Gen ch 4

t B In a three‑point mapping experiment for the genes y‑w‑ec, the following percentages of events are observed: NCO events: 65% SCO events between y and w: 15% SCO events between y and ec: 17% DCO events: 3% What is the map distance between y and ec?

20 map units ( The map distance between any two genes is the sum of the percentages of all detectable recombination events between them, so 17 + 3 = 20.)

For X-linked traits in Drosophila, the male phenotype is determined by the maternally inherited allele.

True (Males inherit only one X chromosome. That chromosome is contributed by the female parent.)

A condition in which one gene pair masks the expression of a nonallelic gene pair is called ________.

epistasis

Genes X, Y, and Z are linked. Crossover gametes between genes X and Y are observed with a frequency of 25%, and crossover gametes between genes Y and Z are observed with a frequency of 5%. What is the expected frequency of double crossover gametes among these genes?

1.25% The probability of a double crossover is the product of the probabilities of the single crossovers: 0.25 x 0.05 = 0.0125, or 1.25%.

Typical ratios resulting from epistatic interactions in dihybrid crosses would be ________.

9:3:4, 9:7

Two genes that are separated by 10 map units show a recombination percentage of 10%. -True -False

true (One map unit is equal to 1% recombination between two genes; 10 map units would be equal to 10% recombination between the genes.)

Assume that the genes from the previous example are located along the chromosome in the order X, Y, and Z. What is the probability of recombination between genes X and Z?

30% Recombination frequencies between linked genes along a chromosome are additive, so the recombination frequency between genes X and Z is 25 + 5 = 30.

If two genes on the same chromosome exhibit complete linkage, what is the expected F2 phenotypic ratio from a selfed heterozygote with the genotype a+b+ ⁄⁄ ab?

3:1 (Each parent produces two types of gametes, a+b+ and ab, giving the simple Mendelian ratio of 3 a+b+ : 1 ab.)

Crossing over during prophase I of meiosis occurs between alleles on sister chromatids

False (Crossing over during meiosis occurs between alleles on nonsister chromatids. )

Which of the following statements about three linked genes that are spaced very close together along a chromosome is most likely to be true? -Interference will be a significant factor in the number of crossovers observed. -Interference and the number of hidden double crossovers will significantly affect the number of crossovers observed. -The number of hidden double crossovers will be very large. -Interference is not likely to affect the number of crossovers observed.

Interference will be a significant factor in the number of crossovers observed. ( Interference effects are more likely when crossovers are confined to a small region.)

Which of the following statements about gamete formation during meiosis is false? -Complete linkage results in the formation of only parental gametes. -Recombinant gametes contain combinations of alleles not found in the parent cell. -Parental gametes contain the same combinations of linked genes as found in the parent cell. -Parental gametes can be formed only if there is no crossing over during meiosis.

Parental gametes can be formed only if there is no crossing over during meiosis. ( If crossing over occurs, half of the gametes formed are parental and the other half are recombinant.)

For linked genes A, B, and C, the fraction of single crossovers for genes A and B is 0.1, the fraction of single crossovers for genes B and C is 0.3, and the fraction of double crossovers is 0.03. Which of the following statements is true? -The distances between A and B and B and C are approximately equal. -Interference has affected the number of double crossovers. -The distance between A and B is greater than the distance between B and C. -The distance between A and B is less than the distance between B and C.

The distance between A and B is less than the distance between B and C. (The frequency of crossover events decreases as the distance between genes decreases. )

ij x kL A_XBXb x A_XBy IV-3 Show diseas IV4=? 1 ProBBlt IV4 in Condt A=1/36 2----------------------------B=1/2

The new information allows you to be certain about two requirements that previously could only be assigned probabilities: You now know that individual III-4 has the genotype X B X b , meaning that she received an X b chromosome from II-4. You also now know that individual IV-4 is a male, meaning that he received a Y chromosome from III-5. This new information changes the probability calculations because you know that these two requirements have been met (thus, they have a probability of one -- certainty). The only requirement that is unchanged is that of III-4 passing on an X b chromosome to IV-4 -- the probability is still 1/2.

In a gene mapping cross, the term Frac NCO refers to the fraction of gametes that have the same genotypes as the parental gametes. True False

True ( Frac NCO refers to the fraction of gametes that have not undergone crossing over and thus their genotypes reflect those of the parental gametes)

Assume that a cross is made between AaBb and aabb plants and that the offspring occur in the following numbers: 106 AaBb, 48 Aabb, 52 aaBb, 94 aabb. These results are consistent with ________.

linkage with approximately 33 map units between the two gene loci

White eye color is an X-linked trait in one line of fruit flies. White eyes is recessive to red eyes. If a red-eyed female and a white-eyed male are crossed, _______. some of their male progeny may have white eyes

some of their male progeny may have white eyes (If the female is heterozygous, approximately half of the male progeny will have white eyes.)

