General Genetics Exam 1 Practice Questions

Briefly describe common misunderstandings or misapplications of the concept heritability.

(1) Heritability is the portion of phenotypic variance due to genetic variance; it doesnot indicate to what extent the phenotype itself is determined by genotype.(2) Heritability applies to populations; it does not apply to individuals. (3) Heritability is determined for a particular population in a particular environmentat a particular time. Heritability determined for one population does not apply to other populations, or even the same population facing different environmental conditions at a different period. (4) A trait with high heritability may still be strongly influenced by environmental factors. (5) High heritability does not mean that differences between populations are due to differences in genotype.

What is a complementation test and what it is used for?

A test designed to determine whether two different mutations are at the same locus (are allelic) or at different loci (are nonallelic). Two individuals that are homozygous for two independently derived mutations are crossed, producing F1 progeny that are heterozygous for the mutations. If the mutations are at the same locus, the F1 will have a mutant phenotype. If the mutations are at different loci, the F1 will have wild-type phenotype.

what are the major results of meiosis?

Four daughter cells that are haploid, which means they each contain half the number of chromosomes of the diploid parent cell.

What are the genetically important results of the cell cycle and mitosis?

-From a single cell, the cell cycle produces two daughter cells that contain the same genetic instructions. The resulting cells are genetically identical with each other and with their parent cell -Each of the cells produced contains a full complement of chromosomes: there is no net reduction or increase in chromosome number. (The genetic material is precisely copied so that the two resulting cells contain the same genetic information.)

List the stages of interphase and the major events that take place in each stage.

-G1 Stage: Growth and development of the cell -S Stage: synthesis of DNA (DNA duplicates) -G2 Stage: Preparation for cell division (Mitosis)

What is the principle of segregation? Why is it important?

-Mendel's first law -states that an organism possess two alleles for any particular trait and that these allele separate during the formation of gametes -it explains how genotypic ratios in the haploid gametes are produced

How does a quantitative characteristic differ from a discontinuous characteristic?

Discontinuous characteristics have only a few distinct (discrete) phenotypes. In contrast, a quantitative characteristic shows a continuous variation in phenotype.

What are the stages of meiosis, and what major events take place in each stage?

-Prophase I: Chromatin begins to condense; Homologous chromosomes synapse; crossing over takes place; nuclear envelope breaks down; mitotic spindle forms -Metaphase I: Homologous pairs of chromosomes line up on the metaphase plate -Anaphase I: The two chromosomes (each with two chromatids) of each homologous pair separate and move toward opposite poles -Telophase I: Chromosomes arrive at opposite spindle poles -Cytokinesis: The cytoplasm divides to produce two cells, each having half the original number of chromosomes -Interkinesis: period between Meiosis I and Meiosis II where no new DNA synthesis takes place, spindle breaks down, nuclear envelope re-forms and chromosomes relax -Prophase II: Chromosome condenses; the spindle forms; nuclear envelope disintegrates -Metaphase II: Individual chromosomes line up on the metaphase plate -Anaphase II: Sister chromatids separate and move as individual chromosomes toward spindle poles -Telophase II: Chromosomes arrive at the spindle poles; the spindle breaks down; nuclear envelope re-forms -Cytokinesis: The cytoplasm divides

List the stages of mitosis and the major events that take place in each stage.

-Prophase: Centrosomes move to opposites; chromatin condense forming distinct/visible chromosomes; nuclear membrane begins to break down -Prometaphase: Nuclear envelope disintegrates, Mitotic spindle (microtubules) forms between centrosomes and kinetochores (located at centromeres); chromosomes move toward metaphase plate; chromosomes continue to condense -Metaphase: Chromosomes align on the metaphase plate; spindle assembly checkpoint -Anaphase: Sister chromatids separate, becoming individual chromosomes that migrate toward opposite poles (to the centrosomes) -Telophase: Chromosomes arrive at spindle poles, the nuclear envelope re-forms, and the condensed chromosomes relax. -Cytokinesis: Cytoplasm divides (forms two daughter cells)

What is the concept of dominance?

-Refers to an allele or phenotype that is expressed in homozygotes (AA) and heterozygotes (Aa) -Controls the trait -The concept of dominance states that when two different alleles are present in a genotype, only the dominant allele is expressed in the phenotype.

Name three essential structural elements of a functional eukaryotic chromosome and describe their functions.

1. Centromere: The central region of a eukaryotic chromosome where the kinetochore is assembled. Centromere is the part of a chromosome, that links sister chromatids. During cell division, it acts as the point of attachment for the spindle fibers. 2. Telomeres: Telomeres play 3 major purposes: a. They help to organize each of our 46 chromosomes in the nucleus of our cells. b. They protect the ends of chromosomes by forming a cap, much like the plastic tip on shoelaces. If the telomeres were not there, chromosomes may end up sticking to other chromosomes. c. They allow the chromosomes to be replicated properly during cell division. d. They may have a role in limiting cell division. 3. Origins of replication (Ori C): It serves as the initiation region of DNA synthesis.

In the Blind Men's Riddle, each blind man buys his own pairs of socks, but the clerk places all the pairs in one bag. Thus, there are two pairs of socks of each color in the bag (two black pairs, two blue pairs, two gray pairs, etc.). What do the two pairs (four socks in all) of each sock represent?

2 pairs of the same color socks=two sets of genetic information (one form each parent/two chromosomes of a homologous pair

What does each part of the Blind Men's Riddle represent in the cell cycle?

2 pairs of the same color socks=two sets of genetic information (one form each parent/two chromosomes of a homologous pair 2 socks of a pair=two chromatids of a chromosome thread that holds the pairs of socks together=cohesion hands of the blind men=Spindle fibers

A horse has 64 chromosomes and a donkey has 62 chromosomes. A cross between a female horse and a male donkey produces a mules, which is usually sterile. How many chromosomes does each mule have? Can you think of any reasons for the fact that most mules are sterile?

32 chromosomes will come from the horse and 31 chromosomes will come from the donkey because sperm and egg are haploid. The mule is sterile because now it contains 63 chromosomes and it is not possible to make haploid gamete in odd number of chromosomes because gametes form from even number of chromosomes

What are the addition and multiplication rules for probability and when should they be used?

ADDITION RULE: -the probability of 2 or more mutually exclusive events occurring -add the probability of each individual event - key word "either/or" *if there is an overlap in events, subtract the overlapping probability so you don't count them twice MULTIPLICATION RULE: -the probability of 2 or more independent events occurring together -multiple the probability of each individual event -key word "and"

what characteristic are exhibited by an X-linked trait?

Allele is located on the X chromosome Males show phenotypes for all x-linked traits because they have one copy of the x chromosome EXAMPLES: Red-green colorblindness; hemophilia; Drosophila

What features are exhibited by a pedigree of a recessive trait? What features are exhibited if the trait is dominant?

Assuming autosomal inheritance and complete penetrance, recessive inheritance is indicated by the presence of affected offspring from two unaffected parents. If the trait is rare, frequently the recessive allele will be passed for a number of generations without the trait appearing in the pedigree. For rare recessive traits, the trait often appears as a result of mating between two close relatives. Finally, if the trait is recessive, all of the offspring of two affected parents will be affected. Assuming autosomal inheritance and complete penetrance, at least one of the parents of affected children should be affected (unless a new mutation has led to creation of a new dominant allele). The trait should not skip generations within a lineage. Two affected parents who are heterozygotes can have unaffected offspring. Unaffected parents do not transmit the trait to their offspring.

How are the events that take place in spermatogenesis and oogenesis similar? How are they different?

BOTH -begins with a diploid germ cell to produce a haploid gamete -involve mitosis and meiosis SPERMATOGENESIS -production of sperm -begins at puberty and continues throughout life -cytoplasm divides evenly -produces four equal-sized haploid spermatids OOGENESIS -production of an egg cell -begins prior to birth and continues through puberty -cytoplasm divides unevenly -produces large mature haploid egg and a small polar body

What is a Barr body? How is it related to the Lyon hypothesis?

Barr body is an inactivated X chromosome that appears as a condensed, darkly staining structure in most cells of female placental mammals Related to the Lyon hypothesis because it is a hypothesis explaining why the phenotypic effect of the X chromosome is the same in the mammalian female which has two X chromosomes as it is in the male which has only one X chromosome: one of each two somatic X chromosomes in mammalian females is selected at random and inactivated early in embryonic development.

