BIS 101 Week 8

Consider the above pedigree. It depicts inheritance of a dominant condition. Affected allele, R. Normal allele, r. The numbers refer to alleles of a marker that may be linked. To explore this possibility, derive a probability for theta = 0.1. Choose the best answer.

0.00092

The gene for albinism (recessive) and the gene for dwarfism (dominant, rare) are linked and 20 m.u. apart. Mating between Meg, a normal woman with no family history of albinism, and Francisco, an albino dwarf, produced a dwarf son, Diego. Diego mates with Rhonda, a normal woman whose father was albino. What is the chance that their child will be albino and not dwarf?

0.05

Blue eyes are recessive to any other color. Mary and John have brown eyes, but each have a blue-eyed father. They have seven children. The probability that four are brown-eyed and three are blue-eyed (in any order) is:

0.15

Alleles at the Y locus (no relation to Y chromosome!) can be genotyped by a lab test and are defined by different numbers (Y1, Y2, Y3, etc). A mother affected by a dominant condition (D) has Y7Y2,Ddgenotype. She marries a man with a Y5Y5,ddgenotype. Assuming no linkage between Y and D, the probability that her child inherits D but not Y7 is:

0.25

You breed a pigmented german shepherd heterozygous for recessive white coat color (Ww) to a female with the same genotype. What is the probability that if they have four pups, two will be white and two pigmented?

27/128

In horses the Red gene (R or r) displays recessive epistasis on the Bay gene (B or b). Horses with the rr genotype are red regardless of their Bay genotype (B_ or bb). Recessive Bay (bb) are black if R_. Crossing a black stallion to a red mare has produced three colts: one bay, one red and one black. What is the P that the next colt from the same pair will be Bay?

0.25

A normal woman with no family history of genetic disease or albinism has a child with a man affected by albinism (recessive) and dwarfism (dominant, rare). The gene for albinism and the gene for dwarfism are linked and 20 m.u. apart. What is the chance that the child is dwarf and carries albinism?

0.5

Marker allele Y7 and allele D are linked at 16 m.u. An affected grandpa that has Y7-D/Y2-d haplotypes. The probability that a grandchild inherits D but not Y7 is: (Hint: this is a challenging problem. Make sure you consider all the possible event paths leading to the final outcome.

~0.07

Alleles at the Y locus can be genotyped by a lab test and have different numbers (Y1, Y2, etc). An affected grandma has Y7Y2,Dd genotype. Assuming no linkage between Y and D, and that no other family member has these alleles, the probability that her grandchild inherits D but not Y7 is: (Hint: this is a challenging problem. Make sure you consider all the possible event paths leading to the final outcome)

~0.19

Marker allele Y7 and allele D are linked in phase at 16 m.u. Consider an individual that has Y7-D/Y2-d haplotypes. The P of a Y7-d gamete is equal to the P of a __________ gamete, and is __________. [a] Y2-D / 0.08 [b] Y2-D / 0.42 [c] Y2-d / 0.25 [d] Y2-d / 0.08 [e] Y2-D / 0.16

A

In the cross AaBbCcDd x AaBbCcDd, what proportion of offspring are expected to be heterozygous for all four genes?

1/16

All sneetches want their children to have stars on their bellies. Sneetches can be black or yellow, and star-bellied or starless. Two loci control these phenotypes: yellow is dominant to black, and star-bellied is dominant to starless. The combination of alleles that make a black, starless sneetch is lethal. If two heterozygous yellow, star-bellied sneetches mate, what is the likelihood their first child will be starless?

1/5

An LOD of 2 means that the likelihood of linkage for the given theta value is __________ more likely than the likelihood of independent assortment.

100

Loci A and B are linked at a 15 m.u distance, with A-B alleles in coupling. The cross AaBb x AaBb produced 1000 progeny. How many aabb do you expect?

180

Two pure breeding lines of pea, one tall and hairy, the other short and smooth, are crossed. The F1 hybrid is tall and hairy. The F2s assort in four phenotypic classes in a 9:3:3:1 ratio. Crossing a short, smooth plant to a short, hairy one will produce __________ phenotypic classes:

2 or 1

Two pure breeding lines of pepper, one dark green and round, the other light green and oval, are crossed. The F1 hybrid is dark green and round. The F2s assort in four phenotypic classes in a 9:3:3:1 ratio. Crossing a light green, oval F2 plant to the F1 will produce __________ phenotypic classes:

4

Forbes' parakeet and the red-crowned parakeet are pure breeding animals. An excaped pair hybridized producing birds of mixed appearance. These F1s intercrossed producing ~1000 birds of ranging phenotypes. Among these, four looks like a pure-bred Forbes and another like a pure-bred red-crowned parakeet. You conclude that __________ genes are likely to control the differences between these "species".

