Bio 288 Achieve HW Exam 2

Suppose that in unicorns, two autosomal loci interact to determine the type of tail. One locus controls whether a tail is present at all. The allele for a tail, T, is dominant over the allele for no tail, t. If a unicorn has a tail, then alleles with an unknown dominance relationship at a second locus determine whether the tail is curly or straight. Define the alleles for tail texture as US for a straight tail and UC for a curly tail. Farmer Baldridge has two unicorns with curly tails. When he crosses them, one‑half of the progeny have curly tails, one‑quarter have straight tails, and one‑quarter do not have a tail. What are the genotypes of the parents at the tail‑presence gene? What are the genotypes of the parents at the tail‑texture gene? Place the unicorn genotype or genotypes below each unicorn phenotype in the pedigree. Some unicorn phenotypes may be the result of more than one genotype.

- Ttx Tt - UsUc x UsUc - Tt UsUc Tt UsUc TTUsUc TT UsUs ttUsUs TtUsUc TtUsUs ttUsUc

Suppose that a rabbit breeder notices two individuals in a litter with large, round noses and names this trait the clown trait. He obtains another rabbit that has a long, thin nose and names it the Pinocchio trait. After breeding the rabbits with unusual nose shapes to wild‑type individuals for several generations, he decides to cross rabbits from families with a history of the clown and Pinocchio traits. The first litter from this cross results in individuals of four phenotypes. The phenotypes are clown, Pinocchio, wild‑type, and heart‑shaped noses. The pedigree illustrates the inheritance of these traits. Individuals 1 and 4 are siblings and are known to come from a true‑breeding wild‑type line. - What is the most likely mode of inheritance for the clown nose? - What is the most likely mode of inheritance for the Pinocchio nose? A careless rabbit sitter leaves several cages open, and a female rabbit mates with an unidentified male. The breeder sends DNA samples from each of the offspring whose fathers are unknown and DNA from individuals IV‑5 through IV‑10 to a genetics lab. The lab staff test the samples for several genes with suspected effects on nose shape. The gel displays the results of genetic testing for the two polymorphic genes detected. The individuals whose father is unknown are labeled A, B, C, and D. - What is the phenotype of individual B?

- X linked recessive - autosomal dominant - Pinocchio

In eukaryotes, extranuclear inheritance occurs when genetic information is transmitted by mechanisms other than through nuclear DNA. Chloroplast DNA (cpDNA) is an example of one mechanism by which extranuclear inheritance can occur. Select the statements that correctly describe cpDNA.

-cpDNA organization is more similar to that of prokaryotes than eukaryotes. - Chloroplast chromosomes contain genes that are involved in photophosphorylation.

Assume that long earlobes in humans are an autosomal dominant trait that exhibits 30% penetrance. A person who is heterozygous for long earlobes mates with a person who is homozygous for normal earlobes. What is the probability that their first child will have long earlobes? Use two decimal places for the answer.

0.15

How many different types of gametes could be produced by the triple mutant male parent?

1

Suppose that allele D produces tiny ears and allele d produces normal ears. A heterozygous male with the Dd genotype mates with a homozygous female with the dd genotype. Tiny ears is an autosomal dominant trait expressed with 20% penetrance. Determine the probability that their offspring will have tiny ears. Report your answer as a percentage without decimal points.

10%

Which of the F2 phenotypic ratios represents dominant epistasis?

12:3:1

Assuming no recombination, how many different types of gametes can be produced by the wild‑type female parent heterozygous for all three mutations?

2

In chickens, the dominant allele Cr produces the creeper phenotype (having extremely short legs). However, the creeper allele is lethal in the homozygous condition. The homozygous recessive genotype results in a normal individual. If two creepers are mated, what will be the phenotypic ratio among the living offspring?

2 creepers : 1 normal

A maternal effect can cause the offspring phenotype ratio to depart from that of classic Mendelian inheritance. In a species of moth, the dominant allele N codes for brown eyes and recessive allele n codes for red eyes. If an Nn female with brown eyes mates with an Nn male, what is the eye color phenotypic ratio of their offspring?