Why do polygenic characteristics have many phenotypes?

Because number of allele combinations increases exponentually with each additional locus.

What is the chromosome theory of heredity? Why was it important?

Boveri and Sutton's chromosome theory of inheritance states that genes are found at specific locations on chromosomes, and that the behavior of chromosomes during meiosis can explain Mendel's laws of inheritance.

What are checkpoints? List some of the important checkpoints in the cell cycle.

Checkpoint allow or prohibit the cell's progression to the next stage in the cell cycle. Checkpoints ensure that all cellular components are present and in good working order, and are necessary to prevent cells with damaged or missing chromosomes from proliferating EXAMPLES: -G1/S Checkpoint: holds the cell in G1 until the cell has all the enzymes necessary for the replication of DNA. (after cell passes this checkpoint, it is committed to divide) -G2/M Checkpoint: Is passed only if the cell's DNA is completely replicated and undamaged.(after cell passes this checkpoint, cell is ready to divided and enters M phase) -Spindle Assembly Checkpoint (SAC): ensures that each chromosome is aligned on the metaphase plate and attached to spindle microtubules from opposite poles

Phenylketonuria (PKU) is a disease that results from a recessive gene. Two normal parents produce a child with PKU. a. what is the probability that a sperm from the father will contain the PKU allele? b. What is the probability that an egg from the mother will contain the PKU allele? c. What is the probability that their next child will have PKU? d. What is the probability that their next child will be heterozygous for the PKU gene?

Consider normal gene is P as diseased gene is p. Two normal parents must be heterozygous recessive in order to produce a diseased son. So father genotype is Pp and mother genotype is Pp. From father both P or p gene can present in the sperm. a. The probability that the sperm from father will contain a PKU allele is 1/2 or 0.5. ( out of P or p any one can present) b. The probability that the egg from mother will contain a PKU allele is also 1/2 or 0.5 ( same as father because sperm or egg both are haploid contains a single gene P or p can be present) c. The probability that their next child will have PKU is 1/4 = 0.25 d. The probability that their next child will be heterozygous for PKU gene is 2/4 = 1/2 = 0.5

Hairlessness in American rat terries is recessive to the presence of hair. Suppose that you have a rat terrier with hair. How can you determine whether this dog is homozygous or heterozygous for the hairy trait?

Cross with a hairless rat terrier. As the hairless rat would be homozygous recessive (hh) genotype as hairlessness is a recessive trait and if the unknown rat is homozygous dominant (HH) for the trait, all the offsprings will have hair (Hh) but if it is heterozygous Hh, half will have hair (Hh) and half of the offsprings will be hairless (hh).

What two processes unique to meiosis are responsible for genetic variation? At what point in meiosis do these processes take place?

Crossing over/Recombination takes place in Prophase I Independent Assortment takes place at metaphase I

Why are the two cells produced by the cell cycle genetically identical?

During DNA synthesis in the S phase create and exact copy of each DNA molecule, giving rise to two genetically identical sister chromatids.

What is the difference between genotype and phenotype?

GENOTPYE -the set of genes possessed by an individual organism on the DNA -the genetic makeup of an individual PHENOTYPE -appearance or manifestation of a characteristic or trait -due to the manifestation of genotype

What is gene interaction? What is the difference between and epistatic gene and a hypostatic gene

Gene interaction is the interaction between genes at different loci that affect the same characteristic. The difference between an epistatic gene and a hypostatic gene is that an epistatic gene can mask the phenotype of a hypostatic gene. EXPLANATION Epistasis is an inereactioon between two genes in which one gene masks the phenotype of the other one. Epistatic gene masks the phenotype of the hypostatic gene. Hypoststic gene's expression gets silenced. It controls the expression of the hypostatic gene. Phenotype is the observable character of an organism.

In the Blind Men's Riddle, what performs the same function as spindle fibers?

Hands of the blind men

In which phase of mitosis and meiosis are the principles of segregation and independent assortment at work?

LAW OF SEGREAGATION This principle can be observed during anaphase of mitosis. LAW OF INDEPENDENT ASSORTMENT This principle can be observed during anaphase I of meiosis.

Direction of shell coiling in the snail Lymnaea peregra results from a genetic maternal effect. An autosomal allele for a right-handed shell (s+), called dextral, is dominant over the allele for a left-handed shell (s), called sinistral. A pet snail called Martha is sinistral and reproduces only as a female (the snails are hermaphroditic). Indicate which of the following statements are true and which are false. Explain your reasoning in each case. a. Martha's genotype must be ss. b. Martha's genotype cannot be s+s+. c. All the offspring produced by Martha must be sinistral. d. At least some of the offspring produced by Martha must be sinistral. e. Martha's mother must have been sinistral. f. All of Martha's brothers must be sinistral

In genetic maternal effects, the genotype of the mother determines the phenotype of the offspring. Because Martha is sinistral, we know her mother must be genotype ss. If Martha's mother is ss, Martha must carry at least one s allele. W have no information about Martha's father. a. False, Martha might have inherited an s+from her father and therefore could be s+s. b. True. Martha must have inherited an s allele from her mother and therefore cannot be s+s+ c. False. The phenotype of Martha's offspring will be determined by Martha's genotype, which we do not know. Martha might have inherited an s+ allele from her father, in which her genotype would be s+s and she would produce all dextral offspring. d. False. Martha's genotype could be s+s, in which case all her offspring would be dextral e. False. Her mother's phenotype is determined by her own mother's genotype, which could have been s+s f. True. Because Martha is sinistral, her mother's genotype is ss and all her offspring should be sinistral like martha

A cell has a circular chromosome and no nuclear membrane. Its DNA is associated with some histone proteins. Does this cell belong to the bacteria, the archaea, or the eukaryotes? Explain your reasoning.

It belongs to arachaea-Eukaryote and arachaea (ancient bacteria) both have DNA wrapped around histones to form chromatin-Eubacterium (true bacteria) do not contain histones-Eukaryotes have multiple linear chromosomes while arachaea have a single circular chromosome

A summer-squash plant that produces disc-shaped fruit is crossed with a summer-squash plant that produces long fruit. All the F1 have disc-shaped fruit. When the F1 are intercrossed, F2 progeny are produced in the following ratio: 9/16 disc-shaped fruit; 6/16 spherical fruit; 1/16 long fruit. Give the genotypes of the F2 Progeny.

Let A and B represent the two loci. The F1 heterozygotes are AaBb. The F2 are A_B_ (disc-shaped), A_bb (spherical), aaB_ (spherical), aabb (long)

List some similarities and differences between mitosis and meiosis. Which differences do you think are most important and why?

MITOSIS - results in two identical diploid daughter cells -no crossing over or genetic variation -On division process in four stages -no synapse -Individual chromosomes line up on metaphase plate in metaphase -Sister chromatids separate in anaphase MEIOSIS -results in four haploid daughter cells -crossing over/genetic variation occurs in meiosis I -two division process in four stages each -synapse in prophase I -Homologous chromosomes line up on metaphase plate in metaphase I and metaphase II has sister chromatids -homologous chromosomes separate in anaphase I

Give the genotypic ratios that may appear among the progeny of simple crosses and the genotypes of the parents that may give rise to each ratio.

MONOHYBRID CROSS -expression of only one characteristic is considered -Genotypic ratio=1 : 2 : 1 (homozygous dominant : heterozygous : homozygous recessive) -phenotypic ratio= 3: 1 (dominant : recessive) -example: RR x rr would produce 1/4 RR, 1/2 Rr, 1/4 rr DIHYBRID CROSS -expression of two characteristics is considered - Genotypic ratio= -Phenotypic ratio= 9:3:3:1 -example: BbEe x BbEe would produce 9/16 R_Y_, 3/16 R_yy, 3/16 rrY_, and 1/16 rryy

Red-green color blindness is an X-linked recessive trait in humans. Polydactyly (extra fingers and toes) is an autosomal dominant trait. Martha has normal fingers and toes and normal color vision. Her mother is normal in all respects, but her father is color blind and polydactylous. Bill is color blind and polydactylous. his mother has normal color vision and normal fingers and toes. If Bill and Martha marry, what types and proportions of children can they produce?

