Module 1 Extended Response

In mice, there is a set of multiple alleles of a gene for coat color. Four of those alleles are as follows: C = full color (wild) cch = chinchilla cd = dilution c = albino Given that the gene locus is not sex-linked and that each allele is dominant to those lower in the list, give the phenotypic ratios expected from each. (a) wild (heterozygous for dilution) X chinchilla (heterozygous for albino) (b) chinchilla (heterozygous for albino) X albino Clearly label (a) and (b), and enter the phenotype and number in ratio format. (ie, 1pink:2yellow:1white)

a) Ccd x cchc b) cch x cc

The following cross AAB x AAB (where the AB allele produces a brown coat color in rats versus the A allele which produce agouti coat color) produces the following phenotypic ratios in the F2 generation: 2/3 brown; 1/3 agouti. What might you conclude about the AB allele?

the AB allele is lethal at homozygosity

A plant with round seeds was crossed with a plant with wrinkled seeds and the following offspring were obtained: 220 round and 180 wrinkled. (a) What is the most probable genotype of each parent? (b) What genotypic and phenotypic ratios are expected, based upon part (a)? (c) Based on the information provided in part (b) above, what are the expected (theoretical) numbers of progeny (400 total) of each phenotypic class?

(a) Assuming that round (W) is dominant to wrinkled (w): Ww X ww (b) 1:1 for either. 1Ww:1ww and therefore, 1 round:1 wrinkled. (c) 200 (200 round + 200 wrinkled = 400, from 1:1 ratio)

Two crosses were done to map genes in the dipteran Drosophila melanogaster. Cross 1: Females homozygous for the recessive mutant characters a, b, and c were crossed with wild-type males from a pure-breeding stock. A testcross was then done by taking the F1 females (all wild-type in appearance) and crossing the with triple mutant males (a, b, c) taken from a pure-breeding stock. The following offspring were counted: Phenotype Number a, b, c 280 a + + 77 b + + 23 c + + 125 a, b + 127 a, c + 25 b, c + 75 wild-type 268 1000 Cross 2: A second cross was done, this time involving the same gene c, as in Cross 1, and a new mutant gene d. Females of phenotype "c" were taken from a pure-breeding line and crossed with males of phenotype "d" from another pure-breeding line. In the F1, the females were all wild-type in appearance, while the males were all of the "c" phenotype. The F1 males and females were crossed together to give F2 offspring. The F2 phenotypes occurred in the following numbers: Males: wild-type, 4; c d, 6; c, 43; d, 47. Females: wild-type, 48; c, 52. (a) Using the information from both of the crosses, map these genes. Calculate recombination frequencies and give map distances. Calculate the interference. (b) Give the genotypes, as chromosome diagrams, for the parents and the F1 flies in both crosses, and for the F2 flies in Cross 2.

(a) Using the information from both of the crosses, map these genes. Calculate recombination frequencies and give map distances. Calculate the interference. DCO are a+c and +b+, so b is in the middle, gene order is abc SCO between a and b are a++ and +bc, (77 +75 + 25 + 23)/1000 = 20 cm SCO between b and c are ab+ and ++c, (125 + 127 + 25 + 23)/1000= 30 cm DCO = 0.048 Expect 0.06 Coefficient of interference is .048/0.06=0.8 Interference is 1- 0.8 = 0.2 Recombination between c and d is (4 + 6)/(4 + 6 + 43 + 47) = 10 cm (b) Give the genotypes, as chromosome diagrams, for the parents and the F1 flies in both crosses, and for the F2 flies in Cross 2. Cross 1: P1: +++/Y x abc/abc F1 +++/abc x abc/Y Cross 2: P1: c+/c+ x +d/y F1: c+/+d x c+/Y

Mutations can be either recessive or dominant to the wild type. For each of the following examples, determine if the mutation would most likely be recessive or dominant. Explain your reasoning. (i) Mutation that disrupts the active site of an enzyme (ii) Over expression of an enzyme

(i) Mutation that disrupts the active site of an enzyme recessive, since the undamaged wild-type copy will still provide function. Function will only be lost of there are two damaged alleles, i.e.recessive. (ii) Over expression of an enzyme there will be increased function in heterozygotes, so the mutation is more likely to be dominant, since it will cause a phenotype in the heterozygous state.

Suppose that in sweet cherry a single gene controls fruit color [red (R-) is dominant to yellow (rr)] and a second gene controls bloom time [early bloom (E-) is dominant to late bloom (ee)]. A true breeding tree with red fruit and early bloom is crossed to a tree that has yellow fruit and blooms late. The resulting F1 is selfed to create a population of 500 F2 trees. The segregation of traits in this F2 population is shown below: Red fruit, early bloom: 341 Red fruit, late bloom: 26 Yellow fruit, early bloom: 32 Yellow fruit, late bloom: 101 (i) What is Mendel's Second Law? (ii) Do a Chi Square analysis to determine whether the results conform to Mendel's Second Law. (iii) If the results do not conform to this law, can you come up with an explanation for the lack of conformity?

(i) What is Mendel's Second Law? Independent assortment, which results in a 9:3:3:1 ratio in a dihybrid cross (ii) Do a Chi Square analysis to determine whether the results conform to Mendel's Second Law. 9/16 of 500 is expected for red/early, 281 3/16 is expected for red/late, 94 3/16 is expected for yellow early, 94 1/16 is expected for yellow late, 31 Chi squared is the (obs-exp)2/exp = (341-281)2/281 + (94-26)2/94 + (94-32)2/94 + (31-101)2/101 = 3600/281 + 4624/94 + 3844/94 + 4900/101 =12.8 + 49 + 40.8 + 48.5= 151 with 3 degrees of freedom the probability is very low that the null hypothesis is correct (iii) If the results do not conform to this law, can you come up with an explanation for the lack of conformity? it is quite possible the two genes exhibit genetic linkage

Suppose that red flower color (RR or Rr) is dominant to white flower color (rr) in Petunia. A friend has a petunia plant with red flowers and wants to determine whether the plant is RR or Rr. In order to determine the genotype, you cross it with a white-flowered petunia (i) What is the name of the type of cross that you performed? (ii) How will this cross help you determine the genotype of the red flowered petunia? That is, how will the results from this cross differ if the red-flowered petunia is RR vs Rr?

(i) testcross (ii) You will observe different segregation in the testcross progeny, depending on the genotype of the red petunia. If the red petunia is RR, then all testcross progeny will be red; If the red petunia is Rr, then of the testcross progeny will be red (Rr) and will be white (rr).