4, brown eyes : 0, red eyes

Alpers' disease is a mitochondrially inherited disease with symptoms that include seizures, dementia, blindness, liver dysfunction, and cerebral degeneration. The pedigree shows the presence of Alpers' disease in three generations. Individual 4 had two daughters and one son, and individual 6 had one daughter and two sons. Which of the individuals indicated are affected by Alpers' disease? Place the correct symbols on the pedigree showing affected and unaffected individuals.

4. black circle 5. black square 7. white circle 10. black square

A recombination frequency of 5% translates to what distance on a genetic map?

5 m.u.

A genetic cross with two genes produces 400 offspring, and 20 of them have recombinant phenotypes. What is the recombination frequency for this cross?

5%

A gene has three alleles. How many different genotypes are possible at this locus in a diploid organism?

6

How many different types of recombinant gametes could be produced by the wild‑type female parent heterozygous for all three mutations?

6

Both genetic and nongenetic factors cause the congenital skeletal abnormality known as clubfoot, which occurs with a worldwide incidence of about 1 in 1000 births. Gurnett et al. identified a family in which clubfoot occurred as an autosomal dominant trait due to a mutation in the PITXI gene. DNA testing revealed that 11 people in the family carried the PITXI mutation, but only eight of these people had clubfoot. Calculate the penetrance of the PITXI mutation in this family

73%

Which of the F2 phenotypic ratios represents recessive epistasis?

9:3:4

Which of the F2 phenotypic ratios represents duplicate recessive epistasis?

9:7

An allelic series determines coat color in rabbits: C (full color), cch (chinchilla, gray color), ch (Himalayan, white with black extremities), and c (albino, all white). The C allele is dominant over all others, cch is dominant over ch and c, ch is dominant over c, and c is recessive to all other alleles. This dominance hierarchy can be summarized as C > cch > ch > c. The rabbits in the table are crossed and produce the progeny shown. Cross Parents Offspring A full color × albino 50% full color, 50% albino B Himalayan × albino50% Himalayan, 50% albino C full color × albino 50% full color, 50% chinchilla D full color × Himalayan 50% full color, 25% Himalayan, 25% albino E full color × full color 75% full color, 25% albino Match the parental genotypes to the letter corresponding to the appropriate cross listed in the table.

A- Cc x cc B- c^h^c x cc C- Cc^ch^ x cc D- Cc x c^h^c E- Cc x Cc

Epistasis often results in modified dihybrid phenotypic ratios. Assume that you obtain one such modified ratio, 9:7, with the gene pairs A and B involved. What would be a possible genotype for a phenotype that would be included with the 9 portion of the modified ratio?

Aa BB

How are mitochondrial genes typically inherited?

An individual inherits their mitochondrial DNA from their mother

As observed in Labrador retriever coat colors in recessive epistasis, the B locus regulates pigment production whereas the E locus regulates pigment deposition to the hair shaft. Which gene is hypostatic?

B gene

Suppose that in goats, an independently sorting autosomal allele that produces a bearded phenotype is dominant in males and recessive in females. The symbol Bb represents the bearded allele and B+ represents the beardless allele. The autosomal allele for a black coat (W), which also independently assorts, is dominant over the allele for white coat (w). Match each set of progeny phenotypes and proportions to the parental cross that produced that set

B+Bb Ww male × B+Bb Ww female males: 9/16 bearded, black; females: 9/16 beardless, black B+Bb Ww male × B+Bb ww female males: 1/8 beardless, white; females: 1/8 bearded, black B+B+ Ww male × BbBb Ww female males: all bearded, 1/4 white; females: all beardless, 3/4 black B+Bb Ww male × BbBb ww female males: all bearded, 1/2 white; females: 1/4 bearded, black

Male‑limited precocious puberty results from a rare, sex‑limited autosomal allele (P) that is dominant over the allele for normal puberty (p) and is expressed only in males. Bill undergoes precocious puberty, but his brother Jack and his sister Beth underwent puberty at the usual time, between the ages of 10 and 14. Although Bill's mother and father underwent normal puberty, his two maternal uncles (his mother's brothers) underwent precocious puberty. All of Bill's grandparents underwent normal puberty. Match the most likely genotypes for each relative in his family. If there is an equal likelihood of multiple genotypes for a relative, place each possible genotype.

