Problem Set for Genetics

8. Brown eyes are dominant over blue. This is NOT a sex-linked trait. If a blue-eyed, red-green colorblind woman marries a normal visioned man who is homozygous for brown eye color, what kind of children might they expect with respect to these two traits? -Genotype of woman -Genotype of man -What two genotypes are possible in the children?

-Genotype of woman? bbXCXC -Genotype of man? BBXY -What two genotypes are possible in the children?BbXCX or BbXCY

What happens if tryptophan levels are high? Put the following list in order (1-4). The trp repressor-tryptophan complex can now bind to the operator of the trp operon Tryptophan does not need to be produced by the trp operon Tryptophan will bind to the repressor protein, changing its conformation RNA Polymerase is blocked from transcribing the genes needed to synthesize tryptophan

3,1,2,4

What is the direction of synthesis of the new strand?

5' -> 3'

7. Wild-type Drosophila flies have red eyes, but a certain mutant type has white eyes. This trait is known to be controlled by a single pair of alleles. A cross between a red-eyed heterozygous female fruit fly and red-eyed male fruit fly yielded 94 red-eyed and 30 whiteeyed progeny. (a) Using a chi-square test check to see if these data are consistent with the hypothesis that white eyes is caused by a recessive allele. (b) 56 of the red-eyed flies were females. Is this consistent? (Use another chi-square test)

Answer: The calculated chi-square value = 0.043, which is less than the critical value of 3.84 (df=1). The data are consistent with the hypothesis of sex-linked recessive. b)The calculated chi-square value = 2.13, which is less than the critical value of 3.84 (df=1). The data are consistent with the predicted female to male ratio among the wild type (red eyed) offspring.

g. removes RNA primer and replaces with DNA

DNA polymerase I

adds DNA nucleotides to new strand

DNA polymerase III

16. A certain rare human disease is caused by a dominant sex-linked allele. Would you expect this disease to be more common among males or females? Explain your answer.

Female, 2 X's, so twice the likelihood of having a mutant X

In alley cats, the coat color is determined by a gene carried on the X chromosome. At the same time, the alleles are expressed as intermediate (nondominance) inheritance. Genotypes and color are as follows:

Females: XbXb = yellow Males: XbY = yellow X+ Xb = calico X+ Y = black X+ X+ = black

Which strand is synthesized towards the replication fork?

Leading strand

f. joins DNA fragments together

Ligase

1. Which sex is more likely have a recessive, sex-linked trait?

Male

holds DNA strands apart

Single-strand binding proteins

Define the origins of replication.

Site where the replication of a DNA molecule begins, consisting of a specific sequence of nucleotides

What do we mean when we say the two strands of DNA are antiparallel?

The two strands are oriented in opposite directions from one another 5'------------------------3' 3'------------------------5'

A certain species of animal has 18 chromosomes in its diploid cells (2n=18). Describe what the metaphase chromosome arrangement would look like for this species in each of the following stages: mitosis, meiosis I, and meiosis II.

There would be 18 individual chromosomes, each with two chromatids, lined up at the metaphase plate. Meiosis I: There would be 9 bivalents lined up, each consisting of 2 paired homologues, each homologue with two chromatids. Meiosis II: There would be 9 individual chromosomes lined up, each with two chromatids (like mitosis, but cells now haploid).

What is the semiconservative model of replication?

Type of DNA replication in which the replicated double helix consists of one old strand, derived from the parental molecule, and one newly made strand

3) Which type of sex chromosome do you find most sex-linked traits on?

X

10. In mice, the gene C for colored fur is dominant over its allele c for white. The gene B for normal behavior (normal) is dominant over b, which causes waltzing. Give the probable genotypes of the parent mice (each was mated repeatedly) that produce the following offspring. Additionally, use a chi-squared analysis to determine if the offspring deviate from the expected ratios. a. Colored, normal mated with white, normal, producing 29 colored, normal and 10 colored waltzers b. Colored, normal mated with colored, normal, produced 38 colored, normal, 15 colored, waltzers, 11 white, normal, and 4 white, waltzers c. Colored, normal mated with a white, waltzer, produced 8 colored, normal, 7 colored waltzers, 9 white, normal, and 6 white, waltzers.

a)Expect a 3:1 ratio. This does not significantly deviate from expected. b)The combined ratio for the two loci is a 9:3:3:1. The genotypes of the parents were CcVv x CcVv. The ratio does not deviate significantly from expected. c)Should be a 1:1:1:1 ratio. X2 = 0.666. p between 0.9 and 0.1. Does not deviate significantly from expected

5. In humans, hemophilia is a X-linked recessive trait. If a female who is a carrier for hemophilia marries a male with normal blood clotting, answer the following questions. a. What fraction of the female children will have hemophilia? b. What fraction of the female children will be carriers? c. What fraction of the male children will have normal blood clotting? d. What fraction of the male children will be carriers? e. What fraction of the male children will have hemophilia?

a. What fraction of the female children will have hemophilia? 0 b. What fraction of the female children will be carriers? 1/2 c. What fraction of the male children will have normal blood clotting? 1/2 d. What fraction of the male children will be carriers? 0 e. What fraction of the male children will have hemophilia?1/2

Conformational changes of DNA and associated histones turn heterochromatin into euchromatin. DNA in active euchromatin is __________________________________(methylated/unmethylated) and histones in heterochromatin are _____________________________ (acetylated/unacetylated).

unmethylated;acetylated

What is the function of the promoter?

• Site of RNA polymerase binding

6. A man whose father was a hemophiliac, but whose own blood clotting time is normal, marries a normal woman with no record of hemophilia in her ancestry. What is the chance of hemophilia in their children?

Man = XY; woman = XX. No chance of hemophilia in their children

Why are cancer cells immortal, while most body cells have a limited life span?

Many cancer cells have telomerase activity. This suggests that in order for cancer cells to divide indefinitely, they need a way to stabilize the telomere length.

13b: In Netherlands dwarf rabbits, a gene showing intermediate inheritance produces three phenotypes. Rabbits that are homozygous for one allele are small rabbits; individuals homozygous for the other allele are deformed and die; heterozygous individuals are dwarf. If two dwarf rabbits are mated, what proportion of their surviving offspring should be dwarf?

2/3 dwarf. 1/3 small

What happens if lactose is present and glucose is scarce? Put the following list in order (1-7). Start with the effects if lactose levels then proceed to the effects of glucose levels. Now that LacI has been removed for the operator, RNA polymerase can proceed with transcription cAMP binds to CAP regulatory protein, causing it to bind to the promoter of the lac operon The enzymes needed for lactose metabolism must be transcribed when lactose is present cAMP levels increase because glucose is scarce (ATP is not being produced through cell respiration) Lactose binds to the LacI repressor, changing LacI's shape and making it fall off the operator CAP binding causes RNA Polymerase to bind to the promoter (higher affinity) and transcribe the gene at a higher level than before The three enzymes involved in the metabolism of lactose are transcribed and expressed at moderate levels

3,6,1,5,2,7,4

. A rare disease which is due to a recessive allele (a) that is lethal when homozygous, occurs within a specific population at a frequency of one in a million. How many individuals in atown with a population of 14,000 can be expected to carry this allele?

Always start with the homozygous recessive percentage if given, which is equal to q 2 . The frequency of the recessive genotype, q 2 is 1 1,000,000 = 0.000001 ∴q = = = 0.001 = the frequency of allele a AND p + q = 1 thus, p = 1 - q = 1 - 0.001 = 0.999 Therefore, the frequency of allele A =p = 0.999 and the frequency of allele a = q =0.001 Carriers are heterozygous and are equal to 2pq. So, 2pq = 2(0.999)(0.001)(14,000) = 27.97 (This should be reported as 28 individuals upon considering the rounding done throughout the mathematics.)

Allele T, for the ability to taste a particular chemical, is dominant over allele t, for the inability to taste the chemical. Four hundred university students were surveyed and 64 were found to be nontasters. Calculate the percentage of heterozygous students. Assume that the population is in H-Wequilibrium.

Calculate the percent of homozygotes recessive individuals 64 400 = 0.16 or 16% ∴ 0.16 = tt = q 2 ∴q = = 0.40 which is the allelic frequency of t Additionally, p + q = 1 thus p = 1 - q or p = 1 - 0.40 = 0.60 which is the allelic frequency of T Solve for the heterozygote (Tt) term, 2pq 2 pq = 2× 0.60× 0.40 = 0.48 or 48% of the population are heterozygous "tasters" and carriers of the recessive allele

The "on switch" controlling gene expression in eukaryotic cells is chemical modification of __________________________________ of chromatin.

DNA and HISTONES

12b. Cytoplasmic traits in certain species of trees are passed from the male plant to all of its progeny. Compare this observation to cytoplasmic inheritance in humans.

In humans, the gamete with the largest cytoplasmic contribution is the egg, so cytoplasmic inheritance is passed from the female parent to all her children. In certain tree species, the male gamete contributes the majority of the cytoplasm to the zygote, so all the mitochondria and chloroplasts in the zygote are inherited from the male parent.

2) Which parent do sons inherit recessive, sex-linked traits from?

Mother

13. In wheat kernel color is determined by a pair of genes in a quantitative way. Each of the two genes can have two alleles (A1, A2, B1, B2). Kernel color ranges from red, when four type 1 alleles (either A1 or B1) are present, to white, when four type 2 alleles (either A2 or B2) are present. Three intermediate colors (dark pink, medium pink, and light pink) can occur depending on the relative numbers type 1 and type 2 alleles (this example was discussed in lecture). Using the forked-line approach, calculate the expected proportions of the five phenotypes that would be produced by a cross between two wheat plants with medium pink kernels that are heterozygous for both genes: A1A2 B1B2 x A1A2 B1B2 .

Phenotype ratio 1/16 red, 1/4 dark pink, 3/8 pink, 1/4 light pink, 1/16 white

synthesizes RNA primer

Primase

Over the past decade, a significant finding in biology has been the identification of miRNAs and siRNAs and their role in regulating the development of many multicellular organisms. Briefly describe the four different ways these small RNAs influence gene expression

Small interfering RNAs and microRNAs regulate gene expression through at least four distinct mechanisms: (1) cleavage of mRNA, (2) inhibition of translation, (3) transcriptional silencing, or (4) degradation of mRNA.

6. In the peppered moth, Biston betularia, there is a gene that determine body color. Three alleles are possible for this gene. The allele for pale color (m) is recessive. A second allele (M'), which is dominant to m, produces a mottled color called insularia. The third allele (M), which is dominant to both of the other two, produces a melanic moth (very dark colored). A pale colored female moth is mated to a melanic male. If half the progeny are melanic and half are insularia, what were the genotypes of the two parents?

The female must be mm because pale color is recessive. The male must have at least one M allele because his phenotype is melanic. From his phenotype alone, his genotype could be MM, MM', or Mm. But because some of the offspring are insularia (M'm), he must be carrying a M' allele. Thus the male parent's genotype is MM'.

2. The French biologist Cuenot crossed wild, gray-colored mice with white (albino) mice. In the first generation, all were gray. He then crossed the F1 mice among themselves. From many litters, he obtained in the F2 generation 198 gray mice and 72 white mice. (a) Propose a hypothesis to explain these results. Diagram both generations (phenotype and genotype) based on your hypothesis. (b)Test your hypothesis using a chi-square test.

The hypothesis is that gray (G) is dominant to (g)

A woman with blood type A marries a man with blood type O. (a) Given only this information, determine all the blood types that are possible for their children? (b) Which blood types are not possible in the children? (c) Would the answers to (a) or (b) change if it was known that both of the woman's parents had blood type AB? Explain

The woman could have genotypes IAIA or IAi. The man must be genotype ii. Therefore, these parents could produce children with blood types A or O. These parents could not produce children with blood types B or AB. Yes. If the woman's parents were both type AB, it would mean that her genotype could only be IAIA, because neither of her parents would carry the i allele. In this case, all her children would be blood type A because they would always inherit a IA allele from their mother.

