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Reika [66]
2 years ago
9

Last week you were wondering what Jessica and Tom's babies might look like with respect to their eyebrows and earlobes (both Tom

and Jessica have thick eyebrows, controlled by gene E, with thick eyebrows dominant to thin eyebrows, and both of them have free earlobes, controlled by gene F, with free earlobes dominant to attached earlobes). You know that Tom's mom has attached earlobes. Additionally, Tom's mom has thin eyebrows, and so does Jessica's dad. No one in Jessica's pedigree going back to her great grandparents are known to have had attached earlobes, so you assume that she must be homozygous for this trait. If your inferences based on their known pedigrees are correct, what is the probability that Jessica's and Tom's first baby will have free earlobes and thin eyebrows?
Biology
1 answer:
mash [69]2 years ago
6 0

Answer:

25%

Explanation:

The possible genotypes and phenotypes are:

  • EE or Ee: thick eyebrows
  • ee: thin eyebrows
  • FF or Ff: free earlobes
  • ff: attached earlobes

Tom and Jessica have thick eyebrows and free earlobes. Tom's mother has attached earlobes, so Tom must have inherited an <em>f</em> allele from her, and because he has free earlobes he must be heterozygous<em> Ff. </em>We assume Jessica is homozygous for free earlobes <em>FF</em>.

Tom's mother and Jessica's father have thin eyebrows, so they are both <em>ee</em> and therefore both Tom and Jessica must be heterozygous <em>Ee</em>.

From the above information, Tom has the genotype EeFf and Jessica is EeFF. If you do a Punnett Square of this cross, you'll get that the probability of having a child with free earlobes and thin eyebrows (eeF_) is 25%.

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Answer: Precise measurements are important for Scientific Accuracy in order to get correct results

Explanation:

Precision is how close the Agreement is between repeated measurements under the same conditions. For example, if a measured value 36.7cm has three digits, or significant figures. Significant figures indicate the precision of a measuring tool that was used to measure a value

Accuracy is how close a measurement is to the correct value for that measurement.

When taking scientific measurements it is very important to be both precise and accurate. Bad equipment, human error can give inaccurate results.

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2 years ago
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Students research unicellular, prokaryotic organisms that live in harsh environments such as volcanic hot springs, brine pools,
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The answer to that would be archaebacteria since they are prokaryotes (no nucleus) but so is Eubacteria, however archaebacteria can live in very harsh conditions.
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Kristoff was recently diagnosed with depression after reporting symptoms such as a loss of interest in activities, poor concentr
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Answer:

The correct answer is option b. "it has high face validity".

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2 years ago
Compared with fraternal twins, identical twins are ________ similar in physical appearance. Compared with unrelated look-alike p
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Answer;

-No more; no more

Compared with fraternal twins, identical twins are no more similar in physical appearance. Compared with unrelated look-alike pairs of individuals, identical twins report no more similar personalities.

Explanation;

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-Since fraternal twins are 2 separate fertilized eggs, they usually develop two separate amniotic sacs, placentas, and supporting structures. Identical, or monozygotic, twins may or may not share the same amniotic sac, depending on how early the single fertilized egg divides into two.

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Alkaptonuria is an infrequent autosomal recessive condition. It is first noticed in newborns when the urine in their diapers tur
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Answer:

See the answer below

Explanation:

Let the disorder be represented by the allele a.

Since the disease is an autosomal recessive one, affected individuals will have the genotype aa and normal individuals will have the genotype Aa or AA.

Since the four adults are carriers, their genotypes would be Aa.

                    Aa     x     Aa

Progeny:    AA    2Aa    aa

Probability of being affected = 1/4

Probability of being a carrier = 1/2

Probability of not being affected = 3/4

(a) The chance that the child second child of Mary and Frank will have alkaptonuria = 1/2

(b) The chance that  the third child of Sara and James will be free of the condition = 3/4

(c)

(d) If someone has no family history of the disorder, their genotype would be AA.

                 AA     x     aa

                        4 Aa

<em>The chance that a child with alkaptonuria will have an offspring with alkaptonuria if the child's mate has no family history </em>= 0

(e)

(f) <em>The chance that a child with alkaptonuria will have an offspring with alkaptonuria if the child's mate has no family history</em> = 0

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