Past papers SU2

How many bivalents would you expect to observe at meiotic metaphase I in a hybrid between species C and species B?

0

How many bivalents would you expect to observe at meiotic metaphase I in a hybrid between species D and species B?

0

What progeny would result from crossing a daughter from the above mating with a brown-eyed male?

100% brown-eyed females 50% ivory-eyed males 50% brown-eyed males

How many univalents would you expect to observe at meiotic metaphase I in a hybrid between species C and species B?

15

How many univalents would you expect to observe at meiotic metaphase I in a hybrid between species D and species B?

15

The haploid number of a species of plant is 10, karyotyping a seedling, reveals that this individual has 21 chromosomes in total.

Aneuploidy

You are a beekeeper and also aspiring breeder. Ivory eye is a recessive characteristic in honeybees; wild-type eyes are brown. What progeny would result from crossing an ivory-eyed queen and a brown-eyed drone?

Brown-eyed females and ivory-eyed males

ABCdef

P (A B C d e f) = 0.40 x 0.05 x 0.35 = 0.007 (probability value) OR 0.7 % (% value) The correct answer is: 0.007 %

AbcDEf

P (A b c D E f) = 0.10 x 0.05 x 0.15 = 0.00075 ~ 0.001 (probability value) OR 0.075 % (% value) The correct answer is: 0.001 %

aBCdef

P (a B C d e f) = 0.10 x 0.05 x 0.35 = 0.00175 ~ 0.002 (probability value) OR 0.2 % (% value) The correct answer is: 0.002 %

abcDeF

P (a b c D e F) = 0.40 x 0.05 x 0.15 = 0.003 (probability value) OR 0.3 % (% value) The correct answer is: 0.003 %

Three genes are on the same chromosome, based on recombination event observations in a mapping experiment, the following distances are estimated: A-B: 15 cM; B-C: 10 cM; and A-C: 20 cM.

Positive interference

An autosomal gene has a wildtype- and mutant allele segregating in a population of sexually reproducing organisms. The interaction between the wildtype and mutant alleles are different depending on the sex of the individual.

Sex Influenced

In a species of fish, a blue stripe on the dorsal fin is observed in males and females. A fish breeder carries out a pair of reciprocal crosses and observes the following results: What is the most likely mode of inheritance for this trait?

Sex linked (with the striped allele dominant)

The following table presents chromosome data on four species of plants and their F1 hybrids Deduce the chromosomal origin of species A.

Species A is an allotetraploid (or amphidiploid) √ resulting from a hybridisation event between species C and D, √ followed by a subsequent diploidisation

The following table presents chromosome data on four species of plants and their F1 hybrids: Deduce the chromosomal origin of species A.

Species A is an allotetraploid with a genome from each of species C and species D

A plant breeder would like to develop a seedless variety of tomato from two existing lines. Line A is a diploid line, and line B is a tetraploid line. Describe the breeding strategy that will produce a seedless line, and support your strategy by describing the results of crosses.

Tetraploids produce diploid gametes √ Diploids produce haploid gametes √ therefore, a hybrid of the tetraploid and diploid will be a triploid. √ The triploid will have either bivalents plus monovalents or all trivalents during synapsis in meiosis I. √ (OR abnormal meioses) It will produce unbalanced (unequal) √, nonviable √ gametes and thus will be seedless.

In humans, Hunter syndrome is known to be an X-linked recessive condition with complete penetrance. For each of the following families, explain the origin of the child indicated in bold-underline. In family A, two phenotypically normal parents have a normal son, a daughter with Hunter and Turner syndromes, and a son with Hunter syndrome.

The daughter with Turner and Hunter syndromes in family A must have received her single X chromosome from her mother, who is heterozygous for the mutation causing Hunter syndrome. The daughter did not receive a sex chromosome from her father because sex chromosome nondisjunction must have occurred during meiosis in his sex cells.