3. Which blood group type of the ABO system is known as the universal donor? Which type is the universal recipient? Explain your answers

Type O is the universal donor. Type O red blood cells have neither A nor B type antigen and so do not stimulate antibody production when given to a recipient with any blood type. Type AB is the universal recipient. AB red blood cells have both A and B antigens, so an AB recipient would not produce antibodies when given any blood type.

3. In a population that is in Hardy-Weinberg equilibrium, the frequency of the recessive homozygote genotype of a certain trait is 0.09. Calculate the percentage of individuals homozygous for the dominant allele.

We know that the frequency of the recessive homozygote genotype is q 2 and equal to0.09. ∴ q = = = 0.30 AND we also know that p + q = 1 Thus, p = 1 - q ∴ p = 1 - 0.30 = 0.70 ∴The homozygote dominants are represented by p 2 = (0.70)2 = 0.49 or 49%

Heterochromatin in transcriptionally ___________________ (active or inactive) and euchromatin is tanscriptionally _______________________ (active or inactive).

inactive;active

Why have genes under regulation?

• Not all proteins are required at all times • Not all proteins are diminished quickly • Waste of the cell's resources if all genes were transcribed and translated at all times • By having some genes under regulation, the cell can manage its resources as needed

What is the function of the operator?

• Site of repressor protein binding, regulates the transcription of the gene

What is the difference between a transcription regulation system that uses induction and a system that uses repression?

(1) Induction: Stimulates expression of a gene in response to a specific substrate. Repression: Prevents expression of a gene until a specific substrate is present. (2) Induction: Transcription is normally off and is stimulated in response to a specific substrate. Repression: Transcription is normally on and is shut off in response to a specific product.

untwists and separates strands

Helicase

An ordinary 6-sided die is rolled repeatedly with the following outcome: 63 ones, 31 twos, 54 threes, 46 fours, 51 fives, and 55 sixes. Would you conclude from this data that it is a fair die? Give statistical support (chi-square test) for your conclusion.

You would expect an equal number of each if this is a fair die. The calculated chi-square value is 11.76 which is slightly greater than the table value of 11.07 (for df=5). The data are not consistent, suggesting that the die is not fair.

In humans, the Rh factor genetic information is inherited from our parents, but it is inherited independently of the ABO blood type alleles. In humans, Rh+ individuals have the Rh antigen on their red blood cells, while Rh− individuals do not. There are two different alleles for the Rh factor known as Rh+ and rh. Assume that a dominant gene Rh produces the Rh+ phenotype, and that the recessive rh allele produces the Rh− phenotype. In a population that is in Hardy-Weinberg equilibrium, if 160 out of 200 individuals are Rh+, calculate the frequencies of both alleles.

Always start with the homozygous recessive percentage if given, which is equal to q 2 . If 160 of 200 individuals are Rh+, then it stands to reason that 40 areRh−. q2 0.38 q2 0.000001 Thus, the frequency of q 2 , the Rh− genotype, is 40 200 = 0.20 ∴ q = 0.20 or q = 0.45 AND p + q = 1 so, p = 1 - q = 1 - 0.45 = 0.55 ∴ Rh+ allele frequency = 0.55 and rh allele frequency =0.45

. In a population that is in Hardy-Weinberg equilibrium, 38 % of the individuals are recessive homozygotes for a certain trait. In a population of 14,500, calculate the percentage of homozygous dominant individuals and heterozygousindividuals.

Always start with the homozygous recessive percentage if given, which is equal to q 2 . ∴ q = ∴ q = ∴ q = 0.616 Solving for p is now straightforward: p + q = 1, thus p = 1 - q = 1 - 0.616 =0.384 The homozygous dominant individuals are representedby p 2 = 0.3842 = 0.147 ∴14,500× 0.0147 = 2,132 homozygous dominant individuals The heterozygotes are represented by the 2pq termand 2 pq = 2× 0.616× 0.384 = 0.473 ∴14,500× 0.473 = 6,859 Double check! 0.38×14.500 = 5,510 5,510 + 2,132 + 6,859 =14,501 (due to some rounding)

PROBLEM #6. A very large population of randomly-mating laboratory mice contains 35% white mice. White coloring is caused by the double recessive genotype, "aa". Calculate allelic and genotypic frequencies for this population.

Answer: 35% are white mice, which = 0.35 and represents the frequency of the aa genotype (or q2). The square root of 0.35 is 0.59, which equals q. Since p = 1 - q then 1 - 0.59 = 0.41. Now that we know the frequency of each allele, we can calculate the frequency of the remaining genotypes in the population (AA and Aa individuals). AA = p2 = 0.41 x 0.41 = 0.17; Aa = 2pq = 2 (0.59) (0.41) = 0.48; and as before aa = q2 = 0.59 x 0.59 = 0.35. If you add up all these genotype frequencies, they should equal 1.

Sickle-cell anemia is an interesting genetic disease. Normal homozygous individuals (SS) have normal blood cells that are easily infected with the malarial parasite. Thus, many of these individuals become very ill from the parasite and many die. Individuals homozygous for the sickle-cell trait (ss) have red blood cells that readily collapse when deoxygenated. Although malaria cannot grow in these red blood cells, individuals often die because of the genetic defect. However, individuals with the heterozygous condition (Ss) have some sickling of red blood cells, but generally not enough to cause mortality. In addition, malaria cannot survive well within these "partially defective" red blood cells. Thus, heterozygotes tend to survive better than either of the homozygous conditions. If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria because they are heterozygous (Ss) for the sickle-cell gene?

Answer: 9% =.09 = ss = q2. To find q, simply take the square root of 0.09 to get 0.3. Since p = 1 - 0.3, then p must equal 0.7. 2pq = 2 (0.7 x 0.3) = 0.42 = 42% of the population are heterozygotes (carriers).

PROBLEM #7. Aftergraduation, you and 19 of yourclosest friends (lets say 10 males and 10 females) chartera plane to go on a round-the-world tour. Unfortunately, you all crash land (safely) on a deserted island. No one findsyou and you start a new population totally isolated from the rest of the world. Two of yourfriends carry (i.e. are heterozygous for) the recessive cystic fibrosisallele (c). Assuming that the frequency of this allele does not change as the population grows, what will be the incidence of cystic fibrosison yourisland?

Answer: There are 40 total alleles in the 20 people of which 2 alleles are for cystic fibrous. So, 2/40 = .05 (5%) of the alleles are for cystic fibrosis. That represents p. Thus, cc or p2 = (.05)2 = 0.0025 or 0.25% of the F1 population will be born with cystic fibrosis.

PROBLEM #10 In a given population, only the "A" and "B" alleles are present in the ABO system; there are no individuals with type "O" blood orwith O alleles in this particularpopulation. If 200 people have type Ablood, 75 have type AB blood, and 25 have type B blood, what are the alleleic frequencies of this population (i.e., what are p and q)?

Answer: To calculate the allele frequencies for A and B, we need to remember that the individuals with type A blood are homozygous AA, individuals with type AB blood are heterozygous AB, and individuals with type B blood are homozygous BB. The frequency of A equals the following: 2 x (number of AA) + (number of AB) divided by 2 x (total number of individuals). Thus 2 x (200) + (75) divided by 2 (200 + 75 + 25). This is 475/600 = 0.792 = p. Since q is simply 1 - p, then q = 1 - 0.792 or 0.208

PROBLEM #4. Within a population of butterflies, the color brown (B) is dominant over the color white (b). And, 40% of all butterflies are white. Given this simple information, which is something that is very likely to be on an exam, calculate the following: A. The percentage of butterflies in the population that are heterozygous. B. The frequency of homozygous dominant individuals.

Answers: The first thing you'll need to do is obtain p and q. So, since white is recessive (i.e. bb), and 40% of the butterflies are white, then bb = q2 = 0.4. To determine q, which is the frequency of the recessive allele in the population, simply take the square root of q2 which works out to be 0.632 (i.e. 0.632 x 0.632 = 0.4). So, q = 0.63. Since p + q = 1, then p must be 1 - 0.63 = 0.37. Now then, to answer our questions. First, what is the percentage of butterflies in the population that are heterozygous? Well, that would be 2pq so the answer is 2 (0.37) (0.63) = 0.47. Second, what is the frequency of homozygous dominant individuals? That would be p2 or (0.37)2 = 0.14.

1. On the soap opera "The Young and the Restless", several individuals suffer from a rapid aging syndrome in which a young child is sent off to boarding school and returns three months later an angry teenager. Victims have been known to age up to two decades in variations of the disorder. In the Newman family, siblings Nicholas and Victoria aged from ages six and eight to sixteen and eighteen within a few months. Their parents, Victor and Nikki, are not affected; in fact, they never seem to age at all. a. What is the mode of inheritance of the rapid aging disorder affecting Nicholas and Victoria? b. How so you know what the mode of inheritance is? c. Draw a pedigree to depict this portion of the Newman family

Autosomal recessive; Parents do not have it but kids do. As long as it is not a new mutation, they must be carriers. If it were a new mutation, it would be unlikely both children would have the trait.;

8. What is the difference between dominance and epistasis? Why didn't any pairs of genes used by Mendel show epistasis?

Dominance means that, for a single gene locus, one allele's expression dominates over another. Epistasis means that an allele at one locus (one gene) influences the expression of another allele at a different locus (two genes are involved). The genes that Mendel chose to use in dihybrid crosses were for such unrelated traits that no gene interaction was observed. Epistasis occurs when more than one gene affects the same trait.

11. The agouti fur color in mice actually results from alternating dark and light bands on each hair. Given this information and what you know about the gene interaction from the previous problem, propose a mechanism that could explain how each of the two genes is actually affecting the overall fur color (agouti, black, or albino).

Given the allele we have been using in the previous problem, one explanation would be: One gene (A- or aa) determines whether any dark pigment is produce (A- : pigment produced, aa : albino). The second gene determines the distribution of the pigment on each hair (B- : alternating dark and light bands - agouti, bb : solid pigmented hair - black). When no pigment is produced (aa) it would not matter if the mouse was B- or bb, so albino is the result. In other words, although both genes affect the same trait (fur color), one actually controls the formation of pigment; the other controls the distribution of the pigment when it is produced.

3. Cystic fibrosis is a genetic disorder in homozygous recessives that causes death during the teenage years. If 4 in 10,000 newborn babies have the disease, what are the expected frequencies of the three genotypes in newborns, assuming the population is at Hardy-Weinberg equilibrium? Why is the assumption not strictly correct?

In a population of 10,000 newborn babies, the expected proportion of babies having the disease is 4/10,000 = 0.0004, which is q2. The frequency of the allele q that causes the disease if the square root of this number, 0.0004 = 0.02. You know that p + q = 1, therefore p = 1 - 0.02 = 0.98. Now, you can calculate the expected frequency of the three genotypes: p2 (homozygote, normal) = 0.9604 2pq (heterozygote, normal) = 0.0392 q2 (recessive, cystic fibrosis) = 0.0004 Assuming Hardy-Weinberg in this case is not correct because there is natural selection acting against the allele that causes the disease, given that the children with the disease die before they reproduce (Remember that Natural Selection is defined as differential reproduction and/or survival among the individuals in a population).

3. The allele y occurs with a frequency of 0.8 in a population of clams. Give the frequency of genotypes YY, Yy, and yy. Show your work.

The allele y has a frequency q = 0.8. You know that p + q = 1, then p = 1 - 0.8 = 0.2. Now you can estimate the frequency of each genotype: YY genotype frequency = p2 = 0.04 Yy genotype frequency = 2pq = 0.32 yy genotype frequency = q2 = 0.64

2. An allele W, for white wool, is dominant over allele w, for black wool. In a sample of 900 sheep, 891 are white and 9 are black. Calculate the allelic frequencies within this population, assuming that the population is in H-W equilibrium.

The allelic frequency of w is represented by the q term and the allelic frequency W is represented by the p term. To calculate the value of q, realize that qq or q 2 represents the homozygous recessive individuals or the black sheep in this case. Since there are 9 black sheep, the frequency of black sheep = # individuals = 9 = 0.01 , thus ww = q 2 =0.01 total individuals 900 ∴ q = = = 0.1 Additionally, p + q = 1 thus p = 1 - q or p = 1 - 0.1 or 0.9 ∴ p = W = 0.9 and q = w =0.1

5. A woman with type O blood, whose father has type A and whose mother has type B has a child with type O. There is a dispute over the identity of the child's father. Two men are possible fathers. One is type AB and the other is type A. (a) What is the mother's genotype? (b) Which man could be the father? (c) If this man is the father, what is his genotype? (d) What are the genotypes of the woman's parents?