Bill: Pp Bill's brother Jack: pp Bill's sister Beth: Pp pp Bill's father: pp Bill's mother: Pp Bill's uncle: Pp Bill's grandfather: pp Bill's grandmother: Pp

In the snail Cepaea nemoralis, an autosomal allele causing a banded shell, u, is recessive to the allele for an unbanded shell, U. Genes at a different locus determine the background color of the shell. The allele for a yellow shell, b, is recessive to the allele for a brown shell, B. A banded, yellow snail is crossed with a homozygous brown, unbanded snail. The F1 progeny are then crossed with banded, yellow snails in a testcross. Predict the percentage of progeny in each phenotype if the two genes are linked with no recombination, if the two genes assort independently, and if the genes are 20 m.u. apart. Place the percentage for each phenotype for each of the scenarios in the table. Percentages can be used more than once.

Brown unbanded: 50%, 25%, 40% Brown banded: 0%,25%, 10% Yellow unbanded: 0%, 25%, 10% Yellow banded: 50%,25%, 40%

A trait, such as height, has high heritability because much of the variation between individuals is the result of genetic variation. However, not all of the variation for height in a population can be attributed to genetic variation alone. Why does genetic variation not always determine the differences in a given trait between individuals?

Changes in the environment can influence the expression of a gene

The dominance pattern of a gene can be determined from the phenotypes of the parents and offspring. In the examples below, assume that each parent is homozygous for the specific allele and that the progeny are heterozygous. Classify each example as either complete dominance, incomplete dominance, or codominance

Complete dominance - A pea plant with all purple flowers and a pea plant with all white flowers produce a pea plant with all purple flowers. Incomplete dominance - A red snapdragon and a white snapdragon produce a pink snapdragon. - A moth with red wings and a moth with yellow wings produce a moth with orange wings. Codominance - A mother with type A blood and a father with type B blood have a daughter with type AB blood. - A white cow and a red bull have a calf that is white with red spots (roan colored).

A woman has blood type A MM. Her child has blood type AB MN. The blood types of the potential fathers are given in the table. Name Blood type George O NN Tom AB MN Bill B MN Claude A NN Henry AB MM Based on their blood types, which of these men could be the father of this child?

Could be the father: Bill Tom Could not be the father: George Claude Henry

Recombination rates between three loci, L2, R, and W2, in corn are shown in the table. Loci Recombination rate ' L2 and R 35% R and W 217% L2 and W218% Use the recombination rates to arrange the three genes on the section of DNA

DNA strand orange section : R yellow section: W2 red section: L2

Although both dominance and epistasis involve masking of gene expression, they are different from each other. What is the major difference?

Dominance involves one allele masking another at the same locus whereas epistasis involves one locus masking a different locus.

Suppose a species of tulip has three alleles for the gene that codes for flower color. The 𝐶𝑅CR allele produces red tulips, the 𝐶𝑝Cp allele produces purple tulips, and the 𝐶𝑤Cw allele produces white tulips. 𝐶𝑅CR is dominant over 𝐶𝑝Cp and 𝐶𝑤Cw , and 𝐶𝑝Cp is dominant over 𝐶𝑤Cw . For each cross, determine the expected phenotype ratio of offspring flower color.

Expected phenotype ratio 𝐶𝑅𝐶𝑝×𝐶𝑝𝐶𝑤- 2 red : 2 purple : 0 white 𝐶𝑅𝐶𝑤×𝐶𝑝𝐶𝑤- 2 red: 1 purple: 1 white

Which statement explains why the recombination frequency between two genes is always less than 50%?

Genes with a recombination frequency near 50% are unlinked and have an equal likelihood of being inherited together or separately.

Suppose a geneticist is using a three‑point test cross to map three linked rabbit morphology and behavioral mutations called br, sf, and H. The gene br is associated with the bristly fur phenotype, and sf is associated with the short‑footed phenotype. Both br and sf are recessive mutations with respect to wild type. H is a dominant mutation that confers the hyper phenotype. The geneticist first crosses true‑breeding hyper rabbits to true‑breeding bristly fur, short‑footed rabbits. Next, the geneticist backcrosses the F1F1 progeny to the bristly fur, short‑footed parents, and obtains the results reported in the table. Place the genes in the correct order in the chromosome.

