Bio 110 - Exam 3 questions

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You cross two true-breeding parents (one with yellow seeds and one with green seeds), grow and self-pollinate the F1 generation, and then count the number of offspring with yellow and green seeds in the F2 generation. If you counted 2,000 offspring and there was a 3:1 ratio of yellow:green seeds, how many offspring would have yellow seeds?

500
667
1999
1,500
1,000
D. When the purple progeny are selfed, both white and purple progeny result.
When Mendel crossed a true-breeding purple-flowered pea plant with a true-breeding white-flowered pea plant, he observed that all of the F1 offspring had purple flowers. When these F1 plants were selfed, he found 3/4 were purple and 1/4 were white. Which of the following observations supports Mendel's conclusion that each progeny can carry two alternative copies of a "factor"?

A. When the white progeny are selfed, only white progeny result.
B. When the purple progeny are crossed with the white progeny, only purple progeny result.
C. When the purple progeny are selfed, only purple progeny result.
D. When the purple progeny are selfed, both white and purple progeny result.
the F1 plants are heterozygous, and tall is dominant to short
You are given two plants. One plant is tall and the other is short. You cross these parental plants and find all F1 progeny are tall. You cross the F1 progeny with one another and find the resulting F2 generation is composed of 770 tall plants and 300 short plants. You should conclude that ___________.

both parental plants are homozygous, and short is dominant to tall
both parental plants are heterozygous, and short is dominant to tall
you can't tell from the information given
the F1 plants are heterozygous, and tall is dominant to short
the F1 plants are homozygous, and tall is dominant to short
3.47:1
In a population of flowering plants, 770 plants have purple flowers and 222 have white flowers. What is the ratio of purple:white flowers in this population?

3.47:1
992:1
548:1
0.29:1
0
Homologous chromosomes separate during Meiosis I.
Which meiotic process results in the Law of Segregation?

Homologous chromosomes separate during Meiosis I.
Homologous chromosomes undergoing crossing over during Meiosis I.
Non-homologous chromosomes segregating into daughter cells independently of each other during Meiosis I.
be homozygous for that trait
A plant that is true-breeding for a particular trait would __________.

be homozygous for that trait
show different traits in the F1 generation when crossed with itself
be heterozygous for that trait
carry two different alleles for that trait
375
You cross two true-breeding parents (one with purple flowers and one with white flowers). All of the F1 progeny (offspring) have purple flowers. You grow and self-pollinate this F1 generation and collect 500 F2 progeny. How many purple flowers should be produced?

167
375
0
500
125
16
The fruit fly Drosophila melanogaster has 3 pairs of autosomal chromosomes and a pair of sex chromosomes. How many different chromosome combinations are produced by independent assortment in D. melanogaster?

3
4
8
16
0.50
A true-breeding pea plant with wrinkled seeds (rr) is crossed with a pea plant that is heterozygous for round seeds (Rr). What portion (expressed as a decimal) of the offspring should have round seeds?

0.50
0
1.00
0.25
0.75
pp
To determine if a plant with purple flowers is homozygous (PP) or heterozygous (Pp) for that trait, one would do a test cross with plants of the _____ genotype.

pp
PP
Pp
p
P
1/8Two F1-hybrid plants with yellow, round seeds (YyRr) are crossed. What is the probability of getting an offspring with the genotype YYRr? 4/8 1/8 1/16 2/8 3/160.25If two F1-hybrid pea plants with purple flowers (Pp) are crossed, what portion (expressed as a decimal) of their offspring would have white flowers (purple flowers are the dominant trait)? 0 0.50 1.00 0.75 0.259/16In a cross of two F1-hybrid plants with purple flowers and round seeds (PpRr), what fraction of the offspring should have purple flowers and round seeds (purple flowers and round seeds are the dominant traits)? 9/16 1/12 1/16 3/16 12/16900You cross two true-breeding parents, one with purple flowers and one with white flowers, grow and self-pollinate the F1 generation, and then count the number of offspring with purple and white flowers in the F2 generation. If you counted 1,200 offspring and there was a 3:1 ratio of purple:white flowers, how many offspring would have purple flowers? 1100 75 300 25 90063Two F1-hybrid plants with purple flowers and round seeds (PpRr) are crossed. If 1,000 randomly chosen offspring from this cross are examined, how many should have white flowers and wrinkled seeds (purple flowers and round seeds are the dominant traits)? You may need to round your answer to the nearest whole number. 83 563 63 750 1881/4A plant with purple flowers and round seeds (PpRr) is crossed with a plant with white flowers and wrinkled seeds (pprr). What fraction of the offspring should have white flowers and wrinkled seeds? (Purple flowers and round seeds are dominant.) 1/5 3/4 1/4 1/3 1/2The parent with green pods had the genotype Gg.You have a pea plant that exhibits the dominant trait of green pods. You cross this plant to a true-breeding plant with the recessive trait of yellow pods. Half of the offspring from this cross have green pods and the other half have yellow pods. Which of the following statements is true? The parent with green pods was true-breeding for green pods. The parent with yellow pods was heterozygous for this character. The parent with green pods had the genotype GG. The parent with green pods had the genotype Gg. The parent with yellow pods had the genotype Gg.the alleles for different characters are randomly allocated to gametes.Mendel's Law of Independent Assortment states that _____. the two alleles for a single character are separated during gamete production his Law of Segregation is not always true offspring inherit traits from their parents homologous chromosomes pair up during Prophase I of meiosis the alleles for different characters are randomly allocated to gametes.Both b and cWhich of the following genotype(s) would result in an individual with the type A blood phenotype? A. ii B. IA IA C. IA I D. Both b and c E. All of the aboveP ... F1 ... F2In his breeding experiments, Mendel first crossed true-breeding plants to produce a second generation, which were