The mother must have genotype ii because that is the only genotype that produces type O blood Only the type A man could be the father. The man with type AB could not carry an i allele and so could not produce type O children. He would have to be heterozygous (IAi). If he were homozygous, he would not be able to produce type O children. The woman's parents would have genotypes IAi and IBi. This is the only way they could have a daughter who is type O

9. In chickens, comb shape is determined by two interacting genes (R/r and P/p). A walnut comb is produced when at least one dominant R allele and at least one dominant P allele are present (R- P-). A rose comb is produced when at least one dominant R allele is present and the second gene is homozygous recessive (R- pp). A pea comb is produced when the first gene is homozygous recessive and at least one dominant P allele is present (rr P-). When both genes are homozygous recessive, a single comb is produced. Predict the phenotype ratios for each of the following matings: (a) RR PP X rr pp (b) Rr pp X Rr pp (c) Rr Pp X rr pp (d) Rr pp X rr Pp (e) Rr Pp X Rr Pp (f) rr pp X rr Pp

a) All are walnut (Rr Pp) (b)3/4 rose (R- pp), 1/4 single (rr pp) (c) 1/4 walnut (Rr Pp), 1/4 rose (Rr pp), 1/4 pea (rr Pp), 1/4 single (rr pp) (d) 1/4 walnut (Rr Pp), 1/4 rose (Rr pp), 1/4 pea (rr Pp), 1/4 single (rr pp) (e) 9/16 walnut (R-P-), 3/16 rose (R-pp), 3/16 pea (rr P-), 1/16 single (rr pp) (f) 1/2 pea (rr Pp), 1/2 single (rr pp)

In an experiment designed to study the inheritance of flower color in four-o'clocks, two plants with pink flowers were crossed. In the progeny from this cross, there were 42 plants with red flowers, 86 with pink flowers, and 39 with white flowers. Using a chi-square test, determine whether those numbers are consistent

Using df = 2, 0.258 is less than the table value of 5.99, so the data are consistent with incomplete dominance.

In corn, kernel color is governed by a dominant allele for white color (W) and by a recessive allele (w). A random sample of 100 kernels from a population that is in H-W equilibrium reveals that 9 kernels are yellow (ww) and 91 kernels are white. (a) Calculate the frequencies of the yellow and white alleles in this population. (b) Calculate the percentage of this population that isheterozygous.

a)

PROBLEM #9. Cystic fibrosis is a recessive condition that affects about 1 in 2,500 babies in the Caucasian population of the United States. Please calculate the following. A. The frequency of the recessive allele in the population. B. The frequency of the dominant allele in the population C. The percentage of heterozygous individuals (carriers) in the population.

a)A. Answer: We know from the above that q2 is 1/2,500 or 0.0004. Therefore, q is the square root, or 0.02. That is the answer to our first question: the frequency of the cystic fibrosis (recessive) allele in the population is 0.02 (or 2%). b)Answer: The frequency of the dominant (normal) allele in the population (p) is simply 1 - 0.02 = 0.98 (or 98%). c)A. Answer: Since 2pq equals the frequency of heterozygotes or carriers, then the equation will be as follows: 2pq = (2)(.98)(.02) = 0.04 or 1 in 25 are carriers.

You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following A. The frequency of the "aa" genotype B. The frequency of the "a" allele. C. The frequency of the "A" allele D. The frequencies of the genotypes "AA" and "Aa." E. The frequencies of the two possible phenotypes if "A" is completely dominant over "a."

a)Answer: 36%, as given in the problem itself. b)Answer: The frequency of aa is 36%, which means that q2 = 0.36, by definition. If q2 = 0.36, then q = 0.6, again by definition. Since q equals the frequency of the a allele, then the frequency is 60%. c)Answer: Since q = 0.6, and p + q = 1, then p = 0.4; the frequency of A is by definition equal to p, so the answer is 40%. d)Answer: The frequency of AA is equal to p2, and the frequency of Aa is equal to 2pq. So, using the information above, the frequency of AA is 16% (i.e. p2 is 0.4 x 0.4 = 0.16) and Aa is 48% (2pq = 2 x 0.4 x 0.6 = 0.48). e)Answers: Because "A" is totally dominate over "a", the dominant phenotype will show if either the homozygous "AA" or heterozygous "Aa" genotypes occur. The recessive phenotype is controlled by the homozygous aa genotype. Therefore, the frequency of the dominant phenotype equals the sum of the frequencies of AA and Aa, and the recessive phenotype is simply the frequency of aa. Therefore, the dominant frequency is 64% and, in the first part of this question above, you have already shown that the recessive frequency is 36%.

PROBLEM #8 You sample 1,000 individuals from a large population for the MN blood group, which can easily be measured since co-dominance is involved (i.e., you can detect the heterozygotes). They are typed accordingly: a)The frequency of each allele in the population B. Supposing the matings are random, the frequencies of the matings. C. The probability of each genotype resulting from each potential cross.

a)Answer: Since MM = p2, MN = 2pq, and NN = q2, then p (the frequency of the M allele) must be the square root of 0.49, which is 0.7. Since q = 1 - p, then q must equal 0.3. b)Answer: This is a little harder to figure out. Try setting up a "Punnett square" type arrangement using the 3 genotypes and multiplying the numbers in a manner something like this:Note that three of the six possible crosses are unique (*), but that the other three occur twice (i.e. the probabilities of matings occurring between these genotypes is TWICE that of the other three "unique" combinations. Thus, three of the possibilities must be doubled. MM x MM = 0.49 x 0.49 = 0.2401 MM x MN = 0.49 x 0.42 = 0.2058 x 2 = 0.4116 MM x NN = 0.49 x 0.09 = 0.0441 x 2 = 0.0882 MN x MN = 0.42 x 0.42 = 0.1764 MN x NN = 0.42 x 0.09 = 0.0378 x 2 = 0.0756 NN x NN = 0.09 x 0.09 = 0.0081 c)Answer: You may wish to do a simple Punnett's square monohybrid cross and, if you do, you'll come out with the following result: MM x MM = 1.0 MM MM x MN = 0.5 MM 0.5 MN MM x NN = 1.0 MN MN x MN = 0.25 MM 0.5 MN 0.25 NN MN x NN = 0.5 MN 0.5 NN NN x NN = 1.0 NN

There are 100 students in a class. Ninety-six did well in the course whereas four blew it totally and received a grade of F. Sorry. In the highly unlikely event that these traits are genetic rather than environmental, if these traits involve dominant and recessive alleles, and if the four (4%) represent the frequency of the homozygous recessive condition, please calculate the following: A.The frequency of the recessive allele. B. The frequency of the dominant allele C. The frequency of heterozygous individuals

a)Answer: Since we believe that the homozygous recessive for this gene (q2) represents 4% (i.e. = 0.04), the square root (q) is 0.2 (20%). b)Answer: Since q = 0.2, and p + q = 1, then p = 0.8 (80%). c)Answer: The frequency of heterozygous individuals is equal to 2pq. In this case, 2pq equals 0.32, which means that the frequency of individuals heterozygous for this gene is equal to 32% (i.e. 2 (0.8)(0.2) = 0.32).

PROBLEM #5. A rather large population of Biology instructors have 396 red-sided individuals and 557 tan-sided individuals. Assume that red is totally recessive. Please calculate the following: A.The allele frequencies of each allele B. The expected genotype frequencies C. The number of heterozygous individuals that you would predict to be in this population. D. The expected phenotype frequencies. E. Conditions happen to be really good this year for breeding and next year there are 1,245 young "potential" Biology instructors. Assuming that all of the Hardy-Weinberg conditions are met, how many of these would you expect to be red-sided and how many tan-sided?

a)Answer: Well, before you start, note that the allelic frequencies are p and q, and be sure to note that we don't have nice round numbers and the total number of individuals counted is 396 + 557 = 953. So, the recessive individuals are all red (q2) and 396/953 = 0.416. Therefore, q (the square root of q2) is 0.645. Since p + q = 1, then p must equal 1 - 0.645 = 0.355. b)Answer: Well, AA = p2 = (0.355)2 = 0.126; Aa = 2(p)(q) = 2(0.355)(0.645) = 0.458; and finally aa = q2 = (0.645)2 = 0.416 (you already knew this from part A above). c)Answer: That would be 0.458 x 953 = about 436. d)Answer: Well, the "A" phenotype = 0.126 + 0.458 = 0.584 and the "a" phenotype = 0.416 (you already knew this from part A above). e)Answer: Simply put, The "A" phenotype = 0.584 x 1,245 = 727 tan-sided and the "a" phenotype = 0.416 x 1,245 = 518 red-sided (or 1,245 - 727 = 518).

Trying to understand the inheritance of the dominant yellow gene in mice, researchers mated two yellow heterozygous mice. A typical result was 56 yellow progeny to 31 wild-type. (a) Use a chi-square test to determine if the outcome of this cross is consistent with the usual 3:1 ratio predicted by Mendelian inheritance for a dominant gene. (b) You will find that the chi-square test done in part (a) indicates the data are not consistent. Now try the hypothesis that the dominant allele is lethal in the homozygous condition. Repeat the chi-square test.

a)Chi-square = 5.25 which is greater than the table value of 3.84 (df=1). Inconsistent b)Chi-square = 0.21 which is less than the table value of 3.84 (df=1). Consistent

2. Achnodroplasia is a common form of hereditary dwarfism that causes very short limbs, stubby hands, and an enlarged forehead. Below are three pedigrees depicting families with this specific type of dwarfism. a. Identify the individuals using Roman numerals for the generation and Arabic numerals to identify the individuals. b. What is the most likely mode of inheritance? Most likely c. Explain your reasoning.

a)I-1, I-4, I-5 II-1, II-2, II-4, II-5, II-9 III-1, III-2 b)Autosomal Dominant (also possible, but less likely: x-linked dominant c)Every generation, both males and females affected. But to rule out x-linked dominant, you would need more information. All males in these pedigrees have daughters that are affected. If there were affected males without affected daughters, then you could rule out x-linked dominant.

10. As discussed in class, one form of fur color in mice is controlled by the interaction of two gene resulting in three phenotypes: agouti, black, and albino. Two agouti mice are crossed repeatedly, and the following offspring are produced: 46 agouti, 16 black, and 23 albino. (a) Based on this cross, which common multiple gene inheritance pattern is operating here? (b) Given your answer to part a, what were the genotypes of the parents? (c) For each of the following crosses, give the possible genotypes of the parents. -Agouti x Agouti produces 37 agouti and 12 black -Black x Albino produces 27 agouti and 24 albino -Black x Black produces 28 black and 10 albino

a)The observed ratio seems to fit the 9:3:4 ratio very closely. This is indicative of the gene interaction pattern recessive epistasis. b)9:3:4 is the expected ratio for recessive epistasis when both parents are heterozygous for each gene, so the parents must be Aa Bb x Aa Bb. c)A- Bb x A- Bb. The B gene must be heterozygous in both parents, because some offspring are black (bb). The exact genotype for the A gene cannot be determined completely, but both parents must have at least one copy of the dominant A allele. Additionally, we know that one or both parents must be homozygous (AA), because there were no albino (aa) offspring.;Aa bb x aa BB. The black parent must be heterozygous for the A gene because some offspring were albino (aa). The albino parent must be homozygous for the B gene, otherwise some offspring would have been black (bb;Aa bb x Aa bb. Both parents must be heterozygous for the A gene, otherwise there could be no albino offspring.