Listed below are blood types for several children and their mothers. Give all possible genotypes and blood types for the father of each child. Child's blood type Mother's blood type a. A A b. O B c. AB A d. B AB

1 IAIA, IAi, ii 2 IAi, ii, IBi 3 IBi, IBIB, IAIB d. IBi, IBIB, IAIBIAIA, IAi, ii

Describe two evolutionarily significant characteristics of meiosis that are not present in mitosis. How do these characteristics increase genetic diversity relative to mitosis?

1) Reshuffling of chromosomes through independent assortment 2) Crossing over exchanges alleles between maternal and paternal chromosomes

The mitochondria within eukaryotic cells have their own genomes. Imagine that a mutation arises on the mitochondrial genome and, at the time of cytokinesis of the host cell, 10% of the mitochondria in that cell have that mutation. In the two daughter cells, what percentage of the mitochondria will possess that mutation? Why?

10% of each daughter cell mitochondria will have the mutation if the mitochondria are distributed randomly.

Tony, who is not diseased, has a sister with cystic fibrosis (CF). Neither of his parents have CF.Tony is expecting a child with Tina. Tina's family history is unknown What is the probability that Tony is heterozygous for the CF gene? Explain your answer.

2/3 Tony's parents are Aa and Aa. Their children's genotypes, with probabilities, are: 1/4 AA; 1/4 Aa; 1/4 aA; 1/4 aa. Tony does not have CF, so he cannot be genotype aa. Of the remaining genotypes, 2/3 are heterozygous.

What will be the probability of observing individuals exhibiting all of the dominant phenotypic traits in a pentahybrid (5 independently assorting gene loci) cross?

243/1024

In the following cross of carriers of the Bombay h allele, what are the expected phenotypic ratios of blood type? AAHh x ABHh

3/8 A; 3/8AB; 2/8 O (or Bombay or "appear to be blood type O) or could be given as 6/16A; 6/16AB; 4/16 O

Two autosomal genes in Drosophila code for eye color and wing shape (the wild type is dominant). The genetic symbols for these traits are for eyes: pr: purple and pr+: red and for wings vg: vestigal and vg+: normal. a) What are the genotypes of a pure-breeding purple vestigial fly and a purebred red normal fly? b) What do we get if we cross these two pure-breeding lines? Give the genotype and the phenotype. c) What would you expect if you test-cross these F1's? d) Which alleles are from the test-cross individual? e) The results that Morgan got were not what you would expect. Instead, he saw fewer of the purple normal (pr pr vg+ vg ) and red vestigial (pr+ pr vg vg). What do these results tell us about the gametes produced by the original heterozygote? f) How did Morgan explain these results?

4 a) What are the genotypes of a pure-breeding purple vestigial fly and a purebred red normal fly? prprvgvg & pr+pr+vg+vg+ b) What do we get if we cross these two pure-breeding lines? Give the genotype and the phenotype. pr+pr vg+vg; red eyes and normal wings c) prpr+vgvg+ x prprvgvg You would expect: pr pr vg vg purple vestigial 1 pr pr vg+ vg purple normal 1 pr+ pr vg vg red vestigial 1 pr+ pr vg+ vg red normal 1 d) all pr and vg. e) There were many more pr vg and pr+vg+ gametes produced by the original heterozygote. f) Morgan proposed that the unusual ratios were a result of linkage and that the genes for eye color and wing shape must be on the same chromosome and usually inherited together.

You may have heard through various media of an animal alleged to be the hybrid of a rabbit and a cat. Given that the cat (Felis domesticus) has a diploid chromosome number of 38 and a rabbit (Oryctolagus cuniculus) has a diploid chromosome number of 44, what would be the expected chromosome number in the somatic tissues of this alleged hybrid? Why?

41

What is a dominant epistatic gene?

A dominant allele that, if present, determines the phenotype of a given trait regardless of which other alleles for the trait are present.

What is the difference between a backcross and a testcross?

A testcross is used to determine genotypes of individuals that may be heterozygous or homozygous for a dominant allele (and therefore express identical phenotypes). The unknown genotypes are exposed by crossing to a "tester" that is known to be homozygous for the recessive allele in question. A backcross is the mating of F1 progeny back to one of their parents. Backcrosses can also be testcrosses if the original parent is homozygous for the recessive allele(s). Backcrosses are typically used to introgress allele(s) of interest, which segregated out in the F1 progeny, back into the genetic background of one of the original parents.

In the above pedigree of Ocular Albinism, a rare X-linked trait: A. The probability that III1 carries the allele B. The probability that III2 carries the allele C. The probability that II5 carries the allele

A. 0 B. 1 C. 1/2

A. In the above pedigree, what are the chances of II4 carrying the allele? B. In the above pedigree, what are the chances of II2 carrying the allele? C. In the above pedigree, what are the chances of the embryo in line III exhibiting the trait?

A. 1 B. 2/3 C. 1/2

In the above pedigree of a rare trait: A. What is the probability of III1 carrying the trait? B. What is the probability of III2 carrying the trait? C. What is the probability of the embryo shown exhibiting the trait?

A. 1/2 or equivalent B. 1/3 or equivalent C. 1/24 or equivalent

Vanessa has obtained two true-breeding strains of mice, each homozygous for an independently discovered recessive mutation that prevents the formation of hair on the body. The discoverer of one of the mutant strains calls his mutation naked, and the other researcher calls her strain hairless. To determine whether the two mutations are simply alleles for the same gene, Vanessa crosses naked and hairless mice with each other. All the offspring are phenotypically wild-type. After intercrossing these F1 mice, however, Vanessa observes 115 wild-type mice and 85 mutant mice in the F2. What is the most likely explanation for the segregation of wild-type and mutant mice in the F2? Are the naked and hairless mutations alleles for the same gene? Using symbols of your own choosing, indicate the genotypes of the parents, the F1, and each of the phenotypic classes of the F2 progeny. Describe any gene interactions in this set of crosses.

