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Check out the Cell Division, Meiosis, and Sex Linkage 1 & Sex Linkage 2 sites at The Biology Project, developed at The University of Arizona.

Also do questions from chapters 11, 13, 14 and 15.

Matching, Fill-ins, True/False etc.

1) Match the description on the left with BEST answer on the right by filling in the blanks. Each answer may be used more than once.

____1 - Chromosomes with two chromatids become chromosomes with one chomatid......A- Metaphase
...........during this phase of meiosis
____2- Chromatin becomes highly supercoiled at the end of this phase of mitosis............B- G0-phase
____3- Crossing over begins in this phase of meiosis.................................................C- Anaphase II
____4- This allows for sorting out of organelles in heteroplasmic cells..........................D- G2-phase
____5- Homologous chromosomes align themselves at the equatorial plane during...........E- Telekinesis
...........this phase of meiosis
____6- Synapsis first occurs in this phase of meiosis..................................................G- S-phase
.........................................................................................................................H- Anaphase I
.........................................................................................................................I- Metaphase I
.........................................................................................................................L- Telophase
.........................................................................................................................M- Cytokinesis
.........................................................................................................................O- G1-phase
.........................................................................................................................P- Anaphase
.........................................................................................................................R- Prophase I
.........................................................................................................................S- Prophase II
.........................................................................................................................T- Metaphase I
.........................................................................................................................U- Prophase

2) Match the description on the left with BEST answer on the right by filling in the blanks. Note: Some answers may have more than 1 answer, and some answers can be used more than once.
____1 - Regions of chromosomes with different staining intensities..............................A- Euchromatin
____2 - Swollen regions on chromatin.....................................................................B- Heterochromatin
____3 - Chromosome composed of MANY dsDNA molecules......................................C- Arms
____4 - Linear dsDNA molecule and the protein attached to it.......................................D- Acrocentric
____5- Ends of a chromosome................................................................................E- Polytene
____6- Areas of active transcription of rRNA.............................................................G- Bands
____7- Primary constrictions..................................................................................H- Nucleoli
____8- Area of active transcription (NOT ONLY rRNA)...............................................I- Chromomeres
____9 - Each acrocentric chromosome in mitotic metaphase has four four... . . . ..............L- Centromere
____10 - Histone found in the nucleosome core..........................................................M- Chromatids
................................................. . . ....................................................................O- Telomeres
..........................................................................................................................P - Golgi
..........................................................................................................................S - H1
..........................................................................................................................T - H2

3) Circle only the true statements below.
A) The cell which divides to give rise to the ascospores is 1n. B) Tetrad analysis determines map distance between a locus and a telomere. C) Ascospores are always formed in multiples of 4. D) An ascus with spores arranged as (AADDAADD) is a recombinant type. E) An ascus with spores arranged as (DDAAAADD) is a recombinant type. G) Histones are negatively charged proteins. H) Cross-overs are most frequent near centromeres. I) Recombination frequencies of greater than 20% overestimate map distance. J) The cell which divides to give rise to the ascospores is 2n. K) Chromosome deletion mutants can be used in physical mapping. L) DNA sequencing is a kind of physical mapping. M) A DNA marker is a DNA sequence which is very closely linked to an allele of interest. N) A three point testcross is an example of physical mapping O) In situ hybridization of chromosomes is an example of genetic mapping. P) Cross-overs are most frequent near telomeres. Q) Every metacentric chromosome, at metaphase of mitosis, has four arms. R) Every acrocentric chromosome, at anaphase II of meiosis, has four arms. S) Chromosomes are composed of two chromatids at the beginning of interphase. T) The telomere of a chromosome joins chromatids together. U) Homeologous chromosomes rarely cross over. 

4) Name the two types of organelles other than nuclei, which contain DNA?

5) Name a histone which is not found in the nucleosome core? ________________________________________

6) Fill-in the phase of cell division which best matches each description.
A) Synapsis first occurs in this phase of meiosis. _________________________________
B) Homologous chromosomes align themselves at the equatorial plane during this phase of meiosis. __________________________________
C) The chromatin becomes highly supercoiled at the end of this phase of mitosis. __________________________________________________
D) Crossing over begins in this phase of meiosis. _________________________________
E) Chromosomes with two chromatids become chromosomes with one chomatid during this meiotic phase. ________________________  

7) Circle the statement which BEST describes our modern view of gene expression?
A) One gene - one enzyme. B) One cistron - one enzyme. C) One cistron - one polypeptide. D) One mutation - one pathway E) One gene - one pathway.