12. In summer squash, there are two pairs of alleles that determine fruit color. The two genes sort independently. Two white-fruited plants are crossed. Both parents are known to be heterozygous for both genes. The cross produces the following offspring: 20 green-fruited plants, 58 yellow-fruited plants, and 218 white-fruited plants. (a) Based on the observed ratio, which common type of epistasis is operating here? That is, which kind of modified dihybrid ratio most closely fits these data? (b) List the four genotype classes in the offspring and give the corresponding phenotype of each. (c) If a doubly heterozygous white plant is crossed with a green plant, what phenotype ratio would you expect in the progeny?

a)The ratio comes close to 12:3:1, which indicates dominant epistasis. b)A-B- = white A-bb = white aaB- = yellow aabb = green c)

3. In the year 2374, humans finally developed the technology necessary for time travels. You are a scientist interested in the population genetics of extinct animals. Taking advantage of this technological advance, you decide to go to the past 8 million years to conduct field work in Venezuela to study a population of Phoberomys pattersoni, the world's largest extinct rodent weighing approximately 700 kg (1500 lb) and looking vaguely like a giant guinea pig. The coat color of this rodent varies between tan (dominant) and brown (recessive). Assume the population is in Hardy-Weinberg equilibrium. You observed 336 tan Phoberomys and 64 brown Phoberomys during your study. a. What is the frequency of the homozygous recessive genotype? b. What is the allelic frequency of the dominant allele in the population? c. Of the animals observed, how many are heterozygous? d.You make another trip to Venezuela and this time you observe 650 animals. How many of the 650 animals would you expect to be tan, assuming the population is still in Hardy-Weinberg equilibrium? e.How many of these tan animals are homozygous for the dominant allele? f. How many of the 650 animals would you expect to be brown, assuming the population is still in Hardy-Weinberg equilibrium? g. As you observe the animals, you count 200 brown Phoberomys and 450 tan. Conduct a chi-square test to determine if your observations are significantly different from what you expect. h. What does your result in (g) mean in terms of the genetic composition of this population of Phoberomys?

a)There are 336 + 64 = 400 animals in the population. 64 are homozygous recessive (brown) Frequency of homozygous recessive = q2 = 64/400 = 0.16 b)Since q2 = 0.16, take the square root to get q = 0.4 Remember that p + q = 1 (formula for allele frequencies) Frequency of the dominant allele p = 0.6 c)Frequency of heterozygote genotypes = 2pq 2pq = (2)*(0.6)*(0.4) = 0.48 Multiply the frequency by the total population size to get the number of animals: (0.48)*(400) = 192 d)If the population is still in H-W equilibrium, then the allele frequencies would be the same: p = 0.6, q = 0.4 The tan phenotype is produced by 2 genotypes, homozygous dominant and heterozygous Frequency of these genotypes is p2 and 2pq p2 = (0.6)2 = 0.36 and 2pq = (2)*(0.6)*(0.4) = 0.48 Multiple each of these frequencies by the total population size (0.36)*(650) = 234, and (0.48)*(650) = 312. Then, 234 + 312 = 546 tan e)We calculated this in the previous problem: p2 = (0.6)2 = 0.36, and then we multiply the genotypic frequency by the total population: (0.36)*(650) = 234 f)Brown animals are homozygous recessive Frequency of brown is q2 = (0.4)2 = 0.16 Multiply the frequency by the total population size (0.16)*(650) = 104 NOTE: You can check your math. Brown + tan should equal 650 g)The calculated X2 is 105.5 There are 2 phenotypes (brown and tan), so there is 1 degree of freedom (2 - 1 = 1) The theoretical X2 for 1 degree of freedom is 3.841, which is much smaller than our calculated one. Therefore, we reject the null hypothesis that the population of 650 is in H-W equilibrium. Our observations are significantly different from our expectation, assuming H-W equilibrium. h)The chi-square test tells us that this population is not in H-W equilibrium with the population that we first observed. This means that evolution has been occurring - at least one of the assumptions of H-W equilibrium is not being met. Perhaps natural selection is occurring, or perhaps there has been some gene flow. We don't know how the H-W equilibrium has been violated, but we do know that the genetic composition of this population is different from that of the original population of 400.

7. In a particular plant, there is a gene with five possible alleles, B1, B2, B3, B4, and B5. (a) Given that two plants with genotypes B2B4 and B1B5 are mated, what types of progeny, and in what proportions, would you expect? (b) For the same gene, if a progeny from a single mating has equal numbers of B1B2 and B2B4 individuals (and no other genotypes), what are the parents' genotypes?

b) The only way to account for the observed offspring is that one parent has genotype B2B2 and the other has genotype B1B4 . (Just construct a 2-box Punnett square to see what the parent gametes must be.)

5. Caleb has a double row of eyelashes, which he inherited from his mother as a dominant trait. His maternal grandfather is the only other relative to have the trait. Veronica, a woman with normal eyelashes, falls madly in love with Caleb, and they marry. Their first child, Polly, has normal eyelashes. Now Veronica is pregnant again and hopes they will have a child who has double eyelashes. a. What chance does a child of Veronica and Caleb have of inheriting double eyelashes? - b. Draw a pedigree of this family.

x-linked recessive mutation. Unless Veronica is heterozygous, it will not appear in the next generation. Caleb can only give the mutant X to his daughters, and a Y to his sons. For his daughters to show the trait, Veronica would also need to give a mutant X to the daughter too. Polly is a carrier of the trait and her sons would have a 50% chance of double eyelashes.

Distinguish between the leading and the lagging strands during DNA replication.

The leading strand is the new complementary DNA strand synthesized continuously along the template strand toward the replication fork in the mandatory 5' -> 3' direction. The lagging strand is a discontinuously synthesized DNA strand that elongates by means of Okazaki fragments, each synthesized in a 5' -> 3' direction away from the replication fork.

Eukaryotic genes can be introduced into bacteria by recombinant DNA techniques. If the introduced gene encodes a protein that is also found in bacteria—for example, a universally used glycolysis enzyme—then, expression of the eukaryotic gene may produce a protein that functions in the bacterial cell. The mouse gene for a glycolysis enzyme is introduced into an E. coli cell that has a mutant gene for the bacterial version of the same enzyme. Even though the mouse enzyme should function in the bacterial cell and restore the cell's ability to perform glycolysis, it does not. Provide two possible reasons why this experiment does not work and propose a solution to overcome one of the problems you suggest.

(1) The prokaryotic RNA polymerase does not recognize the eukaryotic promoter, so there is no transcription. Solution: A eukaryotic gene that will be introduced into bacteria for expression must be engineered to have a prokaryotic promoter. (2) There is no Shine-Dalgarno sequence in the eukaryotic gene promoter, so even if an mRNA is produced, the prokaryotic ribosome will not efficiently recognize the translational start site. Solution: If a prokaryotic promoter is used to express this gene, it should have the Shine-Dalgarno sequence within it. (3) Prokaryotic genes do not have introns, so the cells do not have splicing enzymes and cannot remove introns from RNAs. If the mouse gene has introns, they will not be removed after it is transcribed in E. coli. If the RNA is translated, then the additional intron sequences in the unspliced RNA will result in an altered protein with extra amino acids or, if a stop codon just happens to be in the intron sequence, an incomplete protein. Solution: Expression of the functional protein in bacteria requires engineering a version of the gene that does not have introns.

(b) 56 of the red-eyed flies were females. Is this consistent? (Use another chi-square test)

(62-56) 2 /62 + (62-68) 2 /62 = 0.58 + 0.58 X2 = 1.16 df = 1 p between 0.1 and 0.9 Conforms to expected

15. Wild-type Drosophila flies have red eyes, but a certain mutant type has white eyes. This trait is known to be controlled by a single pair of alleles. A cross between a red-eyed heterozygous female fruit fly and red-eyed male fruit fly yielded 94 red-eyed and 30 white-eyed progeny. (a) Using a chi-square test check to see if these data are consistent with the hypothesis that white eyes is caused by a recessive sex-linked allele.

(93-94)2 /93 + (30-31)2 /31 = 0.011 + 0.0323 X2 = 0.0432 df = 1 p greater than 0.975. Conforms to expected

2. In a particular plant species, 2n = 12. For this organism, how many chromosomes, and how many DNA molecules will be present per cell for each of the following? (a) Leaf cell in G1 12, 12 (b) Leaf cell in G2 12, 24 (c) Root meristem cell in metaphase of mitosis. 12, 24 (d) Root meristem cell in anaphase of mitosis. 12, 24 (e) Root meristem cell in telophase of mitosis (cytokinesis complete) 12, 12 (f) Microspore mother cell (in anther) in prophase I of meiosis. 12, 24 (g) Microspore mother cell in metaphase II of meiosis. 6, 12 (h) Pollen grain (after meiosis and cytokinesis complete) 6, 6

(a) 12, 12 (b)12, 24 (c) 12, 24 (d)12, 24 (e)12, 12 (f)12, 24 (g)6, 12 (h)6, 6

3. The haploid chromosome number in cats is 19. Sex is determined by the XY system as in humans. Answer the following: (a) How many chromosomes are present in a zygote? (b) How many sex chromosomes are present in a sperm cell? (c) How many autosomes are present in a liver cell? (d) How many X chromosomes are present in an egg cell? (e) How many Y chromosomes are present in an egg cell? (f) During meiosis in the female cat, how many bivalents would be present in the primary oocyte?

(a) 38 (b) 1 (c)36 (d) 1 (e) 0 (f) 19

The frequency of two alleles in a gene pool is 0.19 (A) and 0.81(a). Assume that the population is in Hardy-Weinberg equilibrium. (a) Calculate the percentage of heterozygous individuals in the population (b) Calculate the percentage of homozygous recessives in the population.

(a) According to the Hardy-Weinberg Equilibrium equation, heterozygotes are represented by the 2pq term. Therefore, the number of heterozygous individuals (Aa) is equal to 2pq which equals 2 × 0.19 × 0.81 = 0.31 or 31% (b)The homozygous recessive individuals (aa) are represented by the q 2 term in the H-W equilibrium equation which equals 0.81 × 0.81 = 0.66 or 66%

7. Consider the following crosses in Drosophila. The two traits being investigated involve eye color and the presence or absence of wing crossveins. The outcomes of four crosses are shown below. (a) Which eye color is dominant and which is recessive? (b) Which wing vein type is dominant and which is recessive? (c) What is the most likely genotype of each parent for each of the four crosses?

(a) We can consider the two traits separately. First looking at the eye color trait, in cross 2, both parents have red eyes, but offspring can have either red or orange. This means that both parents must be heterozygous, so red is dominant (b) Now looking at just the wing trait-in cross 4, both parents have crossveins, but some offspring are crossveinless. This means that both parents must be heterozygous, so crossveins is dominant. (c) Cross 1: rrVv x rrVv, cross 2: RrVv x Rrvv, cross 3: R-vv x R-Vv (at least one parent must be RR, cross 4: RrVv x RrVv

In fruit flies, eye color is carried on the X chromosome. The allele for red eyes is dominant over its recessive allele, white eyes. Cross a homozygous red-eyed female to white-eyed male. -What is the genotype of the male? -What is the genotype of the female? -How many genotypes are possible among the offspring? -How many phenotypes are possible among the offspring? -What is the probability of getting offspring that are red-eyed males? -What is the probability of getting offspring that are white-eyed males? -What is the probability of getting offspring that are red-eyed females? -What is the probability of getting offspring that are white-eyed females?

-What is the genotype of the male? Xr Y -What is the genotype of the female?XRXR -Genotypes:XRXr XRY -Phenotypes:½ Red eyed female, ½ Red eyed male -How many genotypes are possible among the offspring? 2 -How many phenotypes are possible among the offspring? 2 -What is the probability of getting offspring that are red-eyed males? 1/2 -What is the probability of getting offspring that are white-eyed males? 0 -What is the probability of getting offspring that are red-eyed females? 1/2 -What is the probability of getting offspring that are white-eyed females?0

Make a detailed list of the steps that occur in the synthesis of a new strand.