then allowed to self-pollinate to generate the offspring. How do we name these three generations? Hints F1 ... F2 ... F3 P1 ... P2 ... P3 P1 ... P2 ... F F ... P1 ... P2 P ... F1 ... F2It is homozygous at two loci.Which of the following is true about a plant with the genotype AABbcc? Hints It is heterozygous at two loci. It is homozygous at two loci. It is triploid. It will not express the recessive c allele. It has recessive alleles at three loci.All of the gametes from a homozygote carry the same version of the gene while those of a heterozygote will differ.What is the difference between heterozygous and homozygous individuals? All of the gametes from a homozygote carry the same version of the gene while those of a heterozygote will differ. The homozygote will express the dominant trait and the heterozygote will express the recessive trait. Homozygotes have one chromosome while heterozygotes have two similar chromosomes. Heterozygotes carry two copies of a gene while homozygotes only carry one.gametes, progenyWhen constructing a Punnett square, the symbols on the outside of the boxes represent _______, while those inside the boxes represent _______. progeny, gametes gametes, parents parents, gametes gametes, progenyTrueTrue or false? The same phenotype can be produced by more than one genotype. True FalseFalseTrue or false? In diploid organisms, a dominant phenotype will only be expressed if the individual is homozygous dominant for that trait. Hints True FalseNoneIf an organism with the genotype AaBb produces gametes, what proportion of the gametes would be Bb? Hints None 1/4 3/4 1/225Two mice are heterozygous for albinism (Aa) . The dominant allele (A) codes for normal pigmentation, and the recessive allele (a) codes for no pigmentation. What percentage of their offspring would have an albino phenotype? Hints 100 50 25 75four ... oneConsider pea plants with the genotypes GgTt and ggtt . These plants can each produce how many type(s) of gametes? Hints one ... one one ... two four ... one four ... two two... one1/16Two organisms with genotype AaBbCcDdEE mate. These loci are all independent. What fraction of the offspring will have the same genotype as the parents? Hints 4/3 1/16 9/64 1/4 3/41/64In the cross AaBbCc × AaBbCc, what is the probability of producing the genotype AABBCC? 1/64 1/4 1/8 1/16 1/32the probability that either one of two independent events will occurWhich of the following is the best statement of the use of the addition rule of probability? the probability that either one of two independent events will occur the probability that two or more independent events will both occur in the offspring of one set of parents the probability that two or more independent events will both occur the probability of producing two or more heterozygous offspring the likelihood that a trait is due to two or more meiotic eventsan alternative version of a geneWhat is an allele? the dominant form of a gene a variety of pea plant used by Mendel the recessive form of a gene an alternative version of a gene a type of chromosomeA dihybrid cross involves organisms that are heterozygous for two characters that are being studied, and a monohybrid cross involves organisms that are heterozygous for only one character being studied.What do we mean when we use the terms monohybrid cross and dihybrid cross? A monohybrid cross is performed for one generation, whereas a dihybrid cross is performed for two generations. A dihybrid cross involves organisms that are heterozygous for two characters that are being studied, and a monohybrid cross involves organisms that are heterozygous for only one character being studied. A monohybrid cross involves a single parent, whereas a dihybrid cross involves two parents. A monohybrid cross results in a 9:3:3:1 ratio whereas a dihybrid cross gives a 3:1 ratio.epistasisA gene that affects the expression of a second gene is an example of __________________. pleiotropy codominance epistasis polygenic inheritance incomplete dominancepolygenic inheritance__________ is an example of a character that is controlled by more than one gene (a quantitative character) and demonstrates a continuum of phenotypes. Pleiotropy Polygenic inheritance Epistasis Codominance Incomplete dominance1:2:1, red:orange:yellow earsThe gene controlling ear color in an organism known as a gizmo has two alleles that exhibit incomplete dominance: CR, which codes for red ears; and CY, which codes for yellow ears. Individuals that are homozygous for the CR allele have red ears, whereas the heterozygous genotype produces orange ears. Those that are homozygous for the CY allele have yellow ears. If two individuals with orange ears are crossed, what ratio of phenotypes should be expected in the offspring? 9:3:1, red:yellow:orange ears 3:1, red:yellow ears 1:2:1, red:orange:yellow ears 1:2:1, red:yellow:orange ears 1:3, red:yellow earsbrownMice with the genotypes BBCc and bbcc are crossed. The alleles for brown coat (b) and no pigment (c) are recessive, and black coat (B) is dominant. What color of mice would NOT be produced by this mating? brown white blackincomplete dominanceAn individual that is heterozygous for an allele and that exhibits a phenotype intermediate between that of the two homozygous alternatives is an example of _____. polygenic inheritance pleiotropy codominance incomplete dominance epistasisA, B, and AB onlyWhat possible blood types could offspring have if their two parents have the blood types A and AB, respectively? AB only A, B, AB, and O A and O only A and B only A, B, and AB only1:1, brown:grayIn water snakes, body color is controlled by two alleles that display incomplete dominance. Snakes with the BB genotype are brown, those with the Bb genotype are gray, and individuals with the bb genotype are green. If a brown snake and a gray snake are crossed, what ratio of phenotypes should be expected in the offspring? 1:1, gray:green 1:2:1, brown:gray:green 1:1, brown:gray 1:1:1, brown:gray:green 1:1, brown:greenCodominance.You are examining a trait which shows three distinct phenotypes, with the phenotype of heterozygotes expressing both alleles seen in the homozygotes. This is an example of: Codominance. Complete dominance. Incomplete dominance. Pleiotrophy.did not skip generationsPedigrees reveal that a trait shows a dominant pattern of inheritance if that trait generally _______. skipped generations was lethal did not skip generations was only inherited by malesCystic fibrosis________________________ is an example of a recessive inherited trait. Huntington's disease Cystic fibrosis Brachydactyly. The AB blood type50%If one parent who has Huntington's disease is heterozygous for the trait and the other parent is not affected, what are the chances of their offspring inheriting the trait? 