5

Blue eyes are recessive to any other color. Mary and John have brown eyes, but each have a blue-eyed father. What is the probability that at least one of their three children is brown-eyed? Hint: sometime it is easier to calculate the P of the alternate event

63/64

A and B = chromosomal loci. Dominant and recessive alleles = upper and lower case letters, respectively. The hybrid AaBb is crossed to aabb. The following phenotypic classes and progeny numbers are produce: AB, 25; Ab, 450; aB, 410; ab, 30. Alleles for genes A and B are in __________ phase. The best estimate of map distance between the A and B loci is __________ m.u. [a] repulsion / 6 [b] coupling / 12 [c] repulsion / 12 [d] repulsion / 24 [e] coupling / 6

A

A chocolate male lab mates with a black female lab. The resulting puppies include two golden and one chocolate. Based on the genetics of Q.2 (see also below), the male and female genotypes could be, respectively: __________, __________. from Q.2: Coat colors of Labrador retrievers depend on two genes: one, the inhibitor of coat color pigment (ee) prevents the expression of color alleles at another independently assorting locus, producing a golden coat color. When the dominant condition exists at the inhibitor locus (E_), the alleles of the other locus may be expressed, E_B_ producing black and E_bb producing chocolate. [a] Eebb / EeBb [b] EEbb / EeBb [c] eeBB / EEbb [d] eeBb / Eebb [e] Eebb / EEBb

A

ABO blood type in humans results from this type of genetic system: [a] Three alleles involving dominance and codominance [b] Two alleles with incomplete dominance [c] Two genes involving epistasis [d] Three alleles involving epistasis [e] Two alleles involving a lethal genotype

A

Considering epistatic yellow in retrievers. An aa dog is yellow regardless of the genotype at the B locus. In the presence of the dominant A allele, dogs are black or brown depending on the presence of the dominant B (black) allele or recessive b (brown) allele. If the mating of a brown and a yellow dogs has previously produced puppies of all three colors, what ratio do you expect for black:brown:yellow? [a] 1:1:2 [b] 12:3:1 [c] 9:3:4 [d] 1:1:1 [e] 1:2:1

A

In a pea plant the A and B genes are linked at a 10 m.u. distance. You genotype the pollen grains of this individual. What is the expected ratio of recombinant to parental gametes for the A-B genes? [a] 1:9 [b] 1:1 [c] 2:9 [d] 1:2 [e] 1:5

A

In a trip to Mexico you buy pure-breeding seeds of a tomato with green stripes over a red background, which you call Zebra. You cross zebra to a common red-fruited variety, Early Girl. The F1 is spotted green on red background. In the F2 you observe 33 solid red, 70 spotted, and 29 zebra. The mutation involves change(s) at __________. Individuals with spotted phenotype are __________. [a] A single gene / heterozygous [b] Two genes / has a dominant allele at both [c] A single gene / recessive homozygus [d] Two genes / recessive at all loci [e] A single gene / dominant homozygous

A

In your greenhouse populations you observe 2 mutant plant lines which you call "biteme" mutants. Plants homozygous for a recessive allele you call 'a'are vulnerable to zombie mite bites. The same is true for plants homozygous for another recessive allele you call 'b'. You cross individuals homozygous for each of the mutants and observe that the entire F1 generation remains vulnerable to zombie mite bites. How many loci are manifested by this analysis and what are the genotypes of the parental generation. [a] 1 locus, a and b are two alleles of the same gene) [b] 1 locus, AA x bb [c] 2 loci, AAbb x aaBB [d] 2 loci, AaBb x AaBb [e] None of the other answers

A

Loci A and B are linked at a 25 m.u distance. The cross AaBb x aabb produced 1000 progeny. Of this progeny, approximately __________ are expected to be recombinant (ignore cross-over interference). Alternatively, if the map distance is determined to be 40 m.u., __________ progeny are expected to be NON recombinant. [a] 250 / 600 [b] 250 / 400 [c] 750 / 600 [d] 750 / 400 [e] 600 / 750