.1. Helicase unwinds the parental double helix. 2. Molecules of single-stranded binding protein stabilize the unwound template strands. 3. The leading strand is synthesized continuously in the 5' ->3' direction by DNA polymerase III after being primed by primase. 4. Primase begins synthesis of the RNA primer for the lagging strand. 5. DNA polymerase III synthesizes discontinuously the lagging strand in the 5' -> 3' direction. 6. DNA polymerase I removes all the RNA primer sections and replaces them with DNA nucleotides. 7. The replacement of the primer with DNA leaves the new DNA nucleotides with a free 3' end. DNA ligase joins the free 3' end to its adjacent 5' end, forming a continuous and unbroken strand of DNA on both the leading and lagging strands.

17. In Drosophila there is an autosomal gene that determines body color. Ebony body (e) is recessive to the wild type (gray) body (e+). A second gene, which is X-linked, controls wing size. Normal wing size (m+) is dominant to the mutant miniature wings (m). A female with miniature wings that is heterozygous for the body color gene is mated with an ebony body male (his wings are normal). (a) What are the genotypes of the parents? (b) Using a Punnett square or the forked-line method, determine the expected proportions of each possible progeny phenotype. Since there is a sex-linked gene here, be sure to include consider sex as part of the phenotype.

1) Xm Xm e+e x Xm+ Y e e 2)1:1:1:1 ratio - grey normal wing female, ebony normal wing female, grey miniature wing male, ebony miniature wing male

Explain what is meant by 5' and 3' ends of the nucleotide.

5' end is the exposed phosphate end (attached to the 5th carbon of the sugar) 3' end is the exposed "OH" of the sugar (attached to the 3rd carbon of the sugar)

What is at the 5' end of DNA? What about the 3' end?

5' end is the exposed phosphate end (attached to the 5th carbon of the sugar) 3' end is the exposed "OH" of the sugar (attached to the 3rd carbon of the sugar)

What happens if lactose levels are low? Put the following list in order (1-5). RNA polymerase is blocked from transcribing the genes for the lactose metabolizing enzymes When RNA polymerase binds to the promoter, it cannot get past the LacI repressor protein The enzymes B-galactosidae, B-galacosidae permease, and transacetylase are not required by the cell due to low levels of lactose Lactose does not bind to the repressor protein, LacI LacI, a repressor protein, is bound to the operator, which follows the promoter

5,4,1,2,3

Explain how the Lac Operon is regulated under the following conditions: a.Lactose is absent from the environment b.Lactose is present and glucose is absent c.Lactose and glucose are both present in the environment Include in your answer the roles of the Repressor, CAP, c-AMP, RNA Polymerase and lactose

A. When lactose is absent from the environment, the RNA polymerase will bind to the lacI gene promoter and transcribe the lac repressor mRNA. This mRNA will then be used to produce the lac repressor, which will bind to the Lac operator site. This will inhibit the lac operon from being transcribed. B. When lactose is present and glucose is absent, allolactose will bind to the lac repressor, thereby inhibiting the lac repressor from binding to the operator site. The RNA polymerase can then bind to the lac operon promoter and transcribe an mRNA containing the coding sequence for the lacZ, LacY and LacA genes. These will then be translated into the b-galactosidase, permease and transacetylase proteins. C. When lactose and glucose are both present, it is advantageous to use glucose instead of lactose. Through a positive control method, the lac operon is turned off through catabolite repression involving the CAP protein (catabolite-activating protein). CAP must be bound to the promoter of the lac operon in order for the RNA polymerase to bind to the lac promoter. CAP cannot bind to the lac promoter if the cAMP levels are low. High glucose levels decrease the level of cAMP available through inhibiting the enzyme that converts ATP to cAMP.

A dominant gene "A" governs the production of purple anthocyanin pigment in tomato stems and its recessive allele "a" produces green stem. A dominant gene "H" controls the development of hairy stems and its recessive allele "h" produces hairless stems. A dihybrid purple, hairy plant is test crossed (backcrossed with a parent that is homozygous recessive for both traits) and produces: 73 purple, hairy 12 purple, hairless 75 green, hairless 9 green, hairy A) Show a Punnett square or Branched Diagram for the predicted cross. What is the predicted purple:green ratio? What is the predicted ratio of the hairy:hairless? What is the ratio of both if sorted independently? B) Is the purple:green ratio among the test cross progeny compatible with the expectation for independent segregation? Support your answer using chi-square analysis. (hint: just analyze the cross of the purple/green first) C) Is the hairy:hairless ratio among the test cross progeny compatible with the expectation for independent segregation? Support your answer using chi-square analysis. (hint: just analyze the cross of the hairy/hairless first) D) Do the two genes assort independently (are the observed progeny in line with what is expected if the traits are segregating independently)? Support your answer using chisquare analysis. (cross both traits - dihbrid cross to

AaHh x aahh You would expect each group to have equal numbers. 1:1:1:1 ratio b) c)Because .50 < p < .80, accept null hypothesis, therefore genes segregate normally d)For a test cross, one would expect the four phenotypes to be found in equal numbers. Clearly, that is not the case. A chi-square analysis would result in a very large chi-square value, and hence a very low p-value resulting in a rejection of the null hypothesis. Thus, independent assortment is not occurring and the genes must be linked since they are segregating as expected.

Elizabeth is married to John, and they have four children. Elizabeth has a straight nose (recessive) and is able to roll her tongue (dominant). John is also able to roll his tongue, but he has a convex (Roman) nose (dominant). Of their four children, Ellen is just like her father, and Dan is just like his mother. The other children—Anne, who has a convex nose, and Peter, who has a straight nose—are unable to roll their tongues. Please answer the following questions about this family. a. What are the genotypes of Elizabeth and John? b. Elizabeth's father was a straight-nosed roller, while her mother was a convex-nosed nonroller. What can you figure out about their genotypes? c. John's father was a straight-nosed roller, while his mother was a convex-nosed roller. What can you determine about their genotypes? d. Diagram the three described generations of this family in accepted pedigree form, including the phenotypes for these two traits.

Again, this begins with a simple pedigree analysis problem. a. Using the Rules for Pedigree analysis, we can figure out that Elizabeth is homozygous for straight nose, and heterozygous for tongue rolling (because her children include non-rollers). John is heterozygous for both nose and tongue rolling (because their children include both straight noses and non-rollers). b. Because Elizabeth has a straight nose, her mother must be heterozygous for noses. Elizabeth only got one of her straight nose alleles from her father; the other had to come from her mother. We know Elizabeth is heterozygous for tongue rolling, which means that she got a rolling allele from one parent and a non-rolling allele from the other. Mom had to give her the non-rolling allele, so she got the rolling allele from Dad. However this does not allow us to conclude that her father was homozygous for tongue rolling. Rule #5 tells us that we may not guess about his second allele, and we don't have enough information to figure it out. c. The only thing we can figure out about John's parents is that his father was homozygous for straight nose. We know only that they have at least one dominant allele for the other genes in question, but nothing about John helps us figure out whether they are hiding recessives or not. d. For our little pedigree, we will designate that the left side of each symbol represents nose shape and the right side represents tongue rolling. Open symbols represent the dominant phenotypes and shaded symbols represent recessive phenotypes. As always, circles are females and squares are males. In constructing a pedigree "mating pairs" are connected by a horizontal line. Offspring are connected by a descending vertical line. Siblings are connected by horizontal brackets. So here's our pedigree:

In garden peas, long stems are dominant to short stems, and yellow seeds are dominant to green seeds. 100 long/yellow pea plants, all of which had one short/green parent, are interbred (bred to each other). 1600 progeny result. Please answer the following questions about these progeny. a. Assuming that these two genes are unlinked, about how many long/green pea plants would you expect to find among the offspring? b. What ratio of yellow to green seed color would you expect among the offspring? c. What would you expect the overall phenotypic ratio among the 1600 offspring to be (taking into consideration both traits)?

As is so often true for genetics problems, this begins with pedigree analysis. Assign symbols to the alleles, such as L for long (dominant) and t for short, and Y for yellow (dominant) and y for green. Always make sure you assign legal symbols, meaning that they must be different versions of the same symbol. (In other words, assigning T for tall and s for short is illegal.) As our tall, yellow plant had a short, green parent, it must be heterozygous for both genes. Thus, this is a dihybrid cross. Here's the mating: Parents: LlYy X LlYy Gametes: LY LY Ly Ly lY lY ly ly LY Ly lY ly LY LLYY LLYy LlYY LlYy Ly LLYy LLyy LlYy Llyy lY LlYY LlYy llYY llYy ly LlYy Llyy llYy llyy Genotypic Ratio: 1 LLYY 2 LlYY 1 LLyy 2 Llyy 1 llYY 2 LLYy 1 llyy 2 llYy 4 LlYy Phenotypic Ratio: 9 Long Yellow : 3 Long Green : 3 Short Yellow : 1 Short Green d. Out of 1600 offspring, 300 (3/16) would be expected to be Long Green. e. This question asks about only one of the genes. You would expect a 3 Yellow : 1 Green ratio. f. 9 Long Yellow : 3 Long Green : 3 Short Yellow : 1 Short Green. Note that you would never present the numbers without the descriptions. 9:3:3:1 is never the correct answer to this kind of question.

Mrs. And Mr. Smith both have widow's peaks (dominant). Their first child also has a widow's peak, but their second child doesn't. Mr. Smith accuses Mrs. Smith of being unfaithful to him. Is he necessarily justified? Why or why not? Work the genetics problem predicting the frequencies of the versions of this trait among their prospective children.

Assign symbols. A good choice here would be W for widow's peak and w for no widow's peak. As both parents here have the dominant trait, they must each have at least one W. Their first child obviously inherited this allele from at least one of them, as he also possesses a widow's peak. The second child does not, and thus must be ww. But note that there is no evidence here that prevents us from assuming that both of these parents could be carrying hidden recessive w alleles, so Mr. Smith is not justified. Here's the Punnett's Square, assuming these two are truly the parents of both children.Genotypic ratio predicted: 1 WW : 2 Ww : 1 ww Phenotypic ratio predicted: 3 Widow's Peak : 1 No Widow's Peak

Mr. and Mrs. Anderson both have tightly curled hair. (The hair form gene shows incomplete dominance. There are two alleles, curly and straight. The heterozygote has wavy hair.) The Andersons have a child with wavy hair. Mr. Anderson accuses Mrs. Anderson of being unfaithful to him. Is he necessarily justified? Why or why not?

Assign symbols. Be careful with this one, as your alleles have incomplete dominance. It isn't a good idea to use capital and lower case letters, since that convention predisposes us to think in terms of complete dominance. One good choice would be to us superscripts: C1 for curly; C2 for straight. Because this gene's alleles have incomplete dominance, everyone's genotype is revealed by his or her phenotype, so our curly-haired parents must both be C1 C1 . The child with wavy hair must be C1 C2 . So Mr. Anderson is justified. Of course, there is a small (very small) possibility that Junior's straighthair allele could have arisen as a new mutation, but other hypotheses are more likely.

7. Brown eyes are dominant over blue eyes. This is NOT a sex-linked trait. Red-green color blindness is an X-linked recessive trait. Cross a brown-eyed, red-green colorblind male (whose mother had blue eyes) with a normal blue-eyed female (whose father was red-green colorblind). -Genotype of the male -Genotype of the female -What is the probability of getting offspring that are blue-eyed carrier females? (Remember that a carrier is a female that carries one copy of the sex-linked allele, but does not have the disease.) -What is the probability of getting offspring that are blue-eyed? -What is the probability of getting offspring that are blue-eyed colorblind males? -What is the probability of getting offspring that are brown-eyed carrier females? -What is the probability of getting offspring that are blue-eyed normal males? -What is the probability of getting offspring that are colorblind?

BbXCY Genotype of the male bbXCX Genotype of the female 1/8 What is the probability of getting offspring that are blue-eyed carrier females? (Remember that a carrier is a female that carries one copy of the sex-linked allele, but does not have the disease.) ½ What is the probability of getting offspring that are blue-eyed? 1/8 What is the probability of getting offspring that are blue-eyed colorblind males? 1/8 What is the probability of getting offspring that are brown-eyed carrier females? 1/8 What is the probability of getting offspring that are blue-eyed normal males? 1/2 What is the probability of getting offspring that are colorblind?

