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Mendelian Genetics Practice Problem Solutions

Mendelian Genetics
Practice Problem Solutions


  1. Mice I: Simple Dominance

    a) Cross 1: red X white gives all red - a likely model is that eye color is controlled by one gene which has 2 alleles, red and white; also the red allele's phenotype is likely to be dominant. Therefore, choose appropriate symbols (one letter per gene, capital letter for allele with dominant phenotype):

    R - red allele. Dominant phenotype of red eyes.

    r - white allele. Recessive phenotype of white eyes.

    Cross 1 is then: RR X rr ----> all Rr (all red eyes).

    Cross 2 is then: Rr X Rr ----> 3:1 red:white.

    The predictions agree with the data, so we are done.

    b) Cross 1: long X short gives some long and some short - a likely model is that ear length is controlled by one gene with 2 alleles, long and short. From this data, we cannot determine which allele has the dominant phenotype.

    Cross 2: long X long gives some long and some short - a likely model is that the long allele has the dominant phenotype. Therefore, the appropriate symbols are:

    L - long allele. Dominant phenotype of long ears.

    l - short allele. Recessive phenotype of short ears.

    Cross 1 is then: Ll X ll ----> 1:1 long:short

    Cross 2 is then Ll X Ll ----> 3:1 long:short

    The predictions agree with the data, so we are done.

  2. Flowers: Incomplete dominance

    Cross 1: Blue X white gives all pale blue - a plausible model is that color is controlled by one gene with 2 alleles, blue and white. The color is determined by a semi-dominant mechanism, that is:

    B - blue allele. Semi-dominant blue phenotype.

    b - white allele. Semi-dominant white phenotype.

    Therefore:
    BB - blue
    Bb - pale blue
    bb - white

    Cross 1 is then: BB X bb ----> all Bb, pale blue.

    Cross 2 is then: Bb X Bb ----> 1:2:1 BB:Bb:bb or: blue:pale:white

    The predictions agree with the data, so we are done.

  3. Blood Type: co-dominance

    Blood type is determined by one gene that has three alleles:

    A - allele with co-dominant type-A phenotype

    B - allele with co-dominant type-B phenotype

    O - allele with recessive type-O phenotype

    a) Cross 1 is then: AA X BB ----> all type AB

    Cross 2 is then: AB X AB ----> 1:2:1 types A:AB:B (note skewed ratio with small number of progeny)

    b) Cross 1 is then: AO X BO ----> 1:1:1:1 types A:B:AB:O (note skewed ratio with small number of progeny)

    The predictions agree with the data, so we are done.

    LESSON: In general, given 2 alleles, A1 and A2, each with a different phenotype as a homozygote, there are three formal possibilities for the heterozygote:

        heterozygote phenotype		      type of inheritance
    - neither A1 or A2 (whichever is dominant)	simple dominance
    - neither A1 nor A2				incomplete dominance
    - both A1 and A2				co-dominance
    

  4. Mice II: Lethal Alleles

    Tail-lessness and lethality are controlled by one gene with two alleles:
    T allele - dominant tail-less phenotype; recessive lethal phenotype
    t allele - recessive normal-tail phenotype; dominant viable phenotype

    Therefore:
    TT - tail-less and dead
    Tt - tail-less and alive
    tt - normal-tail and alive

    Cross 1 is then: TT X tt ----> all Tt (tail-less)
    Cross 2 is then: Tt X Tt ----> 1:2:1 tail-less:normal:dead
    The predictions agree with the data, so we are done.

  5. Flies: Sex-Linkage

    There are two different types of notation for sex-linkage:

    X - X chromosome carrying R allele; dominant red-eyed phenotype
    X - X chromosome carrying r allele; recessive white-eyed phenotype
    Y - Y chromosome carrying no eye-color allele

    or: --> equivalent to XX and or simply --> equivalent to XY

    Cross 1 is then: XX cross XY ----> 1:1:1:1

    XX (red female): XX (red female): XY (red male): XY (white male)

    Cross 2 is then of two possible types (and there will be equal #s of each type):

    XY cross XX ----> 1:1:1:1

    XX (red female): XX (white female): XY (red male): XY (white male)

    XY cross XX ----> 1:1 XY(red male):(XX (red female)

    The predictions agree with the data, so we are done.

  6. Pedigrees

    a) sex-linked recessive:
    X - X chromosome with normal allele; dominant phenotype
    X - X chromosome with disease allele; recessive phenotype
    Y - Y chromosome, no disease or normal allele

    b) Autosomal recessive:
    D - normal allele; dominant phenotype
    d - disease allele; recessive phenotype

  7. Mice III: Multiple genes & Simple Dominance

    a) eye color is determined by one gene with two alleles:
    B - dominant blue-eye phenotype
    b - recessive brown eye phenotype
    tooth length is determined by one gene with two alleles:
    S - dominant short-tooth phenotype
    s - recessive long tooth phenotype
    Cross 1 is then: BBss X bbSS ----> all BbSs
    Cross 2 is then: BbSs X BbSs ----> 9:3:3:1 (B_S_): (B_ss): (bbS_): (bbss)

    b) height is determined by one gene with two alleles:
    T - dominant tall phenotype
    t - recessive short phenotype
    color is determined by one gene with two alleles:
    G - dominant green phenotype
    g - recessive yellow phenotype
    Cross 1 is then: TtGG X ttgg ----> 1:1 (TtGg): (ttGg)
    Cross 2 is then: TTGg X ttgg ----> 1:1 (TtGg): (Ttgg)