A

Mr. T has blood type A. Mrs. T has blood type B. They have three kids, whose blood type is, respectively, A, B, and O. The best explanation for the blood phenotypes in this family is: [a] The parents are both heterozygous at the same blood type locus: Mr. T is IA/Io, Mrs. T is IB/Io. [b] The parents are both heterozygous at the same blood type locus: Mr. T is IA/IB, Mrs. T is IA/IB. [c] The parents are homozygous at the same blood type locus: Mr. T is IA/IA, Mrs. T is IB/IB. [d] The parents are both heterozygous at two independent genes controlling blood type. Mr. T is Aabb, Mrs. T is aaBb. [e] The parents are both homozygous at two independent genes controlling blood type. Mr. T is AAbb, Mrs. T is aaBB.

A

Norway spruce can have stiff or droopy branches. You cross two pure breeding varieties, one stiff the other droopy obtaining stiff F1s. Selfing one of these you obtain 178 stiff and 144 droopy F2. The best explanation is two independent loci are involved in [a] double recessive epistasis [b] codominance [c] recessive epistasis [d] multiple alleles [e] dominant epistasis

A

Pursuing the linkage of marker allele G3 to disease allele D, your team obtains the following theta_LOD values series. The first number is theta, the second, separated by _, is the corresponding LOD (some LOD are negative): 0.01_-4, 0.02_-2, 0.03_-1, 0.05_0, 0.07_1, 0.09_2, 0.11_5, 0.14_10, 0.17_15, 0.2_6, 0.25_3, 0.3_1, 0.35_0.2. You conclude that linkage is __________ at __________ m.u. [a] strongly supported / 17 [b] suggestive but not significant / 30 [c] suggestive but not significant / 1 [d] strongly supported / 1 [e] unlikely / any

A

Studying an F2 family, you suspect the action of two unlinked genes because the phenotypic ratio "looks like" 9:3:3:1. To determine whether your guess is reasonable, you subject your observed phenotypic counts to the chi square test obtaining a chi square value of 6.8. The degrees of freedom are __________. You conclude that your guess is __________. [a] 3 / supported [b] 2 / not supported [c] 4 / not supported [d] 3 / unsupported [e] 4 / supported

A

The Ziblets, a family of fairies, display through generations the occasional Pink Nose. This syndrome was first recorded in great-great-grandfather Zibbo. Worried about genetic disease they have since taken great precautions to marry into separate families that have no history of Pink Nose. Alas, Pink Nose pops up with a certain probability among descendants of Zibbo. Children with an affected parent are often normal, but their children can display Pink Nose. The following explanation of the condition is consistent with the observations above: [a] dominant / incompletely penetrant. [b] recessive / variable expressivity [c]dominant / variable expressivity [d] recessive / incomplete penetrance [e] incompletely dominant / fully penetrant

A

Three plant lines each have an independent mutation that causes them to have a curled leaf phenotype. All are true breeding for this phenotype and the wild-type phenotype is straight leafed. When crossing these mutants, you observe the following: Plant 1 x 3 > complementation Plant 1 x 2 > no complementation Given these results, what phenotypes have you observed in the progeny? [a] 1x3 = straight, 1x2 = curled [b] None of the other answers [c] 1x3 = curled, 1x2 = curled [d] 1x3 = curled, 1x2 = straight [e] 1x3 = straight, 1x2 = straight

A

Two normal looking fruit flies were crossed, and in their progeny, there were 202 females and 98 males. Which answers provides the best genetic explanation for this anomaly. [a] X-linked recessive lethal allele [b] None of the other answers [c] X-linked gain of function mutation [d] Y-linked recessive lethal allele [e] Y-linked gain of function mutation

A

A recombination frequency of 45% between the A and B loci suggests that: [a] A and B are on different chromosomes [b] A and B are nearly unlinked: they are distant on the same chromosome. [c] A and B are on different chromosomes, which rarely can undergo pairing and C.O. [d] A and B are linked and in repulsion [e] A and B are tightly linked

B

A tall plant was discovered in a field of pure breeding short sunflowers. Crossing the tall mutant to a true-breeding wild-type plant you obtain 26 tall and 0 short plants. What is the most likely genotype of the tall mutant and the inheritance pattern of the tall phenotype? [a] tt, dominant [b] TT, dominant [c] Tt, dominant [d] Tt, recessive [e] tt, recessive