8) Circle only the true statements below.
A) There are no genes on the human Y chromosome. B) In the XX - XO sex determination system the XO genotype has two sex chromosomes. C) All sex determination is genetic. D) Meiotic crossing over is more common than mitotic crossing over. E) Chromosome walking requires mapping functions. F) Any two genes on the same chromosome are linked. G) Sex influenced traits are X-linked. H) Only females can be carriers of recessive X-linked traits.

9) (Circle the one correct answer) An inactivated X chromosome is called a ( golgi body, chad, barr body, perithecium, periosteum, chromomere, centromere ).

Autosomal 2 Point Problems

1) A mouse of genotype X⁺ b / X^- B is crossed to X^- b / X^- b. If the two loci are 8 cM apart, what is the expected proportion of the progeny that will be X⁺X^- Bb ?

2) Turkeys of genotype Q R⁺/ q R^- are crossed to q R^- / q R⁺. If the two loci are 4 cM apart, what is the expected proportion of the progeny that will be qq R^-R^- ?

3) A rat of genotype e K⁺/ E K^- is crossed to e K⁺ / E K^-.
If the two loci are 12 cM apart, what is the expected proportion of all genotypes and phenotypes if E is codominant to e, and K⁺ is dominant to K^- ? 

4) In crawfish the allele for large claws, L, is dominant to the allele for small claws, S. The allele for high metabolic rate, +, is dominant to the allele for low metabolic rate, met. Crawfish heterozygous for both loci are testcrossed and the results of the cross are shown in the table below.

Phenotypes [Number of individuals]
Large clawed, high metabolism [203], Large clawed, low metabolism [300], Small clawed, high metabolism [229], Small clawed, low metabolism [197]
A) Determine the c² value for the above results when testing for no gene linkage? B) What is the p value for the test? C) Is there significant evidence for gene linkage?

5) In silverfish the allele for light avoidance, A, is dominant to the allele for light indifference, a. The allele for straight line motion, +, is dominant to the allele for zig-zag motion, z. Silverfish heterozygous for both loci are testcrossed to light indifferent, zig-zag moving silverfish and the results of the cross are shown in the table below.

Phenotypes [Number of individuals]
Light indifferent, straight line motion [40], Light avoiding, straight line motion [58], Light indifferent, zig-zag motion [61], Light avoiding, zig-zag motion [42]
A) Determine the c² value for the above results when testing for no gene linkage? B) What is the p value for the test? C) Is there significant evidence for gene linkage?

6) In dragons the allele for green scales, G, is codominant to the allele for red scales, R. The allele for fire-breathing, B, is dominant to the allele for non fire-breathing, b. The autosomal loci for scale color and fire-breathing are 32 cM apart. Red non fire-breathing female dragons are crossed with male red-green dragons heterozygous for the fire-breathing alleles. If the G and B alleles in the male dragons are in trans, what are the expected genotypic and phenotypic ratios in the dragon progeny?

Autosomal 3 Point Problems

1) From the cross and data set given in the following table determine the order of the loci, the map units between the loci, and the alleles on each of the two homologous chromosomes of the trihybrid parent.
Cross Tt Bb +z x tt bb zz. T is dominant to t, B is dominant to b, and + is dominant to z.

Phenotypes observed [Number of individuals]
Tb+ [54], Tbz [498], TBz [9], TB+ [31], tb+ [8], tB+ [478], tBz [60], tbz [29]

2) From the cross and data set given in the following table determine the order of the loci, the map units between the loci, and the alleles on each of the two homologous chromosomes of the trihybrid parent.
Cross Dd Qq Rr x dd qq rr. D is dominant to d, Q is dominant to q, and R is dominant to r.

