Genetics Test 3 Practice

Aneuploidy is caused by _______.

Disjunction

Drosophila melanogaster, like humans, have X and Y chromosomes. However, unlike humans, it is the ratio of the number of X chromosomes to sets of autosomes that determines the sex of the fly. A fly has one X chromosome, two sets of autosomes (a diploid set), and no Y. A ratio of 1 X:A results in a ________.

Female

Drosophila melanogaster, like humans, have X and Y chromosomes. However, unlike humans, it is the ratio of the number of X chromosomes to sets of autosomes that determines the sex of the fly. A fly has one X chromosome, two sets of autosomes (a diploid set), and no Y. A ratio of 0.5 X:A results in a _________ fly.

Male

One form of hemophilia is caused by an X-linked recessive allele of blood clotting factor 8. Assume that a man with hemophilia marries a phenotypically wild-type woman whose father had hemophilia. What is the probability that they will have a daughter with hemophilia? (Notice the phrasing of the question. Calculate the probability they will have a daughter with hemophilia considering they could have either a son or daughter.) a. 1/4 b. 1/16 c. 1/8 d. 3/4 e. 1/2

a. 1/4

The genes A B C occur in that order in fruit flies. A is 10 map units from B, and B is 20 map units from C. If you test crossed triply heterozygous A B C/abc how much recombination would you expect between genes A and B? a. 10% b. 20% c. 30% d. 50% e. 0.2%

a. 10%

Two tomato plant genes A and B are linked and 40 mu apart. A tomato plant homozygous at the A and B loci (AABB) is crossed with an ab plant (aabb). The resulting F1 is two-point testcrossed back to an ab tester. What percent of the progeny is expected to be Ab? a. 20% b. 40% c. 10% d. 60%

a. 20%

Which of the following is an example of interference in gene mapping? a. A crossover in one region inhibits crossovers in nearby regions of a chromosome. b. Double crossovers do not occur. c. Crossovers will not occur near centromeres. d. An actively transcribed gene is less likely to undergo crossover than a silent one.

a. A crossover in one region inhibits crossovers in nearby regions of a chromosome.

In chickens, cock-feathering is a sex-limited recessive trait, where h is the recessive allele. What ratio of offspring would you get from an h h-hen feathered hen X h h-cock feathered rooster? a. All hen-feathered hens and all cock-feathered roosters b. All hen-feathered c. All cock-feathered d. 3:1 hen-feathered hens: cock-feathered roosters e. Unable to predict from this information

a. All hen-feathered hens and all cock-feathered roosters

Usually results from nondisjunction during cell division Example: An individual with Turner syndrome is 45, X or 2n-1 a. Aneuploid b. Metabolism c. Pleitropy d. Euploid

a. Aneuploid

In a plant species, two blue plants are mated and produce 303 blue plants and 98 white plants. These observations can best be explained by a. Complete dominance b. Incomplete dominance c. Codominace d. Haploinsufficiency

a. Complete dominance

Shows as an unpaired loop under the microscope Paralogs can form (more in a minute) Phenotype depends on how much and what information duplicated a. Consequences of a duplication b. Consequences of a replication c. Consequences of a deletion

a. Consequences of a duplication

No gain or loss of genetic information Different gene order along the chromosome Possibility of position effect - change in gene expression because of a change in location Can make two genes come together Might cause a chimeric protein a. Consequences of inversions b. Consequences of diversions c. Consequences of embroyonic lethal d. Consequences of deletions

a. Consequences of inversions

Dextral (D) and sinistral (d) shell coiling in the water snail Limnaea peregra is genetically determined by maternal effect. From a population of mixed sinistral and dextral snails, a sinistral hermaphrodite and a dextral male are chosen and mated. What is the phenotype(s) of their offspring? a. Dd b. dd c. DD

a. Dd b. dd Rationale: Since the maternal parent is sinistral, we know that its genotype must have one d, but could be Dd or dd.The phenotype of the offspring would depend on these two genotypes.Supposing the maternal parent was Dd, then the offspring would be dextral.Supposing the maternal parent was dd, then the offspring would be sinistral

What is the possible genotype for individual III-2? a. Hemizygous with the recessive allele b. Homozygous dominant c. Hemizygous with the dominant allele d. Homozygous recessive

a. Hemizygous with the recessive allele

Xist lncR N A plays a crucial role in X inactivation. The Xist R N A coats the X chromosome that produced it, and whichever X has more Xist, it becomes the inactive X.If the mutant X cannot produce Xist, then ________. a. It can not become coated with Xist RNA. b. It can become coated with Xist DNA. c. It can become coated with Xist carbohydrates. e. It can become coated with Xist protein.

a. It can not become coated with Xist RNA.

What term is best associated to the following: Not just applied to Barr bodies Gene mutations and chromosome mutations happen during development so not every cell in a body has the mutation, and thus the same genotype a. Mosaicism b. Genetic Anticipation c. Wild-type d. Mutant Gene Product

a. Mosaicism

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, but the allele is a recessive lethal. b. The Manx phenotype is a result of heteroplasmy. c. The Manx phenotype is dominant epistatic. d. The Manx genotype exhibits variable expression. e. The Manx phenotype is caused by gene interactions.

