Biology Test 3 Study Questions

5′-TGCAAT-3′

Given a DNA molecule with the sequence of bases 5′-ATTGCA-3′, what would be the sequence of the complementary strand? (It will be helpful to draw the DNA molecule when answering the question.)

messenger RNA (mRNA)

carries information specifying amino acid sequences of polypeptide from DNA to ribosomes

is unchanged by the reactants or products

As a result of its involvement in a chemical reaction, an enzyme

DNA ligase

A biochemist isolates, purifies, and combines in a test tube a variety of molecules needed for DNA replication. When she adds some DNA to the mixture, replication occurs, but each DNA molecule consists of a normal strand paired with numerous segments of DNA a few hundred nucleotides long. What has she probably left out of the mixture?

Each has half the chromosomes and the same amount of DNA of the original cell

A cell goes through S phase, meiosis I, and cytokinesis. As the daughter cells are beginning meiosis II, which of the following is an appropriate description of the contents of each cell?

This organism has the life cycle shown in figure 13.6c. Therefore, it must be a fungus or a protist, perhaps an alga.

A certain eukaryote lives as a unicellular organism, but during environmental stress it produces gametes. The gametes fuse, and the resulting zygote undergoes meiosis

39; 39; 78

A chicken has 78 chromosomes in its somatic cells. How many chromosomes did the chicken inherit from each parent? How many chromosomes are in each of the chicken's gametes? How many chromosomes will be in each somatic cell of the chicken's offspring?

She should clone it. Crossbreeding it with another plant would generate offspring that have additional variation, which she no longer desires now that she has obtained her ideal orchid.

A horticulturalist breeds orchids, trying to obtain a plant with a unique combination of desirable traits. After many years, she finally succeeds. To produce more plants like this one, should she crossbreed it with another plant or clone it? Why?

a sperm

A human cell containing 22 autosomes and a Y chromosome is

A kinetochore connects the spindle ( a motor; note that it has motor proteins) to a chromosome (the cargo it will move)

A kinetochore has been compared to a coupling device that connects a motor to the cargo that it moves. Explain.

1/2

A man has six fingers on each hand and six toes on each foot. His wife and their daughter have the normal number of digits. Remember that extra digits is a dominant trait. What fraction of this couple's children would be expected to have extra digits?

1/2; 1/16

A man with hemophilia (a recessive, sex-linked condition) has a daughter without the condition. She marries a man who does not have hemophilia. What is the probability that their daughter will have hemophilia? Their son? If they have four sons, what is the probability that all will be affected?

Man IAi; woman IBi; child ii. Genotypes for future children are predicted to be 14/IAIB,14/IAi,14/IBi,14/ii.

A man with type A blood marries a woman with type B blood. Their child has type O blood. What are the genotypes of these three individuals? What genotypes, and in what frequencies, would you expect in future offspring from this marriage?

Parents: Ii x ii Genotypic ratio= 1 Ii : 1 ii (2:2 is equivalent) Phenotypic ratio= 1 inflated : 1 constricted (2:2 is equivalent)

A pea plant heterozygous for inflated pods (Ii) is crossed with a plant homozygous for constricted pods (ii). Draw a Punnett square for this cross to predict genotypic and phenotypic ratios. Assume that pollen comes from the ii plant.

Between T and A, 12%; between A and S, 5%

A planet is inhabited by creatures that reproduce with the same hereditary patterns seen in humans. Three phenotypic characters are height (T = tall, t = dwarf), head appendages (A = antennae, a = no antennae), and nose morphology (S = upturned snout, s = downturned snout). Since the creatures are not "intelligent," Earth scientists are able to do some controlled breeding experiments using various heterozygotes in testcrosses. For tall heterozygotes with antennae, the offspring are tall antennae, 46; dwarf antennae, 7; dwarf no antennae, 42; tall no antennae, 5. For heterozygotes with antennae and an upturned snout, the offspring are antennae upturned snout, 47; antennae downturned snout, 2; no antennae downturned snout, 48; no antennae upturned snout, 3. Calculate the recombination frequencies for both experiments.

The black and white alleles are incompletely dominant, with heterozygotes being gray in color. A cross between a gray rooster and a black hen should yield approximately equal numbers of gray and black offspring.

A rooster with gray feathers and a hen of the same phenotype produce 15 gray, 6 black, and 8 white chicks. What is the simplest explanation for the inheritance of these colors in chickens? What phenotypes would you expect in the offspring of a cross between a gray rooster and a black hen?

Arg or (R) 0 Glue or (E) - Pro or (P) - Arg or (R)

A segment in the middle of an mRNA has the sequence 5′-AGAGAACCGCGA-3′. Using the codon table, translate this sequence, assuming the first three nucleotides are a codon.

HT

A 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 for these two genes is possible in a gamete from this organism?

Because the gene for this eye color character is located on the X chromosome, all female offspring will be red-eyed and heterozygous (Xw+Xw); all male offspring will inherit a Y chromosome from the father and be white-eyed (XWY). (Another way to say this is that 1/2 the offspring will be red-eyed heterozygous [carrier] females, and 1/2 will be white-eyed males.)

A white-eyed female Drosophila is mated with a red-eyed (wild-type) male, the reciprocal cross of the one shown in Figure 15.4. What phenotypes and genotypes do you predict for the offspring from this cross?

17%; yes, it is consistent. In Figure 15.9, the recombination frequency was also 17%. (You'd expect this to be the case since these are the very same two genes, and their distance from each other wouldn't change from one experiment to another.)

A wild-type fruit fly (heterozygous for gray body color and normal wings) is mated with a black fly with vestigial wings. The offspring have the following phenotypic distribution: wild-type, 778; black vestigial, 785; black normal, 158; gray vestigial, 162. What is the recombination frequency between these genes for body color and wing size? Is this consistent with the results of the experiment in Figure 15.9?

6%; wild-type heterozygous for normal wings and red eyes×recessive homozygous for vestigial wings and purple eyes

A wild-type fruit fly (heterozygous for gray body color and red eyes) is mated with a black fruit fly with purple eyes. The offspring are wild-type, 721; black purple, 751; gray purple, 49; black red, 45. What is the recombination frequency between these genes for body color and eye color? Using information from problem 3, what fruit flies (genotypes and phenotypes) would you mate to determine the order of the body color, wing size, and eye color genes on the chromosome?

In meiosis, a combined 14-21 chromosome will behave as one chromosome. If a gamete receives the combined 14-21 chromosome and a normal copy of chromosome 21, trisomy 21 will result when this gamete combines with a normal gamete (with its own chromosome 21) during fertilization.

About 5% of individuals with Down syndrome have a chromosomal translocation in which a third copy of chromosome 21 is attached to chromosome 14. If this translocation occurred in a parent's gonad, how could it lead to Down syndrome in a child?

If crossing over did not occur, the two homologs would not be associated in any way; each sister chromatid would be either all maternal or all paternal and would only be attached to its sister chromatid, not to a nonsister chromatid. This might result in incorrect arrangement of homologs during metaphase I and, ultimately, in formation of gametes with an abnormal number of chromosomes.

After the synaptonemal disappears, how would any pair of homologous chromosomes be associated if crossing over did not occur? What affect might this have on gamete formation?

Like histones, the E. coli proteins would be expected to contain many basic (positively charged) amino acids, such as lysine and arginine, which can form weak bonds with the negatively charged phosphate groups on the sugar-phosphate backbone of the DNA molecule.

Although the proteins that cause the E. coli chromosome to coil are not histones, what property would you expect them to share with histones, given their ability to bind to DNA

Fifty percent of the offspring will show phenotypes resulting from crossovers. These results would be the same as those from a cross where A and B were not on the same chromosome, and you would interpret the results to mean that the genes are unlinked. (Further crosses involving other genes on the same chromosome would reveal the genetic linkage and map distances.)

Assume that genes A and B are on the same chromosome and are 50 map units apart. An animal heterozygous at both loci is crossed with one that is homozygous recessive at both loci. What percentage of the offspring will show recombinant phenotypes resulting from crossovers? Without knowing these genes are on the same chromosome, how would you interpret the results of this cross?

(a) For each pair of genes, you had to generate an F1 dihybrid fly; let's use the A and B genes as an example. You obtained homozygous parental flies, either the first with dominant alleles of the two genes (AABB) and the second with recessive alleles (aabb), or the first with dominant alleles of gene A and recessive alleles of gene B (AAbb) and the second with recessive alleles of gene A and dominant alleles of gene B (aaBB). Breeding either of these pairs of P generation flies gave you an F1 dihybrid, which you then testcrossed with a doubly homozygous recessive fly (aabb). You classed the offspring as parental or recombinant, based on the genotypes of the P generation parents (either of the two pairs described above). You added up the number of recombinant types and then divided by the total number of offspring. This gave you the recombination percentage (in this case, 8%), which you can translate into map units (8 map units) to construct your map.

