factor would not be enough

B) Recessive, because the regulated gene would still be turned on by the product of the normal allele.

Which of the following statements concerning mitosis and meiosis are true? A) Sister chromatids separate during mitosis and meiosis I. B) Sister chromatids separate during mitosis and meiosis II. C) Ploidy is reduced (Diploid to haploid) after meiosis II. D) Ploidy is reduced (Diploid to haploid) after mitosis. E) Ploidy is reduced (Diploid to haploid) after meiosis I.

B) Sister chromatids separate during mitosis and meiosis II. E) Ploidy is reduced (Diploid to haploid) after meiosis I.

Feather color in the mythical creature the griffon is determined by the H locus, which is autosomal. The H allele (black feathers) is dominant to the h allele (white feathers), however when two heterozygotes are crossed, the ratio of the progeny is always less than the expected 3:1 ratio (in other words less than 75% of the progeny exhibit the dominant phenotype). Which of the following statements could explain this result given the information given? (CHOOSE ALL THAT APPLY) A) The H allele co-segregates with the h allele in meiosis. B) The HH genotype is embryonic lethal. C) Allele H exhibits incomplete penetrance. D) The trait is sex-linked.

B) The HH genotype is embryonic lethal. C) Allele H exhibits incomplete penetrance.

Which of the following genotypes would not usually be represented in a gamete? (CHOOSE ALL THAT APPLY) A) Ab B) aa C) BB D) ab E) AA F) bb G) aB H) Bb I) Aa J) AB

B) aa C) BB E) AA F) bb H) Bb I) Aa

This is a two-part wrong-word question. In part A you'll choose the wrong word to remove from this sentence, and in part B, you'll choose the correct word to replace it with. Consider the incorrect statement: During meiosis I each pair of homologous chromosomes is pulled apart by the spindle fibers attached to their cohesin, and cell division produces two haploid cells. Part A Which underlined word makes this sentence incorrect? A) haploid B) cohesin C) meiosis I D) homologous chromosomes E) spindle fibers

B) cohesin

In the yawncat (a rare hypothetical animal), the dominant allele R causes solid tail color, and the recessive allele r results in white spots on a colored background. The black coat color allele B is dominant to the brown allele b, but these genes can only be expressed if the animal has an mm genotype at a third gene locus. Animals that are M_ are yellow regardless of which allele from the B locus is present. A mating between a solid yellow-tailed male yawncat and a solid brown-tailed female yawncat produces 16 offspring with the following tail phenotypes: six solid yellow, two spotted yellow, three solid black, one spotted black, three solid brown, and one spotted brown. What is the most likely genotype of the female parent? A) Mm bb rr B) mm bb Rr C) mm Bb rr D) mm Bb Rr E) mm bb rr

B) mm bb Rr

You cross a yellow round (genotype YYRr) with a green wrinkled plant (genotype yyrr). What percent of the progeny will be green wrinkled? A) 100% B) 25% C) 0% D) 75% E) 50%

C) 0% They will all be yellow (Yy). Of those plants, 50% will be round (Rr) and 50% wrinkled (rr).

You run a chi-squared analysis to test whether your data support the null hypothesis. Which of the following p-values (at 95% confidence) would indicate that you reject the null hypothesis? (CHOOSE ALL THAT APPLY) A) 0.1 B) 0.9 C) 0.01 D) 0.1 E) 0.5 F) 0.99

C) 0.01

In the yawncat (a rare hypothetical animal), the dominant allele R causes solid tail color, and the recessive allele r results in white spots on a colored background. The black coat color allele B is dominant to the brown allele b, but these genes can only be expressed if the animal has an mm genotype at a third gene locus. Animals that are M_ are yellow regardless of which allele from the B locus is present. A mating between a solid yellow-tailed male yawncat and a solid brown-tailed female yawncat produces 16 offspring with the following tail phenotypes: six solid yellow, two spotted yellow, three solid black, one spotted black, three solid brown, and one spotted brown. What is the probability of the next offspring from these same two parents having a spotted brown tail? A) 9/16 B) 1/2 C) 1/16 D) 3/16 E) 1/4