  8. Pathways I: Epistasis

    a) 	9 A_B_	purple		b)	9 A_B_	purple
    	3 A_bb	red			3 A_bb	red
    	3 aaB_	white			3 aaB_	white
    	1 aabb	white			1 aabb	red
    or:	9:4:3 purple:white:red		9:4:3 purple:red:white
    

  9. Pathways II: Lethal Allele Combinations

    You'd expect: 	1	AaBb	??? red + blue = purple???
    		1	aaBb	blue
    		1	Aabb	red
    		1 	aabb	white
    
    but the AaBb class is missing - AaBb could be dead (all the precursor used up or red + blue is toxic)

  10. Linkage
    There are 2 genes with 2 alleles each:
    gene 1:
    R - allele with dominant rough phenotype
    r - allele with recessive smooth phenotype
    gene 2:
    H - allele with dominant hard phenotype
    h - allele with recessive soft phenotype
    For crosses with linkage, we need a set of symbols that shows the chromosomal arrangement of the genes and alleles:

    RRhh becomes which means:

    Cross 1 is then: X ---->

    Cross 2 is then: X ---->

    Non-recombinant types: : rough soft and : smooth hard

    Recombinant types: : rough hard and : smooth soft

    So recombination frequency =

  11. Pedigrees

    Since the abnormality appears in women, we can rule out Y-linked inheritance.
    Since it appears more frequently in men than women, it may be sex-linked recessive:
    X - X chromosome with recessive abnormal allele
    X - X chromosome with dominant normal allele

    In this case, we would predict that all the third generation would be affected. This does not match the data.

    If it were sex-linked dominant:
    X - X chromosome with dominant abnormal allele
    X - X chromosome with recessive normal allele

    Therefore: 6 must be XX, 7 must be XY
    If this is so, we would predict that the all the third generation daughters will get an X from 7 and be affected. This does not match the data.

    This leaves autosomal dominant or recessive.
    If it were autosomal recessive:
    a - recessive abnormal allele
    A - dominant normal allele

    Therefore we would predict that all their children would have to be aa and be affected. This does not match the data.

    Try autosomal dominant and see if it works:
    A - dominant abnormal allele
    a - recessive normal allele

    If 1 is AA, then all the children of 1 and 2 would be Aa - affected. This does not match the data, so:
    1 must be Aa
    Using similar logic,
    3 must be Aa
    4 must be aa
    In the second generation, we would predict children with genotypes Aa and aa. All of this are consistent with our model.
    In the third generation, we would predict that the children of 6 and 7 will have genotypes AA, Aa, and aa: These combinations could result from 6 cross 7. All this is consistent with our model
    Note that individual 14 is the key to the problem, without her, many of the models would be indistinguishable.

    The results predicted by this model fit the data. Its one weakness is that it requires two non-blood related individuals in the family to introduce the abnormality. This is an unlikely event, but since no other explanation matches the data, we are forced to accept this model in the absence of other evidence.

  12. The pedigree is as follows:



    It might actually be quite difficult to tell the difference between Y-linked and sex-linked recessive. If you found a female member of the family with the disorder, then you could conclude that it was sex-linked recessive. But if you never found an affected female, you couldn't be sure. You could also look at the sons of the various women in the family - if any of them were affected, then it would have to be sex-linked recessive. (Why?)

  13. Calculate the recombination frequencies in pairwise combinations:

    - l to m: Parental gamete types are: and resulting in progeny:

    	Phenotype	Number
    	+   m   n  	0
    	+   m   + 	899
    	l   +   n	901
    	l   +   +	0
    
    Total parental = 1800
    Recombinant types are: and , resulting in progeny:
    	Phenotype	Number
    	l   m   n	99
    	l   m   +	0
    	+   +   n	0
    	+   +   +	101
    
    Total recombinant = 200
    Therefore l ---> m = 10cM

    - m to n: Parental types are: and resulting in progeny:

    	Phenotype	Number
    	+   m   +       899
    	l   m   +	0
    	+   +   n	0
    	l   +   n	901
    
    Total parental = 1800
    Recombinant types are: and , resulting in progeny:
    		Phenotype	Number
    		l   m   n	99
    		+   m   n	0
    		l   +   +	0
    		+   +   +	101
    
    Total recombinant = 200
    Therefore m ---> n = 10cM

    - l to n: Parental types are: and resulting in progeny:

    	Phenotype	Number
    	+  +   +	101
    	+  m   +	899
    	l  +   n	901
    	l  m   n	99
    
    Total parental = 2000
    Recombinant types are: and , resulting in progeny:
    		Phenotype	Number
    		+   m   n	0
    		+   +   n	0
    		l   +   +	0
    		l   m   +	0
    

    Total recombinant = 0
    Therefore l ---> n < 0.05cM
    This is an unusual map distance, it is too small to be measured with this number of progeny. For this reason, there is no least frequent class to find the map order and the map must remain ambiguous:

    where l and n are 0.05cM apart. (In Drosophila, 0.05cM is roughly 25000 bp of DNA.)

  14. Linkage & Pedigrees Problem

    (a) & (b) The best way to solve this problem is to work out the genotypes of the individuals in the pedigree:



    c) To be type B, they must have gotten an chromosome from mom and a chromosome from dad.

    The recombination frequency (RF) gives the probability of crossover (genetic exchange) between two genes. In the case of individual (3), he can produce 4 types of gametes:

    Recombinant Gametes:
    * probability of gamete =
    * probability of gamete =
    Non-recombinant Gametes
    *probability of gamete =
    * probability of gamete =



    Another way to look at it: in order to be diseased and B, recombination must have occurred and since the chance of recombination is 11%, the chance of being B and diseased is 11%.


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