Assume you are mapping genes A, B, C, and D in Drosophila. You know that these genes are linked on the same chromosome, and you determine the recombination frequencies between each pair of genes to be as follows: A-B, 8%; A-C, 28%; A-D, 25%; B-C, 20%; B-D, 33%. Describe how you determined the recombination frequency for each pair of genes.

Because bananas are triploid, homologous pairs cannot line up during meiosis. Therefore, it is not possible to generate gametes that can fuse to produce a zygote with the triploid number of chromosomes.

Banana plants, which are triploid, are seedless and therefore sterile. Propose a possible explanation.

Both the I A and I B alleles are dominant to the i allele because the i allele results in no attached carbohydrate. The IA and IB alleles are codominant; both are expressed in the phenotype of I A I B heterozygotes, who have type AB blood.

Based on the surface carbohydrate phenotype in (b), what are the dominance relationships among the alleles?

1 9 (Since cystic fibrosis is caused by a recessive allele, Beth and Tom's siblings who have CF must be homozygous recessive. Therefore, each parent must be a carrier of the recessive allele. Since neither Beth nor Tom has CF, this means they each have a 2 3 chance of being a carrier. If they are both carriers, there is a 1 4 chance that they will have a child with CF. 2 3 × 2 3 × 1 4 = 1 9 ); virtually 0 (Both Beth and Tom would have to be carriers to produce a child with the disease, unless a very rare mutation (change) occurred in the DNA of cells making eggs or sperm in a non-carrier that resulted in the CF allele.)

Beth and Tom each have a sibling with cystic fibrosis, but neither Beth nor Tom nor any of their parents have the disease. Calculate the probability that if this couple has a child, the child will have cystic fibrosis. What would be the probability if a test revealed that Tom is a carrier but Beth is not? Explain your answers.

The chance of the fourth child having cystic fibrosis is 14/ , as it was for each of the other children, because each birth is an independent event. We already know both parents are carriers, so whether their first three children are carriers or not has no bearing on the probability that their next child will have the disease. The parents' genotypes provide the only relevant information.

Both members of a couple know that they are carriers of the cystic fibrosis allele. None of their three children has cystic fibrosis, but any one of them might be a carrier. The couple would like to have a fourth child but are worried that he or she would very likely have the disease, since the first three do not. What would you tell the couple? Would it remove some uncertainty from their prediction if they could find out whether the three children are carriers?

The mRNA sequence (5′-UGGUUUGGCUCA-3′) is the same as the nontemplate DNA strand sequence (5′-TGGTTTGGCTCA-3′), except there is a U in the mRNA wherever there is a T in the DNA. The nontemplate strand is probably used to represent a DNA sequence because it represents the mRNA sequence, containing codons. (This is why it's called the coding strand.)

By convention, the nontemplate strand, also called the coding strand, is used to represent a DNA sequence. Write the sequence of the mRNA strand and the nontemplate strand—in both cases reading from 5′ to 3′—and compare them. Why do you think this convention was adopted? (HINT: Why is this called the coding strand?)

G1 phase

Cell A has half as much DNA as cells B, C, and D in a mitotically active tissue. Cell A is most likely in

On both the leading and lagging strands, DNA polymerase adds onto the 3' end of an RNA primer synthesized by primase, synthesizing DNA in the 5′→3′ direction. Because the parental strands are antiparallel, however, only on the leading strand does synthesis proceed continuously into the replication fork. The lagging strand is synthesized bit by bit in the direction away from the fork as a series of shorter Okazaki fragments, which are later joined together by DNA ligase. Each fragment is initiated by synthesis of an RNA primer by primase as soon as a given stretch of single-stranded template strand is opened up. Although both strands are synthesized at the same rate, synthesis of the lagging strand is delayed because initiation of each fragment begins only when sufficient template strand is available.

Compare DNA replication on the leading and lagging strands, including both similarities and differences.

Following mitosis, cytokinesis results in two genetically identical daughter cells in both plant cells and animal cells. However, the mechanism of dividing the cytoplasm is different in animals and plants. In an animal cell, cytokinesis occurs by cleavage, which divides the parent cell in two with a contractile ring of actin filaments. In a plant cell, a cell plate forms in the middle of the cell and grows until its membrane fuses with the plasma membrane of the parent cell. A new cell wall grows inside the cell plate, thus eventually between the two new cells.

Compare cytokinesis and plant cells and animal cells.

The chromosomes are similar in that each is composed of two sister chromatids, and the individual chromosomes are positioned similarly at the metaphase plate. The chromosomes differ in that in a mitotically dividing cell, sister chromatids of each chromosome are genetically identical, but in a meiotically dividing cell, sister chromatids are genetically distinct because of crossing over in meiosis I. Moreover, the chromosomes in metaphase of mitosis can be a diploid set or a haploid set, but the chromosomes in metaphase of meiosis II always consist of a haploid set.

Compare the chromosomes in a cell at metaphase of mitosis with those in a cell at metaphase 2.

Animals and plants both reproduce sexually, alternating meiosis with fertilization. Both have haploid gametes that unite to form a diploid zygote, which then goes on to divide mitotically, forming a diploid multicellular organism. In animals, haploid cells become gametes and don't undergo mitosis, while in plants, the haploid cells resulting from meiosis undergo mitosis to form a haploid multicellular organism, the gametophyte. This organism then goes on to generate haploid gametes. (In plants such as trees, the gametophyte is quite reduced in size and not obvious to the casual observer.)

Compare the life cycles of animals and plants, mentioning their similarities and differences.

During eukaryotic cell division, tubulin is involved in spindle formation and chromosome movement, while actin functions during cytokinesis. In bacterial binary fission, its the opposite: Actin-like molecules are thought to move the daughter bacterial chromosomes to opposite ends of the cell, and tubulin-like molecules are thought to act in daughter cell separation.

Compare the roles of tubulin and actin during eukaryotic cell division with the roles tubulin-like and actin-like proteins during bacterial binary fission.

The processes are similar in that polymerases form polynucleotides complementary to an antiparallel DNA template strand. In replication, however, both strands act as templates, whereas in transcription, only one DNA strand acts as a template.

Compare the use of a template strand during transcription and replication.

In a disorder caused by a dominant allele, there is no such thing as a "carrier," since those with the allele have the disorder. Because the allele is dominant, the females lose any "advantage" in having two X chromosomes, since one disorder-associated allele is sufficient to result in the disorder. All fathers who have the dominant allele will pass it along to all their daughters, who will also have the disorder. A mother who has the allele (and thus the disorder) will pass it to half of her sons and half of her daughters.

Consider what you learned about dominant and recessive alleles in Concept 14.1. If a disorder were caused by a dominant X-linked allele, how would the inheritance pattern differ from what we see for recessive X-linked disorders?

A signal peptide on the leading end of the polypeptide being synthesized is recognized by a signal-recognition particle that brings the ribosome to the ER membrane. There the ribosome attaches and continues to synthesize the polypeptide, depositing it in the ER lumen.

Describe how a polypeptide to be secreted reaches the endomembrane system.

In the context of the ribosome, tRNAs function as translators between the nucleotide-based language of mRNA and the amino-acid-based language of polypeptides. A tRNA carries a specific amino acid, and the anticodon on the tRNA is complementary to the codon on the mRNA that codes for that amino acid. In the ribosome, the tRNA binds to the A site. Then the polypeptide being synthesized (currently on the tRNA in the P site) is joined to the new amino acid, which becomes the new (C-terminal) end of the polypeptide. Next, the tRNA in the A site moves to the P site. After the polypeptide is transferred to the new tRNA, thus adding the new amino acid, the now empty tRNA moves from the P site to the E site, where it exits the ribosome.

Describe how tRNAs function in the context of the ribosome in building a polypeptide.

Much of the chromatin in an interphase nucleus is present as the 30-nm fiber, with some in the form of the 10-nm fiber and some as looped domains of the 30-nm fiber. (These different levels of chromatin packing may reflect differences in gene expression occurring in these regions.) Also, a small percentage of the chromatin, such as that at the centromeres and telomeres, is highly condensed heterochromatin.

Describe the levels of chromatin packing you'd expect to see in an interphase nucleus.

A gene contains genetic information in the form of a nucleotide sequence. The gene is first transcribed into an RNA molecule, and a messenger RNA molecule is ultimately translated into a polypeptide. The polypeptide makes up part or all of a protein, which performs a function in the cell and contributes to the phenotype of the organism.

Describe the process of gene expression, by which a gene affects the phenotype of an organism.

A nucleosome is made up of eight histone proteins, two each of four different types, around which DNA is wound. Linker DNA runs from one nucleosome to the next.

Describe the structure of a nucleosome, the basic unit of DNA packing in eukaryotic cells.