50% 100% 0 25%25%If both parents are carriers of an recessive inherited trait, what are the chances that their child will express that trait? 50% 25% 0 100%phenylketonuria (PKU)The effects of ______________________ can be almost completely overcome by regulating the diet of the individual from infancy. cystic fibrosis sickle-cell disease phenylketonuria (PKU) Huntington's diseasewwThe following question refer to the pedigree chart in the figure below for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle. What is the genotype of individual II-5? ww or Ww ww Ww WW50%The following question refer to the pedigree chart in the figure below for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle. What is the likelihood that the progeny of IV-3 and IV-4 will have the trait? 0% 100% 75% 50%1The following question refer to the pedigree chart in the figure below for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle. What is the probability that individual III-1 is Ww? 3/4 1/4 1 2/4as an autosomal dominantThe figure below shows the pedigree for a family. Dark-shaded symbols represent individuals with one of the two major types of colon cancer. Numbers under the symbols are the individual's age at the time of diagnosis. Males are represented by squares, females by circles. From this pedigree, this trait seems to be inherited _____. as an autosomal dominant as a result of epistasis as an autosomal recessive from mothersThe law of independent assortment requires describing two or more genes relative to one another.Which of the following statements about independent assortment and segregation is correct? The law of independent assortment is accounted for by observations of prophase I. The law of segregation is accounted for by anaphase of mitosis. The law of segregation requires describing two or more genes relative to one another. The law of independent assortment requires describing two or more genes relative to one another.2 and 3In a particular plant, leaf color is controlled by gene locus D. Plants with at least one allele D have dark green leaves, and plants with the homozygous recessive dd genotype have light green leaves. A true-breeding, dark-leaved plant is crossed with a light-leaved one, and the F1 offspring is allowed to self-pollinate. The predicted outcome of the F2 is diagrammed in the Punnett square shown in the figure, where 1, 2, 3, and 4 represent the genotypes corresponding to each box within the square. Which of the boxes marked 1-4 correspond to plants with a heterozygous genotype? 2, 3, and 4 1 2 and 3 1, 2, and 31 and 4 onlyIn a particular plant, leaf color is controlled by gene locus D. Plants with at least one allele D have dark green leaves, and plants with the homozygous recessive dd genotype have light green leaves. A true-breeding, dark-leaved plant is crossed with a light-leaved one, and the F1 offspring is allowed to self-pollinate. The predicted outcome of the F2 is diagrammed in the Punnett square shown in the figure, where 1, 2, 3, and 4 represent the genotypes corresponding to each box within the square. Which of the boxes marked 1-4 correspond to plants that will be true-breeding? 1 only 1 and 4 only 1, 2, 3, and 4 2 and 3 only4Skin color in a certain species of fish is inherited via a single gene with four different alleles. How many different types of gametes would be possible in this system? 8 16 2 4a combination of polygenic inheritance and environmental factorsHeight in humans generally shows a normal (bell-shaped) distribution. What type of inheritance most likely determines height? a combination of polygenic inheritance and environmental factors a combination of complete dominance and environmental factors a combination of epistasis and environmental factors incomplete dominance a combination of multiple alleles and codominanceRed shows incomplete dominance over white.You cross a true-breeding red-flowered snapdragon with a true-breeding white-flowered one. All of the F1 are pink. What does this say about the parental traits? Hints Pink is dominant, and red and white are recessive. Red is completely dominant. Red shows incomplete dominance over white. Both red and white are pleiotropic.roan × roanIn cattle, roan coat color (mixed red and white hairs) occurs in the heterozygous (Rr) offspring of red (RR) and white (rr) homozygotes. Which of the following crosses would produce offspring in the ratio of 1 red:2 roan:1 white? red × roan roan × roan red × white white × roanincomplete dominanceRadish flowers may be red, purple, or white. A cross between a red-flowered plant and a white-flowered plant yields all-purple offspring. The flower color trait in radishes is an example of which of the following? incomplete dominance codominance a multiple allelic system sex linkageB positiveA woman who has blood type A positive has a daughter who is type O positive and a son who is type B negative. Rh positive is a trait that shows simple dominance over Rh negative. Which of the following is a possible phenotype for the father? B positive A negative AB negative O negativeIt is very likely that at least one of Woody Guthrie's parents also have had the allele for Huntington's disease.Folk singer Woody Guthrie died of Huntington's disease, an autosomal dominant disorder. Which statement below must be true? Hints His daughters will die of Huntington's disease but not his sons. All of his children will develop Huntington's disease. It is very likely that at least one of Woody Guthrie's parents also have had the allele for Huntington's disease. His sons will develop Huntington's disease but not his daughters. There is not enough information to answer the question.Regulate the diet of the affected persons to severely limit the uptake of the amino acid.Phenylketonuria (PKU) is a recessive human disorder in which an individual cannot appropriately metabolize the amino acid phenylalanine. This amino acid is not naturally produced by humans. Therefore, the most efficient and effective treatment is which of the following? Regulate the diet of the affected persons to severely limit the uptake of the amino acid. Feed the patients the missing enzymes in a regular cycle, such as twice per week. Feed the patients an excess of the missing product. Feed them the substrate that can be metabolized into this amino acid. Transfuse the patients with blood from unaffected donors.50%The following question refers to the pedigree chart in the figure for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle. What is the likelihood that the progeny of IV-3 and IV-4 will have the trait? 