B

In snapdragons, a true breeding red flowered plant (AA) crossed with a true breeding white flowered plant (aa) produces pink flowered F1 progeny. What phenotypic ratios do you expect from selfing the F1 generation? [a] 3 red: 1 pink [b] 1 red: 2 pink: 1 white [c] None of the other answers [d] 1 red: 2 white: 1 pink [e] 1 pink: 2 red: 1 white

B

No individuals homozygous for BRCA2 mutation have been found. This observation is consistent with: [a] incomplete penetrance [b] lethality [c] dominance [d] variable expressivity [e] combination of wild-type and mutant allele is lethal

B

You prized hybrid cow has the following genotype: Qq,Rr,Tt, Hh. The Q, R, H, and T loci are independent. She can produce __________ egg types, such as __________. [a] 4 / QRT, qrt, QRtHH [b] 16 / QRTH, QrtH, Qrth [c] 8 / QqRT, qRRt, QQrTT [d] 8 / qRt, qrT, QrtH, QRT [e] 8 / qRt, qrT, Qrt, QRT

B

A test cross is performed on a Drosophila (fruit fly) presumed to be B/b, F/f. B = black body, b = brown body. F = forked bristles, f = unforked bristles. The resulting offspring has the following phenotypes: Black, forked = 230, black, unforked = 210, brown, forked = 240, and brown, unforked = 270. The Chi-square value of this data set is __________ and the degrees of freedoms are __________. *Note on genotypic nomenclature: B/b is a equivalent to Bb *Use your textbook instructions, or an Internet site to calculate the chi square [a] 3.75 / 3 [b] 1.5 / 4 [c] 7.9 / 3 [d] 5.75 / 3 [e] 8.25 / 3

C

Considering epistatic yellow in retrievers. An aa dog is yellow regardless of the genotype at the B locus. In the presence of the dominant A allele, dogs are black or brown depending on the presence of the dominant B (black) allele or recessive b (brown) allele. If the mating of two black dogs has previously produced puppies of all three colors, what ratio do you expect for black:brown:yellow? [a] 1:1:1 [b] 1:2:1 [c] 9:3:4 [d] 2:1:1 [e] 12:3:1

C

Curly leaf of pepper is caused by a recessive allele at a single locus. A homozygous curly leaf plant is crossed to a heterozygous, normal plant. In their progeny you count 60 healthy and 15 affected individuals. The best explanation for these results is that the curly leaf trait [a] is negatively affected by non-disjunction [b] has variable expressivity [c] has incomplete penetrance [d] incomplete dominance of [e] X-linked inheritance

C

In Chinese cabbage, wrinkled leaf is dominant, smooth leaf is recessive. You would like to find out if a smooth plant is homozygous or heterozygous. What is your simplest and best strategy? [a] testcross [b] propagate by cuttings and phenotype clonal progeny [c] review your genetics, no need to test [d] cross to another, uknown dark green plant [e] count the chromosomes

C

Shire hobbits living in the villages of Bonning Gate and Newbiggin have curly hair on their feet rather than the normal smooth hair. The curly haired trait is controlled by different genes in the two families. Intervillage matings between the families resulted in all children (F1) having smooth hair. If such children were to intermarry, what would be the probability of their children (F2) having curly haired feet? [a] 0.25 [b] 0.06 [c] 0.44 [d] 0.66 [e] 0.18

C

There are two questions on pineapple genetics. The leaf edges of pineapple can be uniformly spiny spiny on the leaf tip piping, i.e. with nice smooth edges that resemble cloth piping folds Pineapple growers prefer varieties that are piping because harvesting them is safer and easier. The following crosses were carried out using pure breeding varieties whose phenotype is indicated as piping, spiny tip, or spiny. Two pure bred parents are named as W and Z for later reference in the answers. Based on the experimental results, you must conclude that [a] single gene at work, Piping is dominant [b] single gene at work, Spiny is dominant [c] two genes at work, Spiny is double recessive [d] two genes at work, Spiny tip is double recessive [e] two genes at work, Spiny is dominant