The DNA of a eukaryotic cell is packaged into structures called chromosomes. Each chromosomes is a long molecule of DNA, which carries hundreds to thousands of genes, with associated proteins that maintain chromosome structure and help control gene activity. This DNA-protien complex is called chromatin. The chromatin of each chromosome is long and thin when the cell is not dividing. Prior to cell division, each chromosome is duplicated, and the resulting sister chromatids are are attached to each other by proteins at the centromeres and, for many species, all along their lengths (a phenomenon called sister chromatid cohesion.)

Differentiate between chromosomes, chromatin, and chromatids

Because of wobble, the tRNA could bind to either 5'-GCA-3' or 5'-GCG-3', both of which code for alanine (Ala). Alanine would be attached to the tRNA (see diagram, upper right).

Draw a tRNA with the anticodon 3′-CFU-5′. What two different codons could it bind to? Draw each codon on an mRNA, labeling all 5′ and 3′ ends, the tRNA, and the amino acid it carries.

Interphase

During _____, the cell grows and replicates both its organelles and its chromosomes.

four strands

During a single crossover event, how many strands of DNA must break? (Recall that DNA is double-stranded.)

Anaphase

During mitosis, centromeres separate and chromatids become individual chromosomes during which phase?

from the end of S phase in interphase through the end of metaphase in mitosis

During which stages of the cell cycle does a chromosome consist of two identical chromatids?

one low-density and one intermediate-density band

E. coli cells grown on 15N medium are transferred to 14N medium and allowed to grow for two more generations (two rounds of DNA replication). DNA extracted from these cells is centrifuged. What density distribution of DNA would you expect in this experiment?

In these cases, the sex of the parent contributing an allele affects the inheritance pattern. For imprinted genes, either the paternal or the maternal allele is expressed, depending on the imprint. For mitochondrial and chloroplast genes, only the maternal contribution will affect offspring phenotype because the offspring inherit these organelles from the mother, via the egg cytoplasm.

Explain how genomic imprinting and inheritance of mitochondrial and chloroplast DNA are exceptions to standard Mendelian inheritance.

The intracellular receptor, once activated, would be able to act as a transcription factor in the nucleus, turning on genes that may cause the cell to pass a checkpoint and divide. The RTK receptor, when activated by a ligand, would form a dimer, and each subunit of the dimer would phosphorylate the other. This would lead to a series of signal transduction steps, ultimately turning on genes in the nucleus. As in the case of the estrogen receptor, the genes would code for proteins necessary to commit the cell to divide.

Explain how receptor tyrosine kinases and intracellular receptors might function in triggering cell division.

This cell must be undergoing meiosis because the two homologs of a homologous pair are associated with each other at the metaphase plate; this does not occur in mitosis. Also, chiasmata are clearly present, meaning that crossing over has occurred, another process unique to meiosis.

Explain how the cell in question 6 is undergoing meiosis, not mitosis.

In normal hemoglobin, the sixth amino acid is glutamic acid (Glu), which is acidic (has a negative charge on its side chain). In sickle-cell hemoglobin, Glu is replaced by valine (Val), which is a nonpolar amino acid, very different from Glu. The primary structure of a protein (its amino acid sequence) ultimately determines the shape of the protein and thus its function. The substitution of Val for Glu enables the hemoglobin molecules to interact with each other and form long fibers, leading to the protein's deficient function and the deformation of the red blood cell.

Explain how the change of a single amino acid in hemoglobin leads to the aggregation of hemoglobin into long fibers. (Review Figures 5.14, 5.18, and 5.19.)

The RNA polymerase would bind directly to the promoter, rather than being dependent on the previous binding of transcription factors.

Explain how the interaction of RNA polymerase with the promoter would differ if the figure showed transcription initiation for bacteria.

1 2 homozygous dominant (AA), 0 homozygous recessive (aa), and 1 2 heterozygous (Aa)

For any gene with a dominant allele A and recessive allele a, what proportions of the offspring from an AA×Aa cross are expected to be homozygous dominant, homozygous recessive, and heterozygous?

First, during independent assortment in metaphase I, each pair of homologous chromosomes lines up independent of each other pair at the metaphase plate, so a daughter cell of meiosis I randomly inherits either a maternal or paternal chromosome of each pair. Second, due to crossing over, each chromosome is not exclusively maternal or paternal, but includes regions at the ends of the chromatid from a nonsister chromatid (a chromatid of the other homolog). (The nonsister segment can also be in an internal region of the chromatid if a second crossover occurs beyond the first one before the end of the chromatid.) This provides much additional diversity in the form of new combinations of alleles. Third, random fertilization ensures even more variation, since any sperm of a large number containing many possible genetic combinations can fertilize any egg of a similarly large number of possible combinations.

Explain how the processes unique to sexual reproduction generate a great deal of genetic variation

In this cross, the final "3" and "1" of a standard cross are lumped together as a single phenotype. This occurs because in dogs that are ee, no pigment is deposited, thus the three dogs that have a B in their genotype (normally black) can no longer be distinguished from the dog that is bb (normally brown).

Explain the genetic basis for the difference between the ratio (9:3:4) of phenotypes seen in this cross and the 9:3:3:1 ratio seen in Figure 14.8.

Checkpoints allow cellular surveillance mechanisms to determine whether the cell is prepared to go to the next stage. Internal and external signals move a cell past these checkpoints. The G1 checkpoint determines whether a cell will proceed forward in the cell cycle or switch into the G0 phase. The signals to pass this checkpoint are often external, such as growth factors. Passing the G2 checkpoint requires sufficient numbers of of active MPF complexes, which in turn orchestrate several mitotic events, MPF also initiates degradation of its cyclin component, terminating the M phase. The M phase will not begin again until sufficient cyclin is produced during the next S and G2 phases. The signal to pass the M phase checkpoint is not activated until all chromosomes are attached to kinetochore fibers and are aligned at the metaphase plate. Only then will sister chromatid separation occur.

Explain the significance of the G1, G2, and M checkpoints and the go-ahead signals involved in the cell cycle control system.

Genes program specific traits, and offspring inherit their genes from each parent, accounting for similarities in their appearance to one or the other parent. Humans reproduce sexually, which ensures new combinations of genes (and thus traits) in the offspring. Consequently, the offspring are not clones of their parents (which would be the case if humans reproduced asexually).

Explain why human offspring resemble their parents but are not identical to them.

a. 1/64 b. 1/64 c. 1/8 d. 1/32

Flower position, stem length, and seed shape are three characters that Mendel studied. Each is controlled by an independently assorting gene and has dominant and recessive expression as indicated in Table 14.1. If a plant that is heterozygous for all three characters is allowed to self-fertilize, what proportion of the offspring would you expect to be each of the following? (Note: Use the rules of probability instead of a huge Punnett square.) a. homozygous for the three dominant traits b. homozygous for the three recessive traits c. heterozygous for all three characters d. homozygous for axial and tall, heterozygous for seed shape

In each case, the alleles contributed by the female parent (in the egg) determine the phenotype of the offspring because the male in this cross contributes only recessive alleles. Thus, identifying the phenotype of the offspring tells you what alleles were in the egg.

For each type of offspring of the testcross in Figure 15.9, explain the relationship between its phenotype and the alleles contributed by the female parent. (It will be useful to draw out the chromosomes of each fly and follow the alleles throughout the cross.)

Inactivation of an X chromosome in females and genomic imprinting. Because of X inactivation, the effective dose of genes on the X chromosome is the same in males and females. As a result of genomic imprinting, only one allele of certain genes is phenotypically expressed.

Gene dosage—the number of copies of a gene that are actively being expressed—is important to proper development. Identify and describe two processes that establish the proper dosage of certain genes.

No. The order could be A−C−B or C−A−B. To determine which possibility is correct, you need to know the recombination frequency between B and C.

Genes A, B, and C are located on the same chromosome. Testcrosses show that the recombination frequency between A and B is 28% and that between A and C is 12%. Can you determine the linear order of these genes? Explain.

You can't tell which end is the 5' end. You need to know which end has a phosphate group on the 5' carbon (the 5' end) or which end has an —OH group on the 3' carbon (the 3' end).

Given a polynucleotide sequence such as GAATTC, explain what further information you would need in order to identify which is the 5′ end.

He expected that the mouse injected with the mixture of heat-killed S cells and living R cells would survive, since neither type of cell alone would kill the mouse.

Griffith was trying to develop a vaccine for S. pneumonia when he was surprised to discover the phenomenon of bacterial transformation. Look at the second and third panels of Figure 16.2. Based on these results, what result was he expecting in the fourth panel? Explain.

a. 3/4×3/4×3/4=27/64 b. 1−27/64=37/64 c. 1/4×1/4×1/4=1/64 d. 1−1/64=63/64

Hemochromatosis is an inherited disease caused by a recessive allele. If a woman and her husband, who are both carriers, have three children, what is the probability of each of the following? a. All three children are of normal phenotype. b. One or more of the three children have the disease. c. All three children have the disease. d. At least one child is phenotypically normal. (NOTE: It will help to remember that the probabilities of all possible outcomes always add up to 1.)

