Important words and concepts from Chapter 14, Campbell & Reece, 2002 (1/29/2005):
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NOTE THAT YOU HAVE BEEN ASSIGNED THE PROBLEMS FOUND AT THE END OF THIS CHAPTER. IF YOU BLOW OFF DOING THESE PROBLEMS (AND EXPERIENCE SUGGESTS THAT MANY OF YOU WILL), THEN YOU WILL NOT BE TERRIBLY HAPPY WITH YOUR PERFORMANCE COME EXAM TIME.
(1) Chapter title: Mendel and the Gene Idea
(a) "The best way to gain an understanding of genetics is to work with it. The fundamental principles discussed (below) will become clear to you, and you will grasp them more surely, if you carefully think through . . . problems which illustrate the various patterns of inheritance..." (Keeton, 1980, Biological Science third edition, W.W. Norton & Company, p. 621)
WAYS OF THINKING ABOUT GENOTYPE
(a) A gene is a discrete heritable unit
(c) (loci mapped to human chromosome number 3 are shown to the right; M b means mega base and cM means centimorgan) -->
(3) Genotype (see also genotype)
(b) Genotype is subdivided into discrete, heritable units called genes
(c) Genotype is what is passed from parent to offspring on chromosomes
(4) Phenotype (see also phenotype)
(a) Phenotype is specified by genotype
(b) Phenotype is the stuff that you can see and measure about an organism
(c) For example, height, color, number of legs, ability to smell, etc.
(d) Phenotype imperfectly maps onto genotype
(e) That means that there may exist numerous genotypes for any one phenotype
(g) See Figure, Genotype versus phenotype
(b) Often the differences represent only one or a few nucleotides
(c) Two genes that are found at the same locus of a homologous pair of chromosomes, that differ in nucleotide sequence, are referred to as different alleles
(d) Different alleles may or may not elicit (code for) different phenotypes
(e) When a gene has a dominant form and a recessive form, these forms are considered to be separate (i.e., different) alleles which give rise to distinct phenotypes (although it is also possible for two distinct alleles to give rise to the same phenotype)
(f) A diploid organism has up to two alleles present in their genome per locus, or as few as one
(g) See Figure, Alleles, alternative versions of a gene
WAYS OF THINKING ABOUT PHENOTYPE
(a) A character is a heritable feature (phenotype) of an individual
(b) A character can vary
(c) For example, hair color is a character as is ear size, etc.
(d) See Table 14.1, The results of Mendel's F1 crosses for seven characteristics in pea plants
(a) A variant of a character is a trait
(b) Thus, we have genes that are responsible for a given character
(c) And we have different alleles of that gene which are responsible for different traits associated with that character
(c) See Figure, Alleles, alternative versions of a gene
(d) Each of these alleles, whether identical or different, will interact to produce a trait
(e) The interaction between non-identical alleles results in interesting non-correspondences between genotype and phenotype
MORE WAYS OF THINKING ABOUT GENOTYPE
(a) Diploidy also results in interesting patterns of inheritance
(b) The simplest pattern results in all offspring (and their offspring, etc.) always resembling the parents
(d) This would be a total of four identical alleles between the two original parents
(e) Such characters are said to be true breeding because they fail to vary through the generations
(a) True breeding stems from homozygosity
(c) Such an individual is said to be a homozygote
(b) Note that the two alleles defining a heterozygote will segregate into different gametes such that 50% of gametes will posses one allele and the rest (50%) of the gametes will posses the other allele
(c) Thus, at a given locus a flower may contain an allele that codes for the trait purple flowers; the homologous allele, found on a homologous chromosome, might code for white flowers (or purple flowers or whatever), but not plant height, etc. (unless the gene has pleiotropic effects)
(b) This is cytogenetical basis for Mendel's law of segregation
(c) See Figure, Mendel's law of segregation
MORE WAYS OF THINKING ABOUT PHENOTYPE
(b) Such alleles (the former) are said to display dominance
(c) When we abbreviate alleles, typically the dominant allele is capitalized (e.g., A)
(b) When we abbreviate alleles, typically the recessive allele is written in lower case (e.g., a)
(b) That is, at that locus the individual's genotype would be aa where a is the abbreviation for the recessive allele
(b) That is, at that locus the individual's genotype would be AA where A is the abbreviation for the dominant allele
(c) NOTE THAT THERE IS NO SUCH THING AS A "HETEROZYGOUS DOMINANT!"