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

Fur color in Himalayan rabbits is controlled by a temperature-sensitive allele. The encoded protein does not produce brown pigment at high temperatures. If one hot summer, you shaved a small patch from your Himalayan rabbit's brown paw, and then let it grow back while the rabbit played outdoors, what color would the fur grow back? a. White b. Brown c. Light brown d. Depends on the animal's genotype e. Depends on the animal's nutrition

a. White

Two beetles (2n) with patches of green and purple mated and produced 39 green and purple-patched, 18 solid green, and 19 solid purple beetles. These observations can best be explained by a. codominance b. recessive epistasis c. incomplete dominance d. dominant epistasis e. variable expressivity

a. codominance

Part of chromosome missing Unlikely to revert because the missing part is lost Can occur anywhere, not necessarily in the middle of the chromosome a. deletion b. addition c. duplication

a. deletion

Which mutations are generally dominant since one copy in a diploid organism is sufficient to alter the phenotype? a. gain of function b. null c. loss of function d. neutral e. conditional

a. gain of function

Which of the following is an example of codominance? a. in roan cattle, you can see a mix of both red and white furs b. flowers of a certain plant can come in purple, blue, or yellow c. a scorpion's venom is more potent when it's younger d. a certain plant can have jagged or smooth edges on its leaves e. there are many possible mutations that lead to white eyes in flies

a. in roan cattle, you can see a mix of both red and white furs

Drosophila melanogaster, like humans, have X and Y chromosomes so that XX is female and XY is male. However, unlike humans, it is the ratio of the number of X chromosomes to sets of autosomes that determines the sex of the fly. A fly has one X chromosome, two sets of autosomes (a diploid set), and no Y. What is its sex? a. male b. female c. intersex

a. male

Human mitochondrial disorders can be caused by either _________ or _________. a. nuclear mutation b. will smith c. mitochondrial DNA mutation

a. nuclear mutation c. mitochondrial DNA mutation

The Chi-square test involves a statistical comparison between measured (observed) and predicted (expected) values. One generally determines degrees of freedom as ________. a. one less than the number of classes being compared b. ten minus the sum of the two categories c. one more than the number of classes being compared d. the number of categories being compared e. the sum of the two categories

a. one less than the number of classes being compared

Change in the position of a portion of a chromosome a. transversion b. diversion c. caryoversion d. inversion e. cisversion

a. transversion

How many Barr bodies would one expect to see in cells of individuals with Turner syndrome (45, XO) and Klinefelter syndrome (47, XXY)? a. zero and one b. one and one c. one and zero d. zero and two e. zero and zero

a. zero and one

In the mouse, gene A allows pigmentation to be deposited in the individual coat hairs; its allele a prevents such deposition of pigment, resulting in an albino. Gene B gives agouti (wild-type fur); its allele b gives black fur. What would the expected ratio of the progeny be in a cross of a doubly heterozygous agouti mouse mated with a doubly homozygous recessive white mouse? a. 3 (agouti) :4 (black) : 9 (albino) b. 1 (agouti) :1 (black) : 2 (albino) c. 9 (black) : 27(albino) d. 9 (agouti) :3 (black) : 4 (albino) e. 1 (agouti) :1 (black) :1 (albino)

b. 1 (agouti) :1 (black) : 2 (albino)

For a gene that has four alleles in nature, what is the maximum number of genotypes that can exist in a population? a. 4 b. 10 c. 20 d. 16

b. 10

There are five alleles for a gene in a population of diploids. How many genotypes are possible in the population? a. 10 b. 15 c. 25 d. 30

b. 15 Rationale: Use the equation [n(n+1)]/2.

A monohybrid cross with a ________ ratio in the F2s indicates incomplete dominance. a. 1:1:1:1 b. 1:2:1 c. 3:1 d. 2:2:2 e. 9:7

b. 1:2:1

A male observed his stained epithelial cells and saw a Barr body. If a karyotype was performed on his cells, what would you expect to see? a. 46, X Y b. 47 X X Y c. 47 X Y Y d. 46 X X

b. 47 X X Y d. 46 X X Rationale: The presence of a Barr body indicates an inactive X. This is normally seen in XX females, or in any cells with more than one X. Since this is a male, he likely has a Y chromosome, and he must have two X chromosomes (Klinefelter's). His corresponding karyotype would then be 47 chromosomes with 2 X chromosomes and a Y chromosome. Alternatively, he could be XX with a translocation of the SRY locus.