C

There are two questions on pineapple genetics. The leaf edges of pineapple can be uniformly spiny spiny on the leaf tip piping, i.e. with nice smooth edges that resemble cloth piping folds Pineapple growers prefer varieties that are piping because harvesting them is safer and easier. The following crosses were carried out using pure breeding varieties whose phenotype is indicated is indicated as piping, spiny tip, or spiny. Two pure bred parents are named as W and Z for later reference in the answers. You conclude that in regard to the two genes controlling the leaf edge phenotype, variety W and Variety Z have the following genetic relationship. W is dominant for only one, Z dominant for both genes [a] W is recessive for both genes, Z dominant for both [b] each is dominant for one only and not the same one [c] W is dominant for both genes, Z only for one [d] W and Z have the same genotype at both loci

C

Two golden retrievers have normal legs. Three of their nine puppies develop into short-egged dogs. Most likely, the parents are __________. The short-leg condition is __________. [a] one homozygous / the other heterozygous. dominant [b] both homozygous / dominant [c] both heterozygous / recessive [d] both heterozygous / dominant [e] one homozygous / the other heterozygous. recessive

C

You mate two mice with the following genotypes: MmKkLLPpQqRr x mmkkllppqqrr. Assume that each gene controls a different character and that the upper case allele is dominant, how many different genotypes and phenotypes do you expect? [a] 64, 64 [b] 128, 64 [c] 32, 32 [d] 16, 64 [e] 64, 32

C

Consider the epistatic problem of the piping vs spiny pineapple. Piping is epistatic on spiny. If the piping locus "P" has genotype pp, the spiny trait is manifested. P_ plants are always piping. ppS_ = spiny tip, ppss = spiny. Two inbred lines, one piping, one spiny are crossed and produce a piping F1. Selfing of the F1 produces the following F2s: 118 piping, 28 spiny tip and 12 spiny. What ratios, in that order, do you expect if you cross the F1 to the inbred spiny parent? [a] 1:2:1 [b] 12:4:3 [c] 9:3:4 [d] 2:1:1 [e] 1:1:2

D

In German shepherds dogs, white coat (ww) and long hair (ll) are recessive traits controlled by independent genes. The corresponding wild-type alleles are dominant and result in pigmented coat and (medium-)short hair. A pigmented, long coated male is crossed to a pigmented, long hair female. Their pups are as following: 3 pigmented and short-haired, 4 white and short-haired, 2 pigmented and long-haired, 1 white and long-haired. What are the genotypes of the parents (male x female): [a] WWLl x Wwll [b] WWLl x WwLl [c] WwLl x Wwll [d] WwLl x WwLl [e] wwLl x Wwll

D

Investigating a suspicious gambling operation, you observe that the six-faced dice used yields 1 less frequently than expected. You measure its probability of "1" at 0.11. The expected odds are __________. The observed odds are __________. [a] 1:6 / 1:9 [b] 0.17 / 0.11 [c] 1:17 / 1:11 [d] 1:5 / 1:8 [e] 1:16 / 1:11

D

You cross a hairless mouse (aaBB genotype) to a mouse with curly hair (AAbb genotype). All of the F1s have straight hair. You cross two of the F1 mice together. In the F2 generation 18 mice have straight hair, 8 mice are hairless, and 6 have curly hair. What is the phenotype of an aabb mouse in the F2 generation? [a] curly hair [b] straight hair [c] None of the other answers [d] Hairless [e] dead because of lethality

D

You have obtained two true-breeding strains of mice, each homozygous for an independently discovered recessive mutation that prevents the formation of hair on the body. The discoverer of one of the mutant strains calls his mutation naked, and the other researcher calls her strain hairless. You cross naked and hairless mice with each other and all the offspring are phenotypically wild-type. What were the genotypes of the parents? [a] HHNN x hhNN [b] HHNN x hhnn [c] hhnn x HHnn [d] hhNN x HHnn [e] hh x nn (h and n are alleles of the same gene)

D

Choose the best answer: When self-fertilizing plants, why does heterozygosity decrease following generations of successive inbreeding? [a] Self-fertilization reduces allelic frequencies, driving the population towards homozygosity [b] Self-fertilization has no effect on declining heterozygosity rates within a population [c] Self-fertilization increases the frequency of the dominant allele, and decreases the frequency of the recessive allele [d] Heterozygous plants will always produce homozygous offspring, whereas homozygous plants will produce homozygous offspring in a 1:1 ratio [e] Half the progeny of heterozygous plants is homozygou