Phenotypes observed [Number of individuals]
Dqr [40], dqr [55], dQR [47], DQR [53], DQr [289], dQr [9], DqR [11], dqR [307]

Sex Linkage Problems

1) A husband and wife have normal vision, although both of their fathers are red-green color blind, which is inherited as an X-linked recessive condition. What is the probability that their first child will be A) a phenotypically normal daughter? B) a normal son? C) a color-blind son? D) a color-blind daughter?

2) Sue and her brother Sam are normal but their brother Stan expresses a trait for a very, very rare x-linked allele (X^g) which prevents him from solving genetics problems. Sue, Sam, and Stan were adopted at birth and have no knowledge of their biological parents. Unknowingly, Sue marries her first cousin Jake, who has never taken a genetics course. Sue and Jake have three daughters, one of whom has this spine-chilling trait. Note that the symbol for the normal genetics problem solving allele is X^G . A) What are the genotypes of Sue, Sue's mother, and her husband Jake? B) Sue finds her long lost sister, Sissy, who can't solve genetics problems. What is the genotype of Sue's father?

3) Larry has a brother Lars who is addicted to the ingredient in Brand Z tooth paste. Addiction to Brand Z is a very rare recessive X-linked trait. Neither Lars' Mother (Lilly) or Father (Lloyd) shows this addiction. Neither Larry nor Larry's sister Linda has been exposed to Brand Z. A) What is the probability that Larry is susceptible to Brand Z ? B) What is the probability that Linda is susceptible to Brand Z ? C) What is the probability that Linda is a carrier of the recessive allele?

Linda marries Luke. Luke's mother Laura is susceptible to Brand Z but his father Lance is not. D) What is the probability that Linda's first male child is wild type? E) What is the probability that Linda's first female child (Lisa) is susceptible to Brand Z when she grows up to have teeth to brush?

4) In the beetles golden body (go) is a recessive X-linked mutation, and brown eyes (bw) is a recessive autosomal mutation. A female homozygous for golden body and red eyes is mated to a male with brown eyes and a black body. A.) Predict the phenotypes of their F1 offspring. B.) If the F1 progeny are intercrossed, what kinds of progeny will appear in the F2, and in what proportions?

5) A roundworm which parasitizes fish, oysters, and crustaceans, has two loci for host specificity. The autosomal allele F⁺ for fish host specificity is dominant to the allele F^- for the inability to parasitize fish. The X-linked allele ost⁺ for oyster specificity is dominant to the allele ost^- for the inability to parasitize oysters. A male roundworm, removed from an oyster and heterozygous for the fish host locus, is crossed with a female roundworm, unable to parasitize fish and heterozygous for the oyster host locus. What are the expected phenotypes of the resulting progeny?

6) Bookworm males have X and Y chromosomes while females are XX. Wild type bookworms have 125 body segments and short hairs on each segment. An allele S120 is a sex linked recessive allele for 120 body segments while the allele lon is a dominant autosomal allele for long segment hairs. Dr. Ann E. Lida crossed short haired female worms having 120 body segments, from a pure breeding line, with male worms having 125 long haired body segments from a pure breeding line. What are the expected phenotypic ratios of the F1 and F2 generations?

Tetrad Analysis

1) Two strains of Neurospora crassa are crossed with spore phenotypes + and s. The following table shows the number of each type of ascus found. How far is s from the centromere?

Asci type [Number]
+ + s s + + s s [52], s s + + s s + + [65], + + s s s s + + [120], s s + + + + s s [130], + + + + s s s s [198], s s s s + + + + [212]

2) Two strains of Sordaria sp. are crossed with spore phenotypes a and b. The following table shows the number of each type of ascus found. How far is the a/b locus from the centromere?

Asci type [Number]
a a b b b b a a [30], b b a a b b a a [74], a a a a b b b b [245], a a b b a a b b [65], b b a a a a b b [43], b b b b a a a a [323]

Another Problem

1) Propose a mode of inheritance consistent with the pedigree below.

Matching, Fill-ins, True/False etc.