Martha is a carrier of red-green color vision(XcX) (due to X-linked recessive gene from father) and normal fingers/toes(dd). Bill is color blind(XcY) and polydactylous (autosomal dominant Dd). X^cXdd x X^cYDd would produce 1/8 - Daughter is normal vision(carrier of color-blindness) and polydactylous(autosomal dominant) 1/8 - Daughter is normal vision(carrier of color-blindness) and normal finger/toes 1/8 - Daughter is red-green color blind(affected) and polydactylous(autosomal dominant) 1/8 - Daughter is red-green color blind(affected) and normal finger/toes 1/8 - Son has normal vision and polydactylous(autosomal dominant) 1/8 - Son has normal vision and normal finger/toes 1/8 - Son is red-green color blind(affected) and polydactylous(autosomal dominant) 1/8 - Son is red-green color blind(affected) and normal finger/toes

Why was Mendel's approach to the study of heredity so successful?

Mendel used very simple organisms like garden pea which was very easy to control pollination among his experimental groups. He has chosen easily observable characteristics like seed and flower color, shape and height of the plant. He was also very focused to study each trait individually in true breeding plants to understand the pattern of inheritance of allelic genes from parent plant to the offspring. Mendel gave three principle of inheritance: law of dominance, independent assortment and law of segregation to study the transmission of genes and genetic traits which they have already studied and existed in parent plant. As he was a mathematician so, He used algebra to articulate and prove the patterns of inheritance by explaining genotypic and phenotypic ratio result in offspring because of hybrid crosses in parent plants.

Ectrodactyly is a rare condition in which the fingers are absent and the hand is split. This condition is usually inherited as an autosomal dominant trait. Ademar Freire-Maia reported the appearance of ectrodactyly in a family in São Paulo, Brazil, whose pedigree is shown here. Is this pedigree consistent with autosomal dominant inheritance? If not, what mode of inheritance is most likely? Explain your reasoning.

No, this would be an example of autosomal recessive because the trait skips generations.

Joe has a white cat named Sam. When Joe crosses Sam with a black cat, he obtains one-half white kittens and one-half black kittens. When the black kittens are interbred, all the kittens that they produce are black. On the basis of these results, would you conclude that white or black coat color in cats is a recessive trait? Explain your reasoning.

On the basis of these results, we can conclude that the white coat color is dominant over the black coat color. So, black coat color is the recessive trait. Let us call the white coat color as B and the black coat color as b. Then it is said that in the first cross when the white cat was crossed with the black cat, they got half white cats and half black cats. This is only possible if Sam is heterozygous for the white coat (Bb). So the first cross is between Bb and bb where bb is the black cat. In the second cross, they interbred the black cats and got all of the progenies that of black color. Here, the cross is between bb and bb.

What is the principle of independent assortment? How is it related to the principle of segregation?

PRINCIPLE OF INDEPENDENT ASSORTMENT Genes for different characteristics that are at different loci segregate independently to one another PRINCIPLE OF SEGREGATION Indicates that the two alleles at a locus separate randomly RELATIONSHIP BETWEEN BOTH Both related to gamete formation and separation of one allele is not effected by another allele and also not affected by allele of a different gene (separation of alleles is random and an independent task)

What are some genetic differences between Prokaryotic and eukaryotic cells?

PROKARYOTIC CELLS -lack a nucleus -relatively small in diameter (1 to 10 um) -genome is usually one circular DNA molecule -DNA is not complexed with histones in eubacteria; some histones in archaea -relatively small amount of DNA -lacks membrane-bounded organelles -lacks cytoskeleton EUKARYOTIC CELLS -contains nucleus -relatively large in diameter (10 to 100 um) -genome is multiple linear DNA molecules -DNA is complexed with hisones -relatively large amount of DNA -contains membrane-bounded organelles -contains cytoskeleton

In German cockroaches, a curved wing (cv) is recessive to a normal wing (cv+). A homozygous cockroach that has normal wings is crossed with a homozygous cockroach that has curved wings. The F1 are intercrossed to produce the F2. Assume that the pair of chromosomes containing the locus for wing shape is metacentric. Draw this pair of chromosomes as it would appear in the parents, F1, and each class of F2 progeny at metaphase I of meiosis. Assume that no crossing over takes place. At each stage, label a location for the alleles for wing shape (cv and cv+) on the chromosomes.

Page B3 in back of book #27

Coat color in cats is determined by genes at several different loci. At one locus on the X chromosome, one allele (X+) encodes black fur; another allele (Xo) encodes orange fur. Females can be black (X+X+), orange (XoXo), or a mixture of orange and black called tortoiseshell (X+Xo). Males are either black (X+Y) or orange (XoY). Bill has a female tortoiseshell cat named Patches. One night, Patches escapes from Bill's house, spends the night out, and mates with a stray male. Patches later gives birth to the following kittens: one orange male, one black male, two tortoiseshell females, and one orange female. Give the genotypes of Patches, her kittens, and the stray male with which Patches mated with.

Patches: X+Xo Stray male: XoY Kittenes: 1 orange male: XoY 1 black male: X+Y 2 tortoiseshell females: X+Xo 1 orange female: XoXo

A cell in prophase II of meiosis contains 12 chromosomes. How many chromosomes would be present in a cell from the same organism if it were in prophase of mitosis? Prophase I of meiosis?

Prophase of mitosis= 24 chromosomes Prophase I of meiosis=24 chromosomes

Briefly explain why the relation between genotype and phenotype is frequently complex for quantitative characteristic.

Quantitative inheritance is polygenic inheritance meaning it is controlled by many genes and the effects of each gene is added together to produce a phenotype. The phenotype produced varies over a continuous degree within a population and is also influenced by environment.

Outline the process of spermatogenesis and oogenesis in animals.

SPERMATOGENESIS: -Takes place in testes -Each diploid spermatogonia undergo repeated rounds of mitosis to produce numerous spermatogonia -Spermatogonia initiate meiosis and enter Prophase I (Now called primary Spermatocyte=diploid because division hasn't occurred) -Completion of meiosis I gives rise to two haploid secondary spermatocytes that undergo meiosis II -Completion of meiosis II gives rise to four haploid spermatids that mature and develop into sperm OOGENESIS: -starts in embryo -takes place in ovaries -Each diploid Oogonia undergo repeated rounds of mitosis to produce numerous Oogonia or can can enter meiosis I -When Oogonia enter prophase I they are still diploid and are now called primary oocytes -stops in Prophase I and starts up again prior to ovulation -Each primary oocyte completes meiosis I and divides during cytokinesis where most of the cytoplasm is allocated to one of the two haploid cells called the secondary oocyte. -Smaller cell (the one that contains half of the chromosomes but only a small part of cytoplasm) is called the first polar body and it may or may not divide more. -Secondary Oocyte completes meiosis II and cytokinesis is uneven again. Smaller cell is second polar body and the larger cell is the ovum -Only the ovum is capable of being fertilized

Construct a table similar to that in FIGURE 2.10 (page 26) for the different stages of meiosis, giving the number of chromosomes per cell and the number of DNA molecules per cell for a cell that begins with four chromosomes (two homologous pairs) in G1. Include the following stages in your table: G1, S, G2, Prophase I, Metaphase I, Anaphase I, Telophase I (after cytokinesis), Prophase II, Metaphase II, Anaphase II, and Telophase II (after cytokinesis).

STAGE #CHROMOSOMS #DNA G1 4 4 S 4 8 G2 4 8 Prophase I 4 8 Metaphase I 4 8 Anaphase I 4 8 Telophase I (after cytokinesis) 2 4 Prophase II 2 4 Metaphase II 2 4 Anaphase II 2 4 Telophase II (after cytokinesis). 2 2

What is meant by genetic sex determination?

Sex determination in which the sexual phenotype is specified by the genes at one or more loci, but there are no obvious differences in the chromosomes of males and females

How do broad-sense and narrow-sense heritabilities differ?

The broad-sense heritability is the portion of phenotypic variance that is due to all types of genetic variance, including additive, dominance, and genic interaction variances. The narrow-sense heritability is only that portion of the phenotypic variance that is due to additive genetic variance.

