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jonny [76]
3 years ago
10

Four corners vocabulary cards

Biology
1 answer:
gtnhenbr [62]3 years ago
6 0
Are u a lefty? And gimme min to answer

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No stripe on the head (H) is dominant, a black stripe on the head is recessive. If dad's genotype is Hh, what are the odds that
olga55 [171]

The chance of the offsprings with stripe on head is 50%.

Option A.

<h3><u>Explanation:</u></h3>

Here according to the figures, the male is the brown one which has no stripe on his head as mentioned, and the female is the pink one who has a stripe on her head. The gene for the head stripe is denoted by H.

So the genotype of the father which is given as Hh. He is heterozygous.

The genotype of the mother should be hh as she has stripe and she must be homozygous recessive to have it.

So the gametes from father is H and h. Whereas the gametes from mother is only h.

So the genotype of the offsprings are Hh and hh as 50% each.

So the probability of the offspring being striped head is 50%.

7 0
3 years ago
Which of the following is an inference?
kaheart [24]
Inference is d. An observation is something you sense through taste,touch,smell, hear, and see
5 0
2 years ago
Why is dna smaller than rna?
Elis [28]
No it’s the other way around .

RNA is much shorter than DNA. DNA contains the code for making lots and lots of different proteins. Messenger RNA contains the information to make just one single polypeptide chain - in other words for just one protein, or even just a part of a protein if it is made up of more than one polypeptide chain.
7 0
2 years ago
Read 2 more answers
What percentage of Americans use solar power ?
dangina [55]

Answer:

66.7 percent.

Explanation:

5 0
3 years ago
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Consider a cross between a man with an X-linked recessive disorder and a woman that's a carrier for the disorder. Approximately
Vadim26 [7]

Answer:

50%

Explanation:

Let's assume that the X linked allele "a" gives the disorder. Therefore, the genotype of the affected man would be X^aY while the genotype of the carrier woman would be X^aX. A cross between X^aY man and X^aX woman would produce progeny in following phenotype ratio: 1 affected daughter: 1 affected son: 1 normal but carrier daughter: 1 normal son.

Therefore, there are 50% chances that their children can express the disorder.  

6 0
3 years ago
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