E

Coat colors of Labrador retrievers depend upon the action of at least two genes. An inhibitor of coat color pigment (ee) prevents the expression of color alleles at another independently assorting locus, producing golden coats. When the dominant condition exists at the inhibitor locus (E_), the alleles of the other locus may be expressed: E_B_ = black and E_bb= chocolate. When dihybrid (heterozgyous at both loci) black labs are mated together, what are the phenotypic proportions expected in the progeny? [a] 9 golden: 3: brown: 4 black [b] 9 black: 3 golden: 4 chocolate [c] 9 chocolate: 3 golden: 4 black [d] 9 chocolate: 3 black: 4 golden [e] 9 black: 3 chocolate: 4 golden

E

Consider the complementation table below illustrating the analysis of 15 mutants, F to Z. All are recessive and homozygous. When crossed, the phenotype of the F1 is marked as 1 if complementation is seen (wild-type F1), or 0 if the F1 is mutant. How many genes are defined by this analysis and what alleles belong to what gene? [a] 3 / g1=FOQSZMPY / g2=NRX / g3=TVUW [b] 5 / g1=FMNTU / g2=OPRVW / g3=QYX / g4=S / g5=Z [c] 15 / each allele belongs to a different gene [d] 4 / g1=QSYXZ / g2=FOMP / g3=NR / g4=TVUW [e] 5 / g1=FOQSZ / g2=MPY / g3=NRX / g4=TV / g5=UW

E

Flowers of Armenian cyclamens can have three types of petal color: solid dark-blue, red with yellow tips, or solid yellow. You purchase a blue flowered plant, self it, and obtain 230 blue, 64 red and yellow, and 13 yellow. Based on this, the most likely hypothesis is that __________ genes are involved and that the one conferring __________ color displays __________ epistasis. [a] 2 / red / dominant [b] 3 / blue / recessive [c] 2 / red / recessive [d] 2 / yellow / recessive [e] 2 / blue / dominant

E

In ferrets, sable (B) is dominant to black (b) and pointed ear (R) to round (r). A black, pointed ear male is mated to a sable, pointed ear female. Their pups are 3 sable and pointed eared 3 black and round eared 2 sable and round eared 1 black and point eared What are the genotypes of the parents (male x female): [a] bbRR x BbRr [b] bbRr x BBRr [c] BbRr x BbRr [d] BbRR x Bbrr [e] bbRr x BbRr

E

Pursuing the linkage of marker W5 to disease allele D, your team obtains the following LOD maximum at a theta of 0.03 for 4 independent family pedigrees: 2.2, 1.8, 1.3, 2.7. You conclude that linkage between marker alelle W5 and D is __________ fold more likely than no linkage and __________ at __________ m.u. [a] in average 100 / suggestive but not significant / 3 [b] at most 500 / suggestive but not significant / 3 [c] around 10 / suggestive but not significant / 30 [d] at most 2.7 / not significant / 6 [e] 100 million / strongly supported / 3

E

Two pure breeding lines of maize display light green leaves instead of the usual dark green of most varieties. Their F1 is dark green. The F2 consists of 461 dark green and 353 light green. You concludes that __________ gene(s) is/are responsible. Using A, B, etc designation for the gene(s), the dark green and light green F2s would have, respectively, the following generic genotypes __________, __________. [a] one / AA or Aa / while aa is lethal [b] two / aabb / anything else (aaB_, A_B_, A_bb) [c] one / A_ / aa [d] two / A_bb / aaB_ [e] two / A_B_ / anything else (aabb, aaB_, A_bb)

E

What criteria must be met in order for a complementation test to be valid? [a] Mutant lines must have mutations in different genes [b] Mutant lines must show different phenotypes [c] Mutant lines must have mutations in the same gene [d] Mutant lines must have been isolated together [e] Mutant lines must be true breeding

E

You grow sweet peas for a living and know that purple pigment is dominant over white. You also know that two genes are involved in producing pea pigment and display complementary gene interaction (duplicative recessive epistasis). One of your clients wants purple sweet peas for his wedding. Which two sweet pea lines should you cross in order to maximize the number of purple sweet peas in the next generation? [a] AaBb x AAbb [b] AaBb x aaBB [c] AaBb x AaBb [d] AaBb x aaBB [e] AAbb x aaBB

E