H, sf, br

In lizards, the gene for color and the gene for pattern are located on the same chromosome and exhibit recombination. The lined pattern allele (l) is recessive to the solid unpatterned allele (L), and the aqua color allele (b) is recessive to the blue color allele (B). Geneticists cross a homozygous lined aqua lizard with a homozygous solid blue lizard. Then, they carry out a testcross using the F1 progeny. The parental chromosomes for the testcross are given. Identify all possible gametic haplotypes each parent can form.

Heterozygous Parent: BL, Bl, bL, bl Homozygous Parent: bl

Step 1: Identify the characteristics of a genetic system with a maternal effect. Which are characteristics of a genetic maternal effect?

Individuals do not display the phenotypes associated with their own genotypes. Progeny display the phenotypes associated with their mothers' genotypes.

Which statement best describes linkage?

It is a condition in which two or more genes do not show independent assortment

Which statement is the definition of a map unit (centimorgan)?

It is the percent chance of a recombination between two locations on a chromosome

Waxy endosperm (wx), shrunken endosperm (sh), and yellow seedling (v) are encoded by three recessive genes in corn that are linked on chromosome 5. A corn plant homozygous for all three recessive alleles is crossed with a plant homozygous for all the dominant alleles. The resulting F1 are then crossed with a plant homozygous for the recessive alleles in a three‑point testcross. The progeny of the testcross are listed in the table in a random order. wxshV 87 WxShv 94 WxShV 3479 wxshv 3478 WxshV 1515 wxShv 1531 wxSh V292 Wxshv 280 Total 10756 Place each gene in the correct order on the chromosome. Determine the coefficient of coincidence among these genes. Determine the interference among these genes.

Loci 1 wx loci 2 v loci 3 sh Coefficient of coincidence: 0.81 interference: 0.19

Match each definition to the appropriate term.

Locus: the chromosomal site where a specific gene is located Interference: the observed double crossover frequency differs from the expected double crossover frequency Linkage: the tendency for genes located in close proximity on the same chromosome to be inherited together Recombination: the process by which a new pattern of alleles on a chromosome is generated

Now you can use the step‑by‑step method to answer this challenging question. From the pedigree, determine Martha's, her mother's, and her father's genotypes. Place the genotypes next to the individuals.

Martha's genotype: s+s Martha's Mom's genotype: ss Martha's Dad's genotype: s+s

Step 4: Identify the evidence from a pedigree that can be used to determine an individual's genotype in a genetic system with a maternal effect, and determine an individual's genotype from this evidence. Which pieces of information can be used to determine Martha's genotype? What is Martha's genotype?

Martha's mom's genotype is ss. Martha's progeny have dextral shells. s+s

Step 3: Identify the information from the pedigree that helps determine Martha's mom's genotype, and determine Martha's mom's genotype. What information from the pedigree helps determine Martha's mom's genotype? What is Martha's mom's genotype?

Martha's recessive, sinistral phenotype is used to determine her mom's genotype ss

Step 5: Identify the information that can be used to determine an individual's genotype from a pedigree in a system with a genetic maternal effect, and determine the individual's genotype. Which pieces of information can be used to determine Martha's dad's genotype? What is Martha's dad's genotype?

Martha's sister's progeny express the recessive phenotype, sinistral shells. Martha is a heterozygote, s+s. Martha's recessive allele came from her mother. s+s

The genetic map shows the location of three genes on a chromosome. Order the gene pairs based on their likelihood of recombining from most likely to least likely to recombine.

Most likely genes A and C genes A and B genes B and C Least likely

Step 2: Based on the mother's genotype, determine the phenotype expressed in the progeny. Determine the phenotype of the progeny given a specific maternal genotype. Place the expected phenotype of the progeny next to each genotype of the mother

Mothers Genotype: s+s+, s+s, ss Phenotype of progeny: dextral,dextral, sinistral

In 1958, two scientists, R.M. Cooper and J.P. Zubek, conducted an experiment using rats in a maze. They took a group of rats and put them in a maze, recording which were able to solve the maze the quickest. The faster rats were bred together and the slower rats were bred together for a few generations until there were two distinct groups of rats, the bright rats and the dull rats, respectively. Next, groups of the dull and bright rats were raised in different environments. An enriched environment contained toys and colors and a depressing environment was simply a cage with no toys or colors. The different experimental groups were (bright, enriched), (bright, depressing), (dull, enriched), (dull, depressing). After being raised in different environments, the rats were challenged with the maze again yielding the results indicated in the table, where changes were based on the bright and dull rats' initial performance. What genetic phenomena best explains the change in performance of the bright rats raised in a depressing environment?