A protein outside the cell can bind to a receptor protein on the cell surface, causing it to change shape and sending a signal inside the cell

How can a protein outside the cell cause event to happen inside the cell?

The living S cells found in the blood sample were able to reproduce to yield more S cells, indicating that the S trait is a permanent, heritable change, rather than just a one-time use of the dead S cells' capsules.

How did this experiment rule out the possibility that the R cells simply used the dead S cells' capsules to become pathogenic?

A sufficient amount of MPF has to exist for a cell to pass the G2 checkpoint; this occurs through the accumulation of cyclin proteins, which combines with Cdk to form (active) MPF. MPF then phosphorylates other proteins, initiating mitosis.

How does MPF allow a cell to pass the G2 phase checkpoint and enter mitosis?

Such organisms reproduce by mitosis, which generates offspring whose genomes are exact copies of the parents genome (in the absence of mutation)

How does an asexually reproducing eukaryotic organism produce offspring that are genetically identical to each other and to their parents?

In watching a show recorded with a DVR, you watch segments of the show itself (exons) and fast-forward through the commercials, which are thus like introns. However, unlike introns, commercials remain in the recording, while the introns are cut out of the RNA transcript during RNA processing.

How is RNA splicing similar to how you would watch a recorded television show? What would introns be?

1; 1; 2

How many chromosomes are drawn in each part? Ignore the micrograph.

6 chromosomes; they are duplicated; 12 chromatids

How many chromosomes are shown in the figure 12.8? Are they duplicated? How many chromatids are shown?

The radioactivity would have been found in the pellet when proteins were labeled (batch 1) because proteins would have had to enter the bacterial cells to program them with genetic instructions. It's hard for us to imagine now, but the DNA might have played a structural role that allowed some of the proteins to be injected while it remained outside the bacterial cell (thus no radioactivity in the pellet in batch 2).

How would the results have differed if proteins carried the genetic information?

DNA pol III covalently adds nucleotides to new DNA strands and proofreads each added nucleotide for correct base pairing.

Identify two major functions of DNA pol III in DNA replication.

The tube from the first replication would look the same, with a middle band of hybrid 15N−14N DNA, but the second tube would not have the upper band of two light blue strands. Instead, it would have a bottom band of two dark blue strands, like the bottom band in the result predicted after one replication in the conservative model.

If Meselson and Stahl had first grown the cells in 14N-containing medium and then moved them into 15N-containing medium before taking samples, what would have been the result after each of the two replications?

All the males would be color-blind, and all the females would be carriers. (Another way to say this is that ½ the offspring would be color-blind males, and ½ the offspring would be carrier females.)

If a color-blind woman married a man who had normal color vision, what would be the probable phenotypes of their children?

Half of the children would be expected to have type A blood and half type B blood.

If a man with type AB blood marries a woman with type O, what blood types would you expect in their children? What fraction would you expect of each type?

If the segments of the maternal and paternal chromatids that undergo crossing over are genetically identical and thus have the same two alleles for every gene, then the recombinant chromosomes will be genetically equivalent to the parental chromosomes. Crossing over contributes to genetic variation only when it involves the rearrangement of different alleles.

If maternal and paternal chromatids have the same two alleles for every gene, will crossing over leading to genetic variation?

2x

If the DNA content of a diploid cell in the G1 phase of the cell cycle is X, then the DNA content of the same cell at Metaphase of Meiosis 1 will be

Synthesis of the leading strand is initiated by an RNA primer, which must be removed and replaced with DNA, a task that could not be performed if the cell's DNA pol I were nonfunctional. In the overview box in Figure 16.17, just to the left of the top origin of replication, a functional DNA pol I would replace the RNA primer of the leading strand (shown in red) with DNA nucleotides (blue). The nucleotides would be added onto the 3' end of the first Okazaki fragment of the upper lagging strand (the right half of the replication bubble).

If the DNA pol I in a given cell were nonfunctional, how would that affect the synthesis of a leading strand? In the overview box in Figure 16.17, point out where DNA pol I would normally function on the top leading strand.

The two largest classes would still be the offspring with the phenotypes of the true-breeding P generation flies, but now would be gray vestigial and black normal, which is now the "parental type" because those were the specific allele combinations in the P generation.

If the parental (P generation) flies had been true-breeding for gray body with vestigial wings and black body with normal wings, which phenotypic class(es) would be largest among the testcross offspring?

In both cases, the G1 nucleus would have remained in G1 until the time it normally would have entered the S phase. Chromosomes condensation and spindle formation would not have occurred until the S and G2 phases had been completed.

If the progression of phases didn't depend on cytoplasmic molecules and, instead, each phase automatically began with the previous one was complete, how would the results have differed?

It would be packaged in a vesicle, transported to the Golgi apparatus for further processing, and then transported via a vesicle to the plasma membrane. The vesicle would fuse with the membrane, releasing the protein outside the cell.

If this protein were destined for secretion, what would happen to it after its synthesis was completed?

X

If we continue to follow the cell lineage from question 4, then the DNA content of a single cell at metaphase of meiosis II will be

The two members of a homologous pair (which would be the same color) would be associated tightly together at the metaphase plate during metaphase I of meiosis I. In metaphase of mitosis, however, each chromosome would be lined up individually, so the two chromosomes of the same color would be in different places at the metaphase plate.

If you arrested a human cell in metaphase I of meiosis and applied this technique, what would you observe? How would this differ from what you would see in metaphase of mitosis

The ratio would be 1 yellow round : 1 green round : 1 yellow wrinkled : 1 green wrinkled.

If you crossed an F1 plant with a plant that was homozygous recessive for both genes (yyrr), how would the phenotypic ratio of the offspring compare with the 9:3:3:1 ratio seen here?

All offspring would have purple flowers. (The ratio would be 1 purple: 0 white.) The P generation plants are true-breeding, so mating two purple-flowered plants produces the same result as self-pollination: All the offspring have the same trait. If Mendel had stopped after the F1 generation, he could have concluded that the white factor had disappeared entirely and would not ever reappear.

If you mated two purple-flowered plants from the P generation, what ratio of traits would you expect to observe in the offspring? Explain. What might Mendel have concluded if he stopped his experiment after the F1 generation?

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Imagine that you are a genetic counselor, and a couple planning to start a family comes to you for information. Charles was married once before, and he and his first wife had a child with cystic fibrosis. The brother of his current wife, Elaine, died of cystic fibrosis. What is the probability that Charles and Elaine will have a baby with cystic fibrosis? (Neither Charles, Elaine, nor their parents have cystic fibrosis.)

Matings of the original mutant cat with true-breeding noncurl cats will produce both curl and noncurl F1 offspring if the curl allele is dominant, but only noncurl offspring if the curl allele is recessive. You would obtain some true-breeding offspring homozygous for the curl allele from matings between the F1 cats resulting from the original curl×noncurl crosses whether the curl trait is dominant or recessive. You know that cats are true-breeding when curl×curl matings produce only curl offspring. As it turns out, the allele that causes curled ears is dominant.

In 1981, a stray black cat with unusual rounded, curled-back ears was adopted by a family in California. Hundreds of descendants of the cat have since been born, and cat fanciers hope to develop the curl cat into a show breed. Suppose you owned the first curl cat and wanted to develop a true-breeding variety. How would you determine whether the curl allele is dominant or recessive? How would you obtain true-breeding curl cats? How could you be sure they are true-breeding?

In the monohybrid cross involving flower color, the ratio is 3.15 purple: 1 white, while in the human family in the pedigree, the ratio in the third generation is 1 can taste PTC: 1 cannot taste PTC. The difference is due to the small sample size (two offspring) in the human family. If the second-generation couple in this pedigree were able to have 929 offspring as in the pea plant cross, the ratio would likely be closer to 3:1. (Note that none of the pea plant crosses in Table 14.1 yielded exactly a 3:1 ratio.)

In Table 14.1, note the phenotypic ratio of the dominant to recessive trait in the F2 generation for the monohybrid cross involving flower color. Then determine the phenotypic ratio for the offspring of the second-generation couple in Figure 14.15b. What accounts for the difference in the two ratios?

histones

In a nucleosome, the DNA is wrapped around

Recessive

In a research article about alkaptonuria published in 1902, Garrod suggested that humans inherit two "characters" (alleles) for a particular enzyme and that both parents must contribute a faulty version for the offspring to have alkaptonuria. Today, would this disorder be called dominant or recessive?

ribosomal RNA (rRNA)

In a ribosomes, plays a structural role; As a ribozyme, plays a catalytic role (catalyzes peptide bond formation)

A + G = C + T

In analyzing the number of different bases in a DNA sample, which result would be consistent with the base-pairing rules?