In jackalopes, blue eyes are dominant to yellow eyes and having a tail is dominant to being tailless. You cross a true-breeding blue eyed, tailed jackalope parent against a true-breeding yellow eyed, tailless jackalope. As expected, the F1 are all blue eyed and have tails. You perform a test cross with the F1 and score the phenotypes of the F2 as follows: A. Blue and tailed 521 B. Blue and tailless 432 C. Yellow and tailed 423 D. Yellow and tailless 507 Indicate the parental and recombinant phenotypic classes: a. A. Recombinant B. Parental C. Parental D. Recombinant b. A. Parental B. Recombinant C. Recombinant D. Parental c. A. Heterozygous B. Homozygous dominant C. Homozygous recessive D. Heterozygous

b. A. Parental B. Recombinant C. Recombinant D. Parental

Based on the description in the previous question, a mule is an example of a. Endopolyploidy b. Allopolyploidy c. Autopolyploidy d. Trisomy

b. Allopolyploidy

A right-handed coiled snail is the result of a cross between two snails. If this offspring snail is genotypically dd (left-handed genotype), what is the genotype of the snail that provided the egg? a. dd b. Dd c. DD d. either Dd or DD e. either Dd or dd

b. Dd Rationale: In maternal effect, such as shell coiling, the genotype of the parent producing the egg determines the phenotype of the progeny. So the parent donating the egg must have had at least one D (right-handed) allele in order to produce right-handed offspring. However, in order for the F1 snail to be dd, the "mother" must be able to donate a d allele. Therefore, the mother must be Dd.

Two mice with medium tails were mated and produced the following offspring: 24 long tailed mice, 50 medium tailed mice, and 27 short tailed mice. These observations can BEST be explained by a. Complete dominance. b. Incomplete dominance. c. Variation in expressivity. d. A lethal allele with haplosufficiency. e. Epistasis

b. Incomplete dominance.

Which statement best describes wild-type human sex determination?Choose the one(s) that apply to the question above: a. Individuals with a Y chromosome are female. b. Individuals with a Y chromosome are male. c. Individuals without a Y chromosome cannot be male. d. Females are heterogametic. e. Individuals with one X chromosome are male. f. The presence of a Y chromosome does not determine sex. g. Individuals with two X chromosomes are male. h. Individuals with two X chromosomes are female. i. Individuals with at least twice as many X chromosomes as Y chromosomes are female.

b. Individuals with a Y chromosome are male. Rationale: This question is a similar question that is on the test, that is why I added so many choices. But, there is only ONE right answer. (Sorry for the list, it will help you I promise!)

In one strain of mouse, homozygotes for an allele of a gene develop heart defects. In a different strain of mouse, homozygotes for the same allele develop normally. Heterozygotes develop normally in both strains. What is the most likely explanation for the differences between the two strains. a. The allele is recessive. b. The development of the heart defect is determined by more than one locus. c. The allele has pleiotropic effects on development. d. The allele is codominant.

b. The development of the heart defect is determined by more than one locus.

Xist lncR N A plays a crucial role in X inactivation. The Xist R N A coats the X chromosome that produced it, and whichever X has more Xist, it becomes the inactive X. If a mutation inactivated Xist on only one chromosome, what would you predict to occur? a. The mutant chromosome would always be inactivated. b. The mutant chromosome would always be active. c. Both chromosomes would be inactivated. d. Both chromosomes would be active. e. The wild-type Xist would compensate for the mutant, and X inactivation would proceed normally.

b. The mutant chromosome would always be active.

Which of the following is true about human sex chromosomes? a. They have the same gene configuration and same loci. b. They act like homologous chromosomes during meiosis so each gamete will get one sex chromosome. c. They are always metacentric. d. They do not participate in meiosis. e. They are independent during meiosis.

b. They act like homologous chromosomes during meiosis so each gamete will get one sex chromosome.

Sickle-cell disease is caused by a loss-of-function mutation in the hemoglobin gene so that an individual must have two copies of the mutant allele to manifest full sickle cell disease. The first successful case of using CRISPR-modified adult stem cells as a treatment for sickle-cell disease was performed on a woman named Victoria. Victoria was homozygous for mutant hemoglobin and had full sickle cell disease. Doctors harvested some of Victoria's stem cells, corrected the mutation using CRISPR, and reintroduced them into her body. The edited cells then expressed wild-type hemoglobin. The researchers increased the dosage of wild-type hemoglobin in order to correct Victoria's sickle cell disease. What phenomenon does this exemplify? a. Complementation groups b. Threshold effect c. Genetic anticipation d. Epistasis

b. Threshold effect

Dr. Mary Lyon, the scientist who discovered Lyonization, studied variegated mice (patches of white and brown). All of the variegated individuals were female. When she crossed variegated females to white males, she found 50 white females, 50 variegated females, and 50 white males. What is the best explanation of inheritance of this trait? a. X-linked dominant lethal b. X-linked recessive lethal c. Autosomal recessive lethal d. Autosomal dominant lethal

b. X-linked recessive lethal

Which mutations are generally dominant given that one copy in a diploid organism is sufficient to alter the wild-type phenotype? a. conditional b. gain of function c. null d. neutral e. loss of function

b. gain of function

Duchenne muscular dystrophy is caused by a recessive X-linked allele. A man with this disorder _____ a. could have inherited it from either parent b. must have inherited it from his mother c. must have inherited it from both parents d. can pass it along to all of his children e. can pass it along to only his sons