Performance can be attributed to the interaction between the rats' genes and the environment in which they were raised

Suppose a graduate student is studying a loss‑of‑function mutation in the mouse gene zigzag. Whereas wild‑type mice have straight tails, zigzag mutant mice have tails with two sharp kinks, so that the tail looks like the letter Z. To determine how the zigzag phenotype is inherited, he performs the crosses listed in the first column of the table, using parents from true breeding lines. Three possible sets of results from these crosses are shown. Determine the mode of inheritance of the zigzag gene that would yield each result set.

Set A: X- linked recessive inheritance Set B: maternal effect inheritance Set C: genomic imprinting ( maternal allele is inactive)

When two genes are linked but quite far apart, their estimated map distance, based on recombination frequencies, is often an underestimation of their true map distance. What is the explanation for this underestimation?

Some double crossover events go undetected since they do not lead to recombinant progeny.

Suppose in the Magudon people of Glaxoon that red, writhing tentacles are considered remarkably attractive in a mate. Most Magudons have red, floppy tentacles. Red, writhing tentacles have been found to run in families, but the pattern of inheritance has confounded Glaxooni geneticists. The Magudon inheritance of traits is remarkably similar to that of humans, and many traits exhibit classic Mendelian inheritance. Oddly, red, writhing tentacles do not follow a Mendelian inheritance pattern. Earth's first interplanetary geneticists have been asked to bring a new perspective to this intriguing problem. After studying the appearance of the trait in several families, the Earth geneticists determine that the red, writhing tentacles trait is autosomal recessive and exhibits a pattern of genetic imprinting. The maternal copy of the gene is imprinted. A family tree was created for one Magudon family. Both of the parents in generation I are heterozygous for tentacle movement. Individuals 5 and 10 are homozygous for the dominant trait, red, floppy tentacles. The Earth geneticists provide the Glaxooni geneticists with three explanations. Complete each of the statements with the provided terms. Each term may be used more than once or not at all. You are currently in a labeling module. Turn off browse mode or quick nav, Tab to items, Space or Enter to pick up, Tab to move, Space or Enter to drop. Statement 1 Individual 1 is heterozygous for tentacles and displays the recessive trait, red, writhing tentacles. The recessive allele was inherited from Individual 1's , and the maternal copy of the gene has been silenced. Statement 2 The children of individuals 1 and 2 do not show the phenotypic frequencies usually found in an autosomal, heterozygous cross. The frequency of the children from individuals 1 and 2 with the recessive phenotype, red, writhing tentacles, is one‑half. The usual phenotypic frequency of the recessive phenotype from this type of cross is one‑quarter. Some of the children that express the recessive phenotype are likely and received their recessive allele from their . Statement 3 Some of the children of individuals 5 and 6 have red, writhing tentacles. Individual 5 has been identified as homozygous for red, floppy tentacles. In order for any of the children to display red, writhing tentacles, individual 6 must be and must have received the recessive allele from his .

Statement 1: father- dominant Statement 2: heterozygous- father Statement 3: heterozygous - mother

A yellow female Labrador retriever was mated with a brown male. Half of the puppies were brown, and half were yellow. Explain how the same female, when mated with a different brown male, could produce only brown offspring.

The first male was bb Ee, and the second male was bb EE.

n a testcross involving two heterozygous genes, equal numbers of recombinant and nonrecombinant progeny are produced. From this result, what can be concluded?

The genes are not linked.

A three‑point linkage analysis found a coefficient of coincidence value of 1.0. What does this value reveal about the number of observed double crossovers in the progeny?

The number of observed double crossovers was the same as the number of double crossovers predicted by probability.

What information about recombination frequencies enables scientists to create linkage maps?

The recombination frequency is proportional to the distance between the two genes.

Why do calculations of recombination frequencies between loci that are far apart on chromosomes underestimate the true genetic distance between the loci?