25% 50% 0% 75% 100%The disorder may be due to mutation in a single protein-coding gene.Hutchinson-Gilford progeria is an exceedingly rare human genetic disorder in which there is very early senility and death, usually from coronary artery disease, at an average age of approximately 13. Patients, who look very old even as children, do not live to reproduce. Which of the following represents the most likely assumption? Each patient will have had at least one affected family member in a previous generation. Successive generations of a family will continue to have more and more cases over time. All cases must occur in relatives; therefore, there must be only one mutant allele. The disorder may be due to mutation in a single protein-coding gene. The disease is autosomal dominant.1The following question refers to the pedigree chart in the figure for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle. What is the probability that individual III-1 is Ww? 1 1/4 3/4 2/3 2/4as an autosomal dominantUse the following pedigree (the figure) for a family in which dark-shaded symbols represent individuals with one of the two major types of colon cancer. Numbers under the symbols are the individual's age at the time of diagnosis. From this pedigree, how does this trait seem to be inherited? as an autosomal recessive from mothers as an incomplete dominant as a result of epistasis as an autosomal dominant0If a male affected by a recessive sex-linked trait reproduces with a female who is homozygous wild type for that trait, what is the chance that any of their offspring will have the trait? 50% 75% 100% 0 25%by mitosis (simple cell division)In animals that exhibit the haplo/diploid mode of sex determination, the males are fertile and produce sperm. In these animals sperm are produced _______________. via replication of the haploid state to the diploid state, followed by meiosis by mitosis (simple cell division) by meiosisIn the first case, all females will have red eyes and all males will have white eyes. In the second case, there will be equal numbers of white and red eyes.Pull out a piece of paper and pencil because this question will take some thought. First, work out the predicted offspring of a mating between a white-eyed female and a red-eyed male Drosophila. Next, work out the predicted offspring of a heterozygous red-eyed female and a white-eyed male. What is the result of both crosses? In the first case, all females will have white eyes and all males red eyes. In the second case, this is reversed. In the first case, all males will have white eyes. In the second case, this will be reversed. In the first case, all females will have red eyes and all males will have white eyes. In the second case, there will be equal numbers of white and red eyes. In both cases, the result is the same. In the first case, there will be equal numbers of white and red eyes. In the second case, only females will have red eyes.homozyogous for one gene and heterozygous for the otherIn Drosophila, one gene affects the formation of anterior body parts and another gene affects the formation of posterior body parts. These "polarity genes" are designated bcd and nos. A fly designated bcd+/bcd nos+/nos+ would be _________________. heterozygous for both of these dominant mutant alleles homozygous for both genes heterozygous for both genes homozyogous for one gene and heterozygous for the other heterozygous for both of these recessive mutant allelesin this produce market, red eyes and long bristles are wild-type traitsYou are working for a fly geneticist who sends you to a produce market to find new mutations. You rummage through the dumpsters and capture 100 flies. You note that 100% of the flies have red eyes and 90% have long body bristles (10% have short body bristles). Based on these results, you can conclude that __________________. red eyes are dominant and short body bristles are recessive short body bristles and long body bristles are located on different chromosomes red eyes and short body bristles are located on different chromosomes in this produce market, red eyes and long bristles are wild-type traits red eyes and long body bristles are inherited in a dominant mannermales are hemizygous for the X chromosomeMales are more often affected by sex-linked traits than females because _____. female hormones such as estrogen often compensate for the effects of mutations on the X chromosome male hormones such as testosterone often alter the effects of mutations on the X chromosome males are hemizygous for the X chromosome X chromosomes in males generally have more mutations than X chromosomes in femalesa and bThe locations for three genes have been added to this cartoon. For clarity, the locations have only been noted on one chromatid. The lowest frequency of crossovers would occur between genes ____ and ____. b and c a and b a and ca gene region present on the Y chromosome that triggers male developmentSRY is best described as _____. a gene present on the X chromosome that triggers female development a gene region present on the Y chromosome that triggers male development an autosomal gene that is required for the expression of genes on the X chromosome an autosomal gene that is required for the expression of genes on the Y chromosome2 and 3In this example, crossing over could occur between ______. 2 and 3 1 and 2 3 and 4tortoiseshell females; black malesIn cats, black fur color is caused by an X-linked allele; the other allele at this locus causes orange color. The heterozygote is tortoiseshell. What kinds of offspring would you expect from the cross of a black female and an orange male? tortoiseshell females; black males orange females; black males black females; orange males tortoiseshell females; tortoiseshell males1 and 2, 3 and 4; 1 and 3, 1 and 4, 2 and 3, 2 and 4Here is a cartoon of homologous chromosomes. Sister chromatids are represented by _____ and nonsister chromatids are represented by ________. 