b. must have inherited it from his mother

In mice, there is a set of multiple alleles of a gene for coat color. Four of those alleles are as follows:C = full color (wild type)cch = chinchillacd = dilutionc = albinoGiven that the gene locus is not sex-linked and that each allele is dominant to those lower in the list, give the phenotypic ratios expected from the following cross.wild (heterozygous for dilution) × chinchilla (heterozygous for albino) a. 1 full color : 1 chinchilla : 1 dilution : 1 albino b. 1 chinchilla : 2 dilution : 1 albino c. 2 full color : 1 chinchilla : 1 dilution d. 2 full color : 2 chinchilla : 1 dilution e. 1 full color : 1 chinchilla : 2 dilution : 1 albino

c. 2 full color : 1 chinchilla : 1 dilution

G and E are genes in a diploid organism. The cross of genotypes GE/ge × ge/ge produces the following progeny: GE/ge 404; ge/ge 396; gE/ge 97; Ge/ge 103. From these data, one can conclude that there are ________ map units between the G and E loci. a. 10 b. 5 c. 20 d. 25 e. 15

c. 20

Example: 2n=46 human diploid cells, n=23 human haploid (gamete) cells a. Pleitropy b. Metabolism c. Euploid d, Aneuploid

c. Euploid

Double cross over (DCO) does not happen as often as predicted because of ___________. a. Assistance b. Encouragement c. Interference d. Facilitation

c. Interference

Which of the following characteristics would allow you to attribute a disease to organelle inheritance rather than maternal effect? a. It follows Mendelian inheritance patterns b. It does not follow Mendelian inheritance patterns c. Its effects are tempered by heteroplasmy d. Its effects are seen during fetal development e. It is inherited through the ooplasm

c. Its effects are tempered by heteroplasmy

Sometimes a DCO recombinant can look like the _____________. a. Mutant Phenotype b. Neutral Phenotype c. Parental Phenotype d. Wild-Type Phenotype e. Sister Chromosomes

c. Parental Phenotype

In the past, scientists specializing in animal husbandry amplified SRY by PCR to determine whether a mouse was male or female. They then ran the PCR product on an agarose gel. This strategy was problematic since absence of a band could either be caused by a mouse being female or a mistake in the genotyping (PCR) process. (It is hard to interpret the absence of a band.) Since then, sex determination of mice by PCR involves amplification of a repeat region in both the X and Y chromosomes (the genes in this region are called gametologs). During evolution, this region Y has undergone deletion so that one sex has more repeats than the other. Where on the Y chromosome would these gametologs be found? a. The male-specific region b. The pseudo-gene region c. The pseudoautosomal region d. The non-recombining region e. The sex-determining region f. The heterochromatic region

c. The pseudoautosomal region

A true-breeding tomato plant with AB phenotype is two-point testcrossed to determine the relationship of genes A and B. The following F2 progeny are observed.Are A and B linked? Why or why not? Choose all that apply: a. They may be assorting independently b. They may be on the same end of the chromosome. c. They may be assorting dependently. d. They may be on opposite ends of the chromosome

c. They may be assorting dependently. d. They may be on opposite ends of the chromosome Rationale: There is just as much chance (50%) of observing a recombinant as there is a parental (50%).

In a plant species, if the B allele (blue flowers) and the b allele (white flowers) are incompletely dominant (B b is light blue), what offspring ratio is expected in a cross between a blue-flowered plant and a white-flowered plant? a. 1/4 blue:1/2 light blue:1/4 white b. 1/2 blue:1/2 white c. all light blue d. 3/4 blue:1/4 white e. 1/3 blue:1/3 light blue:1/3 white

c. all light blue Rationale: The cross is B B ´ b b

In a Chi-square test, as the value of the χ2 increases, the likelihood of rejecting the null hypothesis ________. a. stays the same b. is doubled c. increases d. decreases

c. increases

Assume that a testcross is made between AaBb and aabb plants and that the offspring occur in the following numbers: 106 AaBb, 48 Aabb, 52 aaBb, 94 aabb. These results are consistent with which of the following? a. linkage with 50% crossing over b. independent assortment c. linkage with approximately 33 map units between the two gene loci d. sex-linked inheritance with 30% crossing over e. 100% recombination

c. linkage with approximately 33 map units between the two gene loci

Small arm of a chromosome. a. a b. b c. p d. g e. q

c. p

Which of the following crosses would indicate that the mutants were in the same complementation group? a. pink eye fly × white eye fly -> all white eyes b. pink eye fly × pink eye fly -> all white eye offspring c. pink eye fly × pink eye fly -> pink eye fly d. red eye fly × red eye fly -> 3 : 1 ratio e. pink eye fly × pink eye fly -> purple eye flies

c. pink eye fly × pink eye fly -> pink eye fly

G and E are genes in a diploid organism. The cross of genotypes GE/ge × ge/ge produces the following progeny: GE/ge 404; ge/ge 396; gE/ge 97; Ge/ge 103. From these data, one can conclude that ________. a. the G and E loci show complete linkage b. the G and E loci assort independently c. the G and E loci are linked d. the G and E loci reside on the same chromosome over 50 map units apart