What information do the mean and variance provide about distribution?

The mean is the center of the distribution. The variance is how broad the distribution is around the mean.

The fruit fly Drosophila Melanogaster (below left) has four pairs of chromosomes, whereas the house fly Musca Domestica (below right) has six pairs of chromosomes. In which species would you expect to see more genetic variation among the progeny of a cross? Explain your answer.

The progeny of the organism whose cells contain the larger number of homologous pairs of chromosomes should be expected to exhibit more variation. The number of different combinations of chromosomes that are possible in the gametes is 2^n, where n is equal to the number of homologous pairs of chromosomes. For the fruit fly, which has 4 chromosomes, the number of possible combinations is 2^4 = 16. For the house fly, which has 6 pairs of chromosomes, the number of possible combinations is 2^6 =64. Thus, the housefly would be the one you would expect to see more genetic variation among the progeny of a cross.

How is the response to selection related to narrow-sense heritability and the selection differential? What information does the response to selection provide?

The response to selection (R) = narrow-sense heritability (h^2 ) × selection differential (S). The value of R predicts how much the mean quantitative phenotype will change with different selection in a single generation.

How is sex determined in humans?

The sex chromosomes determine whether you are female (XX) or male (XY)

Explain why tortoiseshell cats are almost always female and why they have a patchy distribution of orange and black fur.

The tortoiseshell cat has black and orange patches which means that they are heterozygous for this trait on the X chromosome. This genes has two alleles (one that encodes for the black fur and one that encodes for the orange fur) Males only have one x chromosome so they cannot encode for both black fur and orange fur.

How is the chi-square goodness-of-fit test used to analyze genetic crosses? What does the probability associated with a chi-square value indicate about the results of a cross?

Thus the goodness-of-fit chi-square test, is used to evaluate, what is the role chance is playing in causing deviation between expected frequency and observed frequency in the genetic cross, and the probability observed from the chi-square table describes the probability about the random chance that are responsible for producing the deviations between the observed frequency and expected frequency. chi square value is calculated using formula =£ (O-E)2/E £ = sum of all the values . O = observed value , E = expected value P- value indicates the significance level of the result of that cross.

What characteristics of an organism would make it suitable for studies of the principles of inheritance? Can you name several organisms that have these characteristics?

USEFUL CHARACTERISTICS: easy to grow and maintain grows rapidly produces many generations in a short period EXAMPLES OF ORGANISMS: Neurospora (a fungus) Saccharomyces Cerevisiae (a yeast) Arabidopsis (a plant) Caenorhabditis Elegans (a nematode) Drosophila Melanogaster (a fruit fly)

List all the components that contribute to the phenotypic variance and define each component.

VG - Component of variance due to variation in genotype VA - Component of variance due to additive genetic variance VD - Component of variance due to dominance genetic variance VI - Component of variance due to genic interaction variance VE - Component of variance due to environmental differences VGE - Component of variance due to interaction between genes and environment

Many studies have suggested a strong genetic predisposition to migraine headaches, but the mode of inheritance is not clear. L. Russo and colleagues examined migraine headaches in several families, two of which are shown in the following pedigree (Page 72). What is the most likely mode of inheritance for migraine headaches in these families? Explain your reasoning.

X-linked is the mode of inheritance for migraine headaches in these families because Y-linked traits never occur in females. The concept of dominance and recessive do not apply to Y-linked traits either. This trait is X-linked dominant and the mother must be heterozygous because not all of the children have the trait

Red-green color blindness is humans is due to an X-linked recessive gene. Both John and Cathy have normal color vision. After 10 years of marriage to John, Cathy gave birth to a color-blind daughter. John filed for a divorce, claiming he is not the father of the child. Is John justified in his claim of nonpaternity? Explain why. If Cathy had given birth to a color-blind son, would John be justified in claiming nonpaternity?

Yes. He cannot be the father of Cathy's daughter. If Cathy is carrier for the color blindness (mutation on only one allele), she would be normal but will give birth to a colorblind son as the X chromosome comes from the mother. She cannot give birth to a colorblind daughter because for that two alleles should be mutated, in which one allele comes from father and one from mother.

The following Pedigree illustrates the inheritance of Nance-Horan syndrome, a rare genetic condition in which effected persons have cataracts and abnormally shaped teeth. a. On the basis of this pedigree, what do you think is the most likely mode of inheritance for Nance-Horan syndrome? b. If couple III-7 and III-8 have another child, what is the probability that the child will have Nance-Horan syndrome? c. If III-2 and III-7 were to mate, what is the probability their children would have Nance-Horan Syndrome?

a) mode of inheritance is X-linked recessive. b) If III-7 and III-8 have another child, then the probability of the child getting this syndrome = 1/4 or 25% c) If III-2 and III-7 mate, then the probability of the child getting this syndrome = 1/2 or 50%

E. W. Lindstrom crossed two corn plants with green seedlings and obtained the following progeny: 3583 green seedlings, 853 virescent-white seedlings, and 260 yellow seedlings. a. Give the genotypes for the green, virescent-white, and yellow progeny. b. Explain how color is determined in these seedlings. c. Does epistasis take place among the genes that determine color in the corn seedlings? If so, which gene is epistatic and which is hypostatic?

a. The ratio that fits the progeny is 12:3:1 of green:virescent-white:yellow. I arrive to this ratio by dividing 1/16 x 4696=293 and then calculating 3583/293= close to 12 and 853/293)= close to 3. Total progeny is: 4696 1/16 x total is 293 (close to 260) 3/16 x total is 880 close to 853 12/16 x total is 3522 close to 3583 b. Green phenotype when locus 1 is dominant, regardless of the alleles at locus 2. Yellow is seen when loci 1 and 2 are recessive, and virescent-white is seen when locus 1 is homozygous recessive, and locus 2 has a dominant allele. locus 1 can be named G and locus 2 Y after green and yellow. c. Locus 1 is epistatic. It shows dominant epistatis, because locus 2 is masked when there is a dominant allele in locus 1. The locus 2 is hypostatic and the effect of this locus is only seen when locus 1 is homozygous recessive.

The L^M and L^N alleles at the MN blood group locus exhibit codominance. Give the expected genotypes and phenotypes and their ratios in progeny resulting from the following crosses a. LMLM x LMLN b. LNLN x LNLN c. LMLN x LMLN d. LMLN x LNLN e. LMLM x LNLN

a. Genotypes: 1:1 2/4 LMLM; 2/4 LMLN Phenotypes: 1:1 blood group M: blood group MN b. Genotypes: all LNLN Phenotypes: all blood group N c. Genotypes: 1:2:1 1/4 LMLM, 2/4 LMLN, 1/4 LNLN Phenotypes: 1:2:1 blood group M: blood group MN: blood group N d. Genotypes: 1:1 2/4 LMLN; 2/4 LNLN Phenotypes: 1:1 blood group MN: blood group N e. Genotypes: all LMLN Phenotypes: all MN blood group

In cucumbers, dull fruit (D) is dominant over glossy fruit (d), orange fruit (R) is dominant over cream fruit (r), and bitter cotyledons (B) are dominant over non-bitter cotyledons (b). The three characters are encoded by genes located on different pairs of chromosomes. A plant homozygous for dull, orange fruit and bitter cotyledons is crossed with a plant that has glossy, cream fruit and non-bitter cotyledons. The F1 are intercrossed to produce the F2. a. Give the phenotypes and their expected proportions in the F2 b. An F1 plant is crossed with a plant that has glossy, cream fruit and non-bitter cotyledons. Give the phenotypes and expected proportions among the progeny of this cross.