CROSSES INVOLVING ONE LOCUS
(a) A cross is a mating between two individuals
(b) We abbreviate the occurrence of a cross as an x found between two genotypes, e.g., AaBBcc x aaBBcc is a cross between two individuals where we are keeping track of alleles found at three different loci (locus A, locus B, and locus C)
(b) For example, flower color in peas
(c) An example of a monohybrid cross could be abbreviated as Aa x Aa
(d) Note that each individual participating in a monohybrid cross is heterozygotic at the locus in question
(a) P stands for parental
(b) The parents are the first generation to be crossed (i.e., mated)
(c) In an experimental breeding program this first generation is called the P generation
(a) The F1 generation is the product (the offspring) of the parental cross (P generation)
(b) F stands for filial
(a) The F2 generation is the product (the offspring) of the interbreeding of the F1 generation
(a) See Figure, Mendel's law of segregation
(b) AA x aa = P generation
(c) Aa = F1 generation
(d) AA + Aa + Aa + aa = F2 generation
(e) Say A is a dominant allele (e.g., codes for a purple flower color)
(f) Say a is a recessive allele (e.g., cods for a white flower color)
(g) Then the F1 generation will have only purple flowers
(h) The F2 generation will have three purple flowers for every white flower
(c) See Figure, Mendel's law of segregation (see in particular the diamond shaped box associated with the F2 generation)
(d) Note that each progeny organism inherits one allele (and only one allele) from each parent
(e) The Punnett square is limited in its power particularly because introducing multiple alleles (i.e., more than two) and, in particular, multiple loci into a cross results in an enormous mess; to transcend messes such as this we will instead employ probability theory
(c) How do we tell the two apart?
(d) The traditional way is to do a testcross
(e) In a testcross the individual with the unknown genotype is crossed with an individual with a known genotype
(f) What genotype is known?
(g) For the pea flower color example, it is the individuals with the white flowers
(i) In other words, the homozygous recessive condition typically displays a one-to-one mapping of phenotype onto genotype
(j) Examples of a testcross: Aa x aa or AA x aa (the former will give rise to two phenotypes whereas the latter will give rise to only one phenotype, thus the genotype of the first parent in each cross may be determined via this testcross)
(k) See Figure, A testcross
CROSSES INVOLVING MORE THAN ONE LOCUS
(a) Upping the ante of complexity is the dihybrid cross
(b) A dihybrid is, for example, a genotype having an abbreviation of AaBb
(d) A dihybrid cross is AaBb x AaBb
(e) Follow this cross through the F2 generation
(f) See Figure, Testing two hypotheses for segregation in a dihybrid cross
(g) Note the final phenotypic ratio of 9:3:3:1
(h) Make sure you understand this ratio as well as how to generate it
(28) Trihybrid (and higher) crosses [genotype]
(c) A monohybrid cross looks like this: Aa x Aa
(d) A dihybrid cross looks like this: AaBb x AaBb
(e) A trihybrid cross looks like this: AaBbCc x AaBbCc
(f) A tetrahybrid cross looks like this: AaBbCcDd x AaBbCcDd, etc.
(g) To follow the products of these higher-order hybrid crosses it is far easier to employ the tools of probability theory
(a) In considering genetic variation, we have begun to touch upon the theory of probabilities
(b) As we delve further into genetic variation and heredity (i.e., genetics) we will also delve further into probability theory (e.g., Hardy-Weinberg equilibrium)
(c) It has been my experience that students who fail to grasp probability theory to some extent go on to fail to grasp genetics
(d) Probability theory, at its simplest, considers
(ii) A range of probabilities from 0 to 1
(e) Probability theory will be found on exams only in the guise of genetics problems; that is, don't worry about the terms (unless you find them helpful) so much as how to employ probability theory when doing your genetics problems
(a) As a simplifying assumption, we will assume that all events occur independently
(b) This means that the occurrence of one event has no impact on the occurrence of another event
(c) Stated as an example, it means that when you toss two coins, which side one coin lands on does not influence which side the other coin lands on
(d) Note that statistical independence need not always apply but, for now, assume that it does always apply
(a) Probabilities are assigned numerical values
(b) A probability of 0.0 means that an event will never happen
(c) A probability of 1.0 means that an event will always happen
(d) An event with a probability between 0.0 and 1.0 will occur with some proportional likelihood such that the closer the probability is to 1.0, the more likely it will occur
(e) In other words, an event with a probability of 0.5 will occur half the times in which circumstances are such that such an event could occur (e.g., when a coin is tossed, half the time it will come up heads and half the time it will come up tails)
(f) An event with a probability of 0.17 will occur with much less likelihood than an event with a probability of 0.5 (e.g., when tossing dice, the likelihood of a 3 being tossed is 1/6 = 0.17; assuming the dice aren't loaded, a probability of 1/6 should be true for any given number per toss)
(a) Assume that two events are statistically independent
(b) What is the probability that two such events will occur during two circumstances during which the event could occur (e.g., two subsequent tosses of a coin, what is the likelihood that both will be heads?)