c. the G and E loci are linked

You have the goal of creating an ideal corn crop that appeals to OU students' appetites. OU students prefer crimson-colored kernels over those awful burnt-orange kernels that Texas students seem to love. In order to expedite your final mass breeding, you decide it is important to understand the inheritance pattern of kernel color. You perform a monohybrid cross and get 910 burnt-orange to 306 crimson kernels. You thus propose a 3:1 model to test using chi-square analysis. Calculate a chi-square value (total deviation). Choose the closest value to the chi square you've calculated. a. 0.0022 b. 0.0044 c. 0.013 d. 0.018 e. 0.0088

d. 0.018

A cross is made between a female calico cat and a male cat having the allele for black fur on his X chromosome. What fraction of the offspring would one expect to be calico? a. 3/4 b. none c. 1/2 d. 1/4 e. 2/3

d. 1/4

Many of the color varieties of summer squash are determined by several interacting loci: AA or Aa gives white, aaBB or aaBb gives yellow, and aabb produces green. Assume that two fully heterozygous plants are crossed. Give the phenotypes (with frequencies) of the offspring. a. 9 white : 3 yellow : 4 green b. 3 white : 9 yellow : 4 green c. 2 white : 3 yellow : 1 green d. 12 white : 1 yellow : 3 green e. 4 white :9 yellow : 3 green

d. 12 white : 1 yellow : 3 green

Two tomato plant genes A and B are 10 mu apart and genes B and C are 30 mu apart. The order of the genes is A-B-C. A tomato plant homozygous at the A B and C loci (AABBCC) is crossed with an abc plant (aabbcc). The resulting F1 is testcrossed back to an abc tester. What percent of the progeny is expected to be Ac? a. 40% b. 60% c. 80% d. 20%

d. 20% Rationale: A and C are 40 mu apart. Therefore 40% of the progeny is expected to be recombinant. There are two recombinant classes Ac and aC, so divide 40 by 2 = 20%.

A horse is 2n = 64 and a donkey is 2n = 62. If crossed, they produce the hybrid mule. There are _______chromosomes in a mule somatic cell. This cell is _________. a. 2n = 126; aneuploid b. 2n = 126; euploid c. 2n = 63; aneuploid d. 2n = 63; euploid e. either 2n = 64 or 2n = 62, depending on whether the mother was the horse or the donkey; euploid.

d. 2n = 63; euploid

In jackalopes, blue eyes are dominant to yellow eyes and having a tail is dominant to being tailless. You cross a true-breeding blue eyed, tailed jackalope parent against a true-breeding yellow eyed, tailless jackalope. As expected, the F1 are all blue eyed and have tails. You perform a test cross with the F1 and score the phenotypes of the F2 as follows: A. Blue and tailed 521 B. Blue and tailless 432 C. Yellow and tailed 423 D. Yellow and tailless 507 If they are linked, calculate the map units between them. a. 44.3 b. 50.6 c. 45.4 d. 23.4

d. 45.4 Rationale: (432+423)/1883*100=45.4 mU

Both homologous chromosomes missing from an individual Embryonic lethal in humans a. Metabolism b. Pleitropy c. Euploid d. Aneuploid

d. Aneuploid

Uniparental disomy (2n) - Example: Imprinting disorders such as Angelman's syndrome a. Pleitropy b. Metabolism c. Euploid d. Aneuploid

d. Aneuploid

Which of the following statements is true? a. An individual won't show incomplete dominance if one allele is a loss of function allele. b. With both incomplete dominance and codominance, one expects heterozygous and homozygous classes to be phenotypically identical. c. In the threshold effect, one would expect the gain of function mutant to be recessive. d. Even in a seemingly clear-cut example of complete dominance, at the protein or enzyme level, one can detect about half the activity or gene product.

d. Even in a seemingly clear-cut example of complete dominance, at the protein or enzyme level, one can detect about half the activity or gene product.

Displays genetic anticipation because repeats expand Male offspring have a 50% chance of receiving fragile X from a carrier mom Incomplete penetrance: can manifest as intellectual disability or individual can be transmitting male (unaffected) a. Examples of Rabies b. Examples of Elimination c. Examples of Neurodegeneration d. Examples of Fragile Site

d. Examples of Fragile Site

The following pedigree shows the inheritance of X-linked recessive red-green colorblindness in a family. What is the genotype of individual II-3? a. Homozygous for the dominant allele b. Hemizygous for the dominant allele c. Heterozygous d. Hemizygous for the recessive allele e. Homozygous for the recessive allele

d. Hemizygous for the recessive allele

An allele of RNA polymerase in Drosophila demonstrates a recessive lethal inheritance pattern. Which of the following is true? a. Two heterozygotes will have offspring with a 2:1 pattern. b. Homozygous lethal genes will always demonstrate a gain of function pattern. c. The threshold effect is not seen in recessive lethality. d. Homozygous recessive offspring will never be seen. e. Two homozygous recessive parents will have 100% homozygous offspring.

d. Homozygous recessive offspring will never be seen.