1) 1- C- Anaphase II, 2- U- Prophase, 3- R- Prophase I, 4- M- Cytokinesis, 5- I- Metaphase I, 6- R- Prophase I

2) 1 - G- Bands, 2 - I- Chromomeres, 3 - E- Polytene, 4 - M- Chromatids, 5- O- Telomeres, 6- H- Nucleolus, 7- L- Centromeres, 8- A- Euchromatin, 9 - C Arms or O- Telomeres, 10 - T- H2

3) Only the true statements are listed.
C) Ascospores are always formed in multiples of 4. D) An ascus with spores arranged as (AADDAADD) is a recombinant type. E) An ascus with spores arranged as (DDAAAADD) is a recombinant type. J) The cell which divides to give rise to the ascospores is 2n. K) Chromosome deletion mutants can be used in physical mapping. L) DNA sequencing is a kind of physical mapping. M) A DNA marker is a DNA sequence which is very closely linked to an allele of interest. Q) Every metacentric chromosome, at metaphase of mitosis, has four arms. U) Homeologous chromosomes rarely cross over. 

4) Mitochondria, Plastids

5) H1

6) A) Prophase I, B) Metaphase I, C) Prophase, D) Prophase I, E) Anaphase II  

7) C) One cistron - one polypeptide.

8) Only the true statements are listed.
D) Meiotic crossing over is more common than mitotic crossing over. H) Only females can be carriers of recessive X-linked traits.

9) Barr body

Autosomal 2 Point Problems

1) X⁺ b / X^- B X X^- b / X^- b ........ X⁺X^- Bb ?

Gametes
from X⁺ b / X^- B ........................ from X ^- b / X ^- b
nonrecomb. (92%) .................... ALL X ^- b (1.00)
X⁺ b (0.46)
X ^- B (0.46)
recomb. (8%)
X⁺ B (0.04)
X ^- b (0.04)

X⁺X^- Bb = (0.04)(1) = 0.04

2) Q R⁺/ q R^- X q R^- / q R⁺ ........ qq R^-R^- ?

Gametes
from Q R⁺/ q R^- ......................... from q R^- / q R⁺
nonrecomb. (96%) .................... 1/2 q R^- : 1/2 q R⁺
Q R⁺ (0.48)
q R^- (0.48)
recomb. (4%)
q R⁺ (0.02)
Q R^- (0.02)

qq R^-R^- = (0.48)(0.5) = 0.24

3) e K⁺/ E K ^- X e K⁺ / E K ^- ........ all genotypes & phenotypes?

Gametes
from e K⁺/ E K ^- ............................................ from e K⁺ / E K ^-
nonrecomb. (88%) ......................................... nonrecomb. (88%)
e K⁺ (0.44) ...................................................... e K⁺ (0.44)
E K ^- (0.44) ...................................................... E K ^- (0.44)
recomb. (12%) ............................................... recomb. (12%)
e K ^- (0.06) ....................................................... e K ^- (0.06)
E K⁺ (0.06) ..................................................... E K⁺ (0.06)

Genotypes
ee K⁺K⁺ = (0.44)(0.44) = 0.194 . . . . Ee K⁺K^- = (0.44)(0.44) + (0.44)(0.44) + (0.06)(0.06) + (0.06)(0.06) = 0.396 . . . . . ee K⁺K^- = (0.44)(0.06) + (0.06)(0.44) = 0.052 . . . . . Ee K⁺K⁺ = (0.44)(0.06) + (0.06)(0.44) = 0.052 . . . . . EE K^-K^- = (0.44)(0.44) = 0.194 . . . . . Ee K^-K^- = (0.44)(0.06) + (0.06)(0.44) = 0.052 . . . . . EE K⁺K^- = (0.44)(0.06) + (0.06)(0.44) = 0.052 . . . . . ee K^-K^- = (0.06)(0.06) = 0.004 . . . . . EE K⁺K⁺ = (0.06)(0.06) = 0.004

Phenotypes
E K⁺ = 0.052 + 0.004 = 0.056 . . . . E K^- = 0.194 . . . . . e K⁺ = 0.194 + 0.052 = 0.246 . . . . . e K^- = 0.004 . . . . Ee K⁺ = 0.396 + 0.052 = 0.448 . . . . . Ee K^- = 0.194 = 0.052

