Midterm 2 Material: Topic 4- Gene Interaction, Inheritance and phenotype and Topic 5-

LOOK AT GPQ: Complementation Q 6 for table. Consider the complementation table below illustrating the analysis of 15 mutants, F to Z. All are recessive and homozygous. When crossed, the phenotype of the F1 is marked as 1 if complementation is seen (WT F1), or 0 if the F1 is mutant. How many genes are defined by this analysis and what alleles belong to what gene?

5 genes g1=FOQSZ g2=MPY g3=NRX g4=TV g5=UW Note: All mutations that fail to complement (0 interaction) are in the same gene. Start with F and roll into its group all mutants that "intersect" as 0. Then, take the next mutant that does complement with F (gives 1) and start a new group. Continue until you have exhausted all mutants.

Coat colors of Labrador retrievers depend upon the action of at least two genes. An inhibitor of coat color pigment (ee) prevents the expression of color alleles at another independently assorting locus, producing golden coats. When the dominant condition exists at the inhibitor locus (E_), the alleles of the other locus may be expressed: E_B_ = black and E_bb = chocolate. When dihybrid (heterozygous at both loci) black labs are mated together, what are the phenotypic proportions expected in the progeny? A. 9 black: 3 golden: 4 chocolate B. 9 black: 3 chocolate: 4 golden C. 9 golden: 3 brown 4 black D. 9 chocolate: 3 black: 4 golden E. 9 chocolate: 3 golden: 4 black

9 black: 3 chocolate: 4 golden REVIEW Note: This is a typical example of recessive epistasis. Being ee creates golden coat color regardless of the alleles present at the other locus. A typical dihybrid cross produces progeny with the ee phenotype 4 times out of 16, so you would expect 4 golden coat colors.

The gene for albinism (recessive) and the gene for dwarfism (dominant, rare) are linked and 20 m.u. apart. Mating between Meg, a normal woman with no family history of albinism, and Francisco, an albino dwarf, produced a dwarf son, Diego. Diego mates with Rhonda, a normal woman whose father was albino. What is the chance that their child will be albino and not dwarf? A. 0.05 B. 0.1 C. 0.25 D. 0.15 E. 0.2

A. 0.05 Done with tutor check notes for math/explanation. Note: A__ = normal, aa = albino B__ = dwarf, bb = normal Francisco has genotype aaBb and has the linkage phase aB/ab Diego has genotype AaBb and the linkage phase Ab/aB. Rhonda has the genotype Aabb with linkage phase Ab/ab. Note that you can only detect recombination in Diego! The question asks the probability that Diego and Rhonda have a baby with the genotype aabb (ab/ab).

Shire hobbits living in the villages of Bonning Gate and Newbiggin have curly hair on their feet rather than normal smooth hair. The curly haired trait is controlled by different genes in the two families. Intervillage matings between the families resulted in all children (F1) having smooth hair. If such children were in intermarry, what would be the probability of their children (F2) having curly haired feet? A. 0.44 B. 0.18 C. 0.06 D. 0.25 E. 0.66

A. 0.44 HELP Note: We are dealing with two loci because when the families intermate all F1s are "wild-type" or smooth. Thus, AAbb (curly) x aaBB (curly) create F1 AaBb (smooth). Intercrossing the F1 will result in a 9 : 3 : 3 : 1 genotypic ratio. However, the phenotypes will be different than 9 : 3 : 3 :1 because homozygous recessive bb creates the same phenotype as homozygous recessive aa. Therefore, the phenotypic ratio to be 9:7 smooth:curly, where curly can be: A_bb, aaB_, or aabb. Therefore the probability of the F2 having curly hair feet is 7/16, or .44. This is a case of double recessive epistasis.

In snapdragons, a true breeding red flowered plant (AA) crossed with a true breeding white flowered plant (aa) produces pink flowered F1 progeny. What phenotypic ratios do you expect from selfing the F1 generation? A. 1 red: 2 pink: 1 white B. 1 pink: 2 red: 1 white C. 3 red: 1 pink D. None of the other answers E. 1 red: 2 white: 1 pink

A. 1 red: 2 pink: 1 white Note: AA= red, Aa= pink, aa= white This is incomplete dominance. The heterozygous displays an intermediate phenotype distinctly different. So three phenotypes are seen.