a. dull, orange, bitter(D_R_B_)= 3/4x3/4x3/4 =27/64 glossy, orange, bitter(ddR_B_)= 1/4x3/4x3/4= 9/64 dull, cream, bitter(D_rrB_)= 3/4x1/4x3/4= 9/64 dull, orange, non-bitter (D_R_bb)= 3/4x3/4x1/4=9/64 glossy, cream, bitter(ddrrB_)= 1/4x1/4x3/4= 3/64 dull, cream, non-bitter(D_rrbb)= 3/4x1/4x1/4= 3/64 glossy, orange, non-bitter(ddR_bb)= 3/4x3/4x1/4=3/64 glossy, cream, non-bitter(ddrrbb)= 1/4x1/4x1/4= 1/64 b. DdRrBb x ddrrbb (F1 plant x glossy, cream, bitter) will produce dull, orange, bitter (DdRrBb)= 2/4x2/4x2/4=8/64 or 1/8 glossy, orange, bitter (ddRrBb)= 2/4x2/4x2/4=8/64 or 1/8 dull, cream, bitter (DdrrBb)= 2/4x2/4x2/4=8/64 or 1/8 Dull, orange, non-bitter (DdRrbb)= 2/4x2/4x2/4=8/64 or 1/8 glossy, cream, bitter (ddrrBb)= 2/4x2/4x2/4=8/64 or 1/8 dull, cream, non-bitter (Ddrrbb)= 2/4x2/4x2/4=8/64 or 1/8 glossy, orange, non-bitter (ddRrbb)= 2/4x2/4x2/4=8/64 or 1/8 glossy, cream, non-bitter (ddrrbb)= 2/4x2/4x2/4=8/64 or 1/8

What is the probability of rolling two six-sided dice and obtaining the following numbers: a. 2 and 3 b. 6 and 6 c. at least one 6 d. Two of the same number (two 1s, two 2s, or two 3s, etc.) e. an even number on both dice f. an even number on at least one dice

a. 1/18 b. 1/36 c. 11/36 d. 1/6 e. 1/4 f. 3/4

The following two genotypes are crossed: AaBbCcddEe × AabbCcDdEe. What will be the proportion of the genotype AaBbCcDdEe among the progeny of this cross? a. AaBbCcDdEe b. AabbCcddee c. aabbccddee d. AABBCCDDEE

a. 1/2 (Aa) x 1/2 (Bb) x 1/2 (Cc) x 1/2 (Dd) x 1/2 (Ee) =1/32 b. 1/2 (Aa) x 1/2 (bb) x 1/2 (Cc) x 1/2 (dd) x 1/4 (ee) =1/64 c. 1/2 (aa) x 1/2 (bb) x 1/4 (cc) x 1/2 (dd) x 1/4 (ee) =1/256 d. No offspring have this genotype. The AaBbCcddEe parent cannot contribute a D allele, and the AabbCcDdEe parent cannot contribute a B allele. Therefore their offspring cannot be homozygous for the BB and DD gene loci

A cell in G1 of interphase has 12 chromosomes. How many chromosomes and DNA molecules will be found per cell when this original cell progresses to the following stages? a. G2 of interphase b. Metaphase I of meiosis c. Prophase of mitosis d. Anaphase I of meiosis e. Anaphase II of meiosis f. Prophase II of meiosis g. After cytokinesis following mitosis h. After cytokinesis following meiosis

a. 12 chromosomes and 24 DNA molecules b. 12 chromosomes and 24 DNA molecules c. 12 chromosomes and 24 DNA molecules d. 12 chromosomes and 24 DNA molecules e. 12 chromosomes and 12 DNA molecules f. 6 chromosomes and 12 DNA molecules g. 12 chromosomes and 12 DNA molecules h. 6 chromosomes and 6 DNA molecules

All of the following cells, showing in various stages of mitosis and meiosis, come from the same rare species of plant. a. What is the diploid number of chromosomes in this plant? b. Give the names of each stage of mitosis or meiosis. c. Give the number of chromosomes and number of DNA molecules per cell present at each stage.

a. 6 chromosome b. First cell: anaphase I of meiosis; second cell: anaphase of meiosis; third cell: anaphase II of meiosis c. first cell: 6 chromosomes and 12 DNA molecules; second cell: 12 chromosomes and 12 DNA molecules; third cell: 6 chromosomes and 6 DNA molecules

In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The genes for these two characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The F1 are intercrossed to produce the F2. a. What are the phenotypic ratios in the F2? b. If an F1 plant is backcrossed with the bitter, yellow-spotted parent, what phenotypes and proportions are expected in the offspring? c. In an F1 plant is backcrossed with the sweet, non-spotted parent, what phenotypes and proportions are expected in the offspring?

a. 9:3:3:1 bitter fruit/yellow spots (1/9=BBSS, 2/9=BBSs, 2/9=BbSS, 4/9=BbSs) Bitter fruit/no spots (1/3=BBss, 2/3=Bbss) Sweet fruit/yellow spots (1/3=bbSS, 2/3=bbSs) sweet fruit/no spots (1/1=bbss) b. BBSS x BbSs (bitter, yellow-spotted parent x F1 plant) would produce all bitter, yellow-spotted progeny (4/16 BBSS, 4/16 BBSs, 4/16 BbSS, 4/16 BbSs) c. bbss x BbSs (sweet,y non -spotted parent x F1 plant) would produce 4/16 bitter, yellow spotted (BbSs), 4/16 bitter, non-spotted (Bbss), 4/16 sweet, yellow-spotted (bbSs), 4/16 sweet, non-spotted (bbss)

White (w) coat color in guinea pigs is recessive to black (W). In 1909, W. E. Castle and J. C. Phillips transplanted an ovary from a black guinea pig into a white female whose ovaries had been removed. They then mated this white female with a white male. All the offspring from the mating were black in color a. Explain the results of this cross. b. Give the phenotype of the offspring of this cross. c. What, if anything, does this experiment indicate about the validity of the pangenesis and the germ-plasm theories discussed in chapter 1.

a. Although the white female gave birth to the offspring, her eggs were produced by the ovary from the black female guinea pig. The transplanted ovary produced only eggs containing the allele for black coat color. b. The white male guinea pig contributed a w allele, while the white female guinea pig contributed the W allele from the transplants ovary. The offspring are thus Ww. c. The production of black offspring suggests that the allele for black coat color was passed to the offspring from the transplanted ovary, in agreement with the germ-plasm theory. If pangenesis were correct, the whit coat allele from the female's body would have traveled to the transplanted ovary and then into the gametes. The absence of any whit offspring indicates that pangenesis did not occur.

For each of the following pedigrees, give the most likely mode of inheritance, assuming the trait is rare. Carefully explain your reasoning. (Page 72 #38)

a. Autosomal dominant mode of inheritance because a single dominant allele is enough for expression of a trait b. Autosomal recessive mode of inheritance because of the skipping of generation from I to IV.

In this chapter, we considered Joan Barry's paternity suit against Charlie Chaplin and how, on the basis of blood types, Chaplin could not have been the father of her child. a. What blood types are possible for the father of Barry's child? b. If Chaplin had possessed one of these blood types, would that prove that he fathered Barry's child?

a. B or AB b. No. Many other men have these blood types. The results would have meant only that Chaplin could not be eliminated as a possible father of the child

Tatuo Aida investigated the genetic basis of color variation in the medaka (Aplocheilus Latipes), a small fish found in Japan. Aida found that genes at two loci (B, b and R, r) determine the color of the fish: fish with a dominant allele at both loci (B_R_) are brown, fish with a dominant allele at the the B locus only (B_rr) are blue, fish with a dominant allele at the R locus only (bbR_) are red, and fish with recessive alleles at both loci (bbrr) are white. Aida crossed a homozygous brown fish with a homozygous white fish. He then backcrossed the F1 with the homozygous white parent and obtained 228 brown fish, 230 blue fish, 237 red fish, and 222 white fish. a. Give the genotypes of the backcross progeny. b. Use a chi-square test to compare the observed numbers of backcross progeny with the number expected. What conclusion can you make from your chi-square results? c. What results would you expect for a cross between a homozygous red fish and a white fish? d. What results would you expect if you crossed a homozygous red fish with a homozygous blue fish and then backcrossed the F1 with a homozygous red parental fish?

a. BbRr x bbrr (F1 x homozygoys white parent) will produce 1/4 brown fish (BbRr), 1/4 blue fish (Bbrr), 1/4 red fish (bbRr), 1/4 white fish (bbrr) b. c. bbRR x bbrr (homozygous red x homozygous white) will produce all red fish (bbRr) d. bbRR x BbRr (Homozygous red x F1) would produce 1/4 brown (BbRR), 1/4 brown (BbRr), 1/4 red (bbRR), 1/4 red (bbRr)