When one ribosome terminates translation and dissociates, the two subunits would be very close to the cap. This could facilitate their rebinding and initiating synthesis of a new polypeptide, thus increasing the efficiency of translation.

In eukaryotic cells, mRNAs have been found to have a circular arrangement in which proteins hold the poly-A tail near the 5′ cap. How might this increase translation efficiency?

several transcription factors have bound to the promoter.

In eukaryotic cells, transcription cannot begin until

The nucleus on the right was originally in the G1 phase; therefore, it had not yet duplicated its chromosomes. The nucleus on the left was in the M phase, so it had already duplicated its chromosomes.

In figure 12.14, why do the nuclei resulting from experiment 2 contain different amounts of DNA?

Each of the six chromosomes is duplicated, so each contains two DNA double helices. Therefore, there are 12 DNA molecules in the cell. The haploid number, n, is 13. One set is always haploid.

In figure 13.4, how many DNA molecules (double helices) are present? What is the haploid number of this cell? Is a set of chromosomes haploid or diploid?

some substance from pathogenic cells was transferred to nonpathogenic cells, making them pathogenic.

In his work with pneumonia-causing bacteria and mice, Griffith found that

The dominant allele I is epistatic to the P/p locus, and thus the genotypic ratio for the F1 generation will be 9 I-P- (colorless) : 3 I-pp (colorless) : 3 iiP- (purple) : 1 iipp (red). Overall, the phenotypic ratio is 12 colorless : 3 purple : 1 red.

In maize (corn) plants, a dominant allele I inhibits kernel color, while the recessive allele i permits color when homozygous. At a different locus, the dominant allele P causes purple kernel color, while the homozygous recessive genotype pp causes red kernels. If plants heterozygous at both loci are crossed, what will be the phenotypic ratio of the offspring?

At the end of Meiosis 1, the two members of a homologous pair end up in different cells, so they cant pair up and undergo crossing over during Prophase 2.

In prophase I, homologous chromosomes pair up and undergo synapsis and crossing over. Can this also occur during prophase II? Explain.

Self-pollination is sexual reproduction because meiosis is involved in forming gametes, which unite during fertilization. As a result, the offspring in self-pollination are genetically different from the parent. (As mentioned in the footnote near the beginning of Concept 14.1, we have simplified the explanation in referring to the single pea plant as a parent. Technically, the gametophytes in the flower are the two "parents.")

In some pea plant crosses, the plants are self-pollinated. Is self-pollination considered asexual or sexual reproduction? Explain.

In the bubble at the top of the micrograph in (b), arrows should be drawn pointing left and right to indicate the two replication forks.

In the TEM in (b), add arrows in the forks of the third bubble

cells with more than one nucleus

In the cells of some organisms, mitosis occurs without cytokinesis. This will result in

There are 23 pairs of chromosomes and two sets.

In the karyotype shown in Figure 13.3, how many pairs of chromosomes are present? How many sets?

25%, or 1/4, will be cross-eyed; all (100%) of the cross-eyed offspring will also be white.

In tigers, a recessive allele of a particular gene causes both an absence of fur pigmentation (a white tiger) and a cross-eyed condition. If two phenotypically normal tigers that are heterozygous at this locus are mated, what percentage of their offspring will be cross-eyed? What percentage of cross-eyed tigers will be white?

Chromosomes exist as single DNA molecules in G1 of interphase and anaphase and telophase of mitosis. During S phase, DNA replication produces sister chromatids, which persist during G2 phase of interphase and through prophase, pro metaphase, and metaphase of mitosis

In which of the three phases of interphase and the stages of mitosis do chromosomes exist as single DNA molecules?

Incomplete dominance describes the relationship between two alleles of a single gene, whereas epistasis relates to the genetic relationship between two genes (and the respective alleles of each).

Incomplete dominance and epistasis are both terms that define genetic relationships. What is the most basic distinction between these terms?

Heterozygous individuals, said to have sickle-cell trait, have a copy each of the wild-type allele and the sickle-cell allele. Both alleles will be expressed, so these individuals will have both normal and sickle-cell hemoglobin molecules. Apparently, having a mix of the two forms of β-globin has no effect under most conditions, but during prolonged periods of low blood oxygen (such as at higher altitudes), these individuals can show some signs of sickle-cell disease.

Individuals heterozygous for the sickle-cell allele are generally healthy but show phenotypic effects of the allele under some circumstances (see Figure 14.17). Explain in terms of gene expression.

The nuclear lamina is a netlike array of protein filaments that provides mechanical support just inside the nuclear envelope and thus maintains the shape of the nucleus. Considerable evidence also supports the existence of a nuclear matrix, a framework of protein fibers extending throughout the nuclear interior.

Interphase chromosomes appear to be attached to the nuclear lamina and perhaps also the nuclear matrix. Describe these two structures.

primary transcript

Is a precursor to mRNA, rRNA, or tRNA, before being processed; Some intron RNA acts as a ribozyme, catalyzing its own splicing

Joan's genotype is Dd. Because the allele for polydactyly (D) is dominant to the allele for five digits per appendage (d ), the trait is expressed in people with either the DD or Dd genotype. But because Joan's father does not have polydactyly, his genotype must be dd, which means that Joan inherited a d allele from him. Therefore, Joan, who does have the trait, must be heterozygous.

Joan was born with six toes on each foot, a dominant trait called polydactyly. Two of her five siblings and her mother, but not her father, also have extra digits. What is Joan's genotype for the number-of-digits character? Explain your answer. Use D and d to symbolize the alleles for this character.

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Karen and Steve each have a sibling with sickle-cell disease. Neither Karen nor Steve nor any of their parents have the disease, and none of them have been tested to see if they carry the sickle-cell allele. Based on this incomplete information, calculate the probability that if this couple has a child, the child will have sickle-cell disease.

The plant could make eight different gametes (YRI, YRi, YrI, Yri, yRI, yRi, yrI, and yri). To fit all the possible gametes in a self-pollination, a Punnett square would need 8 rows and 8 columns. It would have spaces for the 64 possible unions of gametes in the offspring.

List all gametes that could be made by a pea plant heterozygous for seed color, seed shape, and pod shape (YyRrIi; see Table 14.1). How large a Punnett square would you need to draw to predict the offspring of a self-pollination of this "trihybrid"?

The anticodon on the tRNA is 3′-AAG-5′, so it would bind to the mRNA codon 5′-UUC-3′. This codon codes for phenylalanine, which is the amino acid this tRNA would carry.

Look at the tRNA shown in this figure. Based on its anticodon, identify the codon it would bind to, as well as the amino acid that it would carry.

G2

Measurements of the amount of DNA per nucleus were taken on a large number of cells from a growing fungus. The measured DNA levels ranged from 3 to 6 picograms per nucleus. During the entirety of which stage of the cell cycle did the nucleus contain 6 pg of DNA

sister chromatids separate during Anaphase

Meiosis 2 is similar to Mitosis in that

Each cell contains numerous mitochondria, and in affected individuals, most cells contain a variable mixture of normal and mutant mitochondria. The normal mitochondria carry out enough cellular respiration for survival. (The situation is similar for chloroplasts.)

Mitochondrial genes are critical to the energy metabolism of cells, but mitochondrial disorders caused by mutations in these genes are generally not lethal. Why not?

1/4 (1/2 chance that the child will inherit a Y chromosome from the father and be male ×1/2 chance that he will inherit the X carrying the disease allele from his mother). If the child is a boy, there is a 1/2 chance he will have the disease; a female would have zero chance (but 1/2 chance of being a carrier).

Neither Tim nor Rhoda has Duchenne muscular dystrophy, but their firstborn son does. What is the probability that a second child will have the disease? What is the probability if the second child is a boy? A girl?

altered primary and quaternary structures

Normal hemoglobin is a tetramer, consisting of two molecules of β-globin and two molecules of α-globin; normal hemoglobin molecules do not associate with each other. In sickle-cell disease, the change in a single amino acid results in mutant hemoglobin tetramers, which associate with each other and assemble into large fibers. Based on this information alone, what can we conclude about the changes in the structure of sickle cell hemoglobin as compared to normal hemoglobin?

continue to divide even when they are tightly packed together

One difference between cancer cells and normal cells is that cancer cells

The cells in the vessel with PDGF would would not be able to respond to the growth factor signal and thus would not divide. The culture would resemble that without the added PDGF.

PDGF signal cells by binding to a cell-surface receptor tyrosine kinase. If you added a chemical that blocked phosphorylation, how would the results differ?

According to the law of independent assortment, 25 plants ( 1 16 of the offspring) are predicted to be aatt, or recessive for both characters. The actual result is likely to differ slightly from this value. Punnett square with 16 boxes shows the outcome of a cross between parents uppercase A lowercase a uppercase T lowercase t x uppercase A lowercase a uppercase T lowercase t.