In the garden pea, yellow cotyledon color is dominant to green, and inflated pod shape is dominant to the constricted form. Considering both of these traits jointly in self-fertilized dihybrids, the progeny appeared in the following numbers: 193 green, inflated 184 yellow constricted 556 yellow, inflated 61 green, constricted Do these genes assort independently? Support your answer using Chi-square analysis.

Because .80 < p < .95, accept null hypothesis, therefore genes sort independently

9. You are doing a cross with Drosophila using the following two traits. Curly wings is dominant over straight wings, and round eyes is dominant over elliptical eyes. You cross a female fly that is known to be heterozygous for both genes with a male that is heterozygous for the wing gene but has elliptical eyes. This cross produces 74 flies with curly wings and round eyes, 61 with curly wings and elliptical eyes, 24 with straight wings and round eyes, and 21 with straight wing and elliptical eyes. Calculate the expected phenotype ratios for this cross, then use the chi-square test to see if the observed data are consistent with the expected numbers.

C- = curly wings, cc = straight wings; R- = round eyes, rr = elliptical eyes The cross is CcRr x Ccrr. Calculate the expected probabilities with a forked line. The calculated chi-square value is: (74-67.5)2 /67.5 + (61-67.5)2 /67.5 + (24-22.5)2 /22.5 + (21-22.5)2 /22.5 = 1.45. For df = 3, the critical value is 7.815. 1.45 is less than 7.815, so we accept the tested hypothesis.

What is the connection among DNA methylation, histone deacetylation, and gene regulation in eukaryotes?

DNA methylation and histone deacetylation are associated with more tightly packaged chromatin, which represses transcription. Methylation of DNA recruits histone deacetylases to the modified region. Acetylation neutralizes positive charges of histones, reducing interaction with DNA, so deacetylation strengthens association of histones with DNA.

1. Everyone knows that when you flip a coin, you have a 50 percent chance of heads and a 50 percent chance of tails. This means that out of 100 flips you should get 50 heads and 50 tails. However, if you actually flip a coin 100 times, a 50:50 ratio is only one among many possible outcomes. What if you get a ratio of 48 heads to 52 tails? How can we be certain that what we observe agrees with what we expect taking into account the role that chance plays? Use the χ 2 (chi---square) formula to determine if the observed ratios are due to chance:

Expect 50 heads/50 tails. (48-50)2 /50 + (52-50)2 /50 = 0.08 + 0.08. The χ2 is therefore 0.16. The degrees of freedom is equal to 1. This is expected to occur between 50% and 90% of the time by chance, and therefore it is not a significant difference between what was observed and what was expected.

At what levels can gene expression be controlled in eukaryotes? For each level, provide one example mechanism

Eukaryotic genes can be regulated at the following levels: Gene Structure: CpG islands can repress transcription of adjacent genes Transcription: Regulators, silencers, enhancers. Insulators, response elements mRNA processing: Alternative splicing Regulation of mRNA stability: stability affected by 5'Cap, Poly A tail, 5' UTR, 3' UTR Translation: availability of translational apparatus Post translational protein modification: chemical modification of proteins

19. Women have sex chromosomes of XX, and men have sex chromosomes of XY. Which of a woman's grandparents could not be the source of any of the genes on either of her Xchromosomes?

Father's Father. The father's father contributes only the Y chromosome to his sons, and subsequently to his grandsons.

A calico cat has a litter of eight kittens: one yellow male, two black males, two yellow females, and three calico females. What is the color of the father of the litter?

Female XB Xb x Xb Y (In order to produce a yellow female, the father would have to be Xb Y= yellow) Offspring: Xb Y XB Y Xb Xb XBXb

A man with dark (dominant), curly (see problem 2) hair marries a woman with light, straight hair. Their daughter, who happens to have dark hair, marries a man with light, wavy hair. Answer the following questions about this dark-haired daughter and her family. a. Draw a Punnett's square for this marriage, and predict the phenotypic ratio among the offspring of the daughter and her husband. b. What is the chance that they will have a child with hair just like his or her father's?

First problem—we know the daughter's hair is dark, but is it straight, wavy or curly? The alleles of this gene have incomplete dominance, which means that straight-haired Mom and curly-haired Dad are both homozygous. That means that the daughter must be a heterozygote, and thus have wavy hair. Her husband must also be a heterozygote. For hair color, Dark is dominant to light. That means that anyone in the pedigree with light hair is homozygous. The daughter has a mother with light hair, so she must be a heterozygote. We don't have enough information to figure out whether the father is homozygous or heterozygous. So here's what we've figured out: So the mating is Dd C1C2 X dd C1C2. Next step is to figure out the different kinds of gametes each of these parents can make. Mother Father D C1 d C1 D C2 d C2 d C1 d C2 Note that the father only makes two kinds of gametes because he is homozygous for the D/d gene. The next step is the Punnett's Square: D C1 D C2 d C1 d D2 d C1 Dd C1C1 Dd C1C2 dd C1C1 dd C1C2 d C2 Dd C1C2 Dd C2C2 dd C1C2 dd C2C2 Genotypic Ratio: 1 Dd C1C1: 2 Dd C1C2: 1 Dd C2C2: 1 dd C1C1: 2 dd C1C2: 1 dd C2C2 Remembering that our hair form gene has *incomplete dominance* our phenotypic ratio will be: 1 dark curly : 2 dark wavy : 1 dark straight : 1 light curly : 2 light wavy : 1 light straight. The father's hair is light and wavy, so the chance that a child will have hair like his is 2/8 (or 1/4).

relieves strain caused by unwinding

Gyrase (Topoisomerase)

What type of bond does helicase break?

Hydrogen bonds

What would be the advantage of regulating gene expression at many levels rather than simply regulating at one level (such as at the start of transcription)?

It is wasteful to produce mRNAs or proteins that are not needed. And although it is relatively timeconsuming for the cell to initiate transcription, translation can be executed rapidly if mRNAs are available. Thus, regulation at levels between transcription and translation may be wasteful of mRNAs but not of proteins, and such regulation poises the cell to produce a protein rapidly when it is needed— for example, see the discussion of T lymphocyte activation on page 465. Being able to regulate gene expression at several levels allows great flexibility in producing appropriate levels of proteins as quickly as necessary while stopping unnecessary production of mRNAs or proteins.

What would be the advantage of regulating gene expression at many levels rather than simply regulating at one level (such as at the start of transcription)?

It is wasteful to produce mRNAs or proteins that are not needed. And although it is relatively timeconsuming for the cell to initiate transcription, translation can be executed rapidly if mRNAs are available. Thus, regulation at levels between transcription and translation may be wasteful of mRNAs but not of proteins, and such regulation poises the cell to produce a protein rapidly when it is needed—for example, see the discussion of T lymphocyte activation on page 465. Being able to regulate gene expression at several levels allows great flexibility in producing appropriate levels of proteins as quickly as necessary while stopping unnecessary production of mRNAs or proteins.

Lac Operon: Regulates production of: Number of genes and how they are controlled: What binds to the operator & when does this occur: High levels of what substance affects the operon how?

It regulates the production of B-galactosidase and other proteins involved in the metabolism of lactose It consists of a cluster of three genes under the control of one promoter and one operator The LacI repressor protein binds to the operator when lacose levels are low High levels of lactose induce the operon.

Trp Operon: Regulates production of: Number of genes and how they are controlled: What binds to the operator & when does this occur: High levels of what substance affects the operon how?

It regulates the production of the amino acid tryptophan It consists of a cluster of five genes under the control of one promoter and one operator The corepressor tryptophan binds to the trp repressor protein, and the complex binds to the operator when tryptophan levels are high High levels of tryptophan repress the operon.

3. A sample of mice (all from the same parents) shows 58 Black hair, black eyes | 16 Black hair, red eyes | 19 White hair, black eyes 7 | White hair, red eyes. What is the expected genotypes of the parents? Objective: Perform a chi-square analysis to support your hypothesis.

Looks like a 9/3/3/1 ratio, meaning the parents are HhEe x HhEe Total number of offspring = 100 Expected 9/16 of 100 = 56.25 3/16 of 100 = 18.75 3/16 of 100 = 18.76 1/16 of 100 = 6.25 .05 + .40 + .003 + .09 = .54 = GOOD FIT Df= 3, 0.9

8. In chickens, males are the homogametic sex, ZZ, and females are the heterogametic sex, ZW. In a cross between Z+ W and Z+ Za , where the superscripts indicate the dominant and recessive sexlinked alleles, what proportions of phenotypes would you expect in the progeny? Would the proportions be the same in both sexes?

Male: all dominant phenotype Female: 1/2 dominant phenotype, ½ recessive phenotype

What are Okazaki fragments? How are they welded together?

Okazaki fragments are short segments of DNA synthesized away from the replication fork on a template strand during DNA replication. Many such segments are joined together by the enzyme DNA ligase to make up the lagging strand of newly synthesized DNA.

10. If a husband and wife have a heterozygous girl for colorblindness, a normal boy, a colorblind girl, and a colorblind boy, what would be the genotypes of the parents?

Mom = XCX Dad = XCY Children = XCX XCXC XY XCY

In certain portions of the Jewish population, there is a genetic disease called Tay Sachs disease, which is fatal to infants within the first five years of life. This disease is caused by a recessive allele of a single gene. Why does this disease persist, even though it is invariably fatal long before the afflicted individual reaches reproductive age? (In other words, why doesn't the allele for Tay Sachs disease simply disappear?)

No human who is homozygous for the Tay Sachs allele ever reproduces, but the heterozygotes are just fine. So the reason this allele persists, and doesn't vanish from the population, is that it survived in heterozygotes. The typical frequency pattern for an allele like this is that it decreases if the allele is fairly common, but the more rare it becomes, the slower the decrease becomes. This is because the more rare the alleles is in the population as a whole, the higher the percentage of the alleles will be found in heterozygotes, and the more unusual it will become for two heterozygotes to mate with each other. So the frequency reduction pattern typically looks something like this: Note that the graph tails into what's called an asymptote. This means that it approaches but never reaches zero. The more rare the allele gets, the less selective pressure there is against it (because such a high percentage of the remaining alleles are tied up in heterozygotes) that the background mutation rate is high enough to balance the selection rate, and the overall frequency remains constant.

If a pure-breeding (homozygous) black (dominant), long-haired (recessive) cat is mated to a purebreeding Siamese, short-haired cat, and one of their male offspring is mated to one of their female offspring, what is the chance of producing a Siamese colored, short-haired kitten?

Once again, we begin with some pedigree analysis to figure out everyone's genotypes. A bit of rumination reveals that this problem describes a nearly classic dihybrid cross. "Pure breeding" is a euphemism for "homozygous," so all of the original parents are homozygous. We mate a bbLL cat to a BBll cat, producing a bunch of BbLl kittens as an F1. The problem then instructs us to create an F2 by mating the kittens to each other. Here's this mating: Parents: BbLl X BbLl Gametes: BL BL Bl Bl bL bL bl bl BL Bl bL bl BL BBLL BBLl BbLL BbLl Bl BBLl BBll BbLl Bbll bL BbLL BbLl bbLL bbLl bl BbLl Bbll bbLl bbll Genotypic Ratio: 1 BBLL 2 BbLL 1 BBll 2 Bbll 1 bbLL 2 BBLl 1 bbll 2 bbLl 4 BbLl Phenotypic Ratio: 9 Black Short : 3 Black Long : 3 Siamese Short : 1 Siamese Long The actual question is, "what is the chance of producing a Siamese, short kitten?" The answer is 3/16

2. In the California poppy, an allele for yellow flowers (D) is dominant over an allele for white flowers (d). 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, and the following progeny are produced: 54 yellow and entire 58 yellow and fringed 53 white and entire 10 white and fringed. b. What conclusion can you make from the results of the chi-square test? c. Suggest an explanation for the results

Parents: yellow, entire petals (CC FF) × white, fringed petals (cc ff). F1 (Cc Ff) For the cross of a heterozygous F1 individual (Cc Ff) with a homozygous recessive individual (cc ff) we would expect a phenotypic ratio of 1:1:1:1 for the different phenotypic classes. Phenotype Observed (O) Expected (E) (O - E) 2 / E or (Χ2 ) Yellow, entire 54 43.75 2.40 Yellow, fringed 58 43.75 4.64 White, entire 53 43.75 1.96 White, fringed 10 43.75 26.03 Total 175 175 35.03 Degrees of freedom = 4 - 1 = 3. The chi-square value is greater than 12.838 for a probability value less than 0.005, or 0.5% that random chance produced the observed ratio of California poppies. B.From the chi-square value, we can see that it is unlikely that random variations produced the observed ratio. We reject the hypothesis that the difference we observed between the observed and expected numbers of progeny is due to chance. Some other phenomena must be acting C.The number of plants with the cc ff genotype is much less than expected. Possibly, the cc ff genotype may be sublethal. In other words, California poppies with the homozygous recessive genotypes may be less viable than the other possible genotypes.