A man and a woman who each have wild-type phenotypes have a son with Klinefelter syndrome who has hemophilia.

d. The man is XHY, the woman is XhXH, the child is XhXhY, and nondisjunction occurred in the woman during meiosis II.

What conclusion can you reach based on the X2 analysis.

e. P > 0.05. The deviations are not signicant and due to chance. The genes are not linked.

A man who is colour blind and a woman who is wild type have a son with Jacob syndrome (XYY) who has hemophilia.

f. The man is XcY, the woman is XhXH, the child is XhYY, and nondisjunction must have occurred in the man during meiosis II.

Which sex determination system occurs in this species and explain which allele is dominant?

female is the heterogametic sex (ZW), whereas males are homogametic (ZZ). √Blue stripe is dominant, and non-stripe is recessive. √

In a species of sh, a blue stripe on the dorsal n is observed in males and females. A sh breeder carries out a pair of reciprocal crosses and observes the following results:

reciprocal crosses produced dierent results √ Why does this evidence support the hypothesis that the trait is sex-linked?

A woman has two dominant traits, each caused by a mutation in a different gene: cataracts (an eye abnormality; C), which she inherited from her father, and polydactyly (an extra finger; P), which she inherited from her mother. Her husband has neither trait. The genes for these two traits are 15 cM apart on the same chromosome. What is the woman's genotype?

C + // + P Because she inherited the cataracts mutation from her father and the polydactyly mutation from her mother, the mutant alleles must be on opposite chromosomes:C + // + P

What is the probability that the first child of this couple will have both cataracts and polydactyly? (Give a percentage number, and round to one decimal.)

For a child to inherit both mutant alleles, the woman would have to produce an egg that carried a recombinant chromosome, C P. We can estimate the probability of this event from the distance between the two genes, 15 cM, which, would be equivalent to 15% recombination. However, only half the recombinants will be C P. Thus, the probability that the child will inherit both mutant alleles is (15/2) % - 7.5%

One of the X chromosomes in a particular Drosophila female had a normal order of genes, but carried recessive alleles of the genes for yellow body colour (y), vermilion eye colour (v), and forked bristles (f), as well as the dominant X-linked Bar eye mutation (B). Her other X chromosome carried the wild-type alleles (+) of all four genes, but the region including y+, v+, and f+(but not B+) was inverted with respect to the normal order of genes. This female was crossed to a wild-type male in the cross Why are there no male offspring with the allele combinations of y v f+, y+ v+ f, y v+ f+, or y+ v f (regardless of the allele of the Bar eye gene)?

If a single crossover occurs within the inversion loop of the female inversion heterozygote, a dicentric and an acentric chromosome are formed. Cells containing these types of chromosomes are not viable. The resulting allele combinations from such single crossovers are not recovered. The four phenotypic classes of missing male offspring would be formed by single crossovers between the y and v OR between the v and f genes in the female inversion heterozygote and therefore are not recovered.

Why are there no male offspring with the allele combinations of y v f+; y+ v+ f; y v+ f+; or y+ v f (regardless of the allele of the Bar eye gene)?

If a single crossover√(1/2) occurs within the inversion loop √(1/2) of the female √(1/2) , who is an inversion heterozygote √(1/2); adicentric √(1/2) and an acentric √(1/2) chromosome are formed that creates unbalanced gametes √(1/2) with duplicated √(1/2) and/or deleted √(1/2) genetic information. Zygotes containing these types of chromosomes are not viable √(1/2) (thus, resulting allele combinations from such single crossovers are not recovered). The four phenotypic classes of missing male offspring would be formed by single crossovers between the y and v OR between the v and f genes in the female inversion heterozygote and therefore are not recovered. √(1/2)√(1/2)

Two genes on the same chromosome, 7 cM apart, contradicts Mendel'spostulate on?