Pea plants heterozygous for flower position and stem length (AaTt) are allowed to self-pollinate, and 400 of the resulting seeds are planted. Draw a Punnett square for this cross. How many offspring would be predicted to have terminal flowers and be dwarf?

Small RNA in spliceosome

Play structural and catalytic roles in spliceosomes, the complex of protein and RNA that splice pre-mRNA

To show the mutant phenotype, a male needs to possess only one mutant allele. If this gene had been on a pair of autosomes, the two alleles would both have had to be mutant in order for an individual to show the recessive mutant phenotype, a much less probable situation.

Propose a possible reason that the first naturally occurring mutant fruit fly Morgan saw involved a gene on a sex chromosome and was found in a male.

Recessive; if the disorder were dominant, it would affect at least one parent of a child born with the disorder. The disorder's inheritance is sex-linked because it is seen only in boys. For a girl to have the disorder, she would have to inherit recessive alleles from both parents. This would be very rare, since males with the recessive allele on their X chromosome die in their early teens.

Pseudohypertrophic muscular dystrophy is an inherited disorder that causes gradual deterioration of the muscles. It is seen almost exclusively in boys born to apparently unaffected parents and usually results in death in the early teens. Is this disorder caused by a dominant or a recessive allele? Is its inheritance sex-linked or autosomal? How do you know? Explain why this disorder is almost never seen in girls.

Your classmate would probably point out that the F1 generation hybrids show an intermediate phenotype between those of the homozygous parents, which supports the blending hypothesis. You could respond that crossing the F1 hybrids results in the reappearance of the white phenotype, rather than identical pink offspring, which fails to support the idea of traits blending during inheritance.

QUESTION Suppose a classmate argues that this figure supports the blending hypothesis for inheritance. What might your classmate say, and how would you respond?

The genes for leaf coloration are located in plastids within the cytoplasm. Normally, only the maternal parent transmits plastid genes to offspring. Since variegated offspring are produced only when the female parent is of the B variety, we can conclude that variety B contains both the wild-type and mutant alleles of pigment genes, producing variegated leaves. (Variety A must contain only the wild-type allele of pigment genes.)

Reciprocal crosses between two primrose varieties, A and B, produced the following results: Afemale×Bmale→offspring with all green (nonvariegated) leaves; Bfemale×Amale→offspring with patterned (variegated) leaves. Explain these results.

3/4 yellow x 3/4 round = 9/16 yellow round 3/4 yellow x 1/4 wrinkled = 3/16 yellow wrinkled 1/4 green x 3/4 round = 3/16 green round 1/4 green x 1/4 wrinkled = 1/16 green wrinkled = 9 yellow round : 3 yellow wrinkled : 3 green round : 1 green wrinkled

Redraw the Punnett square on the right side of Figure 14.8 as two smaller monohybrid Punnett squares, one for each gene. Below each square, list the fractions of each phenotype produced. Use the rule of multiplication to compute the overall fraction of each possible dihybrid phenotype. What is the phenotypic ratio?

The physical basis for the law of segregation is the separation of homologs in anaphase I. The physical basis for the law of independent assortment is the alternative arrangements of all the different homologous chromosome pairs in metaphase I.

Review the description of meiosis (see Figure 13.8) and Mendel's laws of segregation and independent assortment (see Concept 14.1). What is the physical basis for each of Mendel's laws?

Yes, this cross would also have allowed Mendel to make different predictions for the two hypotheses, thereby allowing him to distinguish the correct one.

Suppose Mendel had transferred pollen from an F1 plant to the carpel of a plant that was homozygous recessive for both genes. Set up the cross and draw Punnett squares that predict the offspring for both hypotheses. Would this cross have supported the hypothesis of independent assortment equally well?

The transcription factor that recognizes the TATA sequence would be unable to bind, so RNA polymerase could not bind and transcription of that gene probably would not occur.

Suppose X-rays caused a sequence change in the TATA box of a particular gene's promoter. How would that affect transcription of the gene?

The previously presumed pathway would have been wrong. The new results would support this pathway: precursor→citrulline→ornithine→arginine. They would also indicate that class I mutants have a defect in the second step and class II mutants have a defect in the first step.

Suppose the experiment had shown that class I mutants could grow only in MM supplemented by ornithine or arginine and that class II mutants could grow in MM supplemented by citrulline, ornithine, or arginine. What conclusions would the researchers have drawn from those results regarding the biochemical pathway and the defect in class I and class II mutants?

About 3/4 of the F2 offspring would have red eyes and about 1/4 would have white eyes. About half of the white-eyed flies would be female and half would be male; similarly, about half of the red-eyed flies would be female and half would be male. (Note that the homologs with the eye color alleles would be the same shape in the Punnett square, and each offspring would inherit two alleles. The sex of the flies would be determined separately by inheritance of the sex chromosomes. Thus your Punnett square would have four possible combinations in sperm and four in eggs; it would have 16 squares altogether.)

Suppose this eye color gene were located on an autosome. Predict the phenotypes (including gender) of the F2 flies in this hypothetical cross. (HINT: Draw a Punnett square.)

15.10 The two chromosomes below, left, are like the two chromosomes inherited by the F1 female, one from each P generation fly. They are passed by the F1 female intact to the offspring and thus could be called "parental" chromosomes. The other two chromosomes result from crossing over during meiosis in the F1 female. Because they have combinations of alleles not seen in either of the F1 female's chromosomes, they can be called "recombinant" chromosomes. (Note that in this example, the alleles on the recombinant chromosomes, b+vg+ and b vg, are the allele combinations that were on the parental chromosomes in the cross shown in Figures 15.9 and 15.10. The basis for calling them parental chromosomes is that they have the combination of alleles that was present on the P generation chromosomes.)

Suppose, as in the question at the bottom of Figure 15.9, the parental (P generation) flies were true-breeding for gray body with vestigial wings and black body with normal wings. Draw the chromosomes in each of the four possible kinds of eggs from an F1 female, and label each chromosome as "parental" or "recombinant."

No. The child can be either IAIAi or IAii. A sperm of genotype IAIA could result from nondisjunction in the father during meiosis II, while an egg with the genotype ii could result from nondisjunction in the mother during either meiosis I or meiosis II.

The ABO blood type locus has been mapped on chromosome 9. A father who has type AB blood and a mother who has type O blood have a child with trisomy 9 and type A blood. Using this information, can you tell in which parent the nondisjunction occurred? Explain your answer. (See Figures 14.11 and 15.13.)

complementary to the corresponding mRNA codon.

The anticodon of a particular tRNA molecule is

the degradation of cyclin

The decline of MPF activity at the end of mitosis is due to

Without crossing over, independent assortment of chromosomes during meiosis I theoretically can generate 2n possible haploid gametes, and random fertilization can produce 2n×2n possible diploid zygotes. Because the haploid number (n) of grasshoppers is 23 and that of fruit flies is 4, two grasshoppers would be expected to produce a greater variety of zygotes than would two fruit flies.

The diploid number for fruit flies is 8, and the diploid number for grasshoppers is 46. If no crossing over took place, would the genetic variation among offspring from a given pair of parents be greater in fruit flies or grasshoppers? Explain.

cleavage furrow formation and cytokinesis

The drug cytochalasin B blocks the function of actin. Which of the following aspects of the animal cell cycle would be most disrupted by by cytochalasin B?

depends on the action of DNA polymerase.

The elongation of the leading strand during DNA synthesis

Activation of this gene could lead to the production of too much of this kinase. If the kinase is involved in a signaling pathway that triggers cell division, too much of it could trigger unrestricted cell division, which in turn could contribute to the development of a cancer (in this case, a cancer of one type of white blood cell).

The gene that is activated on the Philadelphia chromosome codes for an intracellular tyrosine kinase. Review the discussion of cell cycle control in Concept 12.3, and explain how the activation of this gene could contribute to the development of cancer.

a. 1/256 b. 1/16 c. 1/256 d. 1/64 e. 1/128

The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F2 offspring will have the following genotypes? a. aabbccdd b. AaBbCcDd c. AABBCCDD d. AaBBccDd e. AaBBCCdd

proteins, mRNA, and tRNA

The information contained in DNA is used to make which of the following product(s)?

Recessive. All affected individuals (Arlene, Tom, Wilma, and Carla) are homozygous recessive aa. George is Aa, since some of his children with Arlene are affected. Sam, Ann, Daniel, and Alan are each Aa, since they are all unaffected children with one affected parent. Michael also is Aa, since he has an affected child (Carla) with his heterozygous wife Ann. Sandra, Tina, and Christopher can each have either the AA or Aa genotype.