In both prokaryotes and eukaryotes, groups of genes can be regulated simultaneously (coordinately expressed). However, each group accomplishes this task differently. Explain how coordinate expression differs in prokaryotes compared with eukaryotes.

Prokaryotes have operons, in which structural genes share a single promoter and therefore are expressed together. Eukaryotes have a variety of methods using response elements. A single response-element regulatory sequence may occur at several different genes, meaning that a single stimulus can increase transcription in all. However, a single gene may also be activated in multiple ways; each eukaryotic gene may be activated by a variety of stimuli, whereas the set of genes in a prokaryotic operon are all regulated by the same stimuli.

Two wavy haired people (one male and one female) marry and have eight children. Of these eight, how many would you expect to be curly haired, how many wavy haired and how many straight haired, assuming that the family follows the expected statistically predicted pattern? Suppose you examine the actual children and discover that three of the eight have curly hair. What do you suppose went wrong?

Same gene as above, so use the same symbols you chose there. This is a simple monohybrid cross.Genotypic Ratio: 1 C1 C1 : 2 C1 C2 : 1 C2 C2 Phenotypic Ratio: 1 Curly : 2 Wavy : 1 Straight This phenotypic ratio predicts that ¼ of the offspring should be curly-haired, ½ should be wavy haired and ¼ should be straight-haired. So of the eight children, our prediction would be 2 with curly hair, 4 with wavy hair and 2 with straight hair. The answer to the question, "What went wrong?" is "Nothing." What solving the problem does for us is make a statistical prediction, but every new conception is a new toss of the coins—statistical predictions are only really useful for large sample sizes. So it is very normal for a group as small as a family to show frequency distributions which don't match statistical expectations.

A genetics engineer was attempting to cross a tiger and a cheetah. She predicted a phenotypic outcome of the traits she was observing to be in the following ratio 4 stripes only: 3 spots only: 9 both stripes and spots. When the cross was performed and she counted the individuals she found 50 with stripes only, 41 with spots only and 85 with both. According to the Chi-square test, did she get the predicted outcome?

Set up a table to keep track of the calculations: Expected ratio Observed # Expected # O-E (O-E)2 (O-E)2 /E 4 stripes 50 44 6 36 0.82 3 spots 41 33 8 64 1.94 9 stripes/spots 85 99 -14 196 1.98 16 total 176 total 176 total 0 total Sum = 4.74 4/16 * 176 = expected # of stripes = 44 3/16 * 176 = expected # of spots = 33 9/16 * 176 = expected # stripes/spots = 99 Degrees of Freedom = 3 - 1 = 2 (3 different characteristics - stripes, spots, or both) Since 4.74 is less than 5.991, I can accept the null hypothesis put forward by the engineer.

Explain telomere erosion and the role of telomerase

Telomeres provide their protective function by postponing the erosion of genes located near the ends of DNA molecules. Telomeres become shorter during every round of replication. Telomeric DNA tends to be shorter in dividing somatic cells of older individuals and in cultured cells that have divided many times. Importantly, some cell genomes (such as germ cells) must persist virtually unchanged from an organism to its offspring over many generations. In order to accomplish this, an enzyme called telomerase catalyzes the lengthening of telomeres in eukaryotic germ cells, thus restoring their original length and compensating for the shortening that occurs during DNA replication.

14. Bow-legs is hypothesized to be X-linked recessive trait in Drosophila elanogaster. The P1 virgin females were, once again, homozygous wild type but the males were bow---legged. There were 52 wild type males and 67 wild type females in the F1 generation. The F2 generation contained 30 wild type males, 75 wild type females, 40 bow---legged males and no bow---legged females. Is this what is to be expected? Use chi-square to prove your position.

The chi square for the bow-legs problem approximates 1.5 (depending on how your round your numbers) with two degrees of freedom. This gives a probability between 30% and 50%. Since this is greater than 5% the difference is insignificant, and the hypothesis is accepted.

13. Consider the imaginary trait, bristles---with---split---ends, a trait hypothesized to be X---linked dominant. In a cross, the P1 virgin females were homozygous wild type while the males had bristles- -- with---split---ends. The F1 84 males were all wild type while the 90 females all had split---ends. In addition, the data for the F2 generation revealed 26 wild type males, 35 wild type females, 29 split--- end males and 40 split---end females. Is this what is to be expected? Use chi-square to prove your position.

The chi square for the bristles-with-spit-ends problem approximates 3.7 (depending on how your round your numbers) with three degrees of freedom. This gives a probability of approximately 30%, therefore, difference between what was expected and the data that had been collected was insignificant and the hypothesis is accepted.

11. There are two genes in fruit flies that determine eye color and wing shape. Red eyes is dominant to sepia (brown) eyes, and normal wings is dominant to vestigial (shriveled) wings. Describe a procedure you could use to determine the genotype of a fly that has red eyes and normal wings. This fly may be homozygous or heterozygous for either gene.

The unknown fly could be mated with a fly that has both recessive traits (sepia/vestigial), a typical test cross. Then look for any recessive traits to show up in the offspring. If sepia offspring are produced, then the parent was heterozygous for the eye color gene. If not it was homozygous. If vestigial offspring are produced, then the parent was heterozygous for the wing gene. If not it was homozygous.

6. There are two genes in fruit flies that determine eye color and wing shape. Red eyes is dominant to sepia (brown) eyes, and normal wings is dominant to vestigial (shriveled) wings. Describe a procedure you could use to determine the genotype of a fly that has red eyes and normal wings. This fly may be homozygous or heterozygous for either gene.

The unknown fly could be mated with a fly that has both recessive traits (sepia/vestigial), a typical test cross. Then look for any recessive traits to show up in the offspring. If sepia offspring are produced, then the parent was heterozygous for the eye color gene. If not it was homozygous. If vestigial offspring are produced, then the parent was heterozygous for the wing gene. If not it was homozygous.

When a male pig from a line of true-breeding (homozygous) black, solid-hooved pigs was crossed to a female from a breed (homozygous) of red, cloven-hooved pigs, their several progeny all looked alike with regard to color and hooves. These progeny were all mated to members of the same breed as their red, cloven-hooved mother pig. The offspring from this final cross were: 11 black, cloven-hooved; 8 black, solid-hooved; 14 red, cloven-hooved; and 10 red, solid-hooved. For each of these two genes (coat color and hoof type) determine which allele is the dominant one. Explain your reasoning. What were the phenotypes of the progeny produced by the first mating in this problem?

This is a reasoning problem. The issue is understanding dominance. We begin by mating two truebreeding pigs, one black with solid hooves and one red, with cloven hooves. They produce offspring, which are not described. Note that you know these offspring should all look alike with respect to these traits, because they are the product of the mating of two homozygous parents. These undescribed offspring are then bred back to pigs which are genetically just like the red, cloven parent pig. This is called a "back cross," though strictly speaking a back-cross should literally mean mating back to the parent. The various offspring from the back-cross are described, and you are asked to determine dominance for the alleles of these two genes. The first thing to note here is that the numbers of each offspring class are completely irrelevant. The only thing that matters in this kind of problem is what kinds of offspring result from that back cross. Recall that one of the parents of this second generation of offspring is definitely homozygous for red and homozygous for cloven, just as the original female was. This means that this parent contributed a red allele and a cloven allele to every one of the offspring. And yet, some of the offspring are black and some have solid hooves. This is possible only if black is dominant to red and if solid is dominant to cloven.

In cats, again, black color is dominant to a special, temperature-sensitive albino gene which produces cats with dark legs, faces and tails (Siamese cats, in case you don't recognize it). A short haired (dominant) Siamese colored female is bred to a long-haired black male. They have eight kittens: 2 black, short-haired; 2 black, long-haired; 2 Siamese, short-haired; and 2 Siamese, longhaired. What were the genotypes of the two parents?

This is a simple pedigree analysis problem. We have two genes. For the color gene, black is dominant to Siamese. For the hair length gene, short is dominant to long. Our parents are a short-Siamese female and a long-black male. The only parts of their genotypes in question are her second hair length allele and his second color allele. A Siamese cat is always homozygous, as is a long-haired cat (second rule of pedigree analysis). Their litter of kittens includes babies with all combinations of color and hair length. Note that the numbers (in this case 2 of each kind of kitten) are immaterial. What matters is what colors these parents can produce. Rules 3 and 4 help us figure out the rest of everyone's genotypes. Since they have some kittens who are Siamese (and thus must be homozygous), the black parent must be carrying that recessive allele, so he's heterozygous. Since they have some kittens who are long-haired (and must be homozygous), the short-haired parent must be carrying that recessive allele. Since Mom is homozygous for Siamese and Dad is homozygous for long hair, all of the kittens must carry at least one allele for each of these traits. So the black kittens are heterozygous, and the short-haired kittens are heterozygous as well.

In horses, one which runs best in water (or in wet conditions) is called (WATER), and one which runs best in dry conditions is called (DRY). (WATER) is recessive to (DRY). A horse can also be either a trotter, which we will designate (GAIT) or a pacer, which we will designate (PACE). (PACE) is recessive to (GATE). We have mated two horses, a stallion named Halter-Man and a mare named Erlich-Mane. Halter-Man is a (WATER)(PACER), while Erlich-mane is a (DRY)(GAIT). One of Erlich-Mane's parents was a (WATER)(PACER). What are the chances of Erlich-Mane and HalterMan producing a (WATER)(GAIT) foal (that's a baby horse, in case you didn't know)?

This problem involves two genes. First step: assign symbols for the alleles. First gene: W = Dry (dominant); w = Water Second gene: P = Gate (dominant); p = Pace Using pedigree analysis, we figure out that Halter-Man is wwpp and Erlich-Mane is WwPp. So here's our mating: WwPp X wwpp Gametes: WP Wp wP wp wp Genotypic Ratio of Offspring: 1 WwPp : 1 Wwpp : 1 wwPp : 1 wwpp Phenotypic Ratio of Offspring: 1 Dry-Gate : 1 Dry-Pace : 1 Water-Gate : 1 Water-Pace The question asked is, what is the chance of producing a Water-Gate foal. Answer: 1/4 or 25%.

2. A large ear of corn has a total of 433 grains, including 271 Purple & starchy, 73 Purple & sweet, 63 Yellow & starchy, and 26 Yellow & sweet. Your Tentative Hypothesis: This ear of corn was produced by a dihybrid cross (PpSs x PpSs) involving two pairs of heterozygous genes resulting in a theoretical (expected) ratio of 9:3:3:1. Objective: Test your hypothesis using chi square and probability values.

Total Kernals = 433 Expected Purple and Smooth 9/16 of 433 = 243.6 Purple and Wrinkled 3/16 of 433 = 81.2 Yellow and Smooth 3/16 of 433 = 81.2 Yellow and Wrinkled 1/16 of 433 = 27.1 3.08 + .82 + 4.07 + .04 = 8.01 = chi-square, df=3. P-value: 0.025

The 5' UTR of the trp operon RNA contains several UGG tryptophan codons. What would be the effect on transcription-attenuation gene regulation if the trp codons were converted to UGC cysteine codons?

Translation through the 5 UTR would no longer be dependent on the availability of the amino acid tryptophan. (However, it would now be dependent on levels of cysteine.) Regardless of levels of tryptophan, translation would continue through the 5 UTR, the attenuator hairpin would form, and the antiterminator loop would form. The entire trp operon would not be transcribed (unless levels of cysteine were low).