C) 1/16

During a dihybrid cross involving two linked genes, 18% of the resulting gametes showed a recombinant genotype. These two linked genes are ________ map units apart. A) 0.18 B) 1.8 C) 18 D) 180 E) 32

C) 18

Mendel's studies of garden peas found that the allele causing 'inflated' pods was dominant to the allele causing 'constricted' pods. If you crossed a pure-breeding constricted plant with a pure-breeding inflated plant, and found that the F2 progeny included 650 inflated-pod plants, approximately how many constricted-pod plants would you expect to also have? A) 165 B) 468 C) 217 D) 600

C) 217

Eggs were recovered from a woman who is heterozygous for the allele causing Tay-Sachs disease, and fertilized with sperm from her husband who has the same alleles. A single cell was taken from each embryo at the 8-cell stage and checked for presence of the defective allele. How frequent are embryos containing only wildtype alleles expected to be? A) 67% B) 33% C) 25% D) 12.5% E) 50%

C) 25%

Genes A and B are closely linked. A man with genotype AaBb, who inherited Ab from his mother, has a son with a woman who is homozygous recessive. What is the likelihood that the son is homozygous recessive like his mother? A) more information is needed. B) less than 25% C) 25% D) 0% E) 100% F) greater than 25%

C) 25% (NOT CONFIRMED!)

Which of the following statements regarding pedigree analysis of basic Mendelian traits is true? (CHOOSE ALL THAT APPLY) A) A daughter has the affected trait as does one of her parents. This trait is therefore dominant. B) A son has the affected trait, but neither of his parents have the trait. This trait is therefore dominant. C) A daughter, whose parents both express the trait, is unaffected. This trait is therefore dominant. D) A son has the affected trait as does one of his parents. This trait is therefore recessive.

C) A daughter, whose parents both express the trait, is unaffected. This trait is therefore dominant.

The presence of a beard on some goats is determined by an autosomal gene that is dominant in males and recessive in females. Heterozygous males are bearded, while heterozygous females are beardless. What type of inheritance is exhibited by this trait? A) Autosomal dominant B) Sex-linked C) Sex-influenced D) Autosomal recessive E) Sex-limited

C) Sex-influenced

Which of the following are true statements regarding sex-linked genes? (CHOOSE ALL THAT APPLY) A) Any trait that is specific to one gender is known as a sex-linked gene. B) Some sex-linked genes are contained on autosomes. C) Sex-linked genes are genes that are located on sex chromosomes. D) The probability of expressing a sex-linked trait is different between males and females.

C) Sex-linked genes are genes that are located on sex chromosomes. D) The probability of expressing a sex-linked trait is different between males and females.

Which of the following principles of genetics was not concluded by Mendel? A) The Law of Independent Assortment B) The Law of Dominance C) The Law of Linked Genes D) The Law of Segregation

C) The Law of Linked Genes

Assume that an individual of AB/ab genotype is testcrossed and four classes of testcross progeny are found in equal frequencies. Which of the following statements could be TRUE? (CHOOSE ALL THAT APPLY) A) The genes A and B are between 10 and 20 map units apart on the same chromosome. B) The genes A and B are on the same chromosome and closely linked. C) The genes A and B are located on different chromosomes. D) The genes A and B are on the same chromosome and very far apart.

C) The genes A and B are located on different chromosomes. D) The genes A and B are on the same chromosome and very far apart.

Simple Mendelian genetics often predicts the ratios of offspring by assuming that genes 'assort independently'. Which of the following describes situations where genes will not assort independently? (Choose all that apply.) A) The two loci are on different chromosomes. B) There is no physical connection between the two loci. C) The two loci are close together on the same chromosome.