The pedigree below traces the inheritance of alkaptonuria, a biochemical disorder. Affected individuals, indicated here by the colored circles and squares, are unable to metabolize a substance called alkapton, which colors the urine and stains body tissues. Does alkaptonuria appear to be caused by a dominant allele or by a recessive allele? Fill in the genotypes of the individuals whose genotypes can be deduced. What genotypes are possible for each of the other individuals?

nuclease, DNA polymerase, DNA ligase

The spontaneous loss of amino groups from adenine in DNA results in hypoxanthine, an uncommon base, opposite thymine. What combination of proteins could repair such damage?

Template sequence: 3 prime- A C G A C T G A A-5 prime. mRNA sequence: 5 prime-U G C U G A C U U-3 prime. Translated: Cys-STOP. If the nontemplate sequence could have been used as a template for transcribing the mRNA, the protein translated from the mRNA would have a completely different amino acid sequence and would most likely be nonfunctional. (It would also be shorter because of the UGA stop signal shown in the mRNA sequence above—and possibly others earlier in the mRNA sequence.)

The template strand of a gene contains the sequence 3′-TTCAGTCGT-5′. Imagine that the nontemplate sequence was transcribed instead of the template sequence. Draw the mRNA sequence and translate it using Figure 17.6. (Be sure to pay attention to the 5′ and 3′ ends.) Predict how well the protein synthesized from the nontemplate strand would function, if at all.

No effect: The amino acid sequence is Met-Asn-Arg-Leu both before and after the mutation because the mRNA codons 5'-CUA-3' and 5'-UUA-3' both code for Leu. (The fifth codon is a stop codon.)

The template strand of a gene includes this sequence: 3′-TACTTGTCCGATATC-5′. It is mutated to 3′-TACTTGTCCAATATC-5′. For both wild-type and mutant sequences, draw the double-stranded DNA, the resulting mRNA, and the amino acid sequence each encodes. What is the effect of the mutation on the amino acid sequence?

Meiosis 1

The two homologs of a pair move toward opposite poles of dividing cell during

hydrogen bonds

The two strands of a DNA double helix are held together by _____ that form between pairs of nitrogenous bases.

Due to alternative splicing of exons, each gene can result in multiple different mRNAs and can thus direct synthesis of multiple different proteins.

There are about 20,000 human protein-coding genes. How can human cells make 75,000-100,000 different proteins?

The genotypes that fulfill this condition are ppyyIi, ppYyii, Ppyyii, ppYYii, and ppyyii. Use the multiplication rule to find the probability of getting each genotype, and then use the addition rule to find the overall probability of meeting the conditions of this problem: Fraction of offspring predicted to have at least two recessive traits is 6/16 or 3/8.

Three characters (flower color, seed color, and pod shape) are considered in a cross between two pea plants: PpYyli×ppYyii. What fraction of offspring is predicted to be homozygous recessive for at least two of the three characters?

a plant cell in the process of cytokinesis

Through a microscope, you can see a cell plate beginning to develop across the middle of a cell and nuclei forming on either side of the cell plate. This cell is most likely

450 each of blue oval and white round (parentals) and 50 each of blue round and white oval (recombinants)

Two genes of a flower, one controlling blue (B) versus white (b) petals and the other controlling round (R) versus oval (r) stamens, are linked and are 10 map units apart. You cross a homozygous blue oval plant with a homozygous white round plant. The resulting F1 progeny are crossed with homozygous white oval plants, and 1,000 offspring plants are obtained. How many plants of each of the four phenotypes do you expect?

1 4 BBDD; 1 4 BbDD; 1 4 BBDd; 1 4 BdDd;

Two organisms, with genotypes BbDD and BBDd, are mated. Assuming independent assortment of the B/b and D/d genes, write the genotypes of all possible offspring from this cross and use the rules of probability to calculate the chance of each genotype occurring.

Parents: GgIi x GgIi 9 green inflated : 3 green constricted : 3 yellow inflated : 1 yellow constricted

Two pea plants heterozygous for the characters of pod color and pod shape are crossed. Draw a Punnett square to determine the phenotypic ratios of the offspring.

Looking at any of the DNA strands, we see that one end is called the 5' end and the other the 3' end. If we proceed from the 5' end to the 3' end on the left-most strand, for example, we list the components in this order: phosphate group →5′C of the sugar →3′C→ phosphate →5′C→3′C. Going in the opposite direction on the same strand, the components proceed in the reverse order: 3′C→5′C→ phosphate. Thus, the two directions are distinguishable, which is what we mean when we say that the strands have directionality. (Review Figure 16.5 if necessary.)

Use this diagram to explain what we mean when we say that each DNA strand has directionality.

5' -CTTCGGGAA-3'

Using Figure 17.6, identify a 5′→3′ sequence of nucleotides in the DNA template strand for an mRNA coding for the polypeptide sequence Phe-Pro-Lys.

Between T and S, 18%; sequence of genes is T−A−S

Using the information from problem 4, scientists do a further testcross using a heterozygote for height and nose morphology. The offspring are tall upturned snout, 40; dwarf upturned snout, 9; dwarf downturned snout, 42; tall downturned snout, 9. Calculate the recombination frequency from these data, and then use your answer from problem 4 to determine the correct order of the three linked genes.

Parents pass genes to their offspring; By dictating the production of messenger RNA , the gene programs cells to make specific enzymes and other proteins, whose cumulative action produces an individuals inherited traits

Using what you know of gene expression in a cell, explain what causes the trait of parents (such as hair color) to show up in their offspring.

disruption of mitotic spindle formation

Vinblastine is a standard chemotherapeutic drug used to treat cancer. Because it interferes with the assembly of microtubules, its effectiveness must be related to

Both bacterial and eukaryotic genes have promoters, regions where RNA polymerase ultimately binds and begins transcription. In bacteria, RNA polymerase binds directly to the promoter; in eukaryotes, transcription factors bind first to the promoter, and then RNA polymerase binds to the transcription factors and promoter together.

What are the similarities and differences in the initiation of gene transcription in bacteria and eukaryotes?

Microtubules are made up of subunits of tubulin, and are structures along which substances are transported into the cell.

What best describes micro tubules?

Because the sex chromosomes are different from each other and because they determine the sex of the offspring, Morgan could use the sex of the offspring as a phenotypic character to follow the parental chromosomes. (He could also have followed them under a microscope, as the X and Y chromosomes look different.) At the same time, he could record eye color to follow the eye color alleles.

What characteristic of the sex chromosomes allowed Morgan to correlate their behavior with that of the alleles of the eye color gene?

Each strand in the double helix has polarity; the end with a phosphate group on the 5' carbon of the sugar is called the 5' end, and the end with an —OH group on the 3' carbon of the sugar is called the 3' end. The two strands run in opposite directions, one running 5′→3′ and the other alongside it running 3′→5′. Thus, each end of the molecule has both a 5' and a 3' end. This arrangement is called "antiparallel." If the strands were parallel, they would both run 5′→3′ in the same direction, so an end of the molecule would have either two 5' ends or two 3' ends.

What does it mean when we say that the two DNA strands in the double helix are antiparallel? What would an end of the double helix look like if the strands were parallel?

Phosphorylation changes the shape of the protein, most often activating it

What effect does phosphorylating a protein have on that protein?

In a bacterial cell, part of the RNA polymerase recognizes the gene's promoter and binds to it. In a eukaryotic cell, transcription factors must bind to the promoter first, then the RNA polymerase binds to them. In both cases, sequences in the promoter determine the precise binding of RNA polymerase so the enzyme is in the right location and orientation.

What enables RNA polymerase to start transcribing a gene at the right place on the DNA in a bacterial cell? In a eukaryotic cell?

Both the 5' cap and the 3' poly-A tail help the mRNA exit from the nucleus and then, in the cytoplasm, help ensure mRNA stability and allow it to bind to ribosomes.

What function do the 5′ cap and the poly-A tail serve on a eukaryotic mRNA?

In the mRNA, the reading frame downstream from the deletion is shifted, leading to a long string of incorrect amino acids in the polypeptide, and in most cases, a stop codon will occur, leading to premature termination. The polypeptide will most likely be nonfunctional.

What happens when one nucleotide pair is lost from the middle of the coding sequence of a gene?

A promoter is the region of DNA to which RNA polymerase binds to begin transcription. It is at the upstream end of the gene (transcription unit).

What is a promoter? Is it located at the upstream or downstream end of a transcription unit?

DNA polymerase can join new nucleotides only to the 3′ end of a pre-existing strand, and the strands are antiparallel.

What is the basis for the difference in how the leading and lagging strands of DNA molecules are synthesized?