Answer: The naked and hairless mutations are not the same allele because complementation occurred and the F1 hybrids are phenotypically wild-type. Naked and hairless are instead mutations of two different genes. To explain the phenotypic ratio in the F2, let us first adopt symbols for these mutations and their dominant wild-type alleles: n = naked mutation, N = wild-type allele, h = hairless mutation, H = wild-type allele The genotypes of the true-breeding parental strains are nnHH (naked) and NNhh (hairless). F1 hybrids produced by crossing these strains are therefore NnHh. When these hybrids are intercrossed, we expect several genotypes to appear in the offspring. Each recessive allele, however, when homozygous, prevents the formation of hair on the body. Thus, only mice that are genotypically N-H- will develop hair; all the others—homozygous nn or homozygous hh, or homozygous for both recessive alleles—will fail to develop body hair. We can predict the frequencies of the wild and mutant phenotypes if we assume that the naked and hairless genes assort independently. genotypes phenotypes This gives the following genotype and phenotype ratios: 9 N-H- normal 3 N-hh no hair 3 nnH- no hair 1 nnhh no hair The ratio of normal mice to those with no hair is 9:7. In a sample of 200 F2 progeny, we would expect 200* 9/16 = 112 to be wild-type and 200 * 7/16 = 88 to be mutant. The observed frequencies of 115 wild-type and 85 mutant mice are close to these expected numbers, suggesting that the hypothesis of two independently assorting genes for body hair is, indeed, correct.

A homozygous dominant long haired cat is mated to a short haired cat. They produce a total of 13 kittens. What is the expected ratio of phenotypes under monohybrid inheritance?

Assign genotypes as follows: <br>LL = long hair <br>ll = short hair<br>We perform a cross of LL x ll which will produce Ll kittens with phenotype of long hair.<br>Expected phenotypic ratio will be 1:0 for long hair: short hair

Why are rare X-linked recessive diseases more common in males than in females?

Because males are hemizygous for the X chromosome

Dosage compensation leads to a variety of interesting coat color patterns in certain mammals. For instance, a female cat that is heterozygous for two coat color alleles, say black and orange, will usually have the "calico" or mosaic phenotype. Describe the chromosomal basis for the mosaicism (calico) in the female. Explain why chromosomally normal male cats do not show the mosaic phenotype, but XXY male cats can be "calico."

Because of dosage compensation, one of the X chromosomes randomly "turns off" early in development. Once such a chromosome is inactivated, it remains so in daughter cells. Recessive alleles on the remaining active X chromosome are expressed because their normal allele (on the inactive X chromosome) is not capable of expression. Because males typically have only one X chromosome, X chromosome inactivation does not occur; however, in XXY males that are heterozygous for certain coat color genes, such inactivation and mosaicism is possible.

Discuss the differences, and at least one similarity, between recombination and independent assortment

Both create genetic diversity by shuffling alleles. This is achieved through Independent assortment by making new combinations of chromosomes, while homologous recombination results in new combinations of alleles on the chromosomes themselves.

What is a primary function of the centromeric DNA region found on chromosomes?

Centromeric DNA, specific DNA (tandem repeated sequences) that allows binding by centromeric proteins, which are in turn bound by Kinetochore proteins, which allows attachment to spindle fibers (made of alpha, beta, gamma tubulin), which facilitates chromosome transport to poles

What is the relative gene order based on the results from this three point mapping cross: a B f 404 A b F 395 A B f 13 a b F 17 a B F 150 A b f 191 a b f 154 A B F 139

Correct order: B A F or F A B (small letters are also fine).

Why would it make sense for cyclin proteins to vary in concentration throughout the cell cycle?

Cyclins increase in concentration during that phase in which they are needed to carry out their functions. Once that phase has passed the cyclins are destroyed so they don't do that anymore.

Among dogs, short hair is dominant to long hair and dark coat color is dominant to white (albino) coat color. Assume that these two coat traits are caused by independently segregating gene pairs. For each of the crosses given below, write the most probable genotype (or genotypes if more than one answer is possible) for the parents. It is important that you select a realistic symbol set and define each symbol below. Parental Phenotypes Phenotypes of Offspring Short, Dark Long, Dark Short, Albino Long, Albino (a) dark, short X dark, long 26 24 0 0 (b) albino, short X albino, short 0 0 102 33 (c) dark, short X albino, short 16 0 16 0 (d) dark, short X dark, short 175 67 61 21

D = dark, d = albino L= long, l = short A) DDLl x DDll B) ddll x ddll C) Ddll x ddll D) DdLl x DdLl

List two differences and two similarities between mitosis and meiosis.

Differences: f. Mitosis produces two daughter cells; meiosis produces four daughter cells. g. Mitosis produces identical daughter cells; meiosis produces four different daughter cells. Similarities: c. DNA replication must occur first. d. Cytokinesis usually occurs at the end of each.

Two of the seven different pea traits examined by Mendel involved genes that we now know are linked. Knowing this, can you explain why Mendel was still able to use results from his crossing experiments to develop the principle of independent assortment?

Either peas have lots of chromosomes or just a few very, very big ones. In either case genetic linkage is minimized.

For the purposes of this question, assume that being Rh+ is a consequence of D and that Rh- individuals are dd. The ability to taste phenylthiocarbamide (PTC) is determined by the gene symbolized T (tt are nontasters). A female whose mother was Rh- has the MN blood group, is Rh+ and a nontaster of PTC, and is married to a man who is MM, Rh-, and a nontaster. List the possible genotypes of the children. Assume that all the loci discussed in this problem are autosomal and independently assorting.

F: DdMNtt M: ddMMtt possible offspring: DdMNtt, ddMNtt, DdMMtt, ddMMtt

In peas, gray seed color is dominant to white. For the purposes of this question, assume that Mendel crossed two plants with gray seeds with each other and the following progeny were produced: 320 gray and 80 white. (a) What is the most probable genotype of each parent? (b) What genotypic and phenotypic ratios are expected in the progeny of such a cross?

G = gray, g = white a) Gg x Gg b) 3:1

What is hemophilia and why can females, but not males, be carriers of hemophilia and other X-linked recessive characteristics?

Hemophilia results from mutations in blood clotting factors on the X chromosome. Women can be carriers since they carry the mutation on only one of the two X chromsomes. The other can provide correct genetic information.

What is a homologous pair of chromosomes, and how is it different from a nonhomologous chromosome pair? How do homologous chromosomes differ from sister chromatids?

Homologous chromosomes have the same genes, but may have different alleles for those genes; nonhomologous pairs do not have the same genes. Homologous contain the same genes, but each chromosome may contain different alleles of those genes. Sister chromatids are the genetically identical halves of a single replicated chromosome.

Describe the difference between homologous chromosomes and sister chromatids.

Homologous chromosomes may have different alleles, but carry the same genes. Sister chromatids are duplicates and (except for errors in replication) are identical in sequence.

Describe the difference between meiosis I and meiosis II.

Homologs pair and segregate in meiosis I. Sister chromatids are paired and segregate in meiosis II. Crossing over occurs in meiosis I, but not in meiosis II.