(c) The answer is that the likelihood that two events will occur in two chances is the product of each event occurring during each chance individually (note that this ability to use the law of multiplication fails to work if the two events are not statistically independent)
(d) For example, the likelihood of obtaining two heads in two tosses of a coin are 0.5 x 0.5 = 0.25 (three tosses? 0.5 x 0.5 x 0.5 = 0.125; four? 0.5 * 0.5 * 0.5 * 0.5 = 0.0625 = 0.54)
(e) Note that doing stuff like this can often be much more difficult should one choose to avoid exponential notation; this point can come back to haunt you all through biology
(a) Note that the law of multiplication can be used to calculate the probability of events not happening
(b) To do this, simply calculate the likelihood that an event will happen, then subtract that likelihood from 1.0
(c) For example, what is the likelihood of not rolling a total of exactly three 4's in three rolls of dice? 1.0 -- (0.17 x 0.17 x 0.17) = 1 - (1/6)3 = 0.995
(a) When a complex event (i.e., one requiring more than one step) can occur via more than one series of steps, what is the likelihood that the event will occur, independent of route?
(b) The answer is the sum of the probabilities that it will occur by all of the individual routes
(c) For example, what is the probability that in three subsequent rolls of dice you will roll one 3, one 4, and one 5?
(d) To answer this, first break down the problem into the many possible ways you can roll these numbers, distinguishing them by order: 3-4-5, 4-3-5, 3-5-4, 5-3-4, 5-4-3, 4-5-3
(e) Now, calculate the likelihood for each set: here this works out to (0.17)3 for each
(f) Now add together all of the probabilities of having the event by each possible path: (0.17)3 + (0.17)3 + (0.17)3 + (0.17)3 + (0.17)3 + (0.17)3 = 6 x (0.17)3 = 0.028
(g) The likelihood of not ending up with any of these combinations, by the way, is 1 -- 0.028 = 0.972
Note that the sum of all possibilities must always add up to 1.0; you should always use this fact as a check on your calculations --> if your probabilities do not all add of to 1.0 (and you have included all possibilities in your sum) then you have made some kind of mistake (and/or don't understand the calculation)
(i) WW x ww
(ii) Ww x ww
(iii) Ww x Ww
(b) In peas an allele for tall plants (T) is dominant over the allele for short plants (t). An allele of another independent gene produces smooth peas (S) and is dominant over the allele for wrinkled peas (s). Calculate both phenotypic ratios and genotypic ratios for the results of each of the following crosses:
(i) TtSs x TtSs
(ii) Ttss x ttss
(iii) ttSs x Ttss
(iv) TTss x ttSS
(c) What fraction of the offspring of parents each with the genotype KkLlMm will be:
(l) Multiple alleles
(a) What the organism "looks" like
(f) 9:3:3:1 ratios
(l) Quantitative characters
(m) Norm of reaction
GENOTYPE --> PHENOTYPE COMPLICATIONS
(c) Molecularly, what is happening is that the protein produced by the incompletely dominant allele is not able to make up (completely) for the lack of activity exhibited by the protein made by the recessive allele
(d) See Figure, Incomplete dominance in snapdragon color
(e) White, in the example in the figure, is the product of a non-functional protein, i.e., no color was produced
(a) Complete dominance up to now we have referred to simply as dominance (or dominant)
(c) Alternatively, the effect of the recessive allele may be sufficiently slight that the phenotype associated with the completely dominant allele completely masks that associated with the recessive allele
(d) Example: brown and blue eye color
(e) Example: Black fur versus brown fur in mice
(f) Example: AO and BO in ABO blood group
(a) Codominance occurs when two alleles both express functional proteins
(b) Very often the heterozygote possessing two codominant alleles displays the phenotypic product of both codominant alleles
(e) For example, the IA and IB alleles in the ABO blood group display codominance to each other
(f) In this case, the meaning of codominance is that both the A and the B phenotypes (oligosaccharides) are displayed by red blood cells
(g) Note that, at the molecular level, all dominance relationships are ones of codominance since the very act of describing, for example, one functional and one non-functional protein is describing two very real phenotypes associated with each allele
(h) NOTE THAT PINK FLOWERS ARE NOT AN EXAMPLE OF CODOMINANCE!!!! (if you don't understand why not then you don't understand codominance/haven't learned what codominance means)
ONE LOCUS, MORE THAN TWO ALLELES
(41) Multiple alleles [genotype]
(b) We have been considering the two allele, one locus situations (as well as two alleles per loci, more than one loci situations, e.g., dihybrid cross)
