Cytogenetics Practice Problems Help

Practice 3

Assume that an autotetraploid of genotype AAaa has the A locus 50 ormore map units from the centromere, so that the equivalent of a single crossover always occurs between the centromere and the A locus. In this case, the chromatids will assort independently. Further assume that random assortment of chromatids to the gametes occurs by twos. Determine (a) the expected genotypic ratio of the progeny that results from selfing this autotetraploid, and (b) the expected increase in the incidence of heterozygous genotypes compared with selfed diploids of genotype Aa.

Solution 3

1. Let each of the genes of the duplex tetraploid be labeled as shown in the illustration below. All capital letters represent identical dominant genes; all lowercase letters represent identical recessive alleles.



As in Practice 2, let us first use a Punnett square to determine the kinds and frequencies of different combinations of alleles (in pairs) expected in the gametes of the autotetraploid. Note in the illustration that (for example) alleles A¹ and A², originally on sister chromatids, can enter the same gamete if crossing over occurs between the centromere and the A locus. Likewise, any other pair of alleles could enter a gamete by the same mechanism (chromatid assortment). Random assortment (by twos) of eight chromatids of the autotetraploid during meiosis is shown in the following table. Note that the diagonal of the table represents the nonsense union of any given allele with itself (e.g., A¹ with A¹). The table is symmetrical on either side of the diagonal.



Ignoring the superscript identification of alleles, the expected ratio of possible diploid gametes is 6 AA: 16 Aa: 6 aa or 3 : 8: 3. Using these gametic expectations, we can now construct a zygotic checkerboard to generate the expected progeny. Summary of progeny genotypes: 9 AAAA (quadruplex) : 48 AAAa (triplex) : 82 AAaa (duplex) : 48 Aaaa (simplex) : 9 aaaa (nulliplex).



2. The total of the numbers (3 + 8 + 3) along the top or side of the preceding table is 14. Thus, the total of all crossproducts in the table is 14 × 14 = 196. Nine genotypes are homozygous AAA, and another nine are homozygous aaaa.Thus, all other genotypes (196 – 18 = 178) are heterozygotes. Selfing the autotetraploid produces 178/196 = 91% heterozygous progeny. Selfing a diploid of genotype Aa produces 50% heterozygous progeny. The increase from 50 to 91% is 41/50 = 82%

Practice 4

Pericarp is the outermost layer of the corn kernel and is maternal in origin. A dominant gene B produces brown pericarp, and its recessive allele b produces colorless pericarp. Tissue adjacent to the pericarp is aleurone (triploid). Purple pigment is deposited in the aleurone when the dominant gene C is present; its recessive allele c results in colorless aleurone. Aleurone is actually a single specialized layer of cells of the endosperm. The color of endosperm itself is modified by a pair of alleles. Yellow is governed by the dominant allele Y and white by the recessive allele y. Both C and Y show xenia to their respective alleles. A plant that is bbCcYy is pollinated by a plant of genotype BbCcYy. (a) What phenotypic ratio is expected among the progeny kernels? (b) If the F₁ is pollinated by plants of genotype bbccyy, in what color ratio will the resulting F₂ kernels be expected to appear?

Solution 4

1. If pericarp is colorless, then the color of the aleurone shows through. If aleurone is also colorless, then the color of the endosperm becomes visible. Since the maternal parent is bb, the pericarp on all F₁ seeds will be colorless. Any seeds with C will have purple aleurone. Only if the aleurone is colorless (ccc) can the color of the endosperm be seen.



2. Half of the F₁ embryos is expected to be Bb and will thus lay down a brown pericarp around their seeds in the F₂; the other half is expected to be bb and will envelop its seeds with a colorless pericarp. Thus, half of the seeds of the F₁ plants will be brown. Of the remaining half that has colorless pericarp, we need show only as much of the genotype as is necessary to establish the phenotype.



Summary of F₂ seed colors : 1/2 brown : 1/4 purple: 1/8 yellow: 1/8 white.

Practice 5

In 1931, Curt Stern found two different translocations in Drosophila from which he developed females possessing heteromorphic X chromosomes. One X chromosome had a piece of the Y chromosome attached to it; the other X was shorter and had a piece of chromosome IV attached to it. Two sex-linked genes were used as markers for detecting crossovers; the recessive trait carnation eye color (car) and the dominant trait bar eye (B). Dihybrid bar females with heteromorphic chromosomes (both mutant alleles on the X portion of the X-IV chromosome) were crossed to hemizygous carnation males with normal chromosomes. The results of this experiment provided cytological proof that genetic crossing over involves an actual physical exchange between homologous chromosome segments. Diagram the expected cytogenetic results of this cross showing all genotypes and phenotypes.

Solution 5

The existence of a morphologically normal X chromosome in recombinant male progeny with carnation eyes provides cytological proof that genetic crossing over is correlated with physical exchange between homologous chromosomes in the parents. Similarly, all other phenotypes correlate with the cytological picture. Chromosomal patterns other than the ones shown above may be produced by crossing over outside the inverted region.

