Chapter 2 (test 1)

Describe three advantages of using pea plants as an experimental organism.

1.Pea plants are relatively small and hardy. They produce both pollen and eggs within the same flower. Because a keel covers the flower, self-fertilization is quite easy. 2. In addition, cross-fertilization is possible by the simple manipulation of removing the anthers in an immature flower and later placing pollen from another plant. 3. Finally, peas exist in several variants.

When an abnormal organism has three copies of a chromosome and therefore contains three copies of the genes on that chromosome (instead of the normal number of two copies), the alleles for each gene usually segregate so that a gamete will contain one or two copies of the gene. Let's suppose that an abnormal pea plant has three copies of the chromosome that carries the height gene. Its genotype is TTt. The plant is also heterozygous for the seed color gene, Yy, which is_found on a different chromosome. With regard to both genes, how many types of gametes can this plant make, and in what proportions? (Assume that it is equally likely that a gamete will contain one or two copies of the height gene.)

2 TY, tY, 2 Ty, ty, TTY, TTy, 2 TtY, 2 Tty It may be tricky to think about, but you get 2 TY and 2 Ty because either of the two T alleles could combine with Y or y. Also, you get 2 TtY and 2 Tty because either of the two T alleles could combine with t and then combine with Y or y.

A cross is made between AA Bb Cc Dd and Aa Bb cc dd individuals. Rather than making a very large Punnett square, which statistical operation could you use to solve this problem, and what would be the probability of an offspring that is AA bb Cc dd? A. Product rule, 1/32 B. Product rule, 1/4 C. Binomial expansion, 1/32 D. Binomial expansion, 1/4

A. Product rule, 1/32

In dogs, brown fur color (B) is dominant to white (b). A cross is made between two heterozygotes. If a litter contains six pups, what is the probability that half of them will be white? A. 0.066, or 6.6% B. 0.13, or 13% C. 0.25, or 25% D. 0.26, or 26%

B. 0.13, or 13%

In your own words, describe Mendel's law of segregation. Do not use the word "segregation" in your answer.

Diploid organisms contain two copies of each type of gene. When they make gametes, only one copy of each gene is found in a gamete. Two alleles cannot stay together within the same gamete.

In a cross between a heterozygous tall pea plant and a dwarf plant, predict the ratios of the offspring's genotypes and phenotypes.

Genotypes: 1:1 Tt and tt Phenotypes: 1:1 Tall and dwarf

Would it be possible to deduce the law of independent assortment from a single-factor cross? Explain your answer.

No, the law of independent assortment applies to transmission patterns of two or more genes. In a monohybrid experiment, you are monitoring only the transmission pattern of a single gene.

The term "cross" refers to an experiment in which?

The gametes come from different individuals

Experimental advantages of using pea plants include wof?

They came in several different varieties they were able to self fertilize, easy to cross AOA

With regard to genotypes, what is a true-breeding organism?

a homozygote that has two copies of the same allele

A pea plant has the genotype rrYy. How many different types of gametes can it make and in what proportions?

1 rY : 1 ry

A Tt plant is crossed to a tt plant. What is the expected outcome for the ratio offspring from this cross?

1 tall 1 dwarf

A cross made btw a plant that is RrYy to a plant that is rrYy. What is the predicted outcome of the sedd phenotypes?

3 round, yellow : 1 round, green : 3 wrinkled, yellow : 1 wrinkled green

A true-breeding tall plant was crossed to a dwarf plant. Tallness is a dominant trait. The F individuals were allowed to self-fertilize. What are the following probabilities for the F, generation? A. The first plant is dwarf. B. The first plant is dwarf or tall. C. The first three plants are tall. D. For any seven plants, three are tall and four are dwarf. E. The first plant is tall, and then among the next four, two are tall and the other two are dwarf.

