Microsoft Word - Genetics Problems.doc Genetics 1 Neatly write your answers on a separate sheet of paper. Show your work. You must specify the genotype or phenotype the numbers refer to in your...

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Microsoft Word - Genetics Problems.doc Genetics 1 Neatly write your answers on a separate sheet of paper. Show your work. You must specify the genotype or phenotype the numbers refer to in your ratios. Ex: 1:2:1 is not sufficient. 1AA : 2Aa : 1aa (genotype) or 1 red : 2 pink: 1 white (phenotype) is correct Assume the genes are autosomal unless otherwise specified. 1. In humans, the allele for albinism is recessive to the allele for normal skin pigmentation. If two heterozygotes have children, what is the chance that a child will have normal skin pigment? What is the chance that a child will be albino? If the child is normal, what is the chance that it is a carrier (heterozygous) for the albino allele? 2. In purple people eaters, one-horn is dominant and no horns is recessive. Show the cross of a purple people eater that is heterozygous for horns with a purple people eater that does not have horns. Write the expected genotypic and phenotypic ratios of their offspring. 3. In humans, the brown-eye (B) allele is dominant to the non-brown eye allele (b). (Eye color is actually quite complicated and not well understood yet.) If two heterozygotes have a child, what are the expected genotypic and phenotypic ratios of their offspring? 4. In seals that gene for the length of whiskers has two alleles. The dominant allele (W) codes long whiskers and the recessive allele (w) codes for short whiskers. What percentage of offspring would be expected to have short whiskers from the cross of two long-whiskered seals, one that is homozygous dominant and one that is heterozygous? 5. In pea plants the yellow color allele (Y) is dominant over the green color allele (y) for seed color and the tall allele (T) is dominant over the short allele (t) for plant height. They are independently assorting genes. Parents heterozygous for both traits are cross- pollinated. Using a 16 square Punnett square, determine the frequency for the four- different phenotypes of the offspring. 6. Now let’s try a shortcut way of solving that same dihybrid cross. Because of Mendel’s Law of Independent Assortment, we can work with the color gene and the height gene separately. Set up two separate monohybrid crosses from those same parents. Then use the laws of probability to calculate your frequencies (fraction) of each trait alone and combined (multiply the fractions) and fill in the table. Height Color = Probability Tall, yellow Tall, green Short, yellow Short, green Genetics 2 Follow the directions from Genetics 1. Assume the genes are autosomal unless otherwise specified. 1. In radishes, the gene that controls color exhibits incomplete dominance. True- breeding red radishes crossed with true-breeding white radishes combine to make purple radishes. What are the genotypic and phenotypic ratios when you cross a purple radish with a white radish? 2. Certain breeds of cattle show incomplete dominance in coat color. When true- breeding red cattle are bred with true-breeding white cattle, the offspring are roan (pinkish coat color). Write the expected genotypic and phenotypic ratios of the offspring when a roan cow is mated with a roan bull. 3. The cattle have a second gene for horned vs. hornless cattle. The allele for horns is dominant. If a bull and cow are heterozygous for both the horn and the coat color genes, what is the probability for each possible phenotype? 4. A man with type AB blood marries a woman with type B blood. Her mother has type O blood. Write the expected phenotypic and genotypic ratios of their children. 5. The father of a child has type AB blood. The mother has type A. Which blood type(s) can their children NOT have? Why? 6. A woman with type A blood and a man with type B blood could potentially have offspring with what blood types? 7. The mother has type A blood. Her husband has type B blood. Their child has type O blood. The husband claims that the child can’t be his. Is he right? Explain. 8. The mother has type B blood. Her husband has type AB blood. Their child has type O blood. The husband claims the child can’t be his. Is he right? Explain. 9. The mother has type AB blood. The father has type B blood. His mother has type O blood. What are all the phenotypic possibilities of blood types for their children? 10. Achondroplasia (dwarfism) is caused by a dominant gene. A woman and man both with dwarfism have children. Homozygous achondroplasia results in death of embryos. Write the expected genotypic and phenotypic ratios of live-birth offspring. 11. Hemophilia is a recessive disorder whose defective gene is located on the X chromosome. List the expected genotypic and phenotypic ratios of the children from a man normal for blood clotting and a woman who is a carrier of the defective blood clotting allele. Report the ratios for boys separate from the ratios for girls.
Answered Same DayMay 06, 2021

Answer To: Microsoft Word - Genetics Problems.doc Genetics 1 Neatly write your answers on a separate sheet of...

Malvika answered on May 07 2021
157 Votes
56868
Genetics 1
1) Albinism
    
    P
    p
    P
    PP
    Pp
    p
    Pp
    pp
Parents are heterozygous for normal skin color – Pp
Ratio of children’s genotype:
- 1:2:1 = 1 (PP): 2 (Pp):1 (pp)
Thus, there is 25% chance i.e. one child in four children would be albino and the two heterozygous children would be carriers for the albino allele.
2) One-horn
    
    H
    h
    h
    Hh
    hh
    h
    Hh
    hh
Purple people eater heterozygous for horns – Hh
Purple people eater with no horns – hh
Ratio of offspring’s genotype: - 2:2 = Hh:hh
50% of the offspring would be heterozygous for horns and 50% would have recessive genes and no horns.
3) Brown eye
    
    B
    b
    B
    BB
    Bb
    b
    Bb
    bb
Heterozygous humans for brown eye color – Bb
Phenotypic ratio of children = Brown (3) : Non-brown (1)
Genotypic ratio of children = BB (1) : Bb (2) : bb (1)
4) Whiskers in seal
    
    W
    w
    W
    WW
    Ww
    W
    WW
    Ww
Homozygous dominant seal – WW
Heterozygous dominant seal – Ww
Short whiskers in children – zero
All children would have long whiskers, 2 would be homozygous and 2 would be heterozygous.
5) Yellow pea plants (Y) and Tall pea plants (T)
    
    YT
    Yt
    yT
    yt
    YT
    YYTT
    YYTt
    YyTT
    YyTt
    Yt
    YYTt
    YYtt
    YyTt
    Yytt
    yT
    YyTT
    YyTt
    yyTT
    yyTt
    yt
    YyTt
    Yytt
    yyTt
    yytt
Heterozygous yellow parent – Yy
Heterozygous tall parent – Tt
Offspring phenotype – 9:3:3:1
Yellow and tall – 9, Yellow and short – 3, Green and Tall – 3, Green and short – 1
6) Dihybrid cross in short
    
    Height
    Color
    =...
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