You observe the cells and karyotypes of a novel species of mammal found on an island. No females have a Barr body; however, they all have two X chromosomes. Males have one X and one Y chromosome. Considering dosage compensation, which of the following would you predict? a. Males transcribe their X at half the rate of a female X. b. Females transcribe their X at twice the rate of a male X. c. Transcription rates would not be affected. d. Males transcribe their X at twice the rate of a female X.

d. Males transcribe their X at twice the rate of a female X.

What is the most likely mode of inheritance? a. X-linked recessive b. autosomal recessive c. autosomal dominant d. Mitochondrial e. X-linked dominant

d. Mitochondrial

What term goes best with this example: Nondisjunction during development causes some cells to have a euploid karyotype: (46, XX) or (46, XY) and some cells to have aneuploid (47, 21+) a. Genetic Anticipation b. Wild-type c. Mutant Gene Product d. Mosaicism

d. Mosaicism

Which of the following best describes Lyonization as a process? a. Barr bodies are inactivated X chromosomes. b. The paternal X chromosome of a female is inactivated in an early embryonic stage. c. The maternal X chromosome of a female is inactivated in an early embryonic stage. d. One of the two female X chromosomes is randomly inactivated in an early embryonic stage. e. Each cell of a female randomly inactivates one X chromosome shortly after mitosis.

d. One of the two female X chromosomes is randomly inactivated in an early embryonic stage. Rationale: In most mammals, either the paternal or maternal X chromosome is inactivated at random, and all progeny cells retain the inactivation. This leads to "patchy" expression in females (mosaicism).

A karyotype is done on an individual who has ovaries and no testes, and it is discovered that they are XY. However, the Y chromosome appears to have a deletion. What region is most likely deleted? a. P A R b. M S Y c. centromere d. S R Y e. heterochromatin of M S Y

d. S R Y Rationale: The S R Y (sex-determining region Y) encodes a protein that guides the formation of testes from the gonads. If this is lacking, the individual will develop the female phenotype.

In jackalopes, blue eyes are dominant to yellow eyes and having a tail is dominant to being tailless. You cross a true-breeding blue eyed, tailed jackalope parent against a true-breeding yellow eyed, tailless jackalope. As expected, the F1 are all blue eyed and have tails. You perform a test cross with the F1 and score the phenotypes of the F2 as follows: A. Blue and tailed 521 B. Blue and tailless 432 C. Yellow and tailed 423 D. Yellow and tailless 507 Are the eye color and tail genes linked? Why or why not? a. There are a higher number of recombinants versus parentals, suggesting the genes are linked. b. There are equal numbers of recombinants and parentals, suggesting the genes are unlinked. c. There are no recombinants, suggesting the genes are unlinked. d. There are a lower number of recombinants versus parentals, suggesting the genes are linked.

d. There are a lower number of recombinants versus parentals, suggesting the genes are linked.

You perform a dihybrid cross to test the inheritance of two fruit fly genes: eye color and wing morphology. You propose a 9:3:3:1 model to test using chi-square analysis. You set your alpha to 0.05. Your final p value ends up being 0.09. Which of the following is correct? a. You should reject your null hypothesis. b. You have data to support linkage of the two genes. c. You have proven your null hypothesis. d. There is a 9% chance that if you redid your experiment, you'd get the same or greater amount of observed deviation from the model.

d. There is a 9% chance that if you redid your experiment, you'd get the same or greater amount of observed deviation from the model.

The accompanying figure is a pedigree of a fairly common human hereditary trait; the boxes represent males and the circles represent females. Filled in symbols indicate the abnormal phenotype. Given that one gene pair is involved, what is the inheritance pattern of the trait? a. Autosomal dominant b. Autosomal recessive c. X-linked dominant d. X-linked recessive e. Epistasis

d. X-linked recessive Rationale: More males are affected than females. Carrier mothers give rise to affected sons.

A genetic counselor takes a family history of a disease and creates the above pedigree. What is the most likely mode of inheritance? a. Autosomal dominant b. Autosomal recessive c. X-linked dominant d. X-linked recessive

d. X-linked recessive Rationale: X-linked recessive. Trait skips generations and affects males. It could be autosomal recessive, but unlikely since it affects males so the best choice is X-linked recessive.

In maize, some alleles at the seed coat locus influence our ability to visually determine the endosperm color phenotype. Specifically, an opaque seed coat prevents us from seeing the endosperm within, while we can readily see the endosperm in seeds with transparent coats. The opaque coat color allele thus appears to mask the expression of endosperm color alleles, providing an example of a. codominance. b. pleiotropy. c. polygenic inheritance. d. epistasis. e. maternal effect.

d. epistasis.