20. A female Drosophila of unknown genotype was crossed with a white-eyed male fly, of genotype XwY (w = white eye allele is recessive, + = red-eye allele is dominant.) Half of the male and half of the female offspring were red-eyed, and half of the male and half of the female offspring were white-eyed. What was the genotype of the female fly

XwXw

In peas, axillary flowers (A) is dominant to terminal flowers (a) and colored flowers (R) is dominant to white (r). A true-breeding plant with white, terminal flowers is crossed with a truebreeding plant having colored, axillary flowers. The F1 plants are then allowed to self-pollinate. (a) Using a punnet square to illustrate, determine all possible F2 phenotypes and calculate the probability of each. (b) The F2 generation consisted of 308 plants with axillary, colored flowers; 34 plants with terminal, white flowers; 118 plants with axillary, white flowers; and 132 plants with terminal, colored flowers. Using the chi-square test, determine if this outcome is consistent with the predicted ratios for this cross.

a) b)The calculated chi-square value is 6.53 which is less than the table value of 7.815 for df=3. Data are consistent with the expected ratio

11. In fruit flies, Drosophila, the gene for eye color is carried on the X chromosome. The allele for red eyes is dominant over its recessive allele, white eyes. a) If a white-eyed female is mated with a red-eyed male, what is the appearance of their offspring? -Genotype of white-eyed female -Genotype of red-eyed male -What are the two possible genotypes in the offspring? b) If the daughters from this cross are mated with their father, what types of offspring would be expected and the probability of each? ¼ XRXR ½ Red eyed females ¼ XRXr ¼ red eyed males ¼ XRY ¼ white eyed male ¼ Xr Y

a) XrXr XRY XRXr or XrY b) Genotypes:¼ XRXR, ¼ XRXr, ¼ XRY ¼ XrY Phenotypes:½ Red eyed females, ¼ red eyed males, ¼ white eyed male

If dividing cells are grown in a culture medium containing radioactive thymidine, the thymidine will be covalently incorporated into the cell's DNA during replication. The radioactive DNA can be detected in the nuclei of individual cells by autoradiography (i.e., by placing a photographic film over the cells, radioactive cells will activate the film and show up as black dots when looked at under a microscope). a) During whatphase of the cell cycle will the radioactive thymidine be incorporated?Why? b) Would cells at all phases of the cell cycle contain radioactive DNA immediately after the labeling procedure?Why? c) Why are there initially no mitotic cells that contain radioactive DNA? d) Explain the rise and fall and then rise again of the curve. e) Estimate the length of G2 for these cells f) Estimate the length of the total cell cycle forthese cells

a) Radioactive thymidine will be incorporated during S phase, since this is the phase of the cell cycle where DNA is replicated. b)No, only those cells in S phase during the cell cycle would contain radioactive DNA immediately after the labeling procedure c)At t=0, the only cells that have radioactive DNA are in S phase. The mitotic cells at t=0 contain no radioactive DNA. It takes about two and a half hours before the first labeled mitotic cells appear. d)The only cells that are labeled are those in S phase during the 30 minute treatment. The initial rise of the curve corresponds to cells that were just finishing DNA replication when the radioactive thymidine was added. The curve rises as more of those cells enter mitosis to a peak that corresponds to times when all of the mitotic cells were in the S phase during the time of labeling. The labeled cells then exit mitosis, being replaced by unlabeled mitotic cells that were not yet in S phase during the labeling period. After 20 hours the curve starts rising again, because the cells labeled in the first 30 min pulse are entering a second round of mitosis. e)The initial lag before the cells enter mitosis corresponds to the G2 phase (approximately 2.5 hours), which is the time between the end of S phase and the beginning of mitosis. f)Around 26-28 hours (measure time from 50% of labeled cells entering mitosis for the first time to the time when 50% of labeled cells enter mitosis for the second time).

4. J. A. Moore investigated the inheritance of spotting patterns in leopard frogs (J. A. Moore. 1943. Journal of Heredity 34:3-7). The pipiens phenotype had the normal spots that give leopard frogs their name. In contrast, the burnsi phenotype lacked spots on its back. Moore carried out the following cross, producing the progeny indicated. Parent phenotypes Progeny phenotypes burnsi × burnsi 39 burnsi, 6 pipiens a. What are the likely genotypes of the two parents? Be sure to specify the dominant and the recessive trait. b. 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. One can define the burnsi allele as B and the pipiens allele as b. The burnsi phenotype is dominant over the pipiens phenotype. b.In each of the crosses, we expect that the burnsi allele is dominant to the pipiens allele. By making that assumption, we can make predictions regarding the phenotypic ratios of the offspring and the genotypes of the parents. For the cross of burnsi × burnsi (Bb × Bb), we would expect a phenotypic ratio of 3:1 in the offspring. Phenotype Observed (O) Expected (E) (O-E) 2 /E or (Χ 2 ) burnsi 39 33.75 0.82 pipiens 6 11.25 2.45 Total 45 45.00 3.27 Degrees of freedom (df) = (number of phenotypic classes) - 1. Because there are two phenotypic classes, the degrees of freedom (df) are 1. From the chi-square table, we can see that the calculated chi-square value falls between 2.706 (P of 0.1) and 3.841 (P of 0.05). The probability is sufficient that differences between what we expected and what we observed could have been generated by chance and that our parents are as predicted, (Bb× Bb).

13. The SUC2 gene in yeast encodes an enzyme that converts the sugar, sucrose, into glucose and fructose, which is necessary for yeast to use sucrose as a source of food. In the presence of glucose, SUC2 expression is switched off; but in the absence of glucose, SUC2 expression increases 100-fold. The expression of SUC2 in wild-type yeast and two mutant yeast strains is shown below. The mutations occur in two genes other than SUC2: SNF1 and SSN6. a)Is the SNF1 gene normally important for activation or repression of SUC2 expression? b. Is the SSN6 gene normally important for activation or repression of SUC2 expression? c. If SNF1 and SSN6 work in the same pathway to regulate SUC2 expression, the order in which the genes acts can be one of two possibilities (where indicates activation and indicates inhibition): SSN6 acts to repress SUC2 expression, and SNF1 activates SUC2 by inhibiting the inhibitor, SSN6. SNF1 activates SUC2 expression, and SSN6 represses SUC2 by inhibiting the action of SNF1. Design an experiment to distinguish between these two possible orders. What would you expect the outcome of your experiment to be in each of the two possible case

a. SNF1 is involved in activating SUC2 expression. Because SUC2 is not active when SNF1 is mutated, SNF1 must normally be required to activate SUC2 expression. b. SSN6 is involved in repressing SUC2 expression. Because SUC2 is active even when glucose is present in ssn6 mutant yeast, SSN6 must normally be required to repress SUC2 expression. c. To test the two possible orders the simplest experiment is to generate a yeast strain that is mutant for both genes, an snf1 ssn6 double mutant, and observe the expression of SUC2 in the presence and absence of glucose. If (1) is the correct order, SUC2 will be switched on (100 units in the table above) in the presence and absence of glucose. In this case, the mutant version of snf1 is unable to repress SSN6; however, because ssn6 is also mutated and cannot repress SUC2, SUC2 expression will always be high. If (2) is the correct order, SUC2 will never be switched on (<1 unit in the table above) in the presence and absence of glucose. In this scenario, mutant ssn6 is unable to repress SNF1; but the nonfunctional snf1 is itself unable to switch on SUC2 expression, so SUC2 expression will always be low. Therefore, the phenotype of the snf1 ssn6 double mutant can distinguish between order (1) (SUC2 always on) and order (2) (SUC2 always off

4. In fruit flies, the gene for white eyes is sex-linked recessive. (R) is red and (r) is white. Cross a whiteeyed female with a normal red-eyed male. a. What percent of the males will have red eyes? White eyes? b. What percent of the females will have red eyes? White eyes? c. What total percent of the offspring will be white-eyed? d. What percent of the offspring will be carriers of the white eye trait?

a. What percent of the males will have red eyes? White eyes? 0% of the males will have red eyes. 100 % will have white eyes b. What percent of the females will have red eyes? White eyes? 100 % of females will have red eyes. 0% will have white eyes. c. What total percent of the offspring will be white-eyed? 50% d. What percent of the offspring will be carriers of the white eye trait? 50%

5. Using the same information from question 4, evaluate the following cross: Parent phenotypes Progeny phenotypes burnsi × pipiens 23 burnsi, 33 pipiens c. What are the likely genotypes of the two parents? Be sure to specify the dominant and the recessive trait. d. 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.

c. Using the information gathered from questions one as well as taking into consideration the progeny observed, the only possible genotypes for the parents must be burnsi (Bb) × pipiens (bb) d.For the cross of burnsi × pipiens (Bb X bb), we would expect a phenotypic ratio of 1:1 in the offspring. Phenotype Observed (O) Expected (E) (O-E) 2 /E or (Χ2 ) burnsi 23 28 0.89 pipiens 33 28 0.89 Total 56 56 1.78 Because there are two phenotypic classes, the degrees of freedom (df) are 1. From the chi-square table, we can see that the calculated chi-square value falls between 0.455 (P of 0.5) and 2.706 (P of 0.1). The probability is sufficient that differences between what we expected and what we observed could have been generated by chance and that our parents are as predicted, (Bb X bb).

12. In corn, purple kernels are dominant over yellow kernels, and full kernels are dominant over shrunken kernels. A corn plant having purple and full kernels is crossed with a plant having yellow and shrunken kernels, and the following progeny are obtained: What are the most likely genotypes of the parents and progeny? Test your genetic hypothesis with a chi-square test.

purple, full 112 purple, shrunken 103 yellow, full 91 yellow, shrunken 94 In this cross, purple kernel X yellow kernel produces roughly ½ purple and ½ yellow kernels and similarly examination of simultaneous cross between full and shrunken results in ½ full and ½ shrunken kernels (a 1:1 ratio for each characteristic). P Purple X Yellow Full X shrunken F1 112 + 113 = 215 Purple 112 + 91 = 203 Full 91 + 94 = 185 Yellow 103 + 94 = 197 Shrunken Because purple is dominant, the purple parent must be heterozygous (Pp) and the yellow parent must be homozygous (pp). Similarly, the full parent must be heterozygous (Ff) and the shrunken parent must be homozygous (ff). Using the multiplication rule, you can combine the probability of each characteristic to obtain the overall genotypes and the proportions of each genotype you would expect: P Purple, full X Yellow, Shrunken PpFf X ppff F1 PpFf = 1/2 purple X 1/2 full = 1/4 purple, full PpFf = 1/2 purple X 1/2 shrunken = 1/4 purple, shrunken ppFf = 1/2 yellow X 1/2 full = 1/4 yellow, full ppff = 1/2 yellow X 1/2 shrunken = 1/4 yellow, shrunken Add the total number of progeny (400) and then calculate the expected number for each genotype: Phenotype Observed Expected purple, full 112 1/4 X 400 = 100 purple, shrunken 103 1/4 X 400 = 100 yellow, full 91 1/4 X 400 = 100 yellow, shrunken 94 1/4 X 400 = 100 Using the formula, chi-square = Σ (observed-expected)2 /expected: = 1.44 + 0.09 + 0.81 + 0.36 = 2.70 Because there are four phenotypic classes, the df (n-1) = 4-1 = 3. The calculated chi-square value of 2.7 lies between 2.366 (a probability of 0.5) and 6.251(a probability of 0.1). The probability of (P) associated with the calculated chi-square value is therefore 0.5

List and explain at least two ways that regulation of genes for yeast galactose-metabolizing enzymes is different from regulation of E. coli lactose-utilizing enzymes.

• Yeast genes are not in an operon; E. coli genes are in an operon. • The yeast GAL genes are mainly controlled by a transcriptional activator; E. coli genes are mainly controlled by a transcriptional repressor. • The yeast GAL genes are under positive inducible control; the E. coli genes are under negative inducible control. • Yeast are eukaryotic, so regulation of any genes, including galactose-utilizing enzymes, could include changes in chromatin structure, post-transcriptional modification of RNA, and RNA stability.