For linked genes A, B, and C, the map distance A-B is 5 map units and the map distance B-C is 25 map units. If there are 10 double crossover events out of 1000 offspring, what is the interference?

0.2 ( The coefficient of coincidence is 0.01/0.0125 = 0.8, so the interference is 1 - 0.8 = 0.2.)

Part C - Experimental prediction: Comparing autosomal and sex-linked inheritance You now know that inheritance of eye color in fruit flies is sex-linked: The gene encoding eye color is located on the X chromosome, and there is no corresponding gene on the Y chromosome. How would the inheritance pattern differ if the gene for eye color were instead located on an autosome (a non-sex chromosome)? Recall that for autosomes, both chromosomes of a homologous pair carry the same genes in the same locations. Suppose that a geneticist crossed a large number of white-eyed females with red-eyed males. Consider two separate cases: Case 1: Eye color exhibits sex-linked inheritance. Case 2: Eye color exhibits autosomal (non-sex-linked) inheritance. (Note: In this case, assume that the red-eyed males are homozygous.)

1 Eye color exhibits sex linked inheritance -100 Fm osp,100 red eyes + 0 white - 0 male osp 0 red eyes+ 100 white -100 female Osp 100 red eyes+ 0white -100Male osp 100 red eyes+0 white

Part D - The effect of a fourth gene on fur color In the same mouse species, a fourth unlinked gene (gene P/p) also affects fur color. For mice that are either homozygous dominant (PP) or heterozygous (Pp), the organism's fur color is dictated by the other three genes (A/a, B/b, and C/c). For mice that are homozygous recessive (pp), large patches of the organism's fur are white. This condition is called piebaldism. In a cross between two mice that are heterozygous for agouti, black, color, and piebaldism, what is the probability that offspring will have solid black fur along with large patches of white fur?

Because each gene segregates independently, you need to determine the probability of each genotype independently and then multiply the four probabilities together. The probability of offspring with solid color (aa) is 1/4; the probability of offspring with black fur (BBor Bb) is 3/4; the probability of colored fur (CCor Cc) is 3/4; and the probability of piebald, or white patches (pp), is 1/4. The combined probability is1/4 x 3/4 x 3/4 x 1/4 = 9/256.

Part C - The effect of a third gene on fur color In the same mouse species, a third unlinked gene (gene C/c) also has an epistatic effect on fur color. The presence of the dominant allele C (for color), allows the A/a and B/b genes to be expressed normally. The presence of two recessive alleles (cc), on the other hand, prevents any pigment from being formed, resulting in an albino (white) mouse.

Because the C/c gene is epistatic to both the A/a and B/b genes, any offspring with the cc allele combination will be albino. Otherwise, the A/a and B/b genes are expressed normally. AaBbcc-Albino, AaBBCC-Agouti Black Aabbcc-Albono AAbbCc Agouti Brown aaBbCc-Solid color, Black AABBcc-Albino

Part B - Lethal alleles and epistasis In addition to A and a, the "agouti" gene has a third allele, AY . Here is some information about the inheritance of the AY allele. The AY allele is dominant to both A and a. The homozygous genotype (AYAY ) results in lethality before birth. The heterozygous genotypes (AYA or AYa) result in yellow fur color, regardless of which alleles are present for the B/b gene. (This effect exhibited by the AY allele is known as epistasis--when the expression of one gene masks the expression of a second gene.) In a mating of mice with the genotypes AYaBb x AYaBb , what is the probability that a live-born offspring will have yellow fur?

Because the presence of the AY allele is epistatic to (masks expression of) the B/b gene, the B/b gene does not need to be taken into consideration in this problem. For the AYa x AYa cross, 1/4 of the offspring would have the AYAY genotype, which is lethal before birth. For the live-born offspring, 2/3 would be AYa, and thus have yellow fur.

To construct a mapping cross of linked genes, it is important that the genotypes of some of the gametes produced by the heterozygote can be deduced by examining the phenotypes of the progeny. True False

False (To construct a mapping cross of linked genes, it is important that the genotypes of all of the gametes produced by the heterozygote can be deduced by examining the phenotypes of the progeny, taking into consideration that the homozygote produced only recessive gametes. Gametes and their genotypes can never be observed directly.)

Eye color in Drosophila is an X-linked trait. White eyes is recessive to red eyes. If a Drosophila male has white eyes, which of the following must also be true? -All his male siblings from the same parents must also have white eyes. -His mother had at least one white allele. -Some of his female siblings from the same mating must also have white eyes. -His father must also have had white eyes.