Two genes that are more than 50 map units apart are expected to show _____. a. 25% recombination b. 30% recombination c. 60% recombination d. independent assortment

d. independent assortment

At a kennel, a cross is performed between several black Laborador retrievers known to be heterozygous at two loci. The moms produced 18 black, 6 chocolate, and 8 white pups. These observations can best be explained by a. complete dominance b. dominant epistasis c. codominance d. recessive epistasis e. variable expressivity

d. recessive epistasis

A cross between two heterozygous short-tailed mice results in offspring in the ratio of 67 short-tailed and 34 long-tailed. The best explanation for this result is _____. a. two alleles are codominant b. there are at least three alleles c. there is simple dominance between two alleles d. there is a lethal allele

d. there is a lethal allele

All forms of extranuclear inheritance involve _____. a. DNA carried by organelles other than the nucleus b. endosymbiotic cells carried in the cytoplasm c. gene products carried by the ooplasm d. transmission from the maternal parent to all offspring e. transmission of information not due to nuclear DNA of the individual

d. transmission from the maternal parent to all offspring

Human blood-type is controlled by A, B, and O alleles. A and B are codominant with each other and completely dominant to O. If a mother has type A blood and her son has type O blood, what are the possible blood types of her son's father? a. type O only b. types A or O c. types B or O d. types A, B, or O e. any blood type

d. types A, B, or O Rationale: For the son to be IOIO, he must inherit an IO allele from each parent. Therefore, the father must carry the IO allele, so the father could be IOIO (type O), IAIO (type A), or IBIO (type B), but he could not be IAIB (type AB).

In a Chi-square analysis, what general condition causes one to reject the null hypothesis? Always assume alpha is 0.05 unless otherwise stated. a. usually when the probability value is less than 0.005 b. when observed >> expected c. when observed = expected d. usually when the probability value is less than 0.05 e. usually when the probability value is less than 0.5

d. usually when the probability value is less than 0.05

The N gene controls tail length in widowbirds. The mutant allele n causes shortening of the tail, but not all tails are the same length. This is an example of a. recessive epistasis. b. dominant epistasis. c. codominance. d. variable expressivity. e. lethal allele. f. variable epistasis.

d. variable expressivity.

The genes for purple eyes and curved wings are approximately 21 map units apart on chromosome 2 in Drosophila. A true breeding purple-eyed and wild-type winged female was mated to a true breeding curved wing and wild-type eyed male. The resulting F1 were phenotypically wild-type flies and were mated to purple, curved tester flies. Of 1000 offspring, what would be the expected number of flies with purple eyes and curved wings? a. 790 b. 395 c. 540 d. 210 e. 105

e. 105

The trait for medium-sized leaves in iris is determined by the genetic condition PP′. Plants with large leaves are PP, whereas plants with small leaves are P′P′. The trait for red flowers is controlled by the genes RR, pink by RR′, and white by R′R′. A cross is made between two plants each with medium-sized leaves and pink flowers. If they produce 320 seedlings, what would be the expected phenotypes, and in what numbers would they be expected? Assume no linkage. a. 35 large, red; 35 medium, red; 35 small, red; 35 large, pink; 35 medium, pink; 35 small, pink; 35 large, white; 35 medium, white; 35 small, white b. 80 large, red; 40 medium, red; 20 small, red; 40 large, pink; 20 medium, pink; 40 small, pink; 20 large, white; 40 medium, white; 20 small, white c. 40 large, red; 20 medium, red; 40 small, red; 20 large, pink; 80 medium, pink; 20 small, pink; 40 large, white; 20 medium, white; 40 small, white d. 20 large, red; 40 medium, red; 20 small, red; 80 large, pink; 40 medium, pink; 40 small, pink; 20 large, white; 40 medium, white; 20 small, white e. 20 large, red; 40 medium, red; 20 small, red; 40 large, pink; 80 medium, pink; 40 small, pink; 20 large, white; 40 medium, white; 20 small, white

e. 20 large, red; 40 medium, red; 20 small, red; 40 large, pink; 80 medium, pink; 40 small, pink; 20 large, white; 40 medium, white; 20 small, white

Glucose-6-phosphate dehydrogenase (G6PD) deficiency is inherited as an X-linked recessive gene in humans. A woman whose father suffered from G6PD marries a wild-type man. What percentage of their sons is expected to be G6PD? (Notice that this question is phrased differently than the previous question.) a. 75% b. 25% c. 100% d. 0% e. 50%

e. 50%

An allele of RNA polymerase in Drosophila demonstrates a recessive lethal inheritance pattern. Which of the following is true? a. Homozygous lethal genes will always demonstrate a gain of function pattern. b. The threshold effect is not seen in recessive lethality. c. Two homozygous recessive parents will have 100% homozygous offspring. d. Two heterozygotes will have offspring with a 2:1 pattern. e. Homozygous recessive offspring will never be seen.

e. Homozygous recessive offspring will never be seen.

Colorblindness is an X-linked, recessive trait. The father of a family is colorblind while the mother has wild-type color vision and has no family history of colorblindness. Their daughter displays mild color perception deficits. What might explain her phenotype? a. She is hemizygous for the mutant allele b. She has a de novo mutation in the photoreceptors c. She is intersexed d. She is homozygous for the mutant allele e. She is a mosaic

e. She is a mosaic Rationale: Assuming she is an XX female, some of her cells are expressing the wild-type allele from her mother, while some of her cells are expressing the mutant allele from her father. If she was a female with Turner syndrome, she could have received her X chromosome from her father and would be fully colorblind, similarly to her father.