What is the probability of rolling one six-sided die and obtaining the following numbers: a. 2 b. 1 or 2 c. an even number d. any number but 6

a. Because 2 is found on only one side of a six-sided die, there is a 1/6 chance of rolling a 2. b. The probability of rolling a 1 on a six-sided die is 1/6, and the probability of rolling a 2 on a six sided die is 1/6. Using the addition rule of probability to determine the probability of rolling a 1 or a 2: 1/6+1/6 = 2/6 or 1/3 c. A single die contains three even numbers (2, 4, 6). The probability or rolling and one of these numbers is 1/6. apply the addition rule: 1/6+1/6+1/6 = 3/6 or 1/2 d. The number 6 is found on one side of a six-sided die. The probability or rolling a 6 is therefore 1/6. The probability or rolling any number but six is (1-1/6) = 5/6

In Drosophilia, yellow body color is due to an X-linked gene that is recessive to the gene for gray body color. a. A homozygous gray female is crossed with a yellow male. The F1 are intercrossed to produce F2. Give the genotypes and phenotypes, along with the expected proportions, of the F1 and F2 progeny. b. A yellow female is crossed with a gray male. The F1 are intercrossed to produce F2. Give the genotypes and phenotypes, along with the expected proportions, of the F1 and F2 progeny.

a. F1: 1/2 X+Y (gray males), 1/2 X+X^y (gray females) F2: 1/4 X+Y (gray males, 1/4 X^yY (yellow males), 1/4 X+X^y (gray females), 1/4 X+X+ (gray females) b. F1: 1/2 X^yY (yellow males), 1/2 X+X^y (yellow females) F2: 1/4 X+Y (gray males, 1/4 X^yY (yellow males), 1/4 X+X^y (gray females), 1/4 X^yX^y (gray females)

Miniature wings in Drosophila melanogaster result from an X-linked allele (X^m) that is recessive to the allele for long wings (X+). Sepia eyes are produced by an autosomal allele (s) that is recessive to an allele for red eyes (s+). a. A female fly that has miniature wings and sepia eyes is crossed with a male that has normal wings and is homozygous for red eyes. The F1 are intercrossed to produce the F2. Give the phenotypes and their proportions expected in the F1 and F2 flies from this cross. b. A female fly that is homozygous for normal wings and has sepia eyes is crossed with a male that has miniature wings and is homozygous for red eyes. The F1 are intercrossed to produce the F2. Give the phenotypes and their proportions expected in the F1 and F2 flies from this cross

a. F1: all males have miniature wings and red eyes (XmY,s+s) and all females have long wings and red eyes (X+Xm,s+s) F2: 3/16 male with normal wings and red eyes, 1/16 male with normal wings and sepia eyes, 3/16 male with miniature wings and red eyes, 3/16 male with miniature wings and sepia eyes, 3/16 female with normal wings and red eyes, 1/16 female with normal wings and sepia eyes, 3/16 female with miniature wings and red eyes, 1/16 female with miniature wings and sepia eyes b. F1: all females have long wings and red eyes (X+Xm,s+s) and all males have long wings and red eyes (X+Y,s+s) F2: 3/16 male with long wings and red eyes, 1/16 male with long wings and sepia eyes, 3/16 male with miniature wings and red eyes, 1/16 male with miniature wings and sepia eyes, 3/8 female with long wings and red eyes, 1/8 female with long wings and sepia eyes.

In cats, blood type A results from an allele I^A, which is dominant over an allele i^B. (There is no blood type O as there is in humans). The blood types of males and female cats that were mated and the blood types of their kittens are presented in the following table. Give the most likely genotype for the parents of each litter. MALE P. FEMALE P. KITTENS a. A B A = 4 kittens / B = 3 kittens b. B B B = 6 kittens c. B A A = 8kittens d. A A A = 7 kittens / B = 2 kittens e. A A A = 10 kittens f. A B A = 4 kittens / B = 1 kitten

a. Female parent is iBiB / male parent is IAIA b. Both parents are iBiB c. Male parent is iBiB / female parent is IAIA or possibly IAiB, but a heterozygous female in this mating is unlikely to have produced eight blood-type-A kittens owing to chance alone d. both parents are IAiB e. Either both parents are IAIA or one parents is IAIA and the other is IAiB f. Female parent is iBiB / Male parent is IAiB

The amount of DNA per cell of a particular species is measured in cells found at various stages of meiosis, and the following amounts are obtained: ______3.7 pg ______7.3 pg ______14.6 pg Match the amounts of DNA above with the corresponding stages of meiosis (a through f below). You may use more than on stage for each amount of DNA. a. G1 b. Prophase I c. G2 d. Following Telophase II and cytokinesis e. Anaphase I f. Metaphase II

a. G1 = 7.3 pg b. Prophase I = 14.6 pg c. G2 = 14.6 pg d. Following Telophase II and cytokinesis = 3.7 pg e. Anaphase I = 14.6 pg f. Metaphase II = 7.3 pg EXPLINATION: *The amount of DNA in the cell will be doubles after the completion of the S phase prior to cytokinesis in either mitosis or meiosis I. *At the completion of cytokinesis following meiosis II, the amount of DNA would be halved. a. G1 occurs prior to the S phase and the doubling of the amount of DNA b. The amount of DNA double in the S phase, so during prophase I of meiosis, the amount of DNA in the cell is twice the amount G1 c. G2 takes place directly after the completion of the S phase, so the amount of DNA is twice the amount of G1. d. following cytokinesis associated with meiosis II, each daughter cell will only contain one-half the amount of DNA of a cell found in G1. e. During anaphase I of meiosis, the amount of DNA in the cell is twice the amount in G1. f. In metaphase II of meiosis, the amount of DNA in each cell is the same as G1 because DNA doubled in the S phase but then was reduced by half in the first meiotic division.

In guinea pigs, the allele for black fur (B) is dominant over the allele for brown fur (b). A black guinea pig is crossed with a brown guinea pig, producing five F1 black guinea pigs and six F1 brown guinea pigs. a. How many copies of the black allele (B) will be present in each cell of an F1 black guinea pig at the following stages: G1, G2, Metaphase of mitosis, Metaphase I of meiosis, Metaphase II of meiosis, and after the second cytokinesis following meiosis? Assume that no crossing over takes place. b. How many copies of the brown allele (b) will be present in each cell of an F1 brown guinea pig at the same stages as those listed in part a? assume that no crossing over takes place.

a. In the F1 black guinea pigs (Bb) only one of the two homologous chromosomes possesses the black allele, and so the number of copies present at each stage are as follows: G1 = one black allele, G2 = two black alleles, Metaphase of mitosis = two black alleles, Metaphase I of meiosis = two black alleles, Metaphase II of meiosis, and after the second cytokinesis following meiosis = one black allele but only half of the cells produced by meiosis b. In the F1 brown guinea pigs (bb) both homologs possess the brown allele and so the number of alleles present at each stage are as follows: G1 = two brown alleles, G2 = four brown alleles, Metaphase of mitosis = four brown alleles, Metaphase I of meiosis = four brown alleles, Metaphase II of meiosis = two brown alleles, and after the second cytokinesis following meiosis = one brown allele

A cell has eight chromosomes in G1 of Interphase. Draw a picture of this cell with its chromosomes at the following stages. Indicate how many DNA molecules are present at each stage. a. Metaphase of mitosis b. Anaphase of mitosis c. Anaphase II of meiosis

a. Metaphase of mitosis = 16 DNA molecules b. Anaphase of mitosis = 16 DNA molecules c. Anaphase II of meiosis = 8 DNA molecules

In cucumbers, orange fruit color (R) is dominant to cream fruit color (r). A cucumber plant homozygous for orange fruits is crossed with a plant with cream fruit. The F1 plants are intercrossed to produce an F2 generation. a. Give the genotypes and phenotypes of the parents, the F1, and the F2. b. Give the genotypes and phenotypes of the offspring of a backcross between the F1 and the orange-fruited parent. c. give the genotypes and phenotypes of a backcross between the F1 and the cream-fruited parent.