A. 1/4 B. 1, or 100% C. (3/4)(3/4)(3/4) = 27/64 = 0.42, or 42% D. Use the binomial expansion equation where n = 7, p = 3/4, q = 1/4, x = 3 P = 0.058, or 5.8% E. The probability that the first plant is tall is 3/4. To calculate the probability that among the next four, any two will be tall, we use the binomial expansion equation, where n = 4, p = 3/4, q = 1/4, and x = 2. The probability P equals 0.21. To calculate the overall probability of these two events: (3/4)(0.21) = 0.16, or 16%

Identical twins are produced from the same sperm and egg (which splits after the first mitotic division), whereas fraternal twins are produced from separate sperm and separate egg cells. If two par ents with brown eyes (a dominant trait) produce one twin boy with blue eyes, what are the following probabilities? A. If the other twin is identical, he will have blue eyes. B. If the other twin is fraternal, he or she will have blue eyes. C. If the other twin is fraternal, he or she will transmit the blue eye allele to his or her offspring. D. The parents are both heterozygotes

A. 100% because they are genetically identical. B. Construct a Punnett square. We know the parents are heterozygotes because they produced a blue-eyed child. The fraternal twin is not genetically identical, but it has the same parents as its twin. The answer is 25%. C. The probability that an offspring inherits the allele is 50% and the probability that this offspring will pass it on to his/her offspring is also 50%. We use the product rule: (0.5)(0.5) = 0.25, or 25%. D. Barring a new mutation during gamete formation, the chance is 100% because they must be heterozygotes in order to produce a child with blue eyes.

A true-breeding pea plant with round and green seeds was crossed to a true-breeding plant with wrinkled and yellöw seeds. Round and yellow seeds are the dominant traits. The F plants were allowed to self-fertilize. What are the following probabilities for the F generation? A. An F, plant with wrinkled, yellow seeds. B. Three out of three F, plants with round, yellow seeds. C. Five F, plants in the following order: two have round, yellow seeds; one has round, green seeds; and two have wrinkled, green seeds. D. An F, plant will not have round, yellow seeds.

A. 3/16 B. (9/16)(9/16)(9/16) = 729/4096 = 0.18 C. (9/16)(9/16)(3/16)(1/16)(1/16) = 243/1,048,576 = 0.00023, or 0.023% D. Another way of looking at this is that the probability it will have round, yellow seeds is 9/16. Therefore, the probability that it will not is 1 - 9/16 = 7/16.

Ectrodactyly, also known as "lobster claw syndrome," is a recessive disorder in humans. If a phenotypically unaffected couple produces an affected offspring, what are the following probabilities? A. Both parents are heterozygotes. B. An offspring is a heterozygote. C. The next three offspring will be phenotypically unaffected. D. Any two out of the next three offspring will be phenotypically unaffected.

A. Barring a new mutation during gamete formation, the chance is 100% because they must be heterozygotes in order to produce a child with a recessive disorder. B. Construct a Punnett square. There is a 50% chance of heterozygous children C. Use the product rule. The chance of being phenotypically normal is 0.75 (i.e., 75%), so the answer is 0.75 × 0.75 × 0.75 = 0.422, which is 42.2%. D. Use the binomial expansion equation where n = 3, x = 2, p = 0.75, q = 0.25. The answer is 0.422, or 42.2%.

Huntington disease is a rare dominant trait that causes neurodegeneration later in life. A man in his thirties, who already has three children, discovers that his mother has Huntington disease though his father is unaffected. What are the following probabilities? A. That the man in his thirties will develop Huntington disease. B. That his first child will develop Huntington disease. C. That one out of three of his children will develop Huntington disease.

A. Construct a Punnett square. Because it is a rare disease, we would assume that the mother is a heterozygote and the father is normal. The chances are 50% that the man in his thirties will have the allele. B. Use the product rule: 0.5 (chance that the man has the allele) times 0.5 (chance that he will pass it to his offspring), which equals 0.25, or 25%. C. We use the binomial expansion equation. From part B, we calculated that the probability of an affected child is 0.25. Therefore the probability of an unaffected child is 0.75. For the binomial expansion equation, n = 3, x = 1, p = 0.25, q = 0.75. The answer is 0.422 or 42.2%.

For pea plants with the following genotypes, list the possible gametes that the plant can make: B. TtYYrr D. tt Yy rr

A. T Y R, T y R, T Y r, T y r B. T Y r, t Y r C. T Y R, T Y r, T y R, T y r, t Y R, t Y r, t y R, t y r D. t Y r, t y r

. In humans, the allele for brown eye color (B) is dominant to blue eye color (b). If two heterozygous parents produce children, what are the following probabilities? A. The first two children have blue eyes. B. A total of four children, two with blue eyes and the other two with brown eyes. C. Ihe first child has blue eyes, and the next two have brown eyes

A. Use the product rule: (1/4)(1/4)=1/16 B. Use the binomial expansion equation: n = 4, p = ¼, q = ¾, x = 2 P = 0.21, or 21% C. Use the product rule: (1/4)(3/4)(3/4) = 0.14, or 14%

Based on genes in pea plants that we have considered in this chapter, which statement(s) is/are not correct? A. The gene causing tall plants is an allele of the gene causing dwarf plants. B. The gene causing tall plants is an allele of the gene causing purple flowers. C. The alleles causing tall plants and purple flowers are dominant.