A young boy is diagnosed with hemophilia A, a recessive, X-linked condition. If neither parent has hemophilia A, what are their genotypes? a. XAXA, XaY b. XaXa, XAYA c. Mother could be either XAXA or XAXa, but father is XaY d. cannot predict from the information given e. XAXa, XAY

e. XAXa, XAY Rationale: Since the father does not have hemophilia, he must be hemizygous dominant (XAY), and, since the mother does not have hemophilia, she must be either XAXa or XAXA. However, since sons get their X chromosomes from their mother, the son must have received an Xa from his mother, meaning that the mother must be XAXa.

Rabbits can come in an allelic series where brown > himilayan > chinchilla > albino, where > means dominant over. They also can come in long (recessive) and short fur. You wish to obtain rabbit babies that are albino with long fur. Which of the following crosses would you do? a. homozygous brown with long fur × albino with short fur (heterozygote) b. albino with long fur × brown chinchilla heterozygote with long fur c. albino with long fur × albino with short fur (homozygous) d. albino with long fur × homozygous himilayan with short fur (heterozygote) e. chinchilla heterozygous with albino with short fur (heterozygous) × brown heterozygous with albino with long fur

e. chinchilla heterozygous with albino with short fur (heterozygous) × brown heterozygous with albino with long fur

Larger arm of a chromosome a. a b. b c. p d. g e. q

e. q

A woman has a rare, heritable defect in the bioenergetic function of her mitochondria. So did her father. If her husband does not have the trait, what can you predict from this? a. that all of her children will have the disorder b. that only her sons will have the disorder c. that only her daughters will have the disorder d. that none of her children will have the disorder e. that the trait will follow normal Mendelian inheritance patterns

e. that the trait will follow normal Mendelian inheritance patterns Rationale: Mitochondrial defects can originate in the mitochondrial or nuclear genome. If the woman inherited the trait from her father, it is not in the mitochondrial genome, as mitochondria are only inherited from the mother. Therefore, it is a nuclear mutation that will follow normal Mendelian patterns of inheritance.

G and E are genes in a diploid organism. The cross of genotypes GE/ge × ge/ge produces the following progeny: GE/ge 404; ge/ge 396; gE/ge 97; Ge/ge 103. From these data one can conclude that ________. a. the recombinant progeny are gE/GE and GE/ge b. the recombinant progeny are GE/GE and ge/ge c. the recombinant progeny are Ge/Ge and gE/gE d. the recombinant progeny are GE/ge and GE/ge e. the recombinant progeny are gE/ge and Ge/ge

e. the recombinant progeny are gE/ge and Ge/ge

Which of the following is evidence for the genotypic sex determination over the balance theory of sex determination in humans? a. XO individuals are female b. Most males are XY and most females are XX c. XX males have the SRY locus translocated onto one of their X chromosomes d. XY females have a deactivated SRY e. A and B f. A, C, and D g. A, B, C, and D

f. A, C, and D Rationale: A, C, and D. XY males and XX females could be evidence of either genotypic sex determination OR the balance theory.

In the past, scientists specializing in animal husbandry amplified SRY by PCR to determine whether a mouse was male or female. They then ran the PCR product on an agarose gel. This strategy was problematic since absence of a band could either be caused by a mouse being female or a mistake in the genotyping (PCR) process. (It is hard to interpret the absence of a band.) Since then, sex determination of mice by PCR involves amplification of a repeat region in both the X and Y chromosomes (the genes in this region are called gametologs). During evolution, this region Y has undergone deletion so that one sex has more repeats than the other. Suppose you were genotyping a known female mouse and a known male mouse. Describe what the gel would look like. Which chromosome, X or Y, would yield a larger band on a gel? a. That's not true, the gametolog on the Y chromosome is actually larger than on the X chromosome. b. No, the gametolog on the X chromosome would yield a smaller band. c. Actually, the size of the gametolog on both chromosomes is the same. d. The gametologs on both chromosomes have the same banding pattern. e. The size of the gametologs has no effect on the banding pattern, so the statement is irrelevant. f. Since the Y chromosome gametolog is smaller, the gametolog on the X chromosome would yield a larger band

f. Since the Y chromosome gametolog is smaller, the gametolog on the X chromosome would yield a larger band

The accompanying figure is a pedigree of a fairly common human hereditary trait; the boxes represent males and the circles represent females. Filled in symbols indicate the abnormal phenotype. Given that one gene pair is involved, what is the inheritance pattern of the trait? a. X-linked dominant b. Autosomal aggressive c. X-linked revolution d. epistasis e. X-linked recessive f. autosomal dominant g. X-linked dinosaur e. Epistaxis h. Autosomal sushi i. autosomal recessive

i. autosomal recessive