(c) In addition there exist numerous examples of more than two allele, one locus genetics (and various dominance relationships can exist between different alleles)
(d) This latter case may be referred to as examples of multiple alleles
(e) Note that no matter how many alleles may be found at a given locus within a population of individuals, in a single, diploid individual a maximum of only two alleles may present
(b) The ABO blood groups are controlled by three alleles, IA, IB, and i
(i) IA is codominant to IB
(ii) Both IA and IB are dominant to I
(iii) Note the different notation from that we have employed so far; variation in the notation used to specify genotype or phenotype is very common in genetics
(c) Possible phenotypes associated with their ABO blood groups include
(ii) A = IAIA or IAi (a dominant phenotype)
(iii) B = IBIB or IBi (a dominant phenotype)
(iv) O = ii (the homozygous recessive phenotype)
(d) See Figure, Multiple alleles for the ABO blood groups
MORE GENOTYPE --> PHENOTYPE COMPLICATIONS
(b) Often, however, a single locus will control more than one character
(c) The set of characters controlled by a single locus in this more-complicated case is known as that locus' pleiotropy
(d) Examples: albinism; white, cross-eyed tigers, etc.
(e) Note that typically, at some level of phenotype (e.g., molecular), the effect of an allele is more or less constant, but can have different effects on different systems (cells, tissues, organs) or with regard to different means of measuring the defect
(c) This inter-locus interacting is termed epistasis
(d) Example (from text): Mouse coat color
(i) B = black coat color (dominant)
(ii) b = brown coat color (recessive)
(iii) C = coat is colored (dominant)
(iv) c = coat displays no color (recessive)
(e) Phenotypes (with underlying genotypes) include
(i) Black fur = BBCC, BbCC, BBCc, or BbCc
(ii) Brown fur = bbCC or bbCc
(iii) White fur (albino) = BBcc, Bbcc, or bbcc
(f) See Figure, An example of epistasis
(g) Note that the albino syndrome is actually a pleiotropy since albinism results in many defects in addition to a lack of fur (or hair) color, e.g., lack of skin color, red irises, poor vision, high susceptibility to skin cancer, poor eyesight, etc.
(a) Often one character is controlled by more than one locus (as with epistasis) but where loci display an additive effect (rather than qualitatively different effects as with the mouse example above)
(b) Such characters are termed quantitative and the associated genetics polygenic inheritance
(c) Typically quantitative characters can vary over a significant range controlled by their underlying genetics
(d) For example, height and skin color (in humans) are quantitative characters
(e) Animal and plant breeding typically involves quantitative characters (bigger melons, smaller dogs, higher yields, etc.)
(f) Figure, A simplified model of polygenic inheritance of skin color
(b) In addition, any one genotype may give rise to more than one variation on a character
(d) Additionally, the environment in which the organism develops and lives typically impacts on phenotype
(e) The norm of reaction refers to the degree to which a phenotype associated with a given genotype may vary, particularly as a function of variation in single environmental parameter (e.g., temperature)
(f) Some norms of reaction display no breadth (e.g., ABO blood groups, flower color)
(g) Other norms of reaction display significant breadth (e.g., polygenic characters and behavioral traits)
(h) See Figure, The effect of environment on phenotype
(47) Nature versus nurture [phenotype]
(a) It is from the concept of reaction norm that we derive the concept of nature versus nurture, i.e., which is more important in determining a given phenotype, the underlying genotype or the environment?
(b) The answer to this question is sufficiently complex and ongoingly controversial that we won't even attempt to answer this question here
(a) Many of the advantages of doing genetics using experimental organisms--possessing in particular short generation times--are lost when it comes to human genetics
(b) Humans are long lived, with long generation times, and often prefer to avoid genetic manipulation ("But dear, it's for the sake of science, and Johnny has such interesting traits. All we're asking is that you have a very large family with him.")
(c) On the other hand, some humans with large families have a predilection toward keeping good records of matings and births over decades and even centuries
(d) Large, well-kept pedigrees coupled with genetic diagnoses can be used to infer the genetics (e.g., dominance relationships) of various human traits and diseases
(a) Pedigrees are represented graphically using certain conventions to symbolize individuals, matings, offspring-to-parent relationships, affected individuals, etc.
