help with genetics problems?

Biology: Genetic problems?

Need help on a last minute project, anything would be greatly appreciated! - Thank You

1. How many Barr bodies would you see in the nuclei of persons with the following sex chromosomes?

a. XO

b. XX

c. XY

d. XXY

e. XXX


2. A man with brown teeth mates with a woman with normal white teeth. They have four daughters, all with brown teeth, and three sons, all with white teeth. The sons mate with women with white teeth, and all their children have white teeth. One of the four brown-toothed daughters mates with a man with white teeth, and they have two brown-toothed daughters, one white-toothed daughter, one brown-toothed son, and one white-toothed son.

Explain the inheritance pattern of the brown teeth condition.

(i.e., is the condition autosomal recessive, autosomal dominant, x-linked recessive, or x-linked dominant?)

3. Maple sugar urine disease is a rare inborn error of human metabolism in which the urine of affected individuals smells like maple sugar. If two unaffected individuals have an affected child, what is the probable mode of inheritance of the disease?

(i.e., is the condition autosomal recessive, autosomal dominant, x-linked recessive, or x-linked dominant?)

4.In the ABO blood system in human beings, alleles A and B are codominant and both are dominant to the O allele. In a paternity dispute, a type AB woman claims that one of four men was the father of her type A child.

Which of the following men could be the father of the child on the basis of the evidence given?

a. Type A

b. Type B

c. Type O

d. Type AB

e. All of the above are possibilities

5.You are working with the exotic organism Phobia laboris and are interested in obtaining mutants that work hard. Normal phobes are lazy. Perseverance finally pays off and you successfully isolate a true-breeding line of hard workers. You begin a detailed genetic analysis of this trait. To date you have obtained the following results:

P generation Hard worker x Nonworker

F1 generation All nonworkers of both sexes

You then mate an F1 female with a hard worker male and obtain the following offspring:

½ hard workers: ½ nonworkers of both sexes

From these results, predict the expected phenotypic ratio from crossing two F1 nonworkers

2 Answers

  • Tweek
    Lv 6
    9 years ago
    Favorite Answer

    I don't want to just give you the answers - this is probably too late anyway.

    1. All human cells have exactly one fully functioning X chromosome in it. Any more are scrunched into Barr bodies. Y chromosomes are irrelevant.

    So this is a no-brainer: a. 0, b. 1, c. 0, etc

    2. Again, this is quite simple really. If you have a condition which divides the sons and daughters this clearly, that's just a dead giveaway that this is likely to be (se)x-linked.

    The crucial thing with sex-linked inheritance is that men (normally) have one X chromosome and women have two. If a man has a gene for something on his X chromosome, it makes no difference to him whether it's dominant or recessive because there's nothing to counteract it on the Y chromosome.

    Once the allele passes to a woman however, it makes all the difference. All four of the daughters inherited their dad's brown teeth. That suggests a single allele being sufficient –> dominant.

    The alternative would be that the mother was carrying a hidden recessive brown allele and just happened to pass it on to all and only her daughters. That's unlikely. If you even need to disprove it the brown-toothed daughter having a white toothed son shows she's not homozygous brown - sons have to get their X chromosome from their mums.

    3. Oh come on, you don't need help with this. Gender isn't mentioned and the genes affecting the child are hidden in the parents –> one possible answer.

    4. A and B alleles are dominant to O alleles, so AO and BO yield A and B type just as surely as AA and BB. So if the child could be AO or AA and could have an A from his mum, is there any type which definitely couldn't pass on either an O or a second A?

    5. The separate mutant line is presumably in-bred to be homozygous for its trait and the wild for the other, so F1 tells you which is dominant with the heterozygous case.

    Do a Punnett square to show how you'd work out the results of F1 x F1, but it should be obvious.

  • 4 years ago

    you do no longer say if it rather is intercourse-related, so we''ll assume it rather is easy Mendelian genetics, then it may be D for dominant, extra suitable digits. And d for the classic # of digits. Dad must be Dd or DD. mom is dd. little ones must be Dd : dd, 2/4 extra suitable, 2/4 popular, for the 1st circulate. little ones could be all Dd, 4/4 extra suitable for the 2d circulate. however the daughter is popular, dd. So, Dad is Dd. So it rather is the 1st one, Dd:Dd, the place there's a 50/50 possibility.

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