a. Parents: Homozygous orange fruit color (RR) and homozygous cream fruit color (rr) F1: all will be orange fruit color (Rr) F2: 1/4 homozygous orange, 2/4 heterozygous orange, 1/4 homozygous cream b. RR x Rr (Homozygous orange parent x heterozygous orange from F1) would produce 1/2 homozygous orange (RR) and 1/2 heterozygous orange (Rr) c. rr x Rr (Homozygous cream parent x heterozygous orange from F1) would produce 1/2 heterozygous orange (Rr) and 1/2 homozygous cream (rr)

Alkaptornuria is a metabolic disorder in which affected persons produce black urine. Alkaptornuria results from an allele (a) that is recessive to the allele for normal metabolism (A). Sally has normal metabolism, but her brother has alkaptornuria. Sally's father has alkaptonuria and her mother has normal metabolism. a. Give the genotypes of sally, her mother, her father, and her brother. b. If Sally's parents have another child, what is the probability that this child will have alkaptonuria? c. If Sally marries a man with alkaptonuria, what is the probability that their first child will have alkaptonuria?

a. Sally is Aa, Sally's mother is Aa, Sally's father is aa, and Sally's brother is aa b. 50% c. 50%

When a Chinese hamster with white spots is crossed with another hamster that has no spots, approximately 1/2 of the offspring have white spots and 1/2 have no spots. When two hamsters with white spots are crossed, 2/3 of the offspring possess white spots and 1/3 have no spots. a. What is the genetic basis of white spotting in Chinese hamsters? b. How might you go about producing Chinese hamsters that breed true for white spotting?

a. The 2:1 ratio in the progeny of two spotted hamsters suggests lethality, and the 1:! ratio in the progeny of a spotted hamster and a hamster without spots indicates that spotted is a heterozygous phenotype. If S and s represent alleles at the locus for white spotting, spotted hamsters are Ss and solid-colored hamsters are ss. 1/4 of the zygotes expected from a mating of two spotted hamsters are SS--embryonic lethal---and missing from the progeny, resulting in the 2:1 ratio of spotted to solid progeny. b. Because spotting is a heterozygous phenotype, obtaining Chinese hamsters that breed true for spotting is impossible

Some sweet-pea plants have purple flowers and other plants have white flowers. A homozygous variety of pea that has purple flowers is crossed with a homozygous variety that has white flowers. All the F1 have purple flowers. When these F1 are self- fertilized, the F2 appear in a ratio of 9/16 purple to 7/16 white. a. Give genotypes for the purple and white flowers in these crosses. b. Draw a hypothetical biochemical pathway to explain the production of purple and white flowers in sweet peas.

a. The F2 ratio of 9:7 is a modified dihybrid ratio, indicating two genes interacting. Using A and B as generic gene symbols, we can start with the F1 heterozygotes: F1 AaBb purple self-fertilized F2 9/16 A-B- purple (like F1) 3/16 A-bb white 3/16 aaB- white 1/16 aabb white Now we see that purple requires dominant alleles for both genes, so the purple parent must have been AABB, and the white parent must have been aabb to give all purple F1. b. White precursor 1 white intermediate 2 purple pigment

J. A. Moore investigated the inheritance of spotting patterns in leopard frogs. The pipiens phenotype has the normal spots that give leopard frogs their name. In contrast, the burnsi phenotype lacks spots on its back. Moor carried out the following crosses, producing the progeny indicated. PARENT PHENOTYPES PROGENY PHENOTYPES burnsi x burnsi 39 burnsi / 6 pipiens burnsi x pipiens 23 burnsi / 33 pipiens burnsi x pipiens 196 burnsi / 210 pipiens a. On the basis of these results, what is the most likely mode of inheritance of the burnsi phenotype? b. Give the most likely genotypes of the parent in each cross. (use B for the burnsi allele and B+ for the pipiens allele) c. Use a chi-square test to evaluate the fit of the observed numbers of progeny to the number expected on the basis of your proposed genotypes.

a. The burnsi x burnsi cross produce both burnsi and pipiens offspring , indicating that the parents were heterozygous, with each possessing a burnsi allele and a pipiens allele. This cross indicates that the burnsi allele is dominant over the pipiens allele. The progeny of the burnsi x pipiens cross suggest that each of the crosses was between a homozygous recessive from (pipiens) and a heterozygous dominant frog (burnsi). The results of both crosses are consistent with the burnsi phenotype being dominant to the pipens phenotype. b. Let B represent the burnsi allele and B+ represent the pipiens allele. burnsi (BB+) x burnsi (BB+) burnsi (BB+) x pipiens (B+B+) burnsi (BB+) x pipiens (B+B+) c. For the burnsi (BB+) x burnsi (BB+) cross, we would expect a phenotypic ratio of 3:1 in the offspring. A chi-square test gives a chi-square value of 2.706 with 1 degree or freedom and a probability of between 0.1 and 0.005, indicating that the differences between what we expect and what we observed could have been generated by chance. For the first burnsi (BB+) x pipiens (B+B+) cross, we expect a phenotypic ratio of 1:1. A chi-square test yields X^2=1.78, df=1, P>0.05. For the second burnsi (BB+) x pipiens (B+B+) cross, we expect a phenotypic ration of 1:1. A chi-square test yields x^2=0.46, df=1, P>0.05. Thus, all three crosses are consistent with the prediction that burnsi is dominant over pipens.

Palomino horses have a golden yellow coat, chestnut horses have a brown coat, and cremello horses have a coat that is almost white. A series of crosses between the three different types of horses produced the following offspring: CROSS OFFSPRING palomino x palomino 13 palomino/6 chestnut/5 cremello chestnut x chestnut 16 chestnut cremello x cremello 13 cremello palomino x chestnut 8 palomino / 9 chestnut palomino x cremello 11 palomino / 11 cremello chestnut x cremello 23 palomino a. Explain the inheritance of the palomino, chestnut, and cremello phenotypes in horses. b. Assign symbols for the alleles that determine these phenotypes, and list the genotypes pf all parents and offspring given in the preceding table

a. The results of the crosses indicate that cremello and chestnut are pure-breeding traits (homozygous). Palomino is a hybrid trait (heterozygous) that produces a 2:1:1 ratio when palominos are crosses with each other. The simplest hypothesis consistent with these results is incomplete dominance , with palomino as the phenotype of the heterozygous resulting from chestnuts crossed with cremellos. b. Let C^B=chestnut and C^W=cremello. The parents and offspring of these crosses have the following genotypes: chestnut= C^BC^B, cremello=C^WC^W, Palomino=C^BC^W

J. W. McKay crossed a stock melon plant that produced tan seeds with a plant that produced red seeds and obtained the following results. CROSS: tan ♀ × red ♂ F1: 13 tan seeds F2: 93 tan, 24 red seeds a. Explain the inheritance of tan seeds and red seeds in this plant. b. Assign symbols for the alleles in this cross and give genotypes for all the individual plants.

a. When a cross is made between tan-seed producing plant and red-seed producing plant, F1 generation contains all tan-seed producing progeny. It shows that tan phenotype is dominant to red one. In F2 generation, the ratio is: Tan seed: Red seed 93:24 3.9:1 This ratio is similar but not identical to 3:1 ratio of monohybrid cross. So we can say that F1 parents are heterozygous dominant for tan color. b. Tan color allele= R Red color allele= r tan-seed producing female parent = RR Red seed producing male parent = rr F1 tan seed producing offspring = Rr F2 tan seed producing offspring = RR/Rr F2 red seed producing offspring = rr

How many Barr bodies would you expect to see in a human cell containing the following chromosomes? a. XX b. XY c. XO d. XXY e. XXYY f. XXXY g. XYY h. XXX i. XXXX

a. XX = 1 b. XY = 0 c. XO = 0 d. XXY = 1 e. XXYY = 1 f. XXXY = 2 g. XYY = 0 h. XXX = 2 i. XXXX = 3

In the California poppy, an allele for yellow flowers (C) is dominant over an allele for white flowers (c). At an independently assorting locus, an allele for entire petals (F) is dominant over an allele for fringed petals (f). A plant that is homozygous for yellow and entire petals is crossed with a plant that is white and fringed. A resulting F1 plant is then crossed with a plant that is white and fringed. The following progeny are produced: 54 yellow and entire, 58 yellow and fringed, 53 white and entire, 10 white and fringed. a. Use a chi-square test to compare the observed numbers with those expected for the cross b. What conclusion can you make from the results of the chi-square test? c. Suggest an explication for the results.

a. a heterozygous G1 plant (CcFf) is crossed to a homozygous recessive plant (ccff). We expect a 1:1:1:1 phenotypic ratio in the progeny or 1/4 of each phenotype. A chi-square test yields a chi-square value of 35 with df=3 and p<0.05 b. Th probability that chance produced the differenced between the observed and the expected ratios is very low, indicating that the progeny are not in a 1:1:1:1 ratio c. The number of plants with the ccff genotype is much less than expected. Poppies with the ccff genotype may be less viable than other genotypes.