There is an increased probability of double crossover events with increasing distance, such that a gamete can maintain the parental genotype even after many recombination events

Which statement best describes genetic maps?

They are based on recombination frequencies between genes.

In each of the scenarios, an individual heterozygous for the indicated gene or genes is crossed with an identical heterozygous individual. Predict the number of different phenotypes that would be observed among the progeny of such a cross. Scenario 1 R is completely dominant to r Scenario 2 R is incompletely dominant to r. cenario 3 R and S are completely dominant to r and s, respectively. R and S are linked by 10 m.u. and do not interact with each other. Scenario 4 Genes R and S perform the same function. R and S are completely dominant to r and s, respectively. R and S are unlinked. Scenario 5 R and S are completely dominant to r and s, respectively. R and S are unlinked and show recessive epistasis.

Two Possible Phenotypes: Scenario 1 and 4 Three Possible Phenotypes: Scenario 2 and 5 Four Possible Phenotypes: Scenario 3

Assume that a single crossover occurs between two genes during meiosis. What would be the consequence of this crossover event?

Two of the four chromatids will be recombinant, and two will be nonrecombinant.

In the animation, the original triple‑heterozygous female fly had all of the mutations in a coupled arrangement. If, instead, the original wild‑type, triple‑heterozygous female were the result of two true breeding parents, one being scarlet and ebony and the other being spinless, which of the statements would be true?

Wild‑type and triple‑mutant phenotypes would be the least frequent progeny in the three‑point test cross.

Suppose that a geneticist discovers a new mutation in Drosophila melanogaster that causes the flies to shake and quiver. She calls this mutation quiver, qu, and determines that it is due to an autosomal recessive gene. She wants to determine whether the gene encoding quiver is linked to the recessive gene for vestigial wings, vg. She crosses a fly homozygous for quiver and vestigial traits with a fly homozygous for the wild-type traits, and then uses the resulting F1 females in a testcross. She obtains the flies from this testcross. Phenotype Number of flie s𝑣𝑔+v⁢g+ 𝑞𝑢+q⁢u+ 230 vg qu 224 vg 𝑞𝑢+q⁢u+ 97 𝑣𝑔+v⁢g+ qu 99 Test the hypothesis that the genes quiver and vestigial assort independently by calculating the chi-squared, 𝑋2,X2, for this hypothesis. Provide the 𝑋2X2 to one decimal place Does the X2 value support the hypothesis that the quiver and vestigial genes assort independently? Why or why not? Use the partial table of critical values for X2 calculations to test this hypothesis.

X2= 102.5 No, the 𝑋2=X2= value indicates that the observed progeny are significantly different from what would be expected with independent assortment of the two genes.

Could gene order be determined if rather than eight phenotypes of progeny, as seen in the animation, there were only six phenotypes of progeny observed, representing three reciprocal progeny groups?

Yes, it can be, assuming that the double recombinants are the two types of predicted progeny that are missing and that there are zero of each type.

Which description best defines a haplotype?

a group of alleles in close association on a chromosome that are likely to be inherited together

Penile hypospadias, a birth defect in male humans in which the urethra opens on the shaft instead of at the tip of the penis, results from an autosomal dominant gene in some families. Females who carry the gene show no effects. What type of trait is this birth defect an example of?

a sex‑limited trait because the defect occurs only in males and the gene involved is autosomal

What is the phenomenon called where there is stronger or earlier expression of a genetic trait in succeeding generations?

anticipation

In German cockroaches, bulging eyes, bu, are recessive to normal eyes, bu+, and curved wings, cv, are recessive to straight wings, cv+. Both traits are encoded by autosomal genes that are linked. A cockroach has genotype bu+ bu cv+ cv, and the genes are in repulsion. Which set of genes will be found in the most common gametes produced by this cockroach?

bu cv+

Which statistical test can be used to determine if two genes are linked?

chi‑square test of independence

A boy has blood type MN with a genotype of LMLN. His red blood cells possess both the M antigen and the N antigen. What is the relationship between his two alleles for this gene?

codominance

A man and a woman are both deaf due to being homozygous for a recessive autosomal mutant allele. However, they are homozygous recessive at different gene loci. If all their children have normal hearing, which event has occurred within each child?