2 and 3 1 and 2; 3 and 4 1 and 3; 2 and 4 3 and 4 1 and 2, 3 and 4; 1 and 3, 1 and 4, 2 and 3, 2 and 4XNXn and XNYRed-green color blindness is a sex-linked recessive trait in humans. Two people with normal color vision have a color-blind son. What are the genotypes of the parents? XNXn and XNY XNXN and XnY XNXN and XNY XnXn and XnYThe genes are on the same chromosome and do not assort independently.Of the following statements, which is/are usually true if the genes for two different characters are linked? The alleles for each character won't segregate. The genes are on the same chromosome and do not assort independently. The alleles for the linked genes assort independently from one another. A proportional number of parental phenotypes and recombinant phenotypes are represented among offspring. All of the abovelinkedWhen genes on a particular chromosome do not assort independently in genetic crosses, the genes are _______________. recombinants wild type linked homozygous alleles not linked1:1:1:1What ratio of F1 generation phenotypes would be expected from a test cross of flies differing in two characters (females are heterozygous for both genes, and males are homozygous recessive for both genes) if the two genes assort independently of each other? 8:2:2:1 1:1:1:1 9:3:3:1 3:3:1:1 5:5:1:1They will appear to segregate independently, as if they are located on different chromosomes.You are studying two genes that are located on the same chromosome. They are located 55 map units apart. How will the alleles of these genes behave in a dihybrid cross? They will appear to segregate independently, as if they are located on different chromosomes. Can't say without knowing the recombination frequency. They will stay together.via both crossing over between homologous chromosomes and independent assortmentHow do the events of meiosis I promote the production of new combinations of alleles? via both independent assortment and crossing over between sister chromatids via independent assortment alone via crossing over between homologous chromosomes only via both crossing over between homologous chromosomes and independent assortment via crossing over between sister chromatids onlyEither P-O-N or N-O-PIf the recombination frequency between P and O is 7.4, and between N and O it is 7.9, what is the likely order of these genes on the chromosome if the distance between N and P is 15.7? None of these choices N-O-P O-P-N Either P-O-N or N-O-P O-N-P8 map unitsThe recessive alleles for cinnibar eyes (cn) and vestigial wings (vg) identify two autosomal genes on the third chromosome of Drosophila melanogaster. When females, heterozygous at these two genes (cn+cn vg+vg), were crossed with cinnibar-eyed, vestigial winged males, the following classes and numbers of progeny (n=1000) were obtained: Wild-type eye color, wild-type wings 42 Wild-type eye color, vestigial wings 464 Cinnibar eye color, vestigial wings 38 Cinnibar eye color, wild-type wings 456 Based upon these results, the map distance between the cn and vg genes is estimated to be 4 map units 8 map units 46 map units 92 map unitsThe male gamete determines sex because each male gamete can contribute either an X or a Y chromosome.In humans, what determines the sex of offspring and why? Hints The female gamete determines sex because only the female gametes can have one of two functional sex chromosomes. The male determines sex because the sperm can fertilize either a female egg or a male egg. The female gamete determines sex because only the female gamete provides cytoplasm to the zygote. The chromosome contribution from both parents determines sex because the offspring uses all the parents' chromosomes. The male gamete determines sex because each male gamete can contribute either an X or a Y chromosome.a gene region present on the Y chromosome that triggers male developmentSRY is best described in which of the following ways? a gene required for development, and males or females lacking the gene do not survive past early childhood an autosomal gene that is required for the expression of genes on the Y chromosome a gene region present on the Y chromosome that triggers male development a gene present on the X chromosome that triggers female development an autosomal gene that is required for the expression of genes on the X chromosomeVery rarely: it is rare that an affected male would mate with a carrier female.Duchenne muscular dystrophy is a serious condition caused by a recessive allele of a gene on the human X chromosome. The patients have muscles that weaken over time because they have absent or decreased dystrophin, a muscle protein. They rarely live past their 20s. How likely is it for a woman to have this condition? One-fourth of the daughters of an affected father and a carrier mother could have this condition. Only if a woman is XXX could she have this condition. Women can never have this condition. Very rarely: it is rare that an affected male would mate with a carrier female. One-half of the daughters of an affected man would have this condition.activation of the XIST gene on the X chromosome that will become the Barr bodyAll female mammals have one active X chromosome per cell instead of two. What causes this? crossing over between the XIST gene on one X chromosome and a related gene on an autosome attachment of methyl (CH3) groups to the X chromosome that will remain active activation of the BARR gene on one X chromosome, which then becomes inactive inactivation of the XIST gene on the X chromosome derived from the male parent activation of the XIST gene on the X chromosome that will become the Barr bodya male inherits only one allele of the X-linked gene controlling hair color.Normally, only female cats have the tortoiseshell phenotype because a male inherits only one allele of the X-linked gene controlling hair color. multiple crossovers on the Y chromosome prevent orange pigment production. only males can have Barr bodies. the Y chromosome has a gene blocking orange coloration. the males die during embryonic development.noneRefer to the following information to answer the question below. A man who is an achondroplastic dwarf with normal vision marries a color-blind woman of normal height. The man's father was 6 feet tall, and both the woman's parents were of average height. Achondroplastic dwarfism is autosomal dominant, and red-green color blindness is X-linked recessive. How many of their daughters might be expected to be color-blind dwarfs? half three out of four none one out of four allhalfRefer to the following information to answer the question below. A man who is an achondroplastic dwarf with normal vision marries a color-blind woman of normal height. The man's father was 6 feet tall, and both the woman's parents were of average height. Achondroplastic dwarfism is autosomal dominant, and red-green color blindness is X-linked recessive. What proportion of their sons would be color-blind and of normal height? all none one out of four three out of four half100%Refer to the following information to answer the question below. A man who is an achondroplastic dwarf with normal vision marries a color-blind woman of normal height. The man's father was 6 feet tall, and both the woman's parents were of average height. Achondroplastic dwarfism is autosomal dominant, and red-green color blindness is X-linked recessive. They have a daughter who is a dwarf with normal color vision. What is the probability that she is heterozygous for both genes? 