Independent assortment

Genes a and b are on one chromosome, 20 cM apart; c and d are on a second chromosome, 10 cM apart. Genes e and f are on yet another chromosome and are 30 cM apart. A homozygous A B C D E F individual is crossed to an a b c d e f individual, and the resulting F1 is crossed back to an a b c d e f individual. What are the probabilities of observing individuals of the following phenotypes in the back cross progeny? (give a fractional value, round three decimals)

Infer the recombination frequency between two genes from the map distance between them. For example, a and b are 20 cM apart, so A B/a b will have 20% recombinant (10% each A b, a B) and 80% parental (40% each A B, a b) gametes. Then, since each chromosome pair segregates independently, use the product rule and multiply the probabilities of obtaining alleles from each chromosome. P (A B C D E F) = 0.40 X 0.45 X 0.35 = 0.063 (probability value) OR 6.3 % (% value) The correct answer is: 0.063 %

In Drosophila, the genes bw and st are located on chromosomes 2 and 3, respectively. Flies homozygous for bw mutations have brown eyes, ies homozygous for st mutations have scarlet eyes, and ies homozygous for bw and st mutations have white eyes. Doubly heterozygous males were mated individually to homozygous bw; st females. All but one of the matings produced four classes of progeny: wild-type, brown-eye, scarlet-eye and white-eyed. The single exception produced only wild-type and white-eyed progeny. Explain the nature of this exception in complete detail.

The exceptional male, whose genotype is bw/+ st/+, is heterozygous √ √ for a translocation √ √ between chromosomes 2 and 3. (we cannot distinguish whether it is a balanced or unbalanced translocation) The translocation is either between the two mutant chromosomes √ or between the two wild-type chromosomes √, that is, T(bw; st) or T(+; +) (it is not possible to distinguish with the given data) however, it clearly is not between a mutant chromosome and a wild-type chromosome √, If it were, the progeny would be either brown or scarlet √,

A colour-blind man and a woman who is wild type have a daughter with Turner syndrome who has normal colour vision and normal blood clotting.

The man is XcY, the woman has at least one X chromosome with the wild-type C allele (XCX-), and the child is XCO. Nondisjunction must have occurred in the man at meiosis I or II.

A man who is colour blind and has hemophilia and a woman who is wild type have a daughter with triple X syndrome who has hemophilia and normal colour vision.

The man is XchY, and the woman is XChX-H . The child could be XchXChXCh, where nondisjunction has occurred in the woman during meiosis II; or the child could be XchXchXCh, where nondisjunction has occurred during meiosis II in the man.

In a species of mosquito, a dominant gene R for resistance against insecticide 1 (I1) and a dominant gene D for resistance against insecticide 2 (I2) are known to be on the same chromosome. A cross was carried out between a I1-resistant strain and a I2-resistant strain, and female progeny resistant to both insecticides were testcrossed with wildtype males. Among the progeny, 99 were resistant to both insecticides, 88 were resistant to I1 only, 89 were resistant to I2 only, and 106 were sensitive to both insecticides. Are I1 and I2 alleles of the same gene? How did you conclude this?

The mutant genes are not alleles because they do not show segregation √ ; if they were alleles, all the progeny wouldbe resistant to one insecticide or the other √.

n the tomato, tall (D-) is dominant to dwarf (dd) with reference to plant height; smooth fruit (P-) is dominant to peach fruit (pp); and round fruit shape (O-) is dominant to oblate fruit shape (oo). These three genes are linked on chromosome 1 of tomato in the order dwarf-peach-oblate. There are 12 cM between dwarf and peach and 17 cM between peach and oblate.A trihybrid plant (DPO/dpo) is testcrossed. The accompanying table shows the progeny plants. Identify the mechanism responsible for the resulting data that do not agree with the established genetic map.