What is the maximum recombination frequency observable from a cross? Why is it less than the true genetic distance in some cases?

If a double heterozygote is testcrossed and gametes are produced in a ratio of 1:1:1:1, this can be interpreted as unlinked genes or as genes that are 50 map units apart. Because gametes this far apart or more cannot be distinguished from gametes that are unlinked, it's not possible to ever get a ratio of gametes produced that indicates a distance of greater than 50 map units. Some genes that are farther apart but are still on the same chromosome may appear to be closer because the observed proportion of recombinant gametes neglects to count those double crossover events that occur between the genes. When a double crossover occurs, the parental gametes are formed again and it looks as if no crossovers have occurred at all.

You have found a mutant of the p53 gene product that will affect the G1/S checkpoint. What would you predict would be the effect of this mutation and a potential consequence?

If the G1 checkpoint is abrogated the cell may accumulate DNA damage prior to division. This could lead to unchecked proliferation or an apoptotic response.

Two mice of the same species have different ear shapes. You find that one mouse, having normal shaped ears, was caught in a field in Kenya. The other mouse, with curled ears, was caught in the frozen tundra of Greenland. You have determined that both mice have identical genotypes at the gene loci controlling ear shape. How would you explain the differences in ear shape? Describe an experiment by which you could test your hypothesis. More (correct) detail will increase your likelihood of getting full credit.

If the genotypes are identical than the phenotypes must be due to environmental influences. The most facile way to determine this is to cross the two strains to determine the phenotype. The F1 offspring should then be raised in conditions similar to Greenland or Kenya, to phenocopy the animals there

What is meant by the term epistasis? Distinguish between epistasis and dominance. Do not use examples in answering this question.

In dominance one or the other trait will be seen in heterozygotes. In epistasis a second gene pair can interfere with the first.

How do incomplete and co-dominance differ?

In incomplete dominance there is a third phenotype in the heterozygotes, while codominace can have two different alleles dominant at the same time.

Why is mitosis important within the cell cycle?

It is how cells divide, which is the most important decision a cell can make. Mitosis is used for embryonic development, and in adults for replenishing lost and damaged cells.

You are the proud owner of a calico cat kitten named Buffy. When you take Buffy to the vet for its first visit, you find out that Buffy is a male! Is this unexpected, and if so why? What do you expect the chromosomal karyotype of your cat to look like with respect to X and Y chromosomes? Do you think you will be able to breed Buffy in the future?

It is unexpected because calico cats are generally XX females whose characteristic coat coloration is the result of X-inactivation. I expect the chromosomal karyotype of the sex chromosomes to be XXY, which is analogous to Kleinfelters syndrome in humans, and explains why Buffy is both a calico and male. I do not expect to be able to breed Buffy, because the "Kleinfelter's calico cats" have some of the same phenotypic characteristics as humans, including sterility.

Red-green color-blindness is an X-linked recessive condition. Juliet has a bit of difficulty passing the red-green color distinction test when she tries to get her driver's license. She has two children with a man who is not color blind. Neither Juliet's son (Henry) nor daughter (Roxanne) is color blind. Henry and Roxanne have normal karyotypes. Roxanne has a son who is color blind. Why aren't Henry and Roxanne color blind?

Juliette didn't pass on her X-linked color blindness allele to her son Henry, but she did pass it on to her daughter Roxanne. Roxanne isn't color blind because she has a wild-type allele from her father.

Red-green color-blindness is an X-linked recessive condition. Juliet has a bit of difficulty passing the red-green color distinction test when she tries to get her driver's license. She has two children with a man who is not color blind. Neither Juliet's son (Henry) nor daughter (Roxanne) is color blind. Henry and Roxanne have normal karyotypes. Roxanne has a son who is color blind. What is Juliet's genotype for the color-blindness allele? How would you explain her partial color blindness?

Juliette is a heterozygote for the color blindness allele. She cannot be homozygous as she has a son who isn't color blind. It is likely that many of the cells of her eye that sense color inactivated the wild type X chromosome, leaving them with an active X chromosome with a defective opsin.

What do kinase enzymes do, and why are they important to researchers of diseases such as cancer?

Kinase enzymes add phosphates to proteins. A set of kinases control the cell cycle, making them pivotal in cell cycle disorders like cancer.

Assume that investigators crossed a strain of flies carrying the dominant eye mutation Lobe on the second chromosome with a strain homozygous for the second chromosome recessive mutations smooth abdomen and straw body. The F1 Lobe females were then backcrossed with homozygous smooth abdomen, straw body males, and the following phenotypes were observed: smooth abdomen, straw body 820 Lobe 780 smooth abdomen, Lobe 42 straw body 58 smooth abdomen 148 Lobe, straw body 152 Give the gene order and map units between these three loci.

Lobe is in the middle. smooth abdomen---5---Lobe-----------15-------------straw body (or a description of this is fine)

Most calico cats are female, but you can find male calico cats. How would this be possible?

Male calicos can be XXY individuals, in other words, Kleinfelters cats having the Y chromosome, which would determine maleness, and the 2 X chromosomes producing the color pattern. GRADE NOTE: an alternative answer would be that the male cat SRY region could be translocated to one of the X chromosome, which may also produce this phenotype

Explain why mitosis does not produce genetic variation and how meiosis leads to the production of tremendous genetic variation.

Mitosis results in identical daughter cells, and is carried out to reduce variation. Meiosis involves independent assortment and crossing over, which creates gametes with different combinations of alleles.

Trisomy 21 or Down syndrome occurs when there is a normal diploid chromosomal complement of 46 chromosomes plus one (extra) chromosome #21. Such individuals therefore have 47 chromosomes. Assume that a mating occurs between a female with Down syndrome and a normal 46-chromosome male. What proportion of the offspring would be expected to have Down syndrome? Justify your answer.

One half of the gametes from the Down's individual will be disomic, thus one half of the offspring will exhibit Downs.

For a particular plant, red flowers (A) are dominant over yellow flowers (a). An initial cross was made between a plant that was true-breeding for red flowers, and another plant true-breeding for yellow flowers. F1 progeny, all having red flowers, were allowed to form seeds (through self-fertilization), which were then planted to generate F2 progeny. Pollen from all the resulting F2 plants was pooled and used to fertilize true-breeding yellow plants. What proportion of the progeny resulting from this cross would be expected to have yellow flowers?