C) The two loci are close together on the same chromosome.

One of the traits that Mendel observed was cotyledon color in pea plants (green or yellow). It turns out that at the molecular level, the gene involved codes for an enzyme that converts a green pigment to a yellow pigment. Based on this information, which of the following statements are true (CHOOSE ALL THAT APPLY) A) Green color is dominant to yellow. B) In the biochemical reaction described above that is catalyzed by the enzyme, the green pigment is the product. C) Yellow color is dominant to green. D) The wild-type (normal) allele is dominant to the mutant allele.

C) Yellow color is dominant to green. D) The wild-type (normal) allele is dominant to the mutant allele.

A series of experiments shows that oil content in a diploid grain is influenced by five genes (a through e) with additive alleles. The highest producing strain is 20% oil content; the lowest has close to 0%. A plant of unknown genotype has an oil content of 12%. What is a possible genotype for this plant? (+ = additive alleles) (CHOOSE ALL THAT APPLY) A) a+a+b+bccddee B) a+a+b+b+c+c+d+dee C) aab+bc+cd+d+e+e+ D) a+a+b+b+ ccddee E) a+a+b+b+c+c+d+de+e

C) aab+bc+cd+d+e+e+

If the results of a chi-square test of a given set of data show a P value greater than 0.05, then the null hypothesis A) must be rephrased. B) must be rejected. C) cannot be rejected. D) cannot be accepted.

C) cannot be rejected.

The closer together two genes are on a chromosome, A) the more likely there will be a recombination event between them. B) the greater the chance that a double crossover will occur between them. C) the less likely there will be a recombination event between them. D) the more likely they are to be epistatic. E) the less likely they are to be good genetic markers.

C) the less likely there will be a recombination event between them.

Polydactyly is the condition of having extra fingers or toes. Some polydactylous persons possess extra fingers or toes that are fully functional, whereas others possess only a small tag of extra skin. This is an example of A) complementation. B) independent assortment. C) variable expressivity. D) complete dominance.

C) variable expressivity.

One Parent is a carrier of the sickle cell trait and the other is normal. What is the probability that their child has sickle-cell disease? A) 50% B) 100% C) 25% D) 0% E) 75%

D) 0%

*** ON LAPTOP *** In a species of Drosophila, genes q and r are found on the same chromosome 20 centimorgans apart. A cross was made between the following individuals: What proportion of the offspring would you expect to have the wild-type phenotype (wild-type is denoted by the + alleles and is dominant to the mutant alleles)? A) 0.40 B) 0.20 C) None of these D) 0.10 E) 0.80

D) 0.10

You are studying a gene that controls ossicone (horn) length in giraffes. The wild-type long-ossicone allele (L) is dominant to the mutant short-ossicone (l) allele. However the L allele is only 60% penetrant. You cross two heterozygous giraffes. What proportion of offspring would you expect to exhibit the long ossicone phenotype? Assume the penetrance of L applies equally to both homozygotes and heterozygotes. A) 0.55 B) 0.40 C) 0.6 D) 0.45 E) 0.75

D) 0.45

The color gene locus has two alleles [C (black) is dominant to c (grey)] and the feather gene locus has two alleles [D (shiny) is dominant to d(dull)]. You cross black shiny chickens (CcDd) with grey dull chickens (ccdd) and get the following numbers 10 Black Shiny chickens 10 Grey Dull chickens 90 Black Dull chickens 90 Grey Shiny chickens What is the distance between the color gene locus and the feather gene locus? A) 5 map units B) 100 map units C) 20 map units D) 10 map units E) 50 map units

D) 10 map units

In lilies, white flowers (W) are dominant to purple flowers (w). If two plants that are heterozygous for flower color are mated, the genotypic ratio of the offspring would be A) 1:1. B) 3:1. C) 1:1:1:1. D) 1:2:1. E) 9:3:3:1.

D) 1:2:1.