In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_ P_). A rose comb is produced when at least one dominant gene is present at the first locus and two recessive genes are present at the second locus (genotype R_ pp). A pea comb is produced when two recessive genes are present at the first locus and at least one dominant gene is present at the second (genotype rr P_). If two recessive genes are present at the first and the second locus (rr pp), a single comb is produced. Progeny with what types of combs and in what proportions will result from the following crosses? a. RRPP x rrpp b. RrPp x rrpp c. RrPp x RrPp d. Rrpp x Rrpp e. Rrpp x rrPp f. Rrpp x rrpp

a. all walnut (RrPp) b. 1/4 walnut (RrPp), 1/4 rose (Rrpp), 1/4 pea (rrPp), 1/4 single (rrpp) c. 9/16 walnut (R_P_), 3/16 rose (R_pp), 3/16 pea (rrP_), 1/16 single (rrpp) d. 3/4 rose (R_pp), 1/4 single (rrpp) e. 1/4 walnut (RrPp), 1/4 rose (Rrpp), 1/4 pea (rrPp), 1/4 single (rrpp) f. 1/2 rose (Rrpp), 1/2 single (rrpp)

What will be the phenotypic sex of a human with the following genes or chromosomes or both? a. XY with SRY gene deleted b. XY with SRY gene located on an autosome c. XX with a copy of SRY gene on an autosome d. XO with a copy of the SRY gene on an autosome e. XXY with the SRY gene deleted f. XXYY with one copy of the SRY gene deleted

a. female b. male c. male d. male e. female f. male

In rabbits, an allelic series helps to determine coat color: C (full color), c^ch (chinchilla, gray color), c^h (Himalayan, white with black extremities), and c (albino, all white). The C allele is dominant over all others, c^ch is dominant over c^h and c, c^h is dominant over c, and c is recessive to all the other alleles. This dominance hierchy can be summarized as C > c^ch > c^h > c. The rabbits in the list below are crossed and produce the progeny shown. Give the genotypes of the parents for each cross. PARENT PHENOTYPES OFFSPRING PHENOTYPES a. full color x albino 1/2 full color / 1/2 albino b. Himalayan x albino 1/2 Himalayan / 1/2 albino c. full color x albino 1/2 full color / 1/2 chinchilla d. full color x Himalayan 1/2 full color / 1/4 Himalayan / 1/4 albino e. full color x full color 3/4 full color / 1/4 albino

a. full color x albino (Cc x cc) will produce 1/2 full color (Cc) and 1/2 albino (cc) b. Himalayan x albino (c^hc x cc) will produce 1/2 Himalayan (c^hc) and 1/1 albino (cc) c. full color x albino (Cc^ch x cc) will produce 1/2 full color (Cc) and 1/2 chinchilla (c^chc) d. full color x Himalayan (Cc x c^hc) will produce 1/4 full color (Cc^h), 1/4 full color (Cc), 1/4 himalayan (cc^h), 1/4 albino (cc) e. full color x full color (Cc x Cc) will produce 1/4 full color (CC), 2/4 full color (Cc), 1/4 albino (cc)

Match each of the following terms with its correct definition (parts a through h) phenocopy sex-limited trait pleiotropy genetic maternal effect polygenic trait genomic imprinting penetrance sex-influenced trait a. The percentage of individuals with a particular genotype that express the expected phenotype. b. A trait determined by an autosomal gene that is more easily expressed in one sex. c. A trait determined by an autosomal gene that is expressed in only one sex d. A trait that is determined by an environmental effect and has the same phenotype as a genetically determined trait e. A trait determined by genes at any loci f. The expression of trait affected by the sex of the parent that transmits the gene to the offspring. g. A gene affecting more than one phenotype h. The influence of the genotype of the maternal parent on the phenotype of the offspring.

a. penetrance b. sex-influenced trait c. sex-limited trait d. Phenocopy e. polygenic trait f. genomic imprinting g. pleiotrophy h. genetic maternal effect

In the early 1900s, Lucien Cuenot, a French scientist working at the University of Nancy, studied the genetic basis of yellow coat color in mice. He carried out a number of crosses between two yellow mice and obtained what he thought was a 3:1 ratio of yellow to gray mice in the progeny. The following table gives Cuento's actual results, along with the results of much larger series of crosses carried out by William Castle and Clarence Little. INVESTIGATORS YELLOW NONYELLOW TOTAL Cuenot 263 100 363 Castle and Little 800 435 1235 Both Combined 1063 535 1598 a. Using a chi-square test, determine whether Cuenot's results are significantly different from the 3:1 ratio that he thought he observed. Are they different from a 2:1 ratio? b. Determine whether Castle and Little's results are significantly different from the 3:1 and a 2:1 ratio? c. Combine the Castle and Little results with those of Cuenot and determine whether they are significantly different from the 3:1 and 2:1 ratio. d. Offer and explanation for the different ratios obtained by Cuenot and by Castle and Little

a. reject 3:1, accept 2:! b. reject 3:1, accept 2:1 c. reject 3:1, accept 2:1

Joe has classic hemophilia, an X-linked recessive disease. Could Joe have inherited the gene for this disease from the following persons YES NO a) his mother's mother _______ _______ b) his mother's father _______ _______ c) his father's mother _______ _______ d) his father's father _______ _______

a. yes b. yes c. no d. no

Which of the following statements is an example of a phenocopy? Explain your reasoning? a. Phenylketonuria results from a recessive mutation that causes lights skin as well as intellectual disability. b. Human height is influenced by genes at many different loci. c. Dwarf plants and mottled leaves in tomatoes are caused by genes that are linked. d. Vestigial wings in Dorsophila are produced by a recessive mutation. This trait is also produced by high temperature during development. e. Intelligence in humans is influenced by both genetic and environmental factors

d because it is influenced by environment a is due to genetic basis; hence not phenocopy b is an example of a quantitative trait c is a linked gene and is not an example of phenocopy e is influenced by genotype it is not an example of phenocopy

How do incomplete dominance and codominance differ?

incomplete dominance means the heterozygote expresses an intermediate phenotype between the homozygotes co-dominance get expressed when the heterozygote expresses both the homozygous phenotypes simultaneously.

What is incomplete penetrance and what causes it?

incomplete penetrance occurs in individuals where the genotype does not produce the expected phenotype because of the interaction environmental factor and genetic background of the individual.

A certain species has three pairs of chromosomes: one acrocentric pair and two metacentric pairs. Draw a cell of this species as it would appear in the following stages of meiosis: a. Metaphase I b. Anaphase I c. Metaphase II d. Anaphase II

metacentric (upper and lower halves equal) Submetacentric (upper half smaller than lower half) Acrocentric (lower half longer and upper very small) Telocentric (lower half normal and upper part almost not even there) a. all homologous pairs lined up on metaphase plate anchored by microtubules connected to centrosome at each pole b. homologous pairs separated and pulled by spindle fibers to the poles c. separated chromosomes from metaphase I lined up on metaphase plate d. chromatids separate and are pulled towards poles by spindle fibers

A certain species has three pairs of chromosomes: an acrocentric pair, a metacentric pair, and a submetacentric pair. Draw a cell of this species as it would appear in metaphase of mitosis.

metacentric (upper and lower halves equal) Submetacentric (upper half smaller than lower half) Acrocentric (lower half longer and upper very small) Telocentric (lower half normal and upper part almost not even there) because it it metaphase, all individual chromosomes are lined up on the metaphase plate held together by kinetochore microtubules that are anchored to a centrosomes at each pole