B. This statement is not correct because these are alleles of different genes.

Marfan syndrome is a rare inherited human disorder characterized by unusually long limbs and digits plus defects in the heart (especially the aorta) and the eyes, among other symptoms. Following is a pedigree for this disorder. Affected individuals are shown with filled (black) symbols. What type of inheritance pattern do you think is the most likely

Based on this pedigree, it is likely to be dominant inheritance because an affected child always has an affected parent. In fact, it is a dominant disorder.

Which of the following operations could be used for hypothesis testing?

Chi Square test

How can you determine whether an organism is heterozygous or homozygous for a dominant trait?

Conduct a cross in which the unknown individual is bred to an individual that carries only recessive alleles for the gene in question.

A true-breeding plant with round and green seeds was crossed to a true-breeding plant with wrinkled and yellow seeds. The Fl plants were allowed to self-fertilize. What is the probability of obtaining the following plants in the F, generation: two that have round, yellow seeds; one with round, green seeds; and two with wrinkled, green seeds?

Construct a Punnett square to determine the probability of these three phenotypes. The probabilities are 9/16 for round, yellow; 3/16 for round, green; and 1/16 for wrinkled, green. Use the multinomial expansion equation described in Solved problem S6, where n = 5, a = 2, b = 1, c = 2, p = 9/16, q = 3/16, r = 1/16. The answer is 0.007, or 0.7%, of the time.

. In cocker spaniels, solid coat color is dominant over spotted coat color. If two heterozygous dogs were crossed to each other, what would be the probability of the following combinations of offspring? A. A litter of five pups, four with solid fur and one with spotted fur. B. A first litter of six pups, four with solid fur and two with spotted fur, and then a second litter of five pups, all with solid fur. C. A first litter of five pups, the firstborn with solid fur, and then among the next four, three with solid fur and one with spotted fur, and then a second litter of seven pups in which the firstborn is spotted, the second born is spotted, and the remaining five are composed of four solid and one spotted animal. D. A litter of six pups, the firstborn with solid fur, the second born spotted, and among the remaining four pups, two with spotted fur and two with solid fur.

First construct a Punnett square. The chances are 75% of producing a solid pup and 25% of producing a spotted pup. A. Use the binomial expansion equation where n = 5, x = 4, p = 0.75, q = 0.25. The answer is 0.396 = 39.6% of the time. B. You can use the binomial expansion equation for each litter. For the first litter, n = 6, x = 4, p = 0.75, q = 0.25; for the second litter, n = 5, x = 5, p = 0.75, q = 0.25. Because the litters are in a specified order, we use the product rule and multiply the probability of the first litter times the probability of the second litter. The answer is 0.070, or 7.0%. C. To calculate the probability of the first litter, we use the product rule and multiply the probability of the first pup (0.75) times the probability of the remaining four. We use the binomial expansion equation to calculate the probability of the remaining four, where n = 4, x = 3, p = 0.75, q = 0.25. The probability of the first litter is 0.316. To calculate the probability of the second litter, we use the product rule and multiply the probability of the first pup (0.25) times the probability of the second pup (0.25) times the probability of the remaining five. To calculate the probability of the remaining five, we use the binomial expansion equation, where n = 5, x = 4, p = 0.75, q = 0.25. The probability of the second litter is 0.025. To get the probability of these two litters occurring in this order, we use the product rule and multiply the probability of the first litter (0.316) times the probability of the second litter (0.025). The answer is 0.008, or 0.8%.

A cross was made between a white male dog and two different black females. The first female gave birth to eight black pups, and the second female gave birth to four white and three black pups. What are the likely genotypes of the male parent and the two female parents? Explain whether you are uncertain about any of the genotypes.

If B is the black allele, and b is the white allele, the male is bb, the first female is probably BB, and the second female is Bb. We are uncertain of the genotype of the first female. She could be Bb, although it is unlikely because she didn't produce any white pups out of a litter of eight.