The results show that recombination frequency in the region between dwarf and oblate is much lower than expected. √ This result is consistent with inversion √ of the chromosome region containing peach √. (OR the gene which is located in the middle) and the presence of heterozygosity √ for the inversion in the trihybrid line. Inversion heterozygosity is suppressing √ the appearance of most of the crossover √ The resultant gametes are nonviable √ due to duplications and deletions √ that are created by the crossover within the inverted segment

In family B, two phenotypically normal parents have two phenotypically normal daughters and a son with Hunter and Klinefelter syndromes

The son with Klinefelter syndrome in family B is karyotypically XXY, and both of his X chromosomes carry the mutant allele for Hunter syndrome. This individual must have received two mutant X chromosomes from his heterozygous mother due to X chromosome nondisjunction during the second meiotic division in her sex cells. Y chromosome received from father.

What can you deduce about the genetic positions of I1 and 12?

The two genes are located on the same chromosome and are located more than 50 cM from each other.

Where should crossover(s) have occurred between genes/chromosome regions, relative to the chromosomal abnormality in the female to produce the y v f B+ and y+ v+ f+ B offspring?

The y v f B+ and y+ v+ f+ B offspring are the result of single crossover √(1/2) events outside of the inversion loop √(1/2), between the end of the inversion (just to the right of f on the preceding diagram) √(1/2) and the B gene √(1/2).

What kind of crossovers produced the y v f B+ and y+ v+ f+ B offspring? Can you determine any genetic distance from these classes of progeny?

The y v f B+and y+v+f+B offspring are the result of single crossover events outside of the inversion loop, between the end of the inversion (just to the right of f on the preceding diagram) and the B gene. This region is approximately 16.7 cM in length (19 recombinants out of 114 total progeny).

What kind of crossover/s produced the y+ v f+ B+ and y v+ f B offspring?

The y+v f+B+and y v+f B offspring would result from two crossover events within the inversion loop, one between the y and v genes and the other between the v and f genes.

Where should crossover(s) have occurred between genes/chromosome regions, relative to the chromosomal abnormality in the female to produce the y+ v f+ B+ and y v+ f B offspring?

The y+v f+B+and y v+f B offspring would result from two crossover events √(1/2) within the inversion loop √(1/2), one between the y and v genes √(1/2) and the other between the v and f genes √(1/2).

At one such commercial venture they are investigating the following traits: CBD content (CBD gene) (with the high production allele "C" dominant over the low production allele "c"); THC content (THC gene) (with the low production allele "T" dominant over the high production allele "t"); Bud density (BD gene) (with the dense bud allele "D" dominant over the loose bud allele "d"); and Flowering time (FT gene) (with the late flowering allele "F" dominant over the early flowering allele "f"). Phenotype Count Fenotipe Telling CDTF 423 CDtF 436 cdTf 442 cdtf 439 CdTF 98 CdtF 95 cDTf 96 cDtf 95 CdTf 89 Cdtf 89 cDTF 85 cDtF 84 CDTf 14 CDtf 15 cdTF 16 cdtF 15

Three of the traits are linked (fully or partially), and the other one assorts independently from the rest. Test each gene combination in a pairwise manner, and evaluate if there is a deviation from the 1:1:1:1 ratio (as expected under independent assortment) for the reciprocal two parental and two recombinant phenotypes, e.g.:

A normal female Drosophila produces abnormal eggs that either contain all her autosomes or all her sex chromosomes. She mates with a normal male Drosophila that produces normal sperm. Match the potential chromosomal compositions with the appropriate sexes/phenotype for the offspring produced by this mating.

Two types of egg cells will be formed: either AA or XX. These eggs will be fertilized by two kinds of sperm cells produced in equal proportions: half the sperm will have one X chromosome and one set of autosomes, and the other half will have one Y chromosome and one set of autosomes.

Give genotypes for the parents in cross 1:

ZBZb♂ xZbW♀

The male is the homogametic sex.

ZZ/ZW

Which sex determination system occurs in this species?

ZZ/ZW

Give genotypes for the parents in cross 2:

ZbZb ♂ x ZBW ♀