complementation

In humans, mitochondrial genetic disorders are inherited from only the mother. The severity of such diseases can vary greatly, even within a single family. What form of inheritance does this represent?

cytoplasmic inheritance

In Drosophila melanogaster, black body (b) is recessive to gray body (b+ ), purple eyes (pr) are recessive to red eyes (pr+), and vestigial wings (vg) are recessive to normal wings (vg+). The loci encoding these traits are linked, with the map distances shown in the image. The interference among these genes is 0.5. A fly with a black body, purple eyes, and vestigial wings is crossed with a fly homozygous for a gray body, red eyes, and normal wings. The female progeny are then crossed with males that have a black body, purple eyes, and vestigial wings. If a total of 1000 progeny are examined, determine the expected number of offspring that result from each crossover event.

double crossover: 4 flies single crossover between vg and pr : 126 flies single crossover between b and pr: 56 flies non-recombinant: 814 flies Double crossover: b+ pr vg+ b pr+ vg Single crossover between b and pr b+ pr vg b pr+ vg+ Single crossover between vg and pr b pr vg+ b+ pr+ vg Non‑recombinants b pr vg b+ pr+ vg+

Three‑point testcrosses are often used to map genes. The two least frequent classes from such crosses usually represent which types of progeny?

double‑crossover progeny

Assuming that gene order is as determined in the original animation, what would be the most common three‑point test cross progeny types if, in the triple heterozygous female parent, the scarlet eye and spineless bristle mutations were coupled and the ebony body mutation was in repulsion?

ebony progeny and scarlet, spineless progeny

Assuming that gene order is as determined in the original animation, what would be the least common three‑point test cross progeny types if, in the triple heterozygous female parent, the scarlet eye and spineless bristle mutations were coupled and the ebony body mutation was in repulsion?

ebony, spineless progeny and scarlet progeny

What type of gene action occurs when one gene masks the effect of another gene at a different locus?

epistasis

A gene whose expression is affected by the sex of the transmitting parent demonstrates which of the events?

genomic imprinting

Fur color in a species of mouse is controlled by a single gene pair. BB animals are black and bb animals are white. Bb animals have gray fur and each hair is gray. What type of interaction is being shown by the two alleles in heterozygous animals?

incomplete dominance

A geneticist is using a three‑point testcross to map three linked Drosophila recessive mutations called a, b, and c, where a is associated with anomalous gait, b is associated with buckled wings, and c is associated with curved bristles. She first crosses homozygous anomalous, buckled flies to homozygous curved flies. Next, she testcrosses the F1 progeny to anomalous, buckled, curved flies. She obtains 1000 progeny distributed as shown. From this data, calculate the map distance between b and c. Testcross progeny phenotype Number curved 277 anomalous, buckled283 wild type 4 anamalous, buckled, curved 6 buckled, curved 128 anomalous 132 anomalous, curved 86 buckled 84

map distance: 27 m.u.

Suppose two independently assorting genes are involved in the pathway that determines fruit color in squash. These genes interact with each other to produce the squash colors seen in the grocery store. At the first locus, the W allele codes for a dominant white phenotype, whereas the w allele codes for a colored squash. At the second locus, the allele Y codes for a dominant yellow phenotype, and the allele y codes for a recessive green phenotype. The phenotypes from the first locus will always mask the phenotype produced by the second locus if the dominant allele (W) is present at the first locus. This masking pattern is known as dominant epistasis. A dihybrid squash, Ww Yy, is selfed and produces 160160 offspring. How many offspring are expected to have the white, yellow, and green phenotypes?

number of white offspring: 120 number of yellow offspring: 30 number of green offspring: 10

A female Drosophila fly is heterozygous for three recessive pigmentation mutations called pl, wh, and tp. pl is associated with purple eyes, wh is associated with a white body, and tp is associated with transparent wings. A geneticist crosses this fly to a male purple, white, and transparent fly and obtains the 1000 progeny given in the table. Based on the table, which genes are linked? Calculate the distance between the linked genes. distance between the linked genes:

pl and wh 19 m.u.