(i) Squares represent males
(ii) Circles represent females
(iii) Horizontal lines between individuals represent matings (often, of course, these are marriages)
(iv) Time flows from top to bottom
(v) Vertical lines represent offspring-to-parent relationships
(vi) Various branchings of vertical lines lead to siblings
(vii) Siblings typically are listed left-to-right in order of their births
(viii) Affected individuals (those displaying the trait) are represented by filled symbols
(50) Autosomal dominant traits (see also human autosomal dominant conditions)
(i) No skipping of generations
(ii) Typically only about half of offspring are also affected
(iii) No silent (i.e., not affected) carriers (at least in principle, that is)
(iv) Mutational genesis of early-onset lethal traits
(b) Examples of human autosomal dominant traits include (no need to memorize unless you are interested):
(i) Brown eyes
(ii) Widows peak
(iii) Free (not attached) earlobes
(iv) Huntington's disease
(c) Note that lethal autosomally dominant diseases tend to be very rare because all affected individuals (all who carry the allele) die before passing that allele on to their offspring (evolution in action... literally); this is why such alleles tend to be the products of same-generation mutational genesis
(d) Huntington's disease is excepted because the time of typical onset of symptoms is after child rearing (e.g., 35 to 45 years old)
(e) Consequent to the self-culling effect of early onset, lethal autosomally dominant alleles tend to be very rare
(f) See Figure, Pedigree analysis
(51) Autosomal recessive traits (see also human autosomal recessive conditions)
(i) Skipping generations (essentially same thing as "silent carriers" for early onset diseases)
(ii) Silent carriers (typically both parents)
(iii) One-fourth progeny affected (when both parents are not affected)
(iv) More likely early onset lethal than autosomal dominants
(v) For less serious conditions, parents can be homozygous recessives
(b) Examples of human autosomal recessive traits include (no need to memorize unless you are interested):
(i) Blue eyes
(ii) Lack of widow's peak
(iii) Attached earlobes
(iv) Cystic fibrosis
(v) Tay-Sachs disease
(vi) Sickle-cell disease
(c) For a serious (i.e., life threatening) autosomal recessive trait, the underlying alleles tend to be rare in populations; this is because the homozygous recessives tend to be "culled"; consequently, the majority of such alleles are found, within populations, in individuals who are heterozygous at the loci in question
(d) See Figure, Pedigree analysis
(a) ABO blood group
(k) Dihybrid cross
(n) Dominant alleles
(p) F1 generation
(q) F2 generation
(aa) Homozygous recessive
(bb) Incomplete dominance
(dd) Law of addition
(gg) Monohybrid cross
(hh) Multiple alleles
(jj) Norm of reaction
(kk) P generation
(ll) Pedigree analysis
(rr) Probability theory
(ss) Punnett square
(vv) Reaction norm
(ww) Recessive alleles
(ccc) True breeding
(ddd) 9:3:3:1 ratio
(53) Practice questions [index]
(a) What is an allele?
(b) A variant of a character is __________.
(c) What does it mean for two crossed individuals to each be true-breeding?
(d) What does it mean for an allele to be recessive? See also C:\Documents and Settings\Stephen T. Abedon\Desktop\New Briefcase\Teaching\LECTURES\113 lectures\campbl14 (problems-exam questions).doc for Mendelian genetics problem.
(e) Under what circumstances would you expect the progeny of a monohybrid cross to not be present in a 3:1 phenotypic ratio?
(f) Under what circumstances would you expect the progeny of a dihybrid cross to be present in a 9:4:3 ratio?
(g) Given the following cross: AaBbCCDdEeFf x AABbCCddEeFf, what proportion of the progeny will have the following genotype: AAbbCCddEEff?
(h) If uppercase is dominant to lowercase, then what would be proportion of the most-represented phenotype would be found in the progeny of the following cross: AaBbCCDdEeFf x AABbCCDdEeFf?
(i) In a reaction norm some measure of phenotype is determined as it varies (or does not vary) as a function of __________. In other words, reaction norms are graphs with phenotype on the Y axis and __________ (same term as previous blank) on the X axis.
(j) Given the following cross, IAiRr × IBiRr, what are the expected phenotypic ratios of the progeny? Adapted from
(k) Here is a sample pedigree for eye color. If the people with filled-in (dark) symbols have brown eyes and those with open (light) symbols have blue eyes, can you figure out the genotypes of the people marked with *? Assume that the brown-eye allele (B) is dominant to the blue-eye allele (b).