In rabbits, an allelic series helps to determine coat color: C (full color), c^ch (chinchilla; gray color), c^h (Himalayan; white with black extremities), and c (albino; all white). The C allele is dominant to all others, c^ch is dominant to c^h and c, c^h is dominant to c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C > c^ch > c^h > c. Indicate the phenotypic ratios expected if rabbits with the cross Cc x (c^h)c. A) 1 full color:1 chinchilla B) 3 full color:1 chinchilla C) 1 full color:1 Himalayan D) 2 full color:1 Himalayan:1 albino E) 1 chinchilla:1 Himalayan

D) 2 full color:1 Himalayan:1 albino

How many different offspring genotypes and phenotypes are expected from a cross between F1 parents heterozygous for three traits? (Assume simple dominant/recessive allele relationships.) A) 16 genotypes; 64 phenotypes B) 8 genotypes; 27 phenotypes C) 64 genotypes; 16 phenotypes D) 27 genotypes; 8 phenotypes

D) 27 genotypes; 8 phenotypes

Humans have 23 pairs of homologs, each aligning randomly in Meiosis I with respect to the others. Ignoring the effects of crossovers, how many different combinations of chromosomes are possible in the gametes of a single person? A) 23^2 B) 4 C) 4 x 23 D) 2^23 E) 2 x 23 F) 2

D) 2^23

*** ON LAPTOP *** Which pattern of inheritance best describes the disease associated with the following pedigree? A) X-linked recessive B) Autosomal dominant C) X-linked dominant D) Autosomal recessive E) Y-linked

D) Autosomal recessive

In purple people eaters, purple is dominant to white. A true-breeding white mutant is mated with a different true-breeding white mutant. All of the F1 are purple. When the purple F1 offspring mate with each other, their offspring occur in the ratio of 9 purple:7 white. Which phenomenon explains the purple F1 offspring? A) Dominant epistasis B) Recessive epistasis C) Mutation D) Complementation

D) Complementation

Manx cats have no tails. When two Manx cats are bred together there is always a one third chance that a kitten will have a tail. When a Manx cat is bred to a cat with a normal tail there is a one-half chance that a kitten will have a tail. Which of the following is the best explanation for this? A) The Manx phenotype is dominant epistatic. B) The Manx genotype exhibits variable expression. C) The Manx phenotype is caused by gene interactions. D) The Manx phenotype is dominant, but the allele is a recessive lethal. E) The Manx phenotype is a result of heteroplasmy.

D) The Manx phenotype is dominant, but the allele is a recessive lethal.

*** ON LAPTOP *** What inheritance pattern best explains this pedigree? A) X-linked dominant B) X-linked recessive C) Autosomal recessive D) Y-linked E) Autosomal dominant

D) Y-linked

Plant height is a polygenic trait determined by 4 separate genes. If the minimum plant height is 10 inches and the maximum height is 90 inches. Which of the following genotypes could a 50 inch plant have? (note: + denotes an additive allele, for example 'a+ = additive'; whereas 'a = non-additive') A) a+a+b+b+c+c+d+d+ B) aabbccdd C) a+a+b+b+ccd+d+ D) a+ab+bc+cd+d E) a+abbccd+d

D) a+ab+bc+cd+d Each additive allele contribute 10 inches to the base phenotype so the genotype must have 4 additive alleles for a 50 inch plant 10"[baseline] + 10"[each additive allele contribution] * 4[# of additive alleles] = 50" plant

*** ON LAPTOP *** A series of two-point crosses among fruit flies is carried out between genes for brown eyes (bw), arc wings (a), vestigial wings (vg), ebony body color (e), and curved wings (cv). The following number of nonrecombinant and recombinant progeny were obtained from each cross: Using these data from two-point crosses, what it the best genetic map (in cM) that can be developed? A) bw 5 cv 24 vg 32 a with e assorting independently B) a 5 bw 13 vg 24 e with vg assorting independently C) cv 5 bw 13 a 34 vg with e assorting independently D) bw 5 a 24 cv 13 vg with e assorting independently E) cv 13 bw 5 a 27 vg with e assorting independently