50% 75% 0% 25% 100%how far apart they are on the chromosomeIn general, the frequency with which crossing over occurs between two linked genes depends on what? Hints whether the genes are dominant or recessive the characters the genes code for how far apart they are on the chromosome the phase of meiosis in which the crossing over occurs whether the genes are on the X or some other chromosomeThe two genes are closely linked on the same chromosome.How would one explain a testcross involving F1 dihybrid flies in which more parental-type offspring than recombinant-type offspring are produced? Both of the characters are controlled by more than one gene. Recombination did not occur in the cell during meiosis. The two genes are linked but on different chromosomes. The two genes are closely linked on the same chromosome.All eight possible phenotypes could occur, but a greater proportion of the offspring would have the parental phenotypes.Now, suppose that the three tomato genes from Part A did not assort independently, but instead were linked to one another on the same chromosome. Would you expect the phenotypic ratio in the offspring to change? If so, how? Which statement best predicts the results of the cross MmDdPp x mmddpp assuming that all three genes are linked? Hints Lack of independent assortment means that you cannot predict the frequencies of phenotypes in the offspring. All eight possible phenotypes would occur in equal numbers in the offspring (1:1:1:1:1:1:1:1). Only the parental phenotypes could possibly occur in the offspring. All eight possible phenotypes could occur, but a greater proportion of the offspring would have the parental phenotypes.15A homozygous tomato plant with red fruit and yellow flowers was crossed with a homozygous tomato plant with golden fruit and white flowers. The F1 all had red fruit and yellow flowers. The F1 were testcrossed by crossing them to homozygous recessive individuals and the following offspring were obtained: Red fruit and yellow flowers-41 Red fruit and white flowers-7 Golden fruit and yellow flowers-8 Golden fruit and white flowers-44 How many map units separate these genes? 18.1 17.6 15 17.1b+ svn+ and b svnIn Drosophila, two genes on the fourth chromosome, bent wings (b) and shaven bristles (svn) are completely linked. If a fly with the genotype bb svnsvn is crossed to a fly with the genotype b+b+ svn+svn+, what type(s) of gametes will their offspring produce. b+ svn+ and b svn b svn b+ svn+ b+ svn+, b svn, b+ svn and b svn+They are located close together on the same chromosome.What is the reason that closely linked genes are typically inherited together? Genes align that way during metaphase I of meiosis. Alleles are paired together during meiosis. They are located close together on the same chromosome. The number of genes in a cell is greater than the number of chromosomes.Crossing-over can produce recombinant gametes with the genotypes abC and ABC.This image shows a pair of homologous chromosomes with three linked genes indicated (genes A, B, and C). Which of the following statements is true with respect to these genes? Crossing-over can produce recombinant gametes with the genotypes abC and ABC. The parental arrangement of alleles is abC and ABC. Independent assortment can produce gametes with the genotypes aBC, abC, AbC, ABC. Crossing-over can produce recombinant gametes with the genotypes aBC and AbC.26.0 map unitsThe recessive alleles for yellow body (yb) and cut wings (cw) identify two autosomal genes on the second chromosome of Drosophila melanogaster. When females, heterozygous at these two genes (yb+yb cw+cw), were crossed with yellow-bodied, cut-winged males, the following classes and numbers of progeny (out of 1000) were obtained: wild type body color, wild type wings 120 wild type body color, cut wings 360 yellow body color, cut wings 140 yellow body color, wild type wings 380 Based upon these results, the map distance between the yb and cw genes is estimated to be 74.0 map units 38.0 map units 26.0 map units 12.0 map unitswild type body color, cut wings X yellow body color, wild type wingsThe recessive alleles for yellow body (yb) and cut wings (cw) identify two autosomal genes on the second chromosome of Drosophila melanogaster. When females, heterozygous at these two genes (yb+yb cw+cw), were crossed with yellow-bodied, cut-winged males, the following classes and numbers of progeny (out of 1000) were obtained: wild type body color, wild type wings 120 wild type body color, cut wings 360 yellow body color, cut wings 140 yellow body color, wild type wings 380 Which of the following crosses could have given rise to the females used in the cross described above? Assume that all individuals are true breeding. wild type body color, wild type wings X yellow body color, cut wings wild type body color, cut wings X yellow body color, wild type wings wild type body color, wild type wings X wild type body color, wild type wings yellow body color, cut wings X yellow body color, cut wings38%The recessive alleles for yellow body (yb) and cut wings (cw) identify two autosomal genes on the second chromosome of Drosophila melanogaster. When females, heterozygous at these two genes (yb+yb cw+cw), were crossed with yellow-bodied, cut-winged males, the following classes and numbers of progeny (out of 1000) were obtained: wild type body color, wild type wings 120 wild type body color, cut wings 360 yellow body color, cut wings 140 yellow body color, wild type wings 380 What proportion of the female's gametes is expected to have the genotype (yb cw+)? 