A brown-eyed man whose father was brown-eyed and whose mother was blue-eyed married a blue-eyed woman whose father and mother were both brown-eyed. The couple has a blue-eyed son. For which of the individuals mentioned can you be sure of the genotype? What are their genotypes? What genotypes are possible for the others? Note that blue (b) is recessive to brown (B).
(n) Many different gene pairs control a horse's coat color. A dominant B will give the horse a black color, and a b will give the horse a chestnut color. However, when a dominant W is present in the second gene pair, the horse will be white no matter what. If the second gene pair is ww, then the horse will be the color of whatever the first gene pair is made of. If a heterozygous white (BbWw) mare is crossed with a heterozygous white (BbWw) stallion, what are the offspring's phenotypic ratio? Adapted from
(o) (Bonus) Many different gene pairs control a horse's coat color. A dominant B will give the horse a black color, and a b will give the horse a chestnut color. However, when a dominant W is present in the second gene pair, the horse will be white no matter what. If the second gene pair is ww, then the horse will be the color of whatever the first gene pair is made of. Horses can also be bay in color. This dominant gene allele (A) masks the dominant black color, but not the white gene or the recessive chestnut color. Horses also carry a dilution gene (D). If only one allele is the dominant D, and a dominant A allele and the recessive chestnut color alleles are present, then the horse will be a palomino. If there are two dominant dilution alleles (DD) and a dominant A allele and the recessive chestnut color alleles are present, then the horse will be a pseudo albino. What are the phenotypic ratios of the offspring when a palomino mare (AAbbDdww) is crossed with a palomino stallion (AabbDdww)?
(p) Dennis, 25, and Lucy, 22, go to pre-pregnancy genetic counseling. Dennis had two normal brothers and a normal sister, but also had a brother and a sister who died at ages 22 and 18, respectively, of cystic fibrosis (a recessive condition of malabsorption due to pancreatic insufficiency and recurrent pulmonary infection with progressive respiratory insufficiency, usually leading to death in early adulthood). Dennis is in good health, as is Lucy. Assume, however, that one (but not both) of Lucy's parents is a carrier for cystic fibrosis. Please set up the cross, indicating the odds that Dennis and Lucy are carriers.
(q) Dennis, 25, and Lucy, 22... What are the odds if Dennis and Lucy have ten children--assuming that all you know about Dennis and Lucy's genotypes is that which you derived in the previous problem--that none of those children will be afflicted with cystic fibrosis?
(r) Dennis, 25, and Lucy, 22, go to pre-pregnancy genetic counseling. Dennis had two normal brothers and a normal sister, but also had a brother and a sister who died at ages 22 and 18, respectively, of cystic fibrosis (a recessive condition of malabsorption due to pancreatic insufficiency and recurrent pulmonary infection with progressive respiratory insufficiency, usually leading to death in early adulthood). Dennis is in good health, as is Lucy. Assume, however, that one (but not both) of Lucy's parents is a carrier for cystic fibrosis. What are the odds that the first child of this marriage will have cystic fibrosis?
(54) Practice question answers [index]
(a) An allele is a variant of a gene, a sequence of nucleotides found at specific locus on a chromosome
(c) It means that they are both homozygous at a given locus
(d) It means that the trait associated with that allele is not displayed in the heterozygous state with a dominant allele but instead only when the allele is found in the homozygous state
(e) Given incomplete or co-dominance
(f) This is an example of epistasis that can be found in a practice question in your book; in this case one locus controlled whether there was (or was not) fur color (with fur color dominant to no fur color) and at the other locus black fur was dominant to brown fur such that of 16 progeny of the dihybrid there was an expectation of 9 black, 3 brown, and 4 white individuals
(g) ½ * ¼ * 1 * ½ * ¼ * ¼ = 1/256
(h) ABCDEF occurs with a probability of 1*3/4*1*3/4*3/4*3/4 = 81/256
(j) This is a dihybrid cross with a twist. I will simplify the allelic abbreviations to simplify my typing such that IA = A and IB = B, and not italicizing. In the cross Ai × Bi the expected phenotypic ratios are AB, A, B, and O, 1:1:1:1:. In the cross Rr × Rr the expected phenotypic ratios are 3R to 1r. Therefore the expected phenotypic ratios, assuming independent assortment, are 3ABR, 3AR, 3BR, 3OR, 1ABr, 1Ar, 1Br, and 1Or.