For Mendel's data shown in Figure 2.8, conduct a chi square analysis to determine if the data agree with Mendel's law of independent

If we construct a Punnett square according to Mendel's laws, we expect a 9:3:3:1 ratio. Because a total of 556 offspring were observed, the expected number of offspring are 556 × 9/16 = 313 round, yellow 556 × 3/16 = 104 wrinkled, yellow 556 × 3/16 = 104 round, green 556 × 1/16 = 35 wrinkled, green If we plug the observed and expected values into the chi square equation, we get a value of 0.51. With four categories, our degrees of freedom equal n - 1, or 3. If we look up the value of 0.51 in the chi square table (see Table 2.1), we see that it falls between the P values of 0.80 and 0.95. This means that the probability is a 80% to 95% that any deviation between observed results and expected results was caused by random sampling error. Therefore, we accept the hypothesis. In other words, the results are consistent with the law of independent assortment.

What is the difference between cross-fertilization and self-fertilization?

In the case of plants, cross-fertilization occurs when the pollen and eggs come from different plants while in self-fertilization they come from the same plant.

A true-breeding tall pea plant was crossed to a true-breeding dwarf plant. What is the probability that an Fl individual will be true-breeding? What is the probability that an Fl individual will be a true-breeding tall plant?

It is impossible for the F1 individuals to be true-breeding because they are all heterozygotes.

metabolize the amino acids leucine, isoleucine, and valine. One of the symptoms is that the urine smells like maple syrup. An unaffected couple produced six children in the following order: unaffected daughter, affected daughter, unaffected son, unaffected son, affected son, and unaffected son. The youngest unaffected son marries an unaffected woman and has three children in the following order: affected daughter, unaffected daughter, and unaffected son. Draw a pedigree that describes this family. What type of inheritance (dominant or recessive) would you propose to explain maple syrup urine disease?

It is recessive inheritance. The pedigree is shown here. Affected individuals are shown with filled symbols. The mode of inheritance appears to be recessive. Unaffected parents (who must be heterozygous) produce affected children.

An individual has the genotype Aa Bb Cc and makes an abnormal gamete with the genotype AaBc. Does this gamete violate the law of independent assortment or the law of segregation (or both)? Explain your answer.

It violates the law of segregation because two copies of one gene are in the gamete. The two alleles for the A gene did not segregate from each other

A pea plant is Tt. Which of the following statements is correct?

Its genotype is Tt, and its phenotype is tall

Why did Mendel's work refute the idea of blending inheritance?

Mendel's work showed that genetic determinants are inherited in a dominant/recessive manner. This was readily apparent in many of his crosses. For example, when he crossed two true-breeding plants for a trait such as height (i.e., tall versus dwarf), all the F1 plants were tall. This was not consistent with blending. Perhaps more striking was the result obtained in the F2 generation: 3/4 of the offspring were tall and 1/4 were short. In other words, the F2 generation displayed phenotypes that were like the parental generation. There did not appear to be a blending to create an intermediate phenotype. Instead, the genetic determinants did not seem to change from one generation to the next. C2. Answer: In the case of plants, cross-fertilization occurs when the pollen and eggs come from different plants while in self-fertilization they come from the same plant

Describe the significance of nonparentals with regard to the law of independent assortment. In other words, explain how the appear ance of nonparentals refutes a linkage hypothesis

Offspring with a recombinant (nonparental) phenotype are consistent with the idea of independent assortment. If two different traits were always transmitted together as unit, it would not be possible to get recombinant phenotypic combinations. For example, if a true-breeding parent had two dominant traits and was crossed to a true-breeding parent having the two recessive traits, the F2 generation could not have offspring with one recessive and one dominant phenotype. However, because independent assortment can occur, it is possible for F2 offspring to have one dominant and one recessive trait.

To avoid self fertilization in his pea plants, mendel had to?

Remove the anthers from immature flowers

Honeybees are unusual in that male bees (drones) have only one copy of each gene, but female bees have two copies of their genes. That is because drones develop from eggs that have not been fertilized by sperm cells. In bees, the trait of long wings is dominant over short wings, and the trait of black eyes is dominant over white eyes. If a drone with short wings and black eyes was mated to a queen bee that is heterozygous for both genes, what are the predicted genotypes and phenotypes of male and female offspring? What are the phenotypic ratios if we assume an equal number of male and female offspring?