74% 26% 38% 14%8How many unique gametes could be produced through independent assortment by an individual with the genotype AaBbCCDdEE? 8 16 4 64different possible assortment of chromosomes into gametesThe individual with genotype AaBbCCDdEE can make many kinds of gametes. Which of the following is the major reason? different possible assortment of chromosomes into gametes the tendency for dominant alleles to segregate together recurrent mutations forming new alleles crossing over during prophase IHTA sexually reproducing animal has two unlinked genes, one for head shape (H) and one for tail length (T). Its genotype is HhTt. Which of the following genotypes is possible in a gamete from this organism? HT Hh HhTt Tarrangement of chromosome pairs at the equatorMendel's second law of independent assortment has its basis in which of the following events of meiosis I? arrangement of chromosome pairs at the equator synapsis of homologous chromosomes crossing over separation of cells at telophase1, 2, and 3the figure and the following description to answer the question(s) below. In a particular plant, leaf color is controlled by gene locus D. Plants with at least one allele D have dark green leaves, and plants with the homozygous recessive dd genotype have light green leaves. A true-breeding, dark-leaved plant is crossed with a light-leaved one, and the F1 offspring is allowed to self-pollinate. The predicted outcome of the F2 is diagrammed in the Punnett square shown in the figure, where 1, 2, 3, and 4 represent the genotypes corresponding to each box within the square. Which of the boxes marked 1-4 correspond to plants with dark leaves? 1 only 2 and 3 1, 2, and 3 4 onlyenvironmental factors such as soil pHHydrangea plants of the same genotype are planted in a large flower garden. Some of the plants produce blue flowers and others pink flowers. This can be best explained by which of the following? the allele for blue hydrangea being completely dominant environmental factors such as soil pH the alleles being codominant the knowledge that multiple alleles are involvedpurple and longRadish flowers may be red, purple, or white. A cross between a red-flowered plant and a white-flowered plant yields all-purple offspring. The part of the radish we eat may be oval or long, with long being the dominant trait. If true-breeding red long radishes are crossed with true-breeding white oval radishes, the F1 will be expected to be which of the following? red and long white and long purple and long purple and ovalall sharp-spined progenyGene S controls the sharpness of spines in a type of cactus. Cactuses with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cactuses have dull spines. At the same time, a second gene, N, determines whether or not cactuses have spines. Homozygous recessive nn cactuses have no spines at all. A cross between a true-breeding sharp-spined cactus and a spineless cactus would produce_____. 25% sharp-spined, 50% dull-spined, 25% spineless progeny 50% sharp-spined, 50% dull-spined progeny all sharp-spined progeny It is impossible to determine the phenotypes of the progeny.blue and white offspringFeather color in budgies is determined by two different genes, Y and B, one for pigment on the outside and one for the inside of the feather. YYBB, YyBB, YYBb or YyBb is green; yyBB or yyBb is blue; YYbb or Yybb is yellow; and yybb is white. A blue budgie is crossed with a white budgie. Which of the following results is possible? green and yellow offspring green offspring only blue and white offspring yellow offspring onlyEach parent is either M or MN.A gene for the MN blood group has codominant alleles M and N. If both children are of blood type M, which of the following is possible? Each parent is either M or MN. Both children are heterozygous for this gene. Each parent must be type M. Neither parent can have the N allele.A and BIn humans, ABO blood types refer to glycoproteins in the membranes of red blood cells. There are three alleles for this autosomal gene: IA, IB, and i. The IA allele codes for the A glycoprotein, The IB allele codes for the B glycoprotein, and the i allele doesn't code for any membrane glycoprotein. IA and IB are codominant, and i is recessive to both IA and IB. People with type A blood have the genotypes IAIA or IAi, people with type B blood are IBIB or IBi, people with type AB blood are IAIB, and people with type O blood are ii. If a woman with type AB blood marries a man with type O blood, which of the following blood types could their children possibly have? A, B, and O AB and O A and B A, B, AB, and OQuantitative characters are due to polygenic inheritance, the additive effects of two or more genes on a single phenotypic character. A single gene affected all but one of the pea characters studied by Mendel.Quantitative characters vary in a population along a continuum. How do such characters differ from the characters investigated by Mendel in his experiments on peas? Environment and genes affect quantitative characters, whereas only genes determined the pea characters studied by Mendel. Quantitative characters are due to polygenic inheritance, the additive effects of two or more genes on a single phenotypic character. A single gene affected all but one of the pea characters studied by Mendel. The nature of inheritance of quantitative characters is poorly understood, and Mendel understood the nature of inheritance for the characters he studied in his peas.The chance that an individual taken at random from the F2 generation produces wrinkled seeds is 25% and the chance that the same individual produces yellow seeds is 75%.Look at the Punnett square, which shows the predicted offspring of the F2 generation from a cross between a plant with yellow-round seeds (YYRR) and a plant with green-wrinkled seeds (yyrr). Select the correct statement about wrinkled yellow seeds in the F2 generation. The chance that an individual taken at random from the F2 generation produces wrinkled seeds is 25% and the chance that the same individual produces yellow seeds is 75%. In a set of 16 individuals from the F2 generation, 3 will produce wrinkled yellow seeds. Wrinkled and yellow traits are linked in peas.These homologous chromosomes represent a maternal and a paternal chromosome.Each chromosome in this homologous pair possesses a different allele for flower color. Which statement about this homologous pair of chromosomes is correct? Each of these homologous