(k) The are all heterozygotes
(l) You can be sure of all of the genotypes except for the brown-eyed father; the genotypes you can be sure of are bb for all blue-eyed individuals and Bb for all brown-eyed individuals whose genotype you can be sure of; the genotype of the brown-eyed father is either Bb or BB
(m) The cross is Rr × rr; half of the offspring are pink and the other half are white; the genotypic ratios in the progeny are 1Rr to 1rr
(n) The Bb × Bb cross will yield black to chestnut in a 3:1 ratio while the Ww × Ww cross will yield white to color in a 3:1 ratio; consequently, the 12 white (¾ of 16), 3 black (¾ of ¼ of 16) , and 1 chestnut (¼ of ¼ of 16)
(o) First you have to figure out what the genotypic ratios are, then you have to figure out what are the associated phenotypes. Loci by loci, the ratios are 1AA:1Aa, all bb, 1DD:2Dd:1dd, and all ww. The genotypes therefore will be 1AAbbDDww : 2AAbbDdww : 1AAbbddww : 1AabbDDww : 2AabbDdww : 1Aabbddww. Now one has to figure out the phenotypes (which is the hard part of this problem): The common ww means that the offspring will all have color (i.e., will all not be white as controlled by the W locus). All of the offspring have the dominant A allele; however, none of the offspring posses the B allele, so the presence of A has no impact except with regard to the D locus. That is, ignoring the D locus, we would expect all of the progeny to be chestnut. However, since all are chestnut (ignoring D) and all have the dominant A allele, the phenotypic ratios are determined solely by the D locus: 1pseudo albino (for DD): 2 palomino (for Dd): 1 chestnut (for dd).
(p) The odds that Lucy is a carrier are 0.5; the odds that Dennis is a carrier are 2/3 (since he has the not-afflicted phenotype but is the son of two carriers)
(q) Given that Lucy and Dennis are both carries, then their odds of having ten healthy children are (¾)10, and therefore the odds of having at least one not-healthy child among ten (given that both are carriers) is (1 -- (¾)10) which is multiplied by the odds that both parents are carriers: (1 -- (¾)10) * 1/2 * 2/3; since the latter result are their odds of not having ten healthy children (of ten), their odds of having ten healthy children is equal to 1 -- (1 -- ((¾)10) * 1/2 * 2/3) = 0.685
(r) The odds that Lucy is a carrier are 0.5; the odds that Dennis is a carrier are 2/3 (since he has the not-afflicted phenotype but is the son of two carriers); the odds that the first child will be affected therefore are ½ * 2/3 * ¼ = 1/12
Chapter 14, Bio 113 questions:
(#) A __________ is a discrete heritable unit that generates traits via the coding for the synthesis of proteins.
(#) True or False, there may exist numerous genotypes for any one phenotype.
(#) Give an example of imperfect mapping of phenotype onto genotype.
A: dominant-recessive relationships between alleles.
(#) Two genes that are found at the same locus of a homologous pair of chromosomes, that differ in nucleotide sequence, we refer to as different __________.
(#) A variant of a character is a __________.
(#) In terms of genotype, what does it mean for a character to be true breeding?
A: It means that the genotype is homozygous; the individual (or individuals) carrying the character are homozygous
(#) What is a homologous allele?
A: A homologous allele is an allele that is found at the same locus on a different, homologous chromosome
(#) Show a monohybrid cross resulting in a 3:1 phenotypic ratio, indicating and naming the three generations involved.
A: AA x aa --> Aa --> AA, Aa, Aa, and aa, and separated by the arrows are the P, F1, and F2 generations, respectively
(#) At one time scientists thought that a single gene pair, in a dominant/recessive inheritance pattern, controlled human eye color. The allele for brown eyes was considered dominant over the allele for blue eyes. The genetic basis for eye color is actually far more complex. At the present, three gene pairs controlling human eye color are known. Two of the gene pairs occur on chromosome pair 15 and one occurs on chromosome pair 19. The bey 2 gene, on chromosome 15, has a brown and a blue allele. A second gene, located on chromosome 19 (the gey gene) has a blue and a green allele. A third gene, bey 1, located on chromosome 15, is a central brown eye color gene. Geneticists have designed a model using the bey 2 and gey gene pairs that explains the inheritance of blue, green and brown eyes. In this model the bey 2 gene has a brown and a blue allele. The brown allele is always dominant over the blue allele so even if a person is heterozygous (one brown and one blue allele) for the bey 2 gene on chromosome 15, the brown allele will be expressed. The gey gene also has two alleles, one green and one blue. The green allele is dominant to the blue allele on either chromosome but is recessive to the brown allele on chromosome 15. This means that there is a dominance order among the two gene pairs. If a person has a brown allele on chromosome 15 and all other alleles are blue or green the person will have brown eyes. If there is a green allele on chromosome 19 and the rest of the alleles are blue, eye color will be green. Blue eyes will occur only if all four alleles are for blue eyes. Specify the genotypes of all of my family members, indicating alleles as follows: B for brown, b for blue for the bey 2 locus, G for green, and g for blue for the gey locus. Assume that G, if present, is responsible for the phenotype expressed (i.e., so is not present in people whose eyes are not green). The important phenotypes are as follows. Specify genotypes in the provided blanks. No partial credit for this bonus question.