4) L__ Large ..... SS Small ..........................................+__ High ..... metmet Low
Cross: LS +met X SS metmet
Test 1:1:1:1
Phenotype ..... # observed ..... # expected
Large, High ........ 203 ............ 232.25
Large, Low ........ 300 ............ 232.25
Small, High ....... 229 ............ 232.25
Small, Low ........ 197 ............ 232.25
A) c² = [(203-232.25)² + (300-232.25)² + (229-232.25)² + (197-232.25)²] / 232.25 = 28.84 B) df = 3, p<0.001 C) yes

5) A__ Avoiding ..... aa Indifferent ..........................................+__ Straight line ..... zz zig-zag
Cross: Aa +z X aa zz
Test 1:1:1:1
Phenotype .................. # observed ..... # expected
Indifferent, Straight line ..... 40 .............. 50.25
Avoiding, Straight line ...... 58 .............. 50.25
Indifferent, Zig-zag ............ 61 .............. 50.25
Avoiding, Zig-zag ............. 42 .............. 50.25
A) c² = [(40-50.25)² + (58-50.25)² + (61-50.25)² + (42-50.25)²] / 50.25 = 6.94 B) df = 3, 0.05<p<0.10 C) No

6) GG Green ..... GR Green & Red ..... RR Red ........................................ B__ Fire-breathing ..... bb Non fire-breathing
R b / R b X R B / G b ........ all genotypes & phenotypes?
Gametes
from R b / R b ................................................. from R B / G b
All Rb (1.00) ................................................ nonrecomb. (68%)
.......................................................................... R B (0.34)
.......................................................................... G b (0.34)
....................................................................... recomb. (32%)
.......................................................................... R b (0.16)
.......................................................................... G B (0.16)

Genotypes
RR Bb = (1.00)(0.34) = 0.34 . . . . RG bb = (1.00)(0.34) = 0.34 . . . . . RR bb = (1.00)(0.16) = 0.16 . . . . . RG Bb = (1.00)(0.16) = 0.16

Phenotypes

Red, Fire breathing = 0.34 . . . . Red & Green, non-fire breathing = 0.34 . . . . . Red, non-fire breathing = 0.16 . . . . . Red & Green, fire breathing = 0.16

Autosomal 3 Point Problems

1) Cross: Tt Bb +z x tt bb zz.
From the table of the progeny we notice that the nonrecombinant phenotypes (most abundant) are Tbz and tB+ and the double x-over phenotypes (least abundant) are TBz and tb+. Since the locus in the middle differs between nonrecombinant and double x-over pairs, it means that the B/b locus is in the middle. Therefore the alleles on each chromosome of the trihybrid parent in linkage notation are Tbz / tB+ .

Make two phenotype tables for TB and B+ gene linkage.

Phenotype ............... # .................................................. Phenotype ............... #
Tb ................ 54 + 498 = 552 ........................................ bz ................. 498 + 29 = 527
tB ................. 478 + 60 = 538 ........................................ B+ ................ 31 + 478 = 509
TB .................. 9 + 31 = 40 ........................................... b+ ................... 54 + 8 = 62
tb ....................8 + 29 = 37 ........................................... Bz .................... 9 + 60 = 69

total offspring = 1167

Map distance between loci T and B = [(40 + 37) X 100] / 1167 = 6.60 cM

Map distance between loci B and + = [(62 + 69) X 100] / 1167 = 11.22 cM

Map distance between loci T and + = 6.60 cM + 11.22 cM = 17.82 cM

2) Cross: Dd Qq Rr x dd qq rr.
From the table of the progeny we notice that the nonrecombinant phenotypes (most abundant) are DQr and dqR and the double x-over phenotypes (least abundant) are dQr and DqR. Since the locus in the middle differs between nonrecombinant and double x-over pairs, it means that the D/d locus is in the middle. The alleles on each chromosome of the trihybrid parent in linkage notation are QDr / qdR .

Make two phenotype tables for QD and DR gene linkage.