P1: RR x rr F1: Rr x Rr F2: 1RR, 2Rr, x rr F3: ? 1/2 of the alleles in the F2 are r, thus 1/2 of the offspring in the F3 will be rr, yellow

A couple has one girl and is expecting a second child. What is the probability that this child will be a boy? After several years and several births, this couple now has six daughters and is expecting a seventh. What is the probability that this child will be a boy?

Pboy=1/2

Describe two methods used to study inheritance of human genetic characteristics.

Pedigree analysis, LOD scores (a statistical estimate of whether two loci (the sites of genes) are likely to lie near each other on a chromosome and are therefore likely to be inherited together as a package)

The phenotype of vestigial (short) wings (vg) in Drosophila melanogaster is caused by a recessive mutant gene that assorts independently with respect to a recessive gene for hairy (h) body. Assume that a cross is made between a fly with normal wings and a hairy body and a fly with vestigial wings and normal body hair. The wild-type F1 flies were crossed among each other to produce 1024 offspring. What phenotypes would you expect among the 1024 offspring, and how many of each phenotype would you expect?

Phenotypes: wild, vestigial, hairy, vestigial hairy Numbers expected: wild (576), vestigial (192), hairy (192), vestigial hairy (64)

You are viewing a pedigree and trying to determine the inheritance pattern of a trait. List three characteristics that, if observed, would indicate that the trait is autosomal recessive.

Same number of males and females affected Skips generations Unaffected can have affected children

Provide simple definitions that distinguish segregation and independent assortment

Segregation implies that either allele can be inherited randomly from a heterozygote. independent assortment suggests that genes will not depend on each other for their inheritance if they are on different chromosomes or far apart on the same chromosome

Why is it that human males are so much more likely to express recessive sex-linked traits than females?

Sex-linked traits are carried on the X chromosome. Since males only have one X chromosome if a recessive trait is carried on the X it will always be phenotypically seen in males.

Two pea plants with purple flowers are crossed. Among the offspring, 63 have purple flowers, and 17 have white flowers. With a chi-square test, compare the observed numbers with a 3:1 ratio and determine if the the difference between observed and expected is statistically significant. Note: Chi-square critical value with 1 degree of freedom is 3.84 with alpha equal to 0.05.

Sum (O-E)^2/E = .592; .592 < 3.84, therefore the observed are not statistically significantly different from the expected and differences are due to random chance

In peas, tall is dominant to short. A homozygous tall plant is crossed with a short plant. The F1 are self-fertilized to produce the F2. Both tall and short plants appear in the F2. a. If the tall F2 are self-fertilized, what types of offspring and proportions will be produced? b. If the short F2 are self-fertilized, what types of offspring and proportions will be produced?

T = tall, t = short a) TT x TT; all tall TT xTt; all tall Tt x Tt; 3 tall:1 short b) tt x tt; all short

State Mendel's Law of Independent Assortment. What is the molecular basis of this law (i.e., what happens inside the cell that gives rise to the law)?

That genes do not depend on each other for inheritance, due to the fact that homologous chromosome pairs line up at random across the metaphase plate in meiosis I

If a cyclin protein was produced at a constant level (instead of variably) througout the cell cycle what effect would you predict?

The cell would remain in that portion of the cell cycle that the cyclin and its CDK directed. This would likely force the cell into an apoptotic program.

Describe the centromere and kinetochore.

The centromere is a repeat rich stretch of DNA that binds to a specific set of proteins, while the kinetochore is a set of proteins that binds to the centromeric proteins and to spindle microtubules

What is the chi-square test used for, and what does it tell you?

The chi-square test is a statistical tool for analyzing data generated from genetic experiments to determine if observed results are consistent with a hypothesis proposed to explain them. The calculated chi-square statistic may be used to determine whether or not to accept or reject the proposed hypothesis within pre-determined confidence limits.

In deer mice, red eyes (r) is recessive to normal black eyes (R). Two mice with black eyes are crossed. They produce two offspring, one with red eyes and one with black eyes. Give the genotypes of parents and offspring of this cross.

The parents are heterozygotes, Rr The red-eyed offspring is rr, the black eyed are either RR or Rr

The white-eye gene in Drosophila is recessive and sex-linked. Assume that a white-eyed female is mated to a wild-type male. What would be the phenotypes of the offspring? Explain your answer.

The white-eyed female has the recessive allele on each X chromosome. The red-eyed (wildtype) male has the wildtype allele on his single X chromosome. So for the offspring, the X of the female combined with the X of the male will give wildtype (heterozyous) females (XX). Also, the X of the female combined with the Y of the male (hemizygous) will give white-eyed males (XY).

A color-blind woman with Turner syndrome (XO) has a father who is color-blind. Given that the gene for the color-blind condition is recessive and X-linked, provide a likely explanation for the origin of the color-blind and cytogenetic conditions in the woman. Be sure to describe what occurred in each parent during gametogenesis.

The woman inherited an Xrg chromosome from the father. Nondisjunction in the female (either at meiosis I or II) produced an egg with no X chromosome, which, when fertilized by the Xrg-bearing sperm, produced the Turner syndrome condition.

What are sex-limited and sex-influenced traits and what is the difference between the two?

These are traits encoded by Autosomal genes, but their expression is dependent on the hormonal constitution of the individual, i.e., male or female. The difference between the two is as follows: Sex-limited expression: phenotype is absolutely limited to one sex Sex-Influenced: sex of an individual influences the expression of a phenotype

A geneticist is examining a culture of fruit flies and discovers a single female with strange spots on her legs. The new mutation is named melanotic. When a female melanotic fly is crossed with a normal male, the following progeny are produced: 123 normal females, 125 melanotic females, and 124 normal males. In subsequent crosses with normal males, melanotic females are frequently obtained, but never any melanotic males. Provide a possible explanation for the inheritance of the melanotic mutation. (Hint: The cross produces twice as many female progeny as male progeny.)

These observations can be explained by a single gene locus with two segregating alleles. Note that (i) the trait appears to be very rare and is thus probably a recessive, spontaneously generated mutation, and (ii) the progeny sex ratios are skewed 2:1 (female:male). The most likely explanation for the progeny results indicated is that the melanotic allele is a lethal, recessive, X-linked allele and is therefore lethal in the homozygous or hemizygous condition. Melanotic heterozygotes display the unusual coloring scheme observed. The homozygous condition in females is never produced by the crossing scheme described. Using (m) for the recessive, X-linked, lethal allele: Xm/X+ (melanotic female parent) × X+/y (normal male parent) Xm/X+ (1; melanotic female zygote) Xm/y (1; unviable male zygote) X+/X+ (1; normal female zygote) X+/y (1; normal male zygotes) Note: There are four phenotypic classes in a 1:1:1:1 ratio (with regard to sex and melanotic genotype), which suggests involvement of a single, X-linked locus with two segregating alleles.