1) My son: green eyes __________
2) My daughter: blue eyes __________
3) My wife: blue eyes __________
4) My mother: green eyes __________
5) My father: brown eyes __________
6) My father's mother: brown eyes __________
7) My father's father: blue eyes __________
8) My father's sister: blue eyes __________
9) My brother: green eyes __________
10) My brother's wife: brown eyes __________
11) My brother's son: brown eyes __________
12) My brother's daughter: green eyes __________
13) My eyes green eyes __________
A: from http://www.seps.org/cvoracle/faq/eyecolor.html.
(#) Show the two possibilities of a test cross involving a single locus and two alleles. Indicate any dominance relationships between alleles, phenotypes of all individuals involved, and ratios of resulting phenotypes and genotypes.
A: Aa ´ aa or AA ´ aa; dominant ´ recessive phenotype; 1:1 genotypic and phenotypic ratios for first cross; 1:0 genotypic to phenotypic ratios (all dominant and heterozygous) for second cross
(#) Assume two loci, two alleles per locus, complete independence between loci, and dominance-recessive relationships between alleles at each locus. Starting with two true breeding individuals, one with the dominant-dominant phenotype and the other with the recessive-recessive phenotype, indicate the type and expected ratio of phenotypes present in the grandchildren (F2) generation.
A: 9:3:3:1 of dominant-dominant, recessive-dominant, dominant-recessive, and recessive-recessive
(#) Assume that uppercase alleles are dominant to lowercase alleles within each locus but that loci are independent from one another. Describe two crosses that could give rise to this genotype and only this genotype: AaBbCcDdEE.
A: AABBCCDDEE ´ aabbccddEE plus e.g. aaBBCCDDEE ´ AAbbccddEE, etc.
(#) Give an example of codominance.
A: The easiest one would be the AB blood group, but I suppose that one could get much fancier than that.
(#) Given an dihybrid cross and epistasis, what are the expected phenotypic ratios. Employ the fun example from the book in which B = black coat color (dominant), b = brown coat color (recessive), C = coat color present (dominant), and c = no coat color present (recessive).
(#) Widow's peak is a dominant trait (see exaggerated widow's peak in image to the right). Bob's hairline, though receding, displays a widow's peak whereas both Bob's wife, Mary, and Bob's brother, Bill, have a hairline that does not come to a Widow's peak. All four of their parents display Widow's peaks. What are the odds that Bob and Mary's first son will not display a widow's peak?
A: This question is asking whether the child will be a homozygous recessive. The odds are 0.5 if Bob is a heterozygote. The odds of Bob being a heterozygote is 2/3 since all four parents are heterozygotes and we can disregard Bob being homozygous recessive. Therefore the odds of the Bob and Mary's first son being a homozygous recessive for the Widow's peak locus are 0.5 times 2/3 or 2/6 = 1/3.
(#) Indicate the genotypes of individuals I-1, III-2, and I-3 in the following pedigree for myopia (near sightedness):
A: Mm, mm, and Mm where lowercase indicates recessiveness. From: http://www.atkinson.yorku.ca/~nats1760-2004y/L15%20solving%20GP.htm.
(#) Using standard assumptions for pedigree analysis, indicate and then justify the genotype of individual I-1.
A: Aa since has non-affected children while trait appears to be inherited via a dominant allele. Otherwise would have to assume that individuals I-2, II-1, and II-6 are all heterozygotes. From http://www.geocities.com/CollegePark/Quad/7304/Image2.gif and http://www.geocities.com/CollegePark/Quad/7304/sept2001one.html.
(#) A dominant allele, A, causes yellow color in rats. The dominant allele found at another, independent locus, R, produces black coat color. When the two dominants occur together (A_ R_), they interact to produce gray. Rats of the genotype aa rr are cream-colored. If a gray male and a yellow female, when mated, produce offspring approximately 3/8 of which are yellow, 3/8 gray, 1/8 cream, and 1/8 black, what are the genotypes of the two parents?
A: A/a R/r ´ A/a r/r since there is a 3:1 ratio of Yellow dominant to Yellow recessive and a 1:1 ratio of Black dominant to Black recessive.