D) bw 5 a 24 cv 13 vg with e assorting independently

*** ON LAPTOP *** (This biochemical pathway is also considered in the next question.) Precursors 1 and 2 are both colorless, 3 is blue, and the product 4 is red, so wildtype individuals are red. Precursor 3 can be produced by either enzyme A or enzyme B. What will be the phenotype of an individual with the genotype aabbDD? A) purple B) blue C) red D) colorless

D) colorless

A man has genotype A1A1 G1G2. The A and G genes are on the same chromosome. What gametes will many meioses produce, and in what proportions? A) ½ A1A1, ½ G1G2 B) More information is needed. C) None of the above D) ½ A1G1, ½ A1G2 E) ¼ A1A1, ¼ A1G1, ¼ A1G2, ¼ G1G2 F) ½ A1, ¼ G1A2, ¼ G2

D) ½ A1G1, ½ A1G2

Color blindness is caused by a sex-linked gene. A homozygous normal woman has a daughter with a color blind man. What is the probability that the daughter is color blind? A) 25% B) 50% C) 100% D) 75% E) 0%

E) 0%

Two genes of a flowering plant are on the same chromosome; one controls height and the other controls color. You cross a true-breeding tall plant with white flowers with a true-breeding short plant with purple flowers. The offspring are all tall plants with purple flowers. If the genes are separated by 20 map units, what is the expected frequency of short white-flowered plants in the F2 generation? A) 0.2 B) 0.02 C) 0.1 D) 0 E) 0.01

E) 0.01 If the recombination frequency is 20% (0.20), that means that there is a 20% chance that a gamete will have a recombinant chromosome. 1/2 of those recombinant gametes will have the desired short & white alleles (sp) making the chance of inheriting this gamete from 1 parent (1/2 * 0.20) = 0.10. Because the progeny needs to inherit the sp gamete from both parents you apply the product rule of probabilities and get a frequency of (0.10 * 0.10) = 0.01.

*** ON LAPTOP *** If individual #1 in the following pedigree is a heterozygote for a rare, autosomal disease, what is the probability that individual #7 will be affected by the disease? Assume that #2 and the spouses of #3 and #4 are not carriers. A) 0.75 B) 0.5 C) 0.031 D) 0.25 E) 0.016 F) 0.125 G) 0.063

E) 0.016

A woman has normal vision although her maternal grandfather (her mother's father) had red-green colorblindness, a sex-linked recessive trait. Her maternal grandmother and the woman's own father are assumed to not possess a copy of the mutant allele. The woman marries a man with normal vision although his father was colorblind. What is the probability that the first child of this couple will be colorblind? A) 1/2 B) 1/16 C) 1/4 D) 1/12 E) 1/8

E) 1/8

(See pathway above) What will be the phenotypic ratios among the offspring of a cross AabbDd x AaBbdd? A) all blue B) 9 red: 7 blue C) 13 red: 3 colorless D) 13 red: 3 blue E) 7 red: 7 blue: 2 colorless F) 3 red: 1 blue G) 9 red: 6 blue: 1 colorless H) 12 red: 4 blue I) 1 red: 1 colorless J) 9 red: 3 blue: 3 purple: 1 colorless K) all red

E) 7 red: 7 blue: 2 colorless

Which of the following crosses would produce a 1:1 ratio of phenotypes in the next generation? A) AA x AA B) Aa x Aa C) aa x aa D) AA x aa E) Aa x aa

E) Aa x aa

In dogs, black coat color (B) is dominant over brown (b), and solid coat color (S) is dominant over white spotted coat (s). A cross between a black, solid female and a black, solid male produces only puppies with black, solid coats. This same female was then mated with a brown, spotted male. Half of the offspring from this cross were black and solid, and half of the offspring were black and spotted. Which of the following genotypes could the female have? (Choose all that apply) A) BbSS B) bbss C) BbSs D) BBSS E) BBSs