Practice 6

The centromere of chromosome V in corn is about 7 map units from the end. The gene for light yellow (virescent) seedling (v) is 10 map units from this end, and a gene called brevis (bv) that shortens internode length is 12map units from this end. The break point of a translocation (T) is 20map units from this end. A translocation heterozygote involving chromosomes Vand VIII of genotype + bv t/v + T is pollinated by a normal (nontranslocated, t) plant of genotype v bv t/v bv t. If gametes are formed exclusively by alternate segregation from the ring of chromosomes formed by the translocation heterozygote, predict the ratio of progeny genotypes and phenotypes from this cross (considering multiple crossovers to be negligible).

Solution 6

First let us diagram the effect that crossing over will have between the centromere and the point of translocation. We will label the ends of chromosome V with 1-2, and of chromosome VIII with 3-4. A cross-shaped pairing figure is formed during meiosis.

Alternate segregation produces half functional and half nonfunctional (duplication-deficiency) gametes. Note that the nonfunctional gametes derive only from the crossover chromatids. Thus, recovery of chromatids that experience a crossover between the centromere and the point of translocation is prevented. The combination of genes in this region of the chromosome is prevented from being broken up by crossing over and are thus transmitted as a unit. This situation is analogous to the block of genes within an inversion heterozygote that are similarly held together as a genetic unit. Noncrossover chromatids will form two types of functional gametes with equal frequency: + bv t and v + T. Expected zygotes are: 1/2 + bv t/v bv t = brevis, homozygous for the normal chromosome order and 1/2 v + T/v bv t = virescent, heterozygous for the translocation.

Practice 7

Shrunken endosperm of corn is governed by a recessive gene sh and waxy endosperm by another recessive wx. Both of these loci are linked on chromosome 9. A plant that is heterozygous for a translocation involving chromosomes 8 and 9 and that developed from a plump, starchy kernel is pollinated by a plant from a shrunken, waxy kernel with normal chromosomes. The progeny are

(a) How far is each locus from the point of translocation? (b) Diagram and label the pairing figure in the plump, starchy parent.

Solution 7

1. The point of translocation may be considered as a gene locus because it produces a phenotypic effect, namely, semisterility. The conventio nal symbol for translocation is T, and t is used for the normal chromosome without a translocation. Gene order in the parents must be in order for double crossovers to produce the least frequent phenotypes:
   + + T = plump, starchy, semisterile ear
   sh wx t = shrunken, waxy, normal ear

The map distances are calculated in the usual way for a three-point testcross.

Distance sh – wx = (82 + 49 + 6 + 3)/573 = 24:4 map units
Distance wx – T = (17 + 40 + 6 + 3)/573 = 11:5 map units
Distance sh – T = 24:4 +11:5 = 35:9 map units

2. 

Practice 8

An inversion heterozygote possesses one chromosome in the normal order:

and one in the inverted order:

A four-strand double crossover occurs in the areas f-e and d-c. Diagram and label the first anaphase figures.

Solution 8

A somewhat easier way to diagram the synapsing chromosomes when crossing over is only within the inversion as shown below. This is obviously not representative of the actual pairing figure. Let the crossover in the c-d region involve strands 2 and 3, and the crossover in the e-f region involve strands 1 and 4.

Practice 9

Data from Drosophila studies indicate that noncrossover (NCO) rings are recovered in equal frequencies with NCO rods from ring-rod heterozygotes. What light does this information shed on the occurrence of sister-strand crossing over?

Solution 9

Let us diagram the results of a sister-strand crossover in a rod and in a ring chromosome.

1. Rod chromosome:



2. Ring chromosome:

The double bridge at anaphase will rupture and produce nonfunctional gametes with duplications or deficiences. These would fail to be recovered in viable offspring. The fact that both rings and rods are recovered with equal frequency argues against the occurrence of sister-strand crossing over.

Modern techniques (involving autoradiography with labeled thymidine or fluorescence microscopy of cultured cells that have incorporated 5-bromodeoxyuridine in place of thymine) reveal that some sister-strand exchanges occur by a repair mechanism when DNA is damaged. One of the initiating steps that transforms a normal cell to a cancer cell is DNA damage. Hence, screening chemicals for their ability to induce sister-strand exchanges is one method for detecting potential cancer-inducing agents (carcinogens).

Practice 10

Yellow body color in Drosophila is produced by a recessive gene y at the end of the X chromosome. A yellow male is mated to an attached-X female (X^{^}X) heterozygous for the y allele. Progeny are of two types: yellowfemales and wild-type females. What insight does this experiment offer concerning the stage (two strand or four strand) at which crossing over occurs? (Reminder: the Y chromosome does not determine maleness in Drosophila; see Genic Balance, Chapter 5).

Solution 10

Let us assume that crossing over occurs in the two-strand stage, i.e., before the chromosome replicates into two chromatids.

The yellowmale produces gametes with either a y-bearingXchromosome or one with theYchromosome that is devoid of genetic markers. Trisomic X (X^{^}XX) flies seldom survive (superfemales). Those with (X^{^}XY) will be viable heterozygous wild-type attached-X females. Crossing over fails to produce yellow progeny when it occurs in the twostrand stage.

Let us assume that crossing over occurs after replication of the chromosome, i.e., in the four-strand stage:

The appearance of yellow females in the progeny is proof that crossing over occurs in the four-strand stage.