The drone is sB and the queen is SsBb. According to the laws of segregation and independent assortment, the male can make only sB gametes, while the queen can make SB, Sb, sB, and sb, in equal proportions. Therefore, male offspring will be SB, Sb, sB, and sb, and female offspring will be SsBB, SsBb, ssBB, and ssBb. The phenotypic ratios, assuming an equal number of males and females, will be: Males Females 1 normal wings/black eyes 2 normal wings, black eyes 1 normal wings/white eyes 2 short wings, black eyes 1 short wings/black eyes 1 short wings/white eyes

Explain the technical differences between a cross-fertilization experiment versus a self-fertilization experiment

The experimental difference depends on where the pollen comes from. In self-fertilization, the pollen and eggs come from the same plant. In cross-fertilization, they come from different plants.

Describe the difference between genotype and phenotype. Give three examples. Is it possible for two individuals to have the same phenotype but different genotypes?

The genotype is the type of genes that an individual inherits while the phenotype is the individual's observable traits. Tall pea plants, red hair in humans, and vestigial wings in fruit flies are phenotypes. Homozygous, TT, in pea plants; a heterozygous carrier of the cystic fibrosis allele; and homozygotes for the cystic fibrosis allele are descriptions of genotypes. It is possible to have different genotypes and the same phenotype. For example, a pea plant that is TT or Tt would both have a tall phenotype.

. A pea plant that is dwarf with green, wrinkled seeds was crossed to a true-breeding plant that is tall with yellow, round seeds. The F generation was allowed to self-fertilize. What types of gametes, and in what proportions, would the Fl generation make? What would be the ratios of genotypes and phenotypes of the F, generation?

The genotype of the F1 plants is Tt Yy Rr. According to the laws of segregation and independent assortment, the alleles of each gene will segregate from each other, and the alleles of different genes will randomly assort into gametes. A Tt Yy Rr individual could make eight types of gametes: TYR, TyR, Tyr, TYr, tYR, tyR, tYr, and tyr, in equal proportions (i.e., 1/8 of each type of gamete). To determine genotypes and phenotypes, you could make a large Punnett square that would contain 64 boxes. You would need to line up the eight possible gametes across the top and along the side, and then fill in the 64 boxes. Alternatively, you could use one of the two approaches described in solved problem S3. The genotypes and phenotypes would be: 1 TT YY RR 2 TT Yy RR 2 TT YY Rr 2 Tt YY RR (4 TT Yy Rr 4 Tt Yy RR 4 Tt YY Rr 8 Tt Yy Rr = 27 tall, yellow, round 1 TT yy RR 2 Tt yy RR 2 TT yy Rr 4 Tt yy Rr = 9 tall, green, round 1 TT YY rr 2 TT Yy rr 2 Tt YY rr 4 Tt Yy rr = 9 tall, yellow, wrinkled 1 tt YY RR 2 tt Yy RR 2 tt YY Rr 4 tt Yy Rr = 9 dwarf, yellow, round 1 TT yy rr 2 Tt yy rr = 3 tall, green, wrinkled 1 tt yy RR 2 tt yy Rr = 3 dwarf, green, round 1 tt YY rr 2 tt Yy rr = 3 dwarf, yellow, wrinkled 1 tt yy rr = 1 dwarf, green, wrinkled

A pea plant that is heterozygous with regard to seed color (yel low is dominant to green) is allowed to self-fertilize. What are the predicted outcomes of genotypes and phenotypes of the offspring?

The genotypes are 1 YY : 2 Yy : 1 yy. The phenotypes are 3 yellow : 1 green.

Albinism, a condition characterized by a partial or total lack of skin pigrnent, is a recessive human trait. If a phenotypically unaffected couple produce an albino child, what is the probability that their next child will be albino?

The parents must be heterozygotes, so the probability is 1/4.

Do you know the genotype of an individual with a recessive trait and/or a dominant trait? Explain your answer

The recessive phenotype must be a homozygote. The dominant phenotype could be either homozygous or heterozygous

In a population of wild squirrels, most of them have gray fur, but an occasional squirrel is completely white. If we let P and p represent dominant and recessive alleles, respectively, of a gene that encodes an enzyme necessary for pigment formation, which of the following statements do you think is most likely to be correct?