E) BBSs

In a standard pedigree, an affected male will be represented by which of the following? A) unshaded circle B) an X C) shaded circle D) unshaded square E) shaded square

E) shaded square

A man has genotype A1A2. What gametes will many meioses produce, and in what proportions? A) none of the above B) 1/3 A1, 1/3 A1A2, 1/3 A2 C) ¼ A1A1, ½ A1A2, ¼ A2A2 D) ¼ A1, ½ A1A2, ¼ A2 E) ½ A1, ½ A2 F) More information is needed

E) ½ A1, ½ A2

A red, round, true-breeding tomato is bred with a yellow, oval, true-breeding tomato, and the F1s are testcrossed to a homozygous-recessive tester. This results in the offspring below. Calculate the chi-squared value to test the likelihood of linkage. yellow oval: 113 red round: 115 yellow round: 84 red oval: 88 **This is a multi part question with 2 parts (be sure to answer all of the parts!** What is the Chi-Squared value (enter a numeric value , DO NOT USE fractions). Would you conclude that these genes are linked or unlinked?

What is the Chi-Squared value (enter a numeric value , DO NOT USE fractions). 7.94 Would you conclude that these genes are linked or unlinked? Linked ANSWER: There are 2 acceptable approaches to this question, you could have predicted a 1:1:1:1 outcome of the 4 phenotypes, or you could pool the parentals and recombinants based on the #s and assumed a 1 Parental: 1 Recombinant outcome. Below are the answers for both. 1) There are 400 total offspring, so if the genesare unlinked you would expect 100 for each of the 4 phenotypes. The chi-square value is (113-100)2/100 + (115-100)2/100 + (84-100)2/100 + (88-100)2/100= 7.94. With 3 degree of freedom, the P value isless than 0.05 (between 0.05 & 0.01), so it is unlikely that the resultswere from random assortment and likely that the genes are linked. 2) There are 400 total offspring, so if the genesare unlinked the parentals should equal the recombinants at 200 each. Thechi-square value is (228-200)2/200 + (172-200)2/200 = 7.8. With 1 degree of freedom, the P value isless than 0.05 (between 0.01 & 0.001), so it is unlikely that the resultswere from random assortment and likely that the genes are linked.

A local florist is having trouble with a new species of plant. She received 2 of these rare plants, 1 of which is really short (5 inches) and the other of which is really tall (23 inches). She would like to sell these plants, but would like them to be medium height (14 inches). When she crosses the plants the F1 progeny are all 14 inches tall! She sells out of these plants very quickly and wants to make more, but the parent plants are no longer fertile. She crosses the F1s with one another and to her surprise she gets plants whose sizes ranged in 3 inch intervals from 5 to 23 inches (5, 8, 11, 14, 17, 20, 23). Most were 14 inches, while 1/64 had 5 inches and 1/64 had 23 inches. Explain how the plant heights were inherited by describing a) the mode of inheritance b) indicating how may gene pairs were at work c) and designating the genotypes of the 2 parental plants, and their 14 inch F1 offspring. d) If one of the 14 inch F1 offspring is mated with one of the 5 inch F2 plants, what phenotypic ratio would be predicted if many offspring resulted?

a) Polygenic (quantitative) inheritance b) 3 gene pairs: This can be determined by the formula 1/4^x = extreme phenotype, where x = # of gene pairs. 1/64 = 1/4^x --> x = 3 It can also be determined by the formula 2x+1 = # of phenotypes in polygenic trait. 2x + 1 = 7 --> x = 3 c) Parent 1 = aabbcc : Parent 2 = AABBCC : F1 = AaBbCc d) 1:3:3:1 (5 inch: 8 inch: 11 inch: 14 inch)