Mutations in a gene lead to different versions (alleles) of that gene

What is the original source of variation among the different alleles of a gene?

a. 1 b. 1/32 c. 1/8 d. 1/2

What is the probability that each of the following pairs of parents will produce the indicated offspring? (Assume independent assortment of all gene pairs.) a. AABBCC×aabbcc→AaBbCc b. AABbCc×AaBbCc→AAbbCC c. AaBbCc×AaBbCc→AaBbCc d. aaBbCC×AABbcc→AaBbCc

In the Punnett square, two of the three individuals with normal coloration are carriers, so the probability is 2 3 . (Note that you must take into account everything you know when you calculate probability: You know she is not aa, so there are only three possible genotypes to consider.)

What is the probability that the sister with normal coloration is a carrier of the albinism allele?

In the cell cycle, DNA synthesis occurs during the S phase, between the G1 and G2 phases of interphase. DNA replication is therefore complete before the mitotic phase begins.

What is the relationship between DNA replication and the S phase of the cell cycle?

The cell would divide under conditions where it was inappropriate to do so. If the daughter cells and their descendants also ignored either of the checkpoints and divided, there would soon be an abnormal mass of cells. (This type of inappropriate cell division can contribute to the development of cancer.)

What might be the result if the cell ignored either checkpoint and progressed through the cell cycle?

The single DNA molecule in the chromosome must be replicated

What must happen to a chromosome before a cell starts mitosis?

48 chromosomes (24 from each parent, including one sex chromosome from each.)

What number and types of chromosomes are found in a somatic cell in an animal with a diploid number of 48?

Microtubules made up of tubulin in the cell provide "rails" along which other vesicles and organelles can travel, based on interactions of motor proteins with the tubulin in the microtubules. In muscle cells, actin in microfilaments interactions with myosin filaments to cause muscle contraction.

What other functions do actin and tubulin carry out? Name the proteins with to do so.

A polypeptide made up of 10 Gly (glycine) amino acids

What polypeptide product would you expect from a poly-G mRNA that is 30 nucleotides long?

Complementary base pairing ensures that the two daughter molecules are exact copies of the parental molecule. When the two strands of the parental molecule separate, each serves as a template on which nucleotides are arranged, by the base-pairing rules, into new complementary strands.

What role does complementary base pairing play in the replication of DNA?

First, each aminoacyl-tRNA synthetase specifically recognizes a single amino acid and attaches it only to an appropriate tRNA. Second, a tRNA charged with its specific amino acid binds only to an mRNA codon for that amino acid.

What two processes ensure that the correct amino acid is added to a growing polypeptide chain?

Euchromatin is chromatin that becomes less compacted during interphase and is accessible to the cellular machinery responsible for gene activity. Heterochromatin, on the other hand, remains quite condensed during interphase and contains genes that are largely inaccessible to this machinery.

What two properties, one structural and one functional, distinguish heterochromatin from euchromatin?

When a nucleotide base is altered chemically, its base-pairing characteristics may be changed. When that happens, an incorrect nucleotide is likely to be incorporated into the complementary strand during the next replication of the DNA, and successive rounds of replication will perpetuate the mutation. Once the gene is transcribed, the mutated codon may code for a different amino acid that inhibits or changes the function of a protein. If the chemical change in the base is detected and repaired by the DNA repair system before the next replication, no mutation will result.

What will be the results of chemically modifying one nucleotide base of a gene? What role is played by DNA repair systems in the cell?

Once the mRNA has exited the nucleus, the cap prevents it from being degraded by hydrolytic enzymes and facilitates its attachment to ribosomes. If the cap were removed from all mRNAs, the cell would no longer be able to synthesize any proteins and would probably die.

What would be the effect of treating cells with an agent that removed the cap from mRNAs?

Alternative versions of genes, called alleles, are passed from parent to offspring during sexual reproduction. In a cross between purple- and white-flowered homozygous parents, the F 1 offspring are all heterozygous, each inheriting a purple allele from one parent and a white allele from the other. Because the purple allele is dominant, it determines the phenotype of the F 1 offspring to be purple, and the expression of the white allele is masked. Only in the F 2 generation is it possible for a white allele to exist in a homozygous state, which causes the white trait to be expressed.

When Mendel did crosses of true-breeding purple- and white-flowered pea plants, the white-flowered trait disappeared from the F1 generation but reappeared in the F2 generation. Use genetic terms to explain why that happened.

Crossing over during meiosis I in the heterozygous parent produces some gametes with recombinant genotypes for the two genes. Offspring with a recombinant phenotype arise from fertilization of the recombinant gametes by homozygous recessive gametes from the double-mutant parent.

When two genes are located on the same chromosome, what is the physical basis for the production of recombinant offspring in a testcross between a dihybrid parent and a double-mutant (recessive) parent?

DNA

Which component is not directly involved in translation?

The ABO blood group is an example of multiple alleles because this single gene has more than two alleles ((IA,IBandi). Two of the alleles, IA and IB , exhibit codominance, since both carbohydrates (A and B) are present when these two alleles exist together in a genotype. IAandIB each exhibit complete dominance over the i allele. This situation is not an example of incomplete dominance because each allele affects the phenotype in a distinguishable way, so the result is not intermediate between the two phenotypes. Because this situation involves a single gene, it is not an example of epistasis or polygenic inheritance.

Which genetic relationships listed in the first column of the two tables above are demonstrated by the inheritance pattern of the ABO blood group alleles? For each genetic relationship, explain why this inheritance pattern is or is not an example.

whether or not the cell is compartmentalized by internal membranes

Which of the following clues would tell you if a cell is prokaryotic or eukaryotic?

replication of the DNA

Which of the following does NOT occur in mitosis?

alignment of the chromosomes at the equator

Which of the following events characterizes metaphase of mitosis?

Exons are cut out before mRNA leaves the nucleus.

Which of the following is not true of RNA processing?

It extends from one end of a tRNA molecule.

Which of the following is not true of a codon?

a single nucleotide insertion downstream of, and close to, the start of the coding sequence

Which of the following mutations would be most likely to have a harmful effect on an organism?

DNA to RNA to protein

Which of the following shows the flow of genetic information?

The law of segregation relates to the inheritance of alleles for a single character. The law of independent assortment of alleles relates to the inheritance of alleles for two characters.

Which one of Mendel's laws describes the inheritance of alleles for a single character? Which law relates to the inheritance of alleles for two characters in a dihybrid cross?

The mRNA farthest to the right (the longest one) started transcription first. The ribosome at the top, closest to the DNA, started translating first and thus has the longest polypeptide.

Which one of the mRNA molecules started being transcribed first? On that mRNA, which ribosome started translating the mRNA first?

prophase, prometaphase, metaphase, anaphase, telophase

Which phases make up the stages of mitosis?

In inversions and reciprocal translocations, the same genetic material is present in the same relative amount but just organized differently. In aneuploidy, duplications, deletions, and nonreciprocal translocations, the balance of genetic material is upset, as large segments are either missing or present in more than one copy. Apparently, this type of imbalance is very damaging to the organism. (Although it isn't lethal in the developing embryo, the reciprocal translocation that produces the Philadelphia chromosome can lead to a serious condition, cancer, by altering the expression of important genes.)

Why are inversions and reciprocal translocations less likely to be lethal than are aneuploidy, duplications, deletions, and nonreciprocal translocations?

Males have only one X chromosome, along with a Y chromosome, while females have two X chromosomes. The Y chromosome has very few genes on it, while the X has about 1,000. When a recessive X-linked allele that causes a disorder is inherited by a male on the X from his mother, there isn't a second allele present on the Y (males are hemizygous), so the male has the disorder. Because females have two X chromosomes, they must inherit two recessive alleles in order to have the disorder, a rarer occurrence.

Why are males affected by X-linked disorders much more often than females?

Crossing over results in new combinations of alleles. Crossing over is a random occurrence, and the more distance there is between two genes, the more chances there are for crossing over to occur, leading to new allele combinations.

Why are specific alleles of two distant genes more likely to show recombination than those of two closer genes?

No. Transcription and translation are separated in space and time in a eukaryotic cell, as a result of the eukaryotic cell's nuclear membrane.

Would the coupling of the processes shown in Figure 17.24 be found in a eukaryotic cell? Explain why or why not.

About one-third of the distance from the vestigial wing locus to the brown eye locus

You design Drosophila crosses to provide recombination data for gene a, which is located on the chromosome shown in Figure 15.12. Gene a has recombination frequencies of 14% with the vestigial wing locus and 26% with the brown eye locus. Approximately where is a located along the chromosome?

The primary structure of a protein is the order of amino acids in a polypeptide, as coded for by the DNA of a gene.

Your body contains tens of thousands of different proteins, each with a specific structure and function. The unique three-dimensional shape of each of these diverse proteins is based on several superimposed levels of structure. Which if the following is an accurate statement about proteins?

transfer RNA (tRNA)

serves as translator molecule in protein synthesis; translates mRNA codons into amino acids

One long DNA molecule, associated with many protiens

what is the best description of the structure of a chromosome in the nucleus of a non-dividing cell?