The white squirrels are pp, and the p allele is a loss of function allele

Wooly hair is a rare dominant trait found in people of Scandinavian descent in which the hair resembles the wool of a sheep. A male with wooly hair, who has a mother with straight hair, moves to an island that is inhabited by people who are not of Scandinavian descent. Assuming that no other Scandinavians immigrate to the island, what is the probability that a great-grandchild of this male will have wooly hair? (Hint: You may want to draw a pedigree to help you figure this out.) If this wooly-haired male has eight great-grandchildren, what is the probability that one out of eight will have wooly hair?

The wooly haired male is a heterozygote, because he has the trait and his mother did not. (He must have inherited the normal allele from his mother.) Therefore, he has a 50% chance of passing the wooly allele to his offspring; his offspring have a 50% of passing the allele to their offspring; and these grandchildren have a 50% chance of passing the allele to their offspring (the wooly haired man's greatgrandchildren). Because this is an ordered sequence of independent events, we use the product rule: 0.5 × 0.5 × 0.5 = 0.125, or 12.5%. Because no other Scandinavians are on the island, the chance is 87.5% for the offspring being normal (because they could not inherit the wooly hair allele from anyone else). We use the binomial expansion equation to determine the likelihood that one out of eight great-grandchildren will have wooly hair, where n = 8, x = 1, p = 0.125, q = 0.875. The answer is 0.393, or 39.3%, of the time.

What are the expected phenotypic ratios from the following cross: Tt Rr yy Aa x Tt rr YY Aa, where T = tall, t = dwarf, R = round, r = wrinkled, Y = yellow, y = green, A = axial, a = terminal; T, R, Y, and A are dominant alleles. Note: See solved problem S3 for help in answering this problem.

This problem is a bit unwieldy, but we can solve it using the multiplication rule. For height, the ratio is 3 tall : 1 dwarf. For seed texture, the ratio is 1 round : 1 wrinkled. For seed color, they are all yellow. For flower location, the ratio is 3 axial : 1 terminal. Thus, the product is (3 tall + 1 dwarf)(1 round + 1 wrinkled)(1 yellow)(3 axial + 1 terminal) Multiplying this out, the answer is 9 tall, round, yellow, axial 9 tall, wrinkled, yellow, axial 3 tall, round, yellow, terminal 3 tall, wrinkled, yellow, terminal 3 dwarf, round, yellow, axial 3 dwarf, wrinkled, yellow, axial 1 dwarf, round, yellow, terminal 1 dwarf, wrinkled, yellow, terminal

WOF not be observed in a pedigree if a genetic disorder was inherited in a recessive manner?

Two affected parents have unaffected offspring

How long did it take Mendel to complete the experiment in Figure 2.5?

Two generations would take two growing seasons. About 1 and 1/2 years

A woman with achondroplasia (a dominant form of dwarfism) and a phenotypically unaffected man have seven children, all of whom have achondroplasia. What is the probability of producing such a family if this woman is a heterozygote? What is the probability that the woman is a heterozygote if her eighth child does not have this disorder?

Use the product rule. If the woman is heterozygous, there is a 50% chance of having an affected offspring: (0.5)7 = 0.0078, or 0.78%, of the time. This is a pretty small probability. If the woman has an eighth child who is unaffected, however, she has to be a heterozygote, because it is a dominant trait. She would have to pass a normal allele to an unaffected offspring. The answer is 100%.

A cross is made between a pea plant that has constricted pods (a recessive trait; smooth is dominant) and is heterozygous for seed color (yellow is dominant to green) and a plant that is heterozygous for both pod texture and seed color. Construct a Punnett square that depicts this cross. What are the predicted outcomes of genotypes and phenotypes of the offspring?

c is the recessive allele for constricted pods; Y is the dominant allele for yellow color. The cross is ccYy × CcYy. Follow the directions for setting up a Punnett square, as described in chapter 2. The genotypic ratio is 2 CcYY : 4 CcYy : 2 Ccyy : 2 ccYY : 4 ccYy : 2 ccyy. This 2:4:2:2:4:2 ratio could be reduced to a 1:2:1:1:2:1 ratio. The phenotypic ratio is 6 smooth pods, yellow seeds : 2 smooth pods, green seeds : 6 constricted pods, yellow seeds : 2 constricted pods, green seeds. This 6:2:6:2 ratio could be reduced to a 3:1:3:1 ratio.