Phenotype ............... # .................................................. Phenotype ............... #
qD ................ 40 + 11 = 51 ........................................... dr ................... 55 + 9 = 64
Qd ................. 47 + 9 = 56 ............................................ DR ................ 53 + 11 = 64
QD .............. 53 + 289 = 342 ......................................... dR ................ 47 + 307 = 354
qd ................ 55 + 307 = 362 ......................................... Dr ................ 40 + 289 = 329
total offspring = 811

Map distance between loci Q and D = [(51 + 56) X 100] / 811 = 13.19 cM

Map distance between loci D and R = [(64 + 64) X 100] / 811 = 15.78 cM

Map distance between loci Q and R = 13.19 cM + 15.78 cM = 28.97 cM

Sex Linkage Problems

1)

2)

3)

 

A) Probability of Larry being suseptable = Probability of getting X^z from Mom = 1/2
B) Probability of Linda being suseptable = 0 Linda got a X⁺ from Dad.
C) Probability of Linda being a carrier = probability of getting X^z from Mom = 1/2

D) Probability that Linda's 1st male child is wild type = 1 - (1/2 X 1/2) = 3/4. Linda has a 50% chance of being a carrier and a 50% chance of passing the X^z to her son if she is a carrier. Since we are talking about a male child, Luke must give the male child his Y.
E) Probability that Linda's 1st female child (Lisa) is suseptable = 1/2 X 1/2 = 1/4. Lisa must be X^z X^z to be suseptable. Linda has a 50% chance of being a carrier and a 50% chance of passing X^z to Lisa if Linda is a carrier. Since we are talking about a female child, Luke must give the female child his X^z.

5) Let Bk be the symbol for the black body allele and R be the symbol for the red eye allele.
Cross: X^goX^go RR X YX^Bk bwbw

A) F1: 1/2 golden bodied male : 1/2 black bodied female ..................... all red eyes
1/2 golden bodied, red eye, males : 1/2 black bodied, red eyed females

B) Cross: X^Bk X^go Rbw X YX^go Rbw

1/4 black bodied male : 1/4 golden bodied male : 1/4 black bodied female : 1/4 golden bodied female

3/4 red eyed : 1/4 brown eyed

3/16 black bodied, red eyed males : 1/16 black bodied, brown eyed males : 3/16 golden bodied, red eyed males : 1/16 golden bodied, brown eyed males : 3/16 black bodied, red eyed females : 1/16 black bodied, brown eyed females : 3/16 golden bodied, red eyed females : 1/16 golden bodied, brown eyed females

6) Cross: X^ost+X^ost- F^-F^- X YX^ost+ F⁺F^-

1/4 oyster parasitizing male : 1/4 non oyster parasitizing male : 1/2 oyster parasitizing female

1/2 fish parasitizing : 1/2 non fish parasitizing

1/8 oyster and fish parasitizing male : 1/8 oyster, but not fish, parasitizing male : 1/8 fish, but not oyster, parasitizing male : 1/8 males unable to parasitize oysters or fish (Only parasitizes crustaceans, they have to eat something or its lethal!) : 1/4 oyster and fish parasitizing female : 1/4 oyster, but not fish, parasitizing female

7) Let shr be the symbol for the short hair allele.
P cross: X^S120X^S120 shr shr X YX^S125 lon lon

F1: 1/2 YX^S120 lon shr : 1/2 X^S125X^S120 lon shr
F1: 1/2 males with 120 long haired segments : 1/2 females with 125 long haired segments

F1 cross: X^S125X^S120 lon shr X YX^S120 lon shr

1/4 males with 125 segments : 1/4 males with 120 segments : 1/4 females with 125 segments : 1/4 females with 120 segments

3/4 long haired : 1/4 short haired

F2: 3/16 males with 125 long haired segments : 1/16 males with 125 short haired segments: 3/16 males with 120 long haired segments : 1/16 males with 120 short haired segments : 3/16 females with 125 long haired segments : 1/16 females with 125 short haired segments : 3/16 females with 120 long haired segments : 1/16 females with 120 short haired segments

1) # recombinant asci = 367 ............... total # of asci = 777

Map distance of the S locus from the centromere = 100 X [1/2 X (367 / 777)] = 23.62 cM

2) # recombinant asci = 212 ............... total # of asci = 780

Map distance of the S locus from the centromere = 100 X [1/2 X (212 / 780)] = 13.59 cM

1) The progeny of each generation have only the mother's phenotype so the simplest explanation is maternal (extranuclear) inheritance.