In a plant species, you notice that purple and yellow leaf colors, as well as hairy and smooth stems, segregate. You cross a plant with purple leaves and hairy stems to a plant with yellow leaves and hairy stems, and generate the progeny indicated below: Class Phenotypes 1 68 yellow leaves, hairy stems 2 66 purple leaves, hairy stems 3 22 purple leaves, smooth stems 4 25 yellow leaves, smooth stems Total: 181 a. Indicate the most likely genotypes for each parent. b. Propose a hypothesis to explain the progeny results. Based on this hypothesis, what ratios are expected for each of the four classes of progeny? c. Using the chi-square method, test your hypothesis and indicate whether you accept or reject it.

Using P for the leaf color locus and S for the stem type locus: PpSs × ppSs. b. A possible hypothesis is that each trait is controlled by independently assorting single gene pairs, and that one allele in each pair exhibits complete dominance over the other. Class 1 = 3/8, class 2 = 3/8, class 3 = 1/8, class 4 = 1/8. c. (68 - 68)2/68 + (66 - 68)2/68 + (22 - 23)2/23 + (25 - 23)2/23 = 0.276. This number (the chi-square statistic) is very small; therefore you confidently accept your hypothesis.

Comb shape in chickens represents one of the classic examples of gene interaction. Two gene pairs interact to influence the shape of the comb. The genes for rose comb (R) and pea comb (P) together produce walnut comb. The fully homozygous recessive condition (rrpp) produces the single comb. Assume that a rose-comb chicken is crossed with a walnut-comb chicken and the following offspring are produced: 17 walnut, 16 rose, 7 pea, 6 single. (a) What are the probable genotypes of the parents? (b) Give the genotypes of each of the offspring classes. Some classes may have more than one possible genotype. Separate and clearly label your answers for a-b in the box below.

Walnut crossed with single produces 1 walnut, 1 rose, 1 pea, and 1 single offspring. RrPr x rrpp Rose crossed with pea produces 20 walnut offspring. RRpp x rrPP Pea crossed with single produces 1 single offspring. rrPp x rrpp Rose crossed with pea produces 2 walnut, 1 single, and 1 pea offspring. RrPp x rrpp Rose crossed with single produces 31 rose offspring. RRpp x rrpp Rose crossed with single produces 10 rose and 11 single offspring.

Tony, who is not diseased, has a sister with cystic fibrosis (CF). Neither of his parents have CF.Tony is expecting a child with Tina. Tina's family history is unknown. If the frequency of heterozygotes in the general population is 1/50, what is the probability that Tony and Tina's child will have CF? Explain each factor in your calculation.

With an affected sibling Tony's probability of carrying the allele is 2/3. Tina's is 1/50. The probability that both will have CF alleles is 2/3 x 1/50= 1/75 Even if both have CF alleles only 1/4 of their children will be affected, therefore the probability that they will have affected children is: 1/75 x 1/4 = 1/300

In humans, the presence of the SRY gene, normally on Y, determines maleness. In Drosophila, an X: A (autosome) ratio of 0.5 determines maleness. What is the gender of a human XXY individual? What is the gender of a Drosophila XXY individual? Explain your reasons for each answer.

XXY human is male. The SRY on the Y chromosome determines maleness, in spite of the two X chromosomes. XXY Drosophila is female. Drosophila is diploid, so there are two of each autosome. If there are two X chromosomes, the X:A ratio is 1.0, which is a female.

You are conducting a cross using Drosophila melanogaster. The results of your cross indicate that the recombination frequency is very high. How do you feel about the accuracy of your map unit calculations based on this cross?

You should feel good since lots of numbers makes for statistical confidence. You should determine that the genes are unlinked if the recombination frequency is 50%

While doing field work in Madagascar, you discover a new dragonfly species that has either red (R) or clear (r) wings. Initial crosses indicate that R is dominant to r. You perform three crosses using three different sets of red-winged parents with unknown genotype and observe the following data: Cross Phenotypes 1 72 red-winged, 24 clear-winged 2 4 red-winged 3 96 red-winged a. What is the most likely genotype for each pair of parents? b. Do you think there are a sufficient number of progeny to support each of your answers in the previous question?

a. 1. Rr x Rr 2. RR xRR 3. RR xRR b. The second cross has insufficient numbers. If we hypothesized that the parents were heterozygotes, the chi squared value would be 0.5 with one degree of freedom. That is les thant he value that would allow us to accept the hypothesis.

The red kernel color in wheat is caused by the presence of at least one dominant allele from each of two independently segregating gene pairs (e.g., R-B-). Wheat plants with rrbb genotypes have white kernels, and plants with genotypes R-bb and rrB- have yellow kernels. If you cross a plant true breeding for red kernels with a plant true breeding for white kernels, a. what is the expected phenotype(s) and ratios of the F1 plants? b. what are the relative proportions of the phenotypic classes expected in the F2 progeny after selfing the F1 progeny?

a. All red b. 1/16 white 9/16 red 6/16 yellow

In the yawncat (a rare hypothetical animal), the dominant allele R causes solid tail color, and the recessive allele r results in white spots on a colored background. The black coat color allele B is dominant to the brown allele b, but these genes can only be expressed if the animal has an mm genotype at a third gene locus. Animals that are M- are yellow regardless of which allele from the B locus is present. A mating between a solid yellow-tailed male yawncat and a solid brown-tailed female yawncat produce 16 offspring with the following tail phenotypes: six solid yellow, two spotted yellow, three solid black, one spotted black, three solid brown, and one spotted brown. a. What are the most likely genotypes of the male and female parents? Explain the strategy used to determine the genotypes. b. What is the probability of the next offspring from these same two parents having a spotted brown tail?

a. . Parents are MmBbRr (solid yellow-tailed male) and mmbbRr (solid brown-tailed female). Each must carry recessive alleles for each trait since homozygous recessives for each are seen in the offspring. b. Pbbrr=1/2 x 1/2 = 1/4

The enzyme glucose-6-phosphate dehydrogenase deficiency (G6PD) is inherited as a recessive gene on the X chromosome in humans. A phenotypically normal woman (whose father had G6PD) is married to a normal man. (a) What fraction of their sons would be expected to have G6PD? (b) If the husband had G6PD, would it make a difference in your answer in part (a)?