Consider the set of crosses for hypothetical genes that control eye color and tail length in mice. Diane is working with mice with recessive mutations for the genes that control light eye color (b) and short tail length (t). She knows that these genes display genetic linkage and are found on chromosome III. In her work, she crosses a true‑breeding male with light eyes and a long tail (bbTT) and a true‑breeding female with dark eyes and a short tail (BBtt). She then crosses the resulting heterozygous progeny (BbTt) together in a dihybrid cross. The number of animals of each phenotype of this second cross is shown. dark‑eyed, short‑tailed (𝐵𝐵𝑡𝑡)=24dark‑eyed, short‑tailed (B⁢B⁢t⁢t)=24 dark‑eyed, long‑tailed (𝐵𝑏𝑇𝑡 or 𝐵𝐵𝑇𝑇)=56dark‑eyed, long‑tailed (B⁢b⁢T⁢t or B⁢B⁢T⁢T)=56 light‑eyed, long‑tailed (𝑏𝑏𝑇𝑇)=26light‑eyed, long‑tailed (b⁢b⁢T⁢T)=26 light‑eyed, short‑tailed (𝑏𝑏𝑡𝑡)=3light‑eyed, short‑tailed (b⁢b⁢t⁢t)=3 What is the most likely explanation for why Diane sees light‑eyed, short‑tailed (bbtt) progeny in this cross?

recombination

What is an Ab aB linkage arrangement called?

repulsion linkage

A recessive mutant allele of an autosome gene in a species of mouse results in a shorted tail in males when homozygous. However, when homozygous in females, this genotype has no effect, and the mice have normal tails. What is this genetic phenomenon called?

sex‑limited characteristic

Suppose a geneticist uses a three‑point testcross to map three recessive, linked hummingbird wing‑shape mutations called tn, r, and s, where tn is associated with thin wings, r is associated with round wings, and s is associated with sharp wings. He first crosses homozygous sharp birds to homozygous thin, round birds. Next, he testcrosses the F1 progeny to thin, round, sharp birds. He obtained the results shown. Given this data, label the progeny phenotypic classes as either parental, single crossover (SCO), or double crossover (DCO) recombinant types. Each label may be used multiple times. Two classes have already been filled in.

sharp 801 parental thin, sharp 815 parental round 201 SCO thin, sharp 195 SCO thin 8 DCO round, sharp 8 DCO thin,round, sharp 27 SCO wild type 26 SCO

With genetic maternal effect, the phenotype of an individual is determined by which of the statements?

the nuclear genotype of the maternal parent

What is penetrance?

the percentage of individuals having a particular genotype who express the expected phenotype

A homozygous variety of opium poppy (Papaver somniferum Laciniatum) with lacerate leaves was crossed with another homozygous variety with normal leaves. All the F1 had lacerate leaves (jagged‑edged leaves). Two F1 plants were crossed to produce the F2. Of the F2, 249 had lacerate leaves and 16 had normal leaves. How are lacerate leaves determined in the opium poppy? Give genotypes for all the plants in the P, F1, and F2 generations. Some generations will have more than one genotype.

two genes, with a dominant allele at either or both loci P - AABB, aabb F1- AaBb F2( lacterate)- A_B_ , A_bb, aaB_ F2( normal)- aabb

Suppose a geneticist uses a three‑point testcross to map three linked Drosophila recessive mutations called f, w, and b. Gene f is associated with abnormally fast movement, w is associated with a wavy movement pattern, and b is associated with broad wings. The geneticist first crosses homozygous broad flies to homozygous fast, wavy flies. Next, he testcrosses the F1F1 progeny to fast, wavy, broad parents and obtains the results reported in the table. Based on the data, select the order of the three genes.

w b f f b w

Which statement below defines epistasis?

when one locus affects or covers the outcome of another locus

Suppose a geneticist is using a three‑point testcross to map three linked Drosophila recessive mutations called s, z, and b, where s is associated with abnormally slow movement, z is associated with zigzag movement pattern, and b is associated with broad wings. The geneticist first crosses homozygous broad flies to homozygous slow, zigzag flies. Next, he testcrosses the F1 progeny to slow, zigzag, and broad parents and obtains the results in the table. F1 testcross progeny phenotype Number broad 615 slow, zigzag 633 zigzag 231 slow,broad 223 slow 219 zigzag, broad 237 slow, zigzag, broad 7 wild type 8 Based on this data, select the order of the three genes.

zbs sbz