chromosomes consists of a single chromatid. Therefore, they must come from a haploid cell. These homologous chromosomes represent a maternal and a paternal chromosome. These homologous chromosomes are formed by DNA replication.The gene is present in both males and females.Which of the following is true of an X-linked gene, but not of a Y-linked gene? Hints It is only expressed in female offspring. Sister chromatids separate during mitosis. It is expressed in half of the cells of either male or female. The gene is present in both males and females. It does not segregate like other genes.100%Cinnabar eyes is a sex-linked, recessive characteristic in fruit flies. If a female having cinnabar eyes is crossed with a wild-type male, what percentage of the F1 males will have cinnabar eyes? 100% 25% 50% 0%a male inherits only one allele of the X-linked gene controlling hair colorNormally, only female cats have the tortoiseshell phenotype because _____. the Y chromosome has a gene blocking orange coloration multiple crossovers on the Y chromosome prevent orange pigment production only males can have Barr bodies a male inherits only one allele of the X-linked gene controlling hair colorOne-half of the daughters of an affected father and a carrier mother could have this condition.Duchenne muscular dystrophy is a serious condition caused by a recessive allele of a gene on the human X chromosome. The patients have muscles that weaken over time because they have absent or decreased dystrophin, a muscle protein. They rarely live past their twenties. How likely is it for a woman to have this condition? One-half of the daughters of an affected father and a carrier mother could have this condition. Women can never have this condition. Only if a woman is XXX could she have this condition. One-fourth of the daughters of an affected man would have this condition.how far apart they are on the chromosomeIn general, the frequency with which crossing over occurs between two linked genes depends on what? Hints whether the genes are dominant or recessive whether the genes are on the X or some other chromosome the characters the genes code for the phase of meiosis in which the crossing over occurs how far apart they are on the chromosomeThe closer two genes are on a chromosome, the lower the probability that a crossover will occur between them.Which of the following statements is true of linkage? The observed frequency of recombination of two genes that are far apart from each other has a maximum value of 100%. Linked genes are found on different chromosomes. The closer two genes are on a chromosome, the lower the probability that a crossover will occur between them. All of the traits that Mendel studied-seed color, pod shape, flower color, and others-are due to genes linked on the same chromosome. Crossing over occurs during prophase II of meiosis.A and GThe following is a map of four genes on a chromosome. Between which two genes would you expect the highest frequency of recombination? E and G A and W A and G A and E W and EThe likelihood of a crossover event between these two genes is low.What is the reason that closely linked genes are typically inherited together? Genes align that way during metaphase I of meiosis. The likelihood of a crossover event between these two genes is low. Chromosomes are unbreakable. Alleles are paired together during meiosis. The number of genes in a cell is greater than the number of chromosomes.18%Consider the relative distance between genes on the chromosome below. What is the recombination frequency of genes X and Z? 4% 11% 18% 23%These two genes could either be 50 or more map units apart or could be on different chromosomes.In a dihybrid cross of a female who is heterozygous for body color (yb+ yb) and cut wings (cw+ cw) with a yellow-bodied, cut-winged male (yb yb cw cw), the following classes and numbers of progeny (out of 1000) were obtained: •Wild type body color, wild type wings 248 •Wild type body color, cut wings 252 •Yellow body, cut wings 247 •Yellow body color, wild type wings 253 Which of the following conclusions can you make about the genes for body color and cut wings? These two genes are 25 map units apart. These two genes are 50 map units apart. These two genes are located on different chromosomes. These two genes could either be 50 or more map units apart or could be on different chromosomes.Gametes with the genotypes ab and AB.Based upon what you know about the position of genes and the frequency of crossing-over between them, which gametes would you expect to be produced more often by crossing over? Gametes with the genotypes ab and AB. Gametes with the genotype aB and Ab. Gametes with the genotype CD and cd. Gametes with the genotype Cd and cD.sq+ sq Ze+ ZeYou are studying two new mutations in Drosophila melanogaster. The mutation squiggle produces flies with antennae that are crooked rather than straight. It is inherited as an autosomal recessive. The mutation zebra produces flies that have bodies with black and white stripes. It is inherited as an autosomal dominant. What would be the correct nomenclature for a fly that is heterozygous for both of these mutations ? an+ an B+ B sq+ sq Ze+ Ze sq sq Ze Ze an+ an b+ b3.8 %The recessive alleles for cinnibar eyes (cn) and vestigial wings (vg) identify two autosomal genes on the third chromosome of Drosophila melanogaster. When females, heterozygous at these two genes (cn+cn vg+vg), were crossed with cinnibar-eyed, vestigial winged males, the following classes and numbers of progeny (n=1000) were obtained: •Wild-type eye color, wild-type wings 42 •Wild-type eye color, vestigial wings 464 •Cinnibar eye color, vestigial wings 38 •Cinnibar eye color, wild-type wings 456 The heterozygous females in this cross produce approximately _____________ gametes with the genotype cn vg. 3.8 % 4.2 % 46.4 % 45.6 %The gene for black body (b).This image shows a partial map of chromosome 2 in Drosophila melanogaster. The numbers below the genes indicate the distance of each gene from the gene for short aristae. Which of the genes shown on this map is/are linked to the gene for short aristae at a map distance that can be calculated by doing a dihybrid cross involving the aristae gene? The gene for black body (b). The genes for black body (b) and cinnabar eyes (cn). A map distance can be calculated from the gene for short aristae to all of the genes shown. The genes for black body (b), cinnabar eyes (cn), and vestigial wings (vg).