a. 1/2 b. no, fathers don't give sons X chromosomes

Two gene loci, A and B, are unlinked (and thus assort independently), and alleles A and B are dominant over alleles a and b. Indicate the probabilities of producing the following. a. An AB gamete from an AaBb individual? b. An AB gamete from an AABb individual? c. An AABB zygote from a cross AaBb × AaBb? d. An AaBb zygote from a cross AaBb × AABB? e. An Aabb zygote from a cross AaBb × AAbb? f. An AB phenotype from a cross AaBb × AaBb? g. An AB phenotype from a cross aabb × AABB? h. An aB phenotype from a cross AaBb × AaBB?

a. 1/4 b. 1/2 c. 1/16 d. 1/4 e. 1/4 f. 9/16 g. 1 h. 1/8

The trait of medium-sized leaves in iris is determined by the genetic condition PP'. Plants with large leaves are PP, while plants with small leaves are P'P'. A cross is made between two plants each with medium-sized leaves. (a) If they produce 80 seedlings, how many plants with each type of leaf (large leaves, medium leaves, small leaves) would you expect? (b) What is the term for this allelic relationship? Separate and clearly label your answers for a-b in the box below

a. 20 large, 40 medium, 20 small b. incomplete dominance

A testcross is performed on an individual to examine three linked genes. The most frequent phenotypes of the progeny were Abc and aBC, and the least frequent phenotypes were abc and ABC. a. What was the genotype of the individual tested? b. What is the order of the gene loci? c. Indicate the genotypes of each progeny class.

a. Abc/aBC b. BAC c. Parental: bAc/BaC DCO: bac/bac and BAC/bac SCO: baC and BAC; bAC and Bac

You are studying body color in an African spider and have found that it is controlled by a single gene with four alleles: B (brown), br (red), bg (green), and by (yellow). B is dominant to all the other alleles, and by is recessive to all the other alleles. The bg allele is dominant to by but recessive to br. (a) You cross a pure-breeding brown spider with a pure breeding red spider. Diagram the cross and predict the genotype and phenotype of the progeny. (b) You cross a pure-breeding brown spider with a pure-breeding green spider. Diagram the cross and predict the genotype and phenotype of the progeny. (c) You cross the progeny from parts (a) and (b), Diagram the cross and predict the genotype and phenotype of the progeny. d. You cross the non-brown progeny from part (c) to a pure-breeding yellow spider. Diagram the cross and predict the genotype and phenotype of the progeny.

a. B/B X br / br → B/ br (all brown) b. B/B X bg / bg → B/ bg (all brown) c. B/ br X B/ bg → B/B (brown) Bbg (brown) Bbr (brown) brbg (red) d. brbg X byby → brby (red) bgby (green)

You are doing lab work with a new species of beetle. You have isolated lines that breed true for either blue shells and long antenna, or green shells and short antenna. Crossing these lines yields F1 progeny with blue shells and long antenna. Crossing F1 progeny with beetles that have green shells and short antenna yield the following progeny: blue shell, long antenna 82 green shell, short antenna 78 blue shell, short antenna 37 green shell, long antenna 43 total 240 a. Which shell color and antenna length alleles are dominant? b. Are the shell color locus and antenna length locus linked? c. Check your answer to the previous question with the chi-square test. d. If the genes are linked, calculate the recombination frequency between them.

a. Blue shells and long antenna b. yes c. (82-60)2/60 + (78-60)2/60 + (37-60)2/60 + (43-60)2/60 484/60 + 324/60 + 529/60 + 289/60 8 + 5.4 + 8.8 + 4.8 27, with 1 degree of freedom d. Recombinants 37 + 43/total 240 = 37.2 cm

In rabbits, an allelic series helps to determine coat color: C (full color), cch (chinchilla; gray color), ch (Himalayan; white with black extremities), and c (albino; all white). The C allele is dominant to all others, cch is dominant to ch and c, ch is dominant to c, and c is recessive to all the other alleles. This dominance hierarchy can be summarized as C>cch>ch>c. Give the genotypic and phenotypic ratios expected if rabbits with the following genotypes are crossed: a. Ccch × Cch b. Cch × chc c. Cch × cc d. cchch × chc e. Cc × chc

a. Ccch x Cch 1 CC (full color); 2 Ccch (full color); 1 cch cch (1 chinchilla) 3 full color:1 chinchilla b. Cch x chc 1 Cch (full color); 1 Cc (full color); 1 ch ch (Himalayan); 1 ch c (Himalayan) 1 full color: 1 Himalayan c. Cch x cc 1 Cc (full color):1 ch c (Himalayan) 1 full color: 1 Himalayan d. cchch x chc 1 cch ch (chinchilla); 1 cch c (chinchilla); 1 ch ch (Himalayan); 1 ch c (Himalayan) 1 chinchilla:1 Himalayan e. Cc x chc 1 Cch (full color); 1 Cc (full color); 1 cch (Himalayan); 1 cc (albino) 2 full color: 1 Himalayan; 1 albino

Meiosis involves three events that are unique and are not seen in mitosis. List these events and the stage of meiosis in which they occur. All the events occur during meiosis I.

a. Homologous chromosomes come into contact during prophase I of meiosis and crossing over of non-sister chromatids occurs. b. At metaphase I homologous pairs line up on the metaphase plate. c. During anaphase 1 of meiosis the sister chromatids remain attached and go to the same end of the cell.

In Drosophila melanogaster, cut wings (ct) is recessive to normal wings (ct+), sable body (s) is recessive to gray body (s+), and vermilion eyes (v) is recessive to red eyes (v+). All three recessive mutations are X-linked. A female fly with cut wings, sable body, and vermilion eyes is crossed to a male with normal wings, gray body, and red eyes. The F1 females produced by this cross were mated with cut, sable, vermilion males in a testcross. The following are the progeny resulting from the testcross. v ct s 510 v+ ct s 1 v+ ct+ s 14 v+ ct+ s+ 500 v+ ct s+ 73 v ct s+ 20 v ct+ s 81 v ct+ s+ 1 Total → 1200 a. Determine the order of these genes on the chromosome. b. Calculate the map distances between the genes. c. Determine the coefficient of coincidence and the interference among these genes.

a. Identify progeny classes representing the double crossover gametes (i.e., the least frequent class; v ct+ s+), and parental gametes (i.e., the most frequent class; v+ ct+ s+). Next, identify the locus whose alleles, if switched, will convert from DCO to parental class, and vice versa. For example, switching the mutant allele (v) in the DCO class to the parental allele v+ converts the DCO phenotype to the parental phenotype. The (v) locus therefore lies between the (ct) and (s) loci, and the relative order of this linkage group is (ct+ v s+) or (s+ v+ ct+). Both orders are correct because, from this experiment alone, we have no way of orienting (ct) or (v) relative to either end of the chromosome containing this linkage group. All that a single 3-factor cross can do is fix the position of the middle marker, in this case the (v) locus, relative to the other two loci that flank it. In practice, many 3-factor crosses using other markers within and located outside this linkage group are performed, and the relative positions of increasing numbers of loci are established (i.e., more genes are mapped). b. s - v: (14 + 20 + 2)/(1200) = 0.030 = 3.0% = 3.0 cM v - ct: (81 + 73 + 2)/(1200) = 0.13 = 13.0% = 13.0 cM c. The expected number of double crossovers (DCO EXP) equals the product of the number of observed single crossovers (SCO). Use the calculated map distances for each of the two SCO regions, and apply the multiplicative rule to calculate the expected number of double crossovers: DCO EXP = (0.03)(0.13) = 0.004 (0.4%). Therefore, 0.4% of the 1200 total progeny, or (0.004)(1200) = 4.8 progeny are expected to be DCO progeny (DCO EXP). The observed number of double crossover progeny (DCO OBS) = 2. The coefficient of coincidence (CC) = DCOOBS /DCO EXP = 2/4.8 = 0.417; and the interference (I) = 1 - CC, therefore I = 1 − 0.417 = 0.583.

You just bought two black guinea pigs from the pet store that are known to be heterozygous (Bb). You also know that black fur (BB) is dominant over white fur (bb), and that a lethal recessive allele is located only one cM away from the recessive b allele. You decide to start raising your own guinea pigs, but after mating these animals several times, you discover they produce only black progeny. a) How would you explain this result? b) If the original black guinea pigs produce an average of 10 offspring per mating, how many matings would you have to make with these same parents before you'd expect to see a white guinea pig? c) Indicate the most likely genotype of the white offspring.

a. any bb homozygotes are dying because of the closely linked lethal allele b. you would expect to see ten litters before you got your first white pig c. the white offspring is bb in which the lethal allele has been recombined away

A mutant gene that produces brown eyes (bw) is located on chromosome #2 of Drosophila melanogaster, whereas a mutant gene producing bright red eyes, scarlet (st), is located on chromosome #3. Phenotypically wild-type flies (with dull red eyes), whose mothers had brown eyes and whose fathers had scarlet eyes, were mated. The 800 offspring possessed the following phenotypes: wild type (dull red), white, scarlet (bright red), and brown. Most of the 800 offspring had wild-type eyes, whereas those with white eyes were the least frequent. (a) The cross for the P generation is: bw/bw; st+/st+ X bw+/bw+; st/st (brown-eyed mothers X scarlet-eyed fathers) Using standard symbolism, like that in the P generation cross above, diagram the cross for the F1 generation. Be certain to provide the alleles of the mutant genes. (b) From the information presented above, how many white-eyed flies would you expect in the F2 generation? Separate and clearly label your answers for a-b in the box below.

a. bw+/bw;st+/st x bw+/bw;st+/st b. 1/16 of 800, or 50

In the fruit fly, Drosophila melanogaster, a spineless (no wing bristles) female fly is mated to a male that is claret (dark eyes) and hairless (no thoracic bristles). Phenotypically wild-type F1 female progeny were mated to fully homozygous (mutant) males, and the following progeny (1000 total) were observed: Phenotypes Number Observed spineless 321 wild 38 claret, spineless 130 claret 18 claret, hairless 309 hairless, claret, spineless 32 hairless 140 hairless, spineless 12 (a) Which gene is in the middle? (b) With respect to the three genes mentioned in the problem, what are the genotypes of the homozygous parents used in making the phenotypically wild F1 heterozygote? (c) What are the map distances between the three genes? Separate and clearly label answers for a-d in the box below.

a. hairless is in the middle b. sp + + and + hr cl c. 9cm between spineless and hairless 34cm between hairless and claret

In snapdragons, the allele for red flowers is incompletely dominant over the allele for white flowers, and thus heterozygotes have pink flowers. What ratios of snapdragon flower colors would you expect to see among progeny generated from the following crosses? a. red × white b. red × pink c. white × pink d. white × white e. pink × pink f red × red

a. red × white 100% pink b. red × pink 50% pink; 50% red c. white × pink 50% pink; 50% white d. white × white 100% white e. pink × pink 25% red; 50% pink; 25% white f. red × red 100% red

What are three of the characteristics of an autosomal dominant trait observed in a pedigree analysis?

affects both sexes equally never skips generations half the children of affected parents are affected

You observe continuous variation in tail length in a wild population of rats. Describe an experiment to determine whether this variation is an example of variable expressivity or continuous variation?

breed long tailed animals with short tailed, then interbreed their offspring. If the differences are caused by allelic differences the F1 will be intermediate and the F2 will have many different tail lengths.

What is the fundamental difference with respect to the chromatids between metaphase of mitosis and metaphase I of meiosis? How does the resulting cell differ in mitosis vs. meiosis I?

in mitosis single chromosomes line up at the metaphase plate. In meiosis I chromosome pairs line up at the metaphase plate. After mitosis daughter cells are diploid, meiotic daughter cells are haploid.

What events during sexual reproduction are significant in contributing to genetic diversity?

independent assortment of homologous chormosomes meiotic crossing over random fertilization of eggs by sperm

The ability to curl one's tongue into a U-shape is a genetic trait that is inherited in a standard Mendelian fashion. Curlers always have at least one curler parent but noncurlers can have one or both parents who are curlers. Using C and c to symbolize this trait, what is the genotype of a noncurler?

non curlers are cc

What are two features that distinguish a prokaryotic cell from a eukaryotic cell?

nucleus, histones, organelles

You are viewing a pedigree and trying to determine the inheritance pattern of a trait. List three characteristics that, if observed, would indicate that the trait is X-linked recessive.

occurs in males more than females affected fathers never have affected sons skips generations

In what type of cell—in humans—would you expect to find meiosis occuring?

primary spermatogonia and primary oogonia

Assume that the genes for tan body and bare wings are 15 map units apart on chromosome II in Drosophila. Assume also that a tan-bodied, bare-winged female was mated to a wild-type male and that the resulting F1 phenotypically wild-type females were mated to tan-bodied, bare-winged males. Of 1000 offspring, what would be the expected phenotypes, and in what numbers would they be expected?

wild type = 425; tan-bare = 425; tan = 75; bare = 75