161The Gene Book 1.7) In the following case of disputed paternity, determine the

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161The Gene Book 1.7) In the following case of disputed paternity, determine the probable parent (OR NEITHER). Mother is type B, child is type AB. Father #1 is A; father #2 is B. A. father #1 B. father#2 C. neither 1.8) What proportion of the plants from the following cross would be tall with yellow seeds? T-TALL, Y-YELLOW ttYy x ttyy B. D. 0 1.9) A child is born with cystic fibrosis. The pediatrician explains that this is an inherited genetic A. Make a Punnett square to show the parents how this disorder can be inherited since B. Use the Punnett square to explain to the parents the likelihood that their next child will be C. Based on these observations, is cystic fibrosis a dominant of recessive disorder? How can disorder. neither parent has the disorder. affected with cystic fibrosis, also. you tell? 1.10) A family comes into your office for genetic counseling. The husband has the allele for Huntington's Disease, a rare dominant neurodegenerative disorder that produces symptoms beginning in midlife. The wife and husband are both Rh+. The wife has Type O blood and the husband has type AB blood. Their son has Type O blood and is Rh+. Based on this information, what can you tell the family about the likelihood that their son may have inherjted Huntington's Disease? Where do we go from here? Well, let's see... it looks like we're going to be exploring family trees and playing genetic dice! We hurry off to CHAPTER 2 REFERENCES

Explanation / Answer

1.7) Answer: A; father #1 is the original parent,

If the child is AB group that means he got heterozygotic allie from both of his parents, if one parent (mother) is B, then the other parent must be A.

Mother: iB iB or iB i (B group) x iB iB or  iB i or iB iA (B or AB group)

1.

Then only the iA iB child (AB group) is possible.

1,8) Answer: D '0'.

in the cross ttYy x ttyy

the possible types of gametes are

ty tY x ty (gametes)

then the possible genotypes would be ttyy and ttYy (short and green, short and yellow).

So yellow tall phenotype is not possible from the cross.

1.9) A) If neither of the parents affected then they must be carriers and their phenotype must be

Ff and Ff

they produce two types of gametes F and f

1.10) the possibility of the child having huntington's decease is 50%.

If the father is deceased then his genotype must be either dominant or heterozygous

DD or Dd

mother unaffected means her genotype must be recessive

dd

then

parental cross DD X dd or Dd x dd

children: Dd (all affected)or Dd (50% affected) and dd (50% unaffected)

  

So the phenotypic ratio is 3(non deceased) : 1 (cystic fibrosis)

genotypic ratio is 1 (FF) : 2 (Ff) : 1 (ff)

B) The possibility of next child having deceased phenotype is

so the possibility of any child having deceased phenotype is 1/4.

C) based on observations of cystic fibrosis is a recessive disorder. The unaffected parents having affected children, so there is a skipping of phenotype in generations, and heterozygotic individuals here are normal. So these are the factors suggest the studied disease as a recessive trait.

1.10) Answer: The probability of a child with deceased phenotype varies from 50- 100%. Because it is a dominant disease the genotype of the father is either DD or Dd

If father here is a homozygous dominant with the genotype of DD, mother anyway is unaffected so the genotype is dd

DD X dd

children: Dd (100% deceased)

If the father is a heterozygotic dominant with the genotype of (Dd) and mother anyway is a recessive (dd)

Dd X dd

children; Dd (affected 50%) dd (unaffected 50%).

parents F f F FF(normal) Ff(carrier) f Ff(carrier) ff(cystic fibrosis)

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