His mother had at least one white allele. (Because this male had white eyes, he must have inherited a white allele from his mother.)

Part A - Experimental technique: Reciprocal crosses When Gregor Mendel conducted his genetic experiments with pea plants, he observed that a trait's inheritance pattern was the same regardless of whether the trait was inherited from the maternal or paternal parent. Mendel made these observations by carrying out reciprocal crosses: For example, he first crossed a female plant homozygous for yellow seeds with a male plant homozygous for green seeds, and then crossed a female plant homozygous for green seeds with a male plant homozygous for yellow seeds. Unlike Mendel, however, Morgan obtained very different results when he carried out reciprocal crosses involving eye color in his fruit flies. The diagram below shows Morgan's reciprocal cross: He first crossed a homozygous red-eyed female with a white-eyed male, and then crossed a homozygous white-eyed female with a red-eyed male.

When Morgan crossed a homozygous red-eyed female with a white-eyed male, all of the offspring had red eyes. However, in the reciprocal cross (homozygous white-eyed female with a red-eyed male), all of the females had red eyes while all of the males had white eyes.

Part B - Experimental results: The F2 generation In one of Morgan's experiments, he crossed his newly discovered white-eyed male with a red-eyed female. (Note that all of the females at that time were homozygous for red eyes because the allele for white eyes had not yet propagated through Morgan's flies.) All of the F1 flies produced by this cross (both males and females) had red eyes.

When a homozygous red-eyed female was crossed with the white-eyed male (w+w+ × wY), the resulting F1 females were w+w and the F1 males were w+ Y. Crossing the F1 males and F1 females would yield these results: All the F2 females would have red eyes, although some would be homozygous (w+w+ ) and others would be heterozygous (w+w). Half the F2 males would have red eyes (w+ Y), and half would have white eyes (wY).

In mice, agouti fur is a dominant trait resulting in individual hairs having a light band of pigment on an otherwise dark hair shaft. A mouse with agouti fur is shown here, along with a mouse with solid color fur, which is the recessive phenotype (A = agouti; a = solid color). A separate gene, which is not linked to the agouti gene, can result in either a dominant black pigment or a recessive brown pigment (B = black; b = brown). A litter of mice from the mating of two agouti black parents includes offspring with the following fur colors: solid color, black solid color, brown (sometimes called chocolate) agouti black agouti brown (sometimes called cinnamon) What would be the expected frequency of agouti brown offspring in the litter?

Because the two traits are determined by unlinked genes, they assort independently. As a result, you need to use the multiplication rule to calculate the probability of agouti brown offspring (A_ bb) from AaBb parents. The probability of A_offspring is 3/4, and the probability of bb offspring is 1/4. The combined probability is therefore 3/4 x 1/4 = 3/16.

A=Autosomal recessive B X-linked Recessive

Begin a pedigree analysis by determining the inheritance mode of the condition. Is the condition recessive or dominant? Dominant conditions require that an affected individual have at least one affected parent. An affected offspring of two unaffected parents indicates a recessive condition. (This is the case for both conditions A and B.) Is the condition autosomal or X-linked? An affected female offspring of two unaffected parents indicates an autosomal recessive condition. (This is the case for condition A.) If the condition were X-linked, the male parent would be affected. To distinguish between rare X-linked and autosomal recessive conditions, choose the simpler mode for the observed pattern—that is, the mode that requires fewer unrelated individuals to carry a rare allele. In this example, the autosomal recessive mode would require two unrelated horses (II-1 and II-5) to carry a rare recessive allele for condition B, whereas the X-linked recessive mode requires only one. Therefore, the X-linked recessive mode is a better choice for condition B.

Pics from 8 ii4a=Aa, 4b=XBXb ii5c-A_, 5d-XBY II7e-A_, 7f-XBY ii8g-AA, 8h-XBXB III4i-A_, 4J-XBX_ III5k-A_ , 5I-XBY

Use these rules when assigning genotypes for X-linked recessive conditions: Males have only one allele for every X-linked gene. Affected males have the recessive allele, and non-affected males have the dominant (wild-type) allele. Unaffected females with affected sons are heterozygous (carriers). Unaffected females with affected fathers are heterozygous. All sons of affected females will be affected. All daughters of affected females will be heterozygous. Use these rules when assigning genotypes for autosomal recessive conditions: Unaffected parents with an affected offspring are both heterozygous. Unaffected offspring of one affected parent will be heterozygous. Unaffected offspring of two heterozygous parents may be homozygous dominant or heterozygous.