--R---10cM---D------20cM-----Y-- Three genes of corn, R, D, and Y, reside one chromosome 9. In a testcross RDY/rdy x rdy/rdy, how many RdY/rdy plants are expected in a progeny of 1000 if there was no interference? A. 10 B. 5 C. 0 D. 15 E. 7

A. 10 Note: Look at notes for explanation

An LOD of 2 means that the likelihood of linkage for the given theta value is ______ more likely than the likelihood of independent assortment. A. 100 B. 3.5 C. 1000 D. 5 E. 2

A. 100 Note: LOD is the log base 10 of the odds (likelihood of linkage / likelihood of independent assortment) 10 ^2 = 100

Considering epistatic yellow retrievers. An aa dog is yellow regardless of the genotype at the B locus. In the presence of the dominant A allele, dogs are black or brown depending on the presence of the dominant B (black) allele or recessive b (brown) allele. If the mating of a brown and a yellow dogs has previously produced puppies of all three colors, what ratio do you expect for black:brown:yellow? A. 1:1:2 B. 12:3:1 C. 9:3:4 D. 1:1:1 E. 1:2:1

A. 1:1:2 Notes: Check out notes for math/explanation

A recombination frequency of 45% between the A and B loci suggests that: A. A and B are nearly unlinked: they are distant on the same chromosome. B. A and B are on different chromosomes, which rarely can undergo pairing and C.O. C. A and B are tightly linked D. A and B are on different chromosomes

A. A and B are nearly unlinked: they are distant on the same chromosome Note: 45% suggests they are on the same chromosome but sufficiently distant that C.O. scrambles them nearly as efficiently as being on different chromosomes.

The Ziblets, a family of fairies, display through generations the occasional Pink Nose. This syndrome was first recorded in great-great-grandfather Zibbo. Worried about genetic disease they have since taken great precautions to marry into separate families that have no history of Pink Nose. Alas, Pink Nose pops up with a certain probability among descendants of Zibbo. Children with an affected parents are often normal, but their children can display Pink Nose. The following explanation of the condition is consistent with the observations above: A. Dominant, incompletely penetrant B. Recessive, variable expressivity C. Recessive, incomplete penetrance D. Dominant, variable expressivity E. Incompletely dominant, fully penetrant

A. Dominant, incompletely penetrant THIS ONE TRICKY SO PAY ATTENTION Note: The condition can skip generations but it is not eliminated by outcrossing to healthy lines. Thus, it cannot be recessive. Incomplete penetrance makes sense. Not all individuals carrying the allele manifest the condition.

A chocolate male lab mates with a black female lab. The resulting puppies include two golden and one chocolate. Based on the genetics of Q2, the male and female genotypes could be respectively:_______ , _____. A. Eebb, EeBb B. eeBb, Eebb C. EEbb, EeBb D. eeBB, EEbb E. Eebb, EEBb

A. Eebb, EeBb Note: ee__ produces golden regardless of what the other B allele is. E__B__ produces black. E__bb produces chocolate. So if two children are golden this means it is ee__ and if one child is chocolate this means it is E__bb. In order to get "ee" the parent has to heterozygous for this allele and there has to be a little b next to the other big B.

Same question background as flashcard before. You conclude that in regard to the two genes controlling the leaf edge phenotype, variety W and Variety Z have the following genetic relationship. A. W is dominant for both genes, Z only for one B. W is recessive for both genes, Z dominant for both C. W is dominant for only one, Z dominant for both genes D. W and Z have the same genotype at both loci E. each is dominant for one only and not the same one

A. W is dominant for both genes, Z only for one NEED HELP IN UNDERSTANDING THIS ONE

Norway spruce can have stiff or droopy branches. You cross two pure breeding varieties, one stiff the other droopy obtaining stiff F1s. Selfing one of these you obtain 178 stiff and 144 droopy F2. The best explanation is two independent loci are involved in A. double recessive epistasis B. Co-dominance C. recessive epistasis D. multiple alleles E. dominant epistasis

A. double recessive epistasis

A and B = chromosomal loci. Dominant and recessive allele = upper and lower case letters, respectively. The hybrid AaBb is crossed to aabb. The following phenotypic classes and progeny numbers are produced: AB, 25; Ab, 450; aB, 410; ab, 30. Alleles for genes A and B are in __________phase. The best estimate of map distance between the A and B loci is ____________ m.u. A. repulsion, 6 B. coupling, 12 C. repulsion, 12 D. coupling, 6 E. repulsion, 24

A. repulsion, 6 Notes: The genes must be in repulsion phase because the most observed progeny represent the parental class, and in this example, the parental classes are gametes Ab or aB. Map distance is estimated by recombinants/total offspring = (25 + 30) / (25 + 30 +410 +450) = 0.06 or 6% or 6 M.U.

Two pure breeding lines of maize display light green leaves instead of the usual dark green of most varieties. Their F1 is dark green. The F2 consists of 461 dark green and 353 light green. You concludes that _______ gene(s) is/are responsible. Using A, B, etc designation for the gene(S), the dark green and light green F2s would have, respectively, the following generic genotypes _______, ________. A. two, A_B_ , anything else (aabb, aaB_ , A_bb) B. two, A_bb, aaB_ C. two, aabb, anything else (aaB_ , A_B_ , A_bb) D. one, A_ , aa E. one, AA or Aa, while aa is lethal

A. two, A_B_ , anything else (aabb, aaB_ , A_bb) REVIEW Note: The numbers resemble 1:1 ratio, but the data fail the chi square test (chi-sq= 14, well past the critical value for one degree of freedom). The numbers fit well a 9:7 ratio, consistent with two genes. There are 9 A_B_ genotypes, and 7 for all other combinations. The product of two genes is needed for wild type leaves. A_bb and aaB_ are equally defective.

A normal woman with no family history of genetic disease or albinism has a child with a man affected by albinism (recessive) and dwarfism (dominant, rare). The gene for albinism and the gene for dwarfism are linked and 20 m.u. apart. What is the chance that the child is dwarf and carries albinism? A. 0.75 B. 0.5 C. 0.05 D. 0.1 E. 0.66

B. 0.5 Note: A__ =normal pigment, dd= normal D__ = dwarfism, dd= normal Mom is AAdd. Dad is aaDd. The question asks the P of a child with genotype aAdD. The P of getting Ad from mom is 1 because she is AAdd. Dad can only produce two types of gametes: aD and ad. Note that recombination does not affect the probability of gametes since the A locus is homozygous. So, 0.5 for aD gamete and aD/Ad child.

In a three point test cross, 15 of 1000 progeny had phenotypes indicating inheritance of a double cross-over recombinant chromosome. The distance between the three genes was 25 m.u. and 15 m.u. Calculate cross-over interference between the genes. A. 0.4 B. 0.6 C. 1 D. 0.8 E. 0.2

B. 0.6 Note: Check out notes for math.

Three plant lines each have an independent mutation that causes them to have a curled leaf phenotype. All are true breeding for this phenotype and the WT phenotype is straight leafed. When crossing these mutants, you observe the following: - Plant 1 x 3-> complementation - Plant 1 x 2-> no-complem. Given these results, what phenotypes have you observed in the progeny? A. 1 x 3 = curled, 1 x 2 = straight B. 1 x 3 = straight, 1 x 2 = curled C. 1 x 3 = curled, 1 x 2 = curled D. None of the other answers E. 1 x 3 = straight, 1 x 2 = straight

B. 1 x 3 = straight, 1 x 2 = curled Note: This question connects phenotype with the complementation test. Since 1 x 3 complement you will get WT, which is straight phenotype. Since 1 x 2 do not complement you will get mutant which is curled phenotype.

In a pea plant the A and B genes are linked at a 10 M.U. distance. You genotype the pollen grains of this individual. What is the expected ratio of recombinant to parental gametes for the A-B genes? A. 1:1 B. 1:5 C. 1:2 D. 1:9 E. 2:9

D. 1:9 Note: Since the map distance between A and B is 10 m.u., you would expect recombination between A and B to occur 10% of the time. The answer 1:9 means that you expect one recombinant for every 9 parental gametes observed (or 10%).

All sneetches want their children to have stars on their bellies. Sneetches can be black or yellow, and star-bellied or starless. Two loci control these phenotypes: yellow is dominant to black, and star-bellied is dominant to starless. The combination of alleles that make a black, starless sneetch is lethal. If two heterozygous yellow, star-bellied sneetches mate, what is the likelihood their first child will be starless? A. 3/8 B. 1/5 C. 3/16 D. 1/2 E. 1/4

B. 1/5 Note: Y_=yellow, yy=black S_=starred, ss=starless ss yy=lethal YySs x YySs -> p of __ss = 1/4, but we must consider two loci. Normally: Y_S_: 9 yyS_: 3 Y_ss: 3 yyss: 1 Lethality of yyss makes the ratio 3/15 instead in 4/16. Thus, 1/5 The question is at first sight a bit disconcerting. It deals with a lethal interaction; yyss is lethal. If two heterozygous parents are mated, what is the P of yyss? This genotype is taken out of the count. So what is left? You must factor both genes in the calculation.

No individuals homozygous for BRCA2 mutation have been found. This observation is consistent with: A. Incomplete penetrance B. Lethality C. Combination of WT and mutant allele is lethal D. Dominance E. Variable expressivity

B. Lethality

Two normal looking fruit flies were crossed, and in their progeny, there were 202 females and 98 males. Which answers provides the best genetic explanation for this anomaly. A. Y-linked gain of function mutation B. X-linked recessive lethal allele C. None of the other answers D. Y-linked recessive lethal allele E. X-linked gain of function mutation

B. X-linked recessive lethal allele Note: X-Linked recessive will have more affected males than females. All sons of affected females will be affected. Skips a generation. When lethal the pair will lead to death. X^Le X^Le, X^LeY, X^leX^Le= Live X^le X^le, X^le Y = dead X^le X^Le x X^Le Y -> X^Le X^Le, X^le X^Le , X^Le Y, X^le Y in 1:1:1:[dead] ratio. So twice as many females as males

e^+ w^+ j 442 e w j+ 435 e^+ w j+ 12 e w^+ j 11 e^+ w j 52 e w^+ j^+ 46 e^+ w^+ j^+ 1 e w j 1 Total = 1000 The above progeny were obtained in a test cross of a female Drosophilia heterozygous for three linked genes. What genes are in coupling phase in the parental hapltypes? A. None are in coupling phase B. e^+ and w^+ C. j^+ and w^+ D. e^+, w^+ and j^+ E. e^+ and j^+

B. e^+ and w^+ Review Definitions Note: The non-recombinants are the most frequent haplo-types (e^+w^+j , ewj^+). They reflect the parental chromosomes. The + alleles of e and w are in coupling, and both are in repulsion with j.

Investigating a suspicious gambling operation, you observe that the six-faced dice used yields 1 less frequently than expected. You measure its probability of "1" at 0.11. The expected odds are ________. The observed odds are _______. A. 0.17, 0.11 B. 1:6, 1:9 C. 1:17, 1:11 D. 1:5, 1:8 E. 1:16, 1:11

C. 1:5, 1:8 Note: fair die: 1 chance of getting a "one" : five chances of not getting a "one" other die: 0.11:(1-0.11) = 0.11:0.89 = 1:8

A wild-type fruit flies is heterozygous for genes A, B, and D. Genes A, B, and D are linked. Assuming no recombination, how many different gamete combinations of these genes can this fruit fly produce? How many could theoretically be produced assuming recombination? A. 2, 10 B. 2, 6 C. 2, 8 D. 2, 4 E. 2, 12

C. 2, 8 Note: Regardless of the linkage phase, if there is no recombination between the loci, only two types of gametes can be produced. For instance if the linkage phase is AbC / aBc, then the only gametes produced would contain either AbC or aBc. With recombination, then a cross over can happen in region 1, which gives us two combinations. A recombination in region 2 can happen, which gives us another two 2 combinations. And a double crossover can happen which gives us another 2 combinations and of course we still have the parental which is two combinations. With a total of 8 recombinations that can have with recombinants. For example gametes could be ABC, abc, Abc, aBC, AbC, aBc, abC, ABc.

Considering epistatic yellow in retrievers. An aa dog is yellow regardless of the genotype at the B locus. In the presence of the dominant A allele, dogs are black or brown depending on the presence of the dominant B (black) allele or recessive b (brown) allele. If the mating of two black dogs has previously produced puppies of all three colors, what ratio do you expect for black: brown: yellow? A. 1:1:1 B. 1:2:1 C. 9:3:4 D. 2:1:1 E. 12:3:1

C. 9:3:4 Note: Check out work in notes for explanation

You grow sweet peas for a living and know that purple pigment is dominant over white. You also know that two genes are involved in producing pea pigment and display complementary gene interaction (duplicative recessive epistasis). One of your clients wants purple sweat peas for his wedding. Which two sweet pea lines should you cross in order to maximize the number of purple sweat peas in the next generation? A. AaBb x aaBB B. AaBb x aaBB C. AAbb x aaBB D. AaBb x AAbb E. AaBb x AaBb

C. AAbb x aaBB REVIEW Note: Any pea plant that is homozygous recessive for EITHER gene A or B results in white pea plants (based on your knowledge this is duplicate recessive epistasis). AAbb x aaBB crosses yield all purple plants (AaBb). No other answer will give you all purple flowers in the F1 generation.

You cross a hairless mouse (aaBB genotype) to a mouse with curly hair (AAbb genotype). All of the F1s have straight hair. You cross two of the F1 mice together. In the F2 generation 18 mice have straight hair, 8 mice are hairless, and 6 have curly hair. What is the phenotype of an aabb mouse in the F2 generation? A. straight hair B. curly hair C. Hairless D. dead because of lethality E. None of the other answers

C. Hairless REVIEW Note: The F1 is AaBb and straight haired (i.e. wild-type). In the F2 we expect the following ratio: A_B_ : A_bb : aaB_ : aabb in a 9:3:3:1. aaB_ mice are hairless, A_bb mice have curly hair. The 9:4:3 ratio suggests recessive epistasis with 9 straight haired, 4 hairless, and 3 curly. From this ratio and the aaB_ phenotype, we infer that aabb must be hairless.

The following is not true about mapping functions. RF = recombination frequency CO = crossing over A. Different mapping functions may or may not consider CO interference B. Mapping functions will correct the map distance between any two genes for which RF can be measured C. Mapping function allowed precise measurement of RF between any two loci D. Mapping functions are based on theoretical assumptions E. The greater the RF between two loci, the larger the correction applied by the function.

C. Mapping function allowed precise measurement of RF between any two loci Review canvas notes for definitions. Note: The mapping function attempts to correct for double CO events, but it cannot predict whether they happen and how frequently. It is based on theoretical assumptions. It uses RF as an input. Very small corrections are applied to closely linked genes, but increasingly larger to distantly linked genes.

ABO blood type in human results from this type of genetic system: A. Three alleles involving epistasis B. Two alleles with incomplete dominance C. Three alleles involving dominance and co-dominance D. Two genes involving epistasis E. Two alleles involving a lethal genotype

C. Three alleles involving dominance and co-dominance. Note: I^A and I^B are dominant to i. I^A and I^B are co-dominant with respect to each other.

e^+ w^+ j 442 e w j+ 435 e^+ w j+ 12 e w^+ j 11 e^+ w j 52 e w^+ j^+ 46 e^+ w^+ j^+ 1 e w j 1 Total = 1000 The above progeny were obtained in a testcross of a female Drosphila heterozygous for three linked genes. Construct a genetic map with the data. A. w - 2.5cM - j - 10cM - e B. w - 2.5cM - e - 10cM - j C. e - 2.5cM - j - 10cM - w D. j - 2.5cM - w - 10cM - e E. e - 2.5cM - w - 10cM - j

C. e - 2.5cM - j - 10cM - w Note: Check notes on how to do this. Review with tutor

You have 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 discover of one of the mutant strain calls his mutation naked, and the other researcher calls her strain hairless. You cross naked and hairless mice with each other and all the offspring are phenotypically wild-type. What were the genotypes of the parents? A. HHNN x hhnn B. hh X nn (h and n are alleles of the same gene) C. hhNN x HHnn D. HHNN xhhNN E. hhnn x HHnn

C. hhNN x HH nn Notes: Question is focused on complementation. If the mutation occurs in the same gene, crossing two mutants together would yield F1 progeny that still display the mutant hairless trait. This is because each parent donates a mutant allele, and no WT allele is present. If the mutation is in a separate gene, then the parent will each contribute one bad allele of one gene and a functional copy of the other. Having one functional copy of each gene produces the WT phenotype in the F1.

There are two questions on pineapple genetics. Start with this one first. The leaf edges of pineapple can be: 1. uniformly spiny 2. spiny on the leaf tip 3. piping, i.e. with nice smooth edges that resemble cloth piping folds Pineapple growers prefer varieties (Links to an external site.)Links to an external site. that are piping because harvesting them is safer and easier. The following crosses were carried out using pure breeding varieties whose phenotype is indicated as piping, spiny tip, or spiny. Two pure bred parents are named as W and Z for later reference in the answers. ( have to look at question 6 picture to make sense) A. single gene at work, piping is dominant B. two genes at work, Spiny tip is double recessive C. two genes at work, Spiny is double recessive D. two genes at work, Spiny is dominant E. Single gene at work, Spiny is dominant

C. two genes at work, Spiny is double recessive HELP ON HOW TO KNOW IF TWO GENES ARE PRESENT Note: The results do not fit simple Mendelian genetics rules. At first sight they are confusing. Focus first on the F2 with the most phenotypic classes. There are two that have three classes. Epistasis could be at work! What is the least frequent phenotype? Piping>Spiny Tip>Spiny. It is Spiny! What about ratios? In both the three-phenotype F2s the ratio matches fairly well 12:3:1 ratio for Piping>Spiny Tip>Spiny. This is indicative of dominant epistasis. We conclude that the spiny individuals are double homozygous recessive.

In your greenhouse populations you observe 2 mutant plants lines which you call "biteme" mutants. Plants homozygous for a recessive allele you call 'a' are vulnerable to zombie mite bites. The same is true for plants homozygous for another recessive allele you call 'b'. You cross individuals homozygous for each of the mutants and observe that the entire F1 generation remains vulnerable to zombie mite bites. How many loci are manifested by this analysis and what are the genotypes of the parental generation. A. None of the other answers B. 1 locus, AA x bb C. 2 loci, AaBb x AaBb D. 1 locus, a and b are two alleles of the same gene E. 2 loci, AAbb x aaBB

D. 1 locus, a and b are two alleles of the same gene Note: Complementation. If the mutation occurs in the same gene, crossing two mutants together would yield F1 progeny that still display the vulnerable-to-zombie-bites phenotype. This is because each parent donates a mutant allele, and no WT allele is present. If the mutation b were in a separate gene, then each parent would contribute a bad allele of one gene and a functional copy of each gene would produce the WT phenotype in the F1. Each mutant is homozygous recessive. Since complementation does not occur, we know the mutations are in the same gene: we are dealing with one locus. The parental cross is aa x bb and the F1 then has the genotype ab. The F1 genotype has two mutated alleles at this locus and is thus vulnerable to zombie bites.

Consider the epistatic problem of the piping vs spiny pineapple. Piping is epistatic on spiny. If the piping locus "P" has genotype pp, the spiny trait is manifested. P__ plants are always piping. ppS__ = spiny tip, ppss = spiny. Two inbred lines, on piping, one spiny are crossed and produce a piping F1. Selfing of the F1 produces the following F2s: 118 piping, 28 spiny tip and 12 spiny. What ratios, in that order, do you expect if you cross the F1 to the inbred spiny parent? A. 1:2:1 B. 12:4:3 C. 9:3:4 D. 2:1:1 E. 1:1:2

D. 2:1:1 Note: The ratio is the classical 12:3:1 for dominant epistasis. Piping is the epistatic trait that hides the two types of spiny. Genotype -> Phenotype P_, __ = Piping pp, S_ = Spiny tip pp, ss = Spiny The two original lines must have been PPSS x ppss, since the F1 yields the ratio expected from a PpSs. Crossing PpSs x ppss will yield four identical genotypic classes, shown here with the resulting phenotype: PpSs = Piping Ppss = Piping ppSs = Spiny tip ppss = Spiny Therefore, the ratio will be 2 Piping : 1 Spiny tip : 1 Spiny However this can also be done thru conditional probability, tutor should me how.

In a trip to Mexico you buy pure-breeding seeds of a tomato with green stripes over a red background, which you call Zebra. You cross zebra to a common red-fruited variety, Early Girl. The F1 is spotted green on red background. In the F2 you observe 33 solid red, 70 spotted, and 29 zebra. The mutation involves change(s) at _ _ _. Individuals with spotted phenotype are _ _ _ _. A. A single gene, dominant homozygous B. A single gene, recessive homozygous C. Two genes, has a dominant allele at both D. A single gene, heterozygous E. Two genes, recessive at all loci

D. A single gene, heterozygous NEED HELP WITH EXPLANATION.

In a cross between AaBbCc and aabbcc, the most frequent gamete classes produced by the hybrid are ABc and abC. The least frequent are aBb and AbC. The parental haplotype and gene order is: A. AcB and aCb B. CBa and cbA C. BAC and bac D. BAc and baC

D. BAc and baC Note: Look at notes for explanation

What criteria must be met in order for a complementation test to be valid? A. Mutant lines must have been isolated together. B. Mutant lines must have mutations in the same gene C. Mutant lines must have mutations in different genes D. Mutant lines must be true breeding E. Mutant lines must show different phenotypes

D. Mutant lines must be true breeding Note: Can not interpret results without them being true breed. A. does not make sense because the complementation test is often used on independently discovered mutants.

Curly leaf of pepper is caused by a recessive allele at a single locus. A homozygous curly leaf plant is crossed to a heterozygous, normal plant. In their progeny you count 60 healthy and 15 affected individuals. The best explanation for these results is that the curly leaf trait. A. incomplete dominance of B. is negatively affected by non-disjunction C. X-linked inheritance D. has low penetrance E. has variable expressivity

D. has low penetrance Note: Penetrance is the fraction of individuals with given genotype that actually express the expected phenotype. Think of penetrance as value determined by "yes" or "no"

In horses the Red gene (R or r) displays recessive epistasis on the Bay gene (B or b). Horses with the rr genotype are red regardless of their Bay genotype (B_ or bb). Recessive Bay (bb) are black if R_. Crossing a black stallion to a red mare has produced three colts: one bay, one red and one black. What is the P that the next colt from the same pair will be Bay? A. 0.375 B. 0.5 C. 0.0625 D. 0.5625 E. 0.25

E. 0.25 Note: Check out notes for math/explanation

Loci A and B are linked at a 15 m.u. distance, with A-B alleles in coupling. The cross AaBb x AaBb produced 1000 progeny. How many aabb do you expect? A. 6 B. 105 C. 62 D. 12 E. 180

E. 180 Notes: Check out notes for explanation/math.

Flowers of Armenian cyclamens can have three types of petal color: solid dark-blue, red with yellow tips, or solid yellow. You purchase a blue flowered plant, self it, and obtain 230 blue, 64 red and yellow, and 13 yellow. Based on this, the most likely hypothesis that ____ genes are involved and that the one conferring ____ color displays _____ epistasis. A. 2 red, dominant B. 3, blue, recessive C. 2, red, recessive D. 2, yellow, recessive E. 2, blue, dominant

E. 2, blue, dominant Still Iffy on this Note:

Loci A and B are linked at a 25 m.u. distance. The cross AaBb x aabb produced 1000 progeny. Of this progeny, approximately ________ are expected to be recombinant (ignore cross-over interference). Alternatively, if the map distance is determined to be 40 m.u., ________ progeny are expected to be NON recombinant. A. 600, 750 B. 750, 600 C. 250, 400 D. 750, 400 E. 250, 600

E. 250, 600 Note: 0.25 x 1000 = 250 recombinant 0.4 x 1000 = 400 recombinant 1000-400 = 600 NON recombinant

Mr.T has blood type A. Mrs. T has blood type B. They have three kids, whose blood type is, respectively, A, B, and O. The best explanation for the blood phenotypes in this family is: A. The parents are both heterozygous at two independent genes controlling blood type. Mr.T is Aabb, Mrs.T is aaBb. B. The parents are homozygous at the same blood type locus: Mr.T is I^A/I^A, Mrs.T is I^B/I^B. C. The parents are both heterozygous at the same blood type locus: Mr.T is I^A/I^B, Mrs.T is I^A/I^B. D. The parents are both homozygous at two independent genes controlling blood type. Mr.T is AAbb, Mrs.T is aaBB. E. The parents are both heterozygous at the same blood type locus: Mr.T is I^A/I^o, Mrs.T is I^B/I^o.

E. The parents are both heterozygous at the same blood type locus: Mr.T is I^A/I^o, Mrs.T is I^B/I^o. Note: IAi x IBi --> all type progeny possible:O, A, B and AB from, respectively, ii, IAi, IBi, IAIB

GPQ: